NCERT Solutions Class 12 Mathematics Chapter 9

Q.1 Determine order and degree (if defined) of differential equation

$\frac{{\mathrm{d}}^{\mathrm{4}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{4}}}\hspace{0.17em}+\mathrm{\hspace{0.17em}}\sin (\mathrm{y}”’)\hspace{0.17em}=\hspace{0.17em}0$

Ans.

$\begin{array}{l}\frac{{\mathrm{d}}^{\mathrm{4}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{4}}}+\sin \left(\mathrm{y}”’\right)=0\Rightarrow \mathrm{y}””\mathrm{\hspace{0.17em}}+\sin \left(\mathrm{y}”’\right)=0\\ \mathrm{The}\mathrm{heighest}\mathrm{order}\mathrm{derivative}\mathrm{present}\mathrm{in}\mathrm{the}\mathrm{differential}\mathrm{equation}\\ \mathrm{is}\frac{{\mathrm{d}}^{\mathrm{4}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{4}}}.\mathrm{Therefore}\mathrm{its}\mathrm{order}\mathrm{is}4\mathrm{.}\\ \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{polynomial}\mathrm{equation}\mathrm{in}\mathrm{y}‘\mathrm{and}\mathrm{degree}\\ \mathrm{of}\mathrm{such}\mathrm{a}\mathrm{differential}\mathrm{equation}\mathrm{can}\mathrm{not}\mathrm{be}\mathrm{defined}\mathrm{.}\end{array}$

Q.2

\begin{array}{l}\text{Determine order and degree}\left(\text{if defined}\right)\text{of differential}\text{equation:}\\ \text{y’ + 5y}\text{=}\text{0}\end{array}

Ans.

\begin{array}{l}\text{y’ + 5y=0}\\ \text{The heighest order derivative present in the differential equation}\\ \text{is y’}\text{. Therefore its order is 1}\text{.}\\ \text{The given equation is a polynomial equation in y’ so the heighest}\\ \text{power raised by y’ is one}\text{. Thus, degree of the differential eqution}\\ \text{is 1}\text{.}\end{array}

Q.3

$\begin{array}{l}\text{Determine order and degree}\left(\text{if defined}\right)\text{of differential}\text{equation:}\hfill \\ {\left(\frac{\text{ds}}{\text{dt}}\right)}^{\text{4}}\text{+ 3s}\frac{{\text{d}}^{\text{2}}\text{s}}{{\text{dt}}^{\text{2}}}\text{=0}\hfill \end{array}$

Ans.

\begin{array}{l}{\left(\frac{\text{ds}}{\text{dt}}\right)}^{\text{4}}\text{+ 3s}\frac{{\text{d}}^{\text{2}}\text{s}}{{\text{dt}}^{\text{2}}}\text{=0}\\ \text{The}\text{heighest oder derivative present in differential equation is}\\ \frac{{\text{d}}^{\text{2}}\text{s}}{{\text{dt}}^{\text{2}}}\text{. Therefore, its order is 2}\text{.}\\ \text{Since, given differential equation is a polynomial in}\frac{{\text{d}}^{\text{2}}\text{s}}{{\text{dt}}^{\text{2}}}\text{and}\frac{\text{ds}}{\text{dt}}\text{.}\\ \text{The power raised by}\frac{{\text{d}}^{\text{2}}\text{s}}{{\text{dt}}^{\text{2}}}\text{is 1, so the degree of equation is 1}\text{.}\end{array}

Q.4

\begin{array}{l}\text{Determine order and degree}\left(\text{if defined}\right)\text{of differential}\text{equation:}\\ {\left(\frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\right)}^{\text{2}}\text{+}\text{cos}\left(\frac{\text{dy}}{\text{dx}}\right)\text{=}\text{0}\end{array}

Ans.

\begin{array}{l}{\left(\frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\right)}^{\text{2}}\text{+cos}\left(\frac{\text{dy}}{\text{dx}}\right)\text{=0}\\ \text{The}\text{heighest oder derivative present in differential equation is}\\ \frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\text{. Therefore, its order is 2}\text{.}\\ \text{Since, given differential equation is not a polynomial in}\frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\\ \text{and}\frac{\text{dy}}{\text{dx}}\text{.}\\ \text{So, its degree is not defined}\text{.}\end{array}

Q.5

\begin{array}{l}\text{Determine order and degree}\left(\text{if defined}\right)\text{of differential}\text{equation:}\\ \frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\text{=}\text{cos3x + sin3x}\end{array}

Ans.

\begin{array}{l}\frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\text{=cos3x + sin3x}\Rightarrow \frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}-\text{cos3x}-\text{sin3x=0}\\ \text{The}\text{heighest oder derivative present in differential equation is}\\ \frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\text{. Therefore, its order is 2}\text{.}\\ \text{Since, given differential equation is a polynomial in}\frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\text{.}\\ \text{The power raised by}\frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\text{is 1, so the degree of equation is 1}\text{.}\end{array}

Q.6

\begin{array}{l}\text{Determine order and degree}\left(\text{if defined}\right)\text{of differential equation:}\\ {\left(\text{y”’}\right)}^{\text{2}}\text{+}{\left(\text{y”}\right)}^{\text{3}}\text{+}{\left(\text{y’}\right)}^{\text{4}}\text{+}{\left(\text{y}\right)}^{\text{5}}\text{=}\text{0}\end{array}

Ans.

\begin{array}{l}{\left(\text{y”’}\right)}^{\text{2}}\text{+}{\left(\text{y”}\right)}^{\text{3}}\text{+}{\left(\text{y’}\right)}^{\text{4}}\text{+}{\left(\text{y}\right)}^{\text{5}}\text{=0}\\ \text{The heighest order derivative in given differential equation is y”’}\text{.}\\ \text{So, its order is 3}\text{.}\\ \text{This differential equation is a polynomial in y”’, y”, y’ and y}\text{.}\\ \text{The heighest power raised by y”’ is 2, therefore degree of}\\ \text{differential equation is 2}\text{.}\end{array}

Q.7

\begin{array}{l}\text{Determine order and degree}\left(\text{if defined}\right)\text{of differential equation:}\\ \text{y”’ + 2y”}\text{+}\text{y’}\text{=}\text{0}\end{array}

Ans.

$\begin{array}{l}\mathrm{y}”’\; +\; 2\mathrm{y}”+\mathrm{y}‘=0\\ \mathrm{The}\mathrm{heighest}\mathrm{order}\mathrm{derivative}\mathrm{in}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{y}”’\mathrm{.}\\ \mathrm{So},\mathrm{its}\mathrm{order}\mathrm{is}3\mathrm{.}\\ \mathrm{This}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{in}\mathrm{y}”’,\mathrm{y}”\mathrm{and}\mathrm{y}‘\mathrm{.}\\ \mathrm{The}\mathrm{heighest}\mathrm{power}\mathrm{raised}\mathrm{by}\mathrm{y}”’\mathrm{is}1,\mathrm{therefore}\mathrm{degree}\mathrm{of}\\ \mathrm{differential}\mathrm{equation}\mathrm{is}1\mathrm{.}\end{array}$

Q.8

\begin{array}{l}\text{Determine order and degree}\left(\text{if defined}\right)\text{of differential equation:}\\ \text{y’}\text{+ y}\text{=}{\text{e}}^{\text{x}}\end{array}

Ans.

\begin{array}{l}{\text{y’+ y=e}}^{\text{x}}{\text{Þy’+ y-e}}^{\text{x}}\text{=0}\\ \text{The heighest order derivative in given differential equation is y’}\text{.}\\ \text{So, its order is 1}\text{.}\\ \text{This differential equation is a polynomial in y’ and y}\text{.}\\ \text{The heighest power raised by y’ is 1, therefore degree of}\\ \text{differential equation is 1}\text{.}\end{array}

Q.9

\begin{array}{l}\text{Determine order and degree}\left(\text{if defined}\right)\text{of differential equation:}\\ \text{y” +}{\left(\text{y’}\right)}^{\text{2}}\text{+}\text{2y}\text{=}\text{0}\end{array}

Ans.

\begin{array}{l}\text{y” +}{\left(\text{y’}\right)}^{\text{2}}\text{+2y=0}\\ \text{The heighest order derivative in given differential equation isy”}\text{.}\\ \text{So, its order is 2}\text{.}\\ \text{This differential equation is a polynomial in y”, y’ and y}\text{.}\\ \text{The heighest power raised by y” is 1, therefore degree of}\\ \text{differential equation is 1}\text{.}\end{array}

Q.10

\begin{array}{l}\text{Determine order and degree}\left(\text{if defined}\right)\text{of differential equation:}\\ \text{y”}\text{+ 2y’}\text{+ siny}\text{=}\text{0}\end{array}

Ans.

\begin{array}{l}\text{y”+ 2y’+ siny=0}\\ \text{The heighest order derivative in given differential equation is}\text{y”}\text{.}\\ \text{So, its order is 2}\text{.}\\ \text{This differential equation is a polynomial in y” and y’}\text{.}\\ \text{The heighest power raised by y” is 1, therefore degree of}\\ \text{differential equation is 1}\text{.}\end{array}

Q.11

\begin{array}{l}\text{The degree of the differential equation}\\ {\left(\frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\right)}^{\text{3}}\text{+}{\left(\frac{\text{dy}}{\text{dx}}\right)}^{\text{2}}\text{+ sin}\left(\frac{\text{dy}}{\text{dx}}\right)\text{+1}\text{=}\text{0}\text{is}\\ \left(\text{A}\right)\text{3}\left(\text{B}\right)\text{2}\left(\text{C}\right)\text{1}\left(\text{D}\right)\text{not}\text{‹}\text{defined}\end{array}

Ans.

\begin{array}{l}{\left(\frac{d^{2}y}{d{x}^2}\right)}^3+{\left(\frac{dy}{dx}\right)}^2+\sin \left(\frac{dy}{dx}\right)+1=0\\ \text{S}\text{i}\text{n}\text{c}\text{e}\text{given equation is not a polynomial equation in y’}\text{. Therefore,}\\ \text{its degree is not defined}\text{.}\\ \text{Hence, the correct answer is D}\text{.}\end{array}

Q.12

\begin{array}{l}\text{The order of the differential equation}\\ {\text{2x}}^{\text{2}}\frac{{\text{d}}^{\text{2}}\text{y}}{{\text{dx}}^{\text{2}}}\text{–}\text{3}\frac{\text{dy}}{\text{dx}}\text{+}\text{y}\text{=}\text{0}\text{is}\\ \left(\text{A}\right)\text{3}\left(\text{B}\right)\text{2}\left(\text{C}\right)\text{0}\left(\text{D}\right)\text{not}\text{‹}\text{defined}\end{array}

Ans.

\begin{array}{l}2x^2\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+y=0\\ \text{T}\text{h}\text{e}\text{highest derivative of given differential equation is}\frac{d^{2}y}{d{x}^2}.\\ \text{T}\text{h}\text{e}\text{r}\text{e}\text{f}\text{o}\text{r}\text{e},\text{the order of given differential equation is 2}\text{.}\\ \text{Thus, correct option is B}\text{.}\end{array}

Q.13

\begin{array}{l}\text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ {\text{y = e}}^{\text{x}}\text{+ 1}\text{:}\text{y” – y’ = 0}\end{array}

Ans.

\begin{array}{l}{\text{y = e}}^{\text{x}}\text{+ 1}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ {\text{y’ = e}}^{\text{x}}\mathrm{\dots }\left(\text{i}\right)\\ \text{Differentiating equation}\left(\text{i}\right)\text{w}\text{.r}\text{.t}\text{. x, we get}\\ {\text{y” = e}}^{\text{x}}\\ \text{Substituting values of y’ and y” in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y” – y’ = 0, we get}\\ {\text{y” – y’ =e}}^{\text{x}}{\text{– e}}^{\text{x}}\text{= 0 = R}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.14

\begin{array}{l}\text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ {\text{y = x}}^{\text{2}}\text{+2x + C}\text{:}\text{y’ –2x –2 = 0}\end{array}

Ans.

\begin{array}{l}{\text{y = x}}^{\text{2}}\text{+2x + C}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y’ = 2x + 2}\\ \text{Substituting values of y’ in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y’ –2x –2 = 0, we get}\\ \text{y’ – 2x –2 = 2x + 2 –2x –2}\\ \text{= 0 = R}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.15

\begin{array}{l}\text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ \text{y = cos x + C}\text{:}\text{y’ + sinx = 0}\end{array}

Ans.

\begin{array}{l}\text{y = cos x + C}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y’ = –sinx}\\ \text{Substituting values of y’ in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y’ + sinx = 0, we get}\\ \text{y’ + sinx = –sinx + sinx = 0 = R}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.16

\begin{array}{l}\text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ \text{y =}\sqrt{{\text{1 + x}}^{\text{2}}}\text{:}\text{y’ =}\frac{\text{xy}}{{\text{1 + x}}^{\text{2}}}\end{array}

Ans.

\begin{array}{l}\text{y =}\sqrt{{\text{1 + x}}^{\text{2}}}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y’ =}\frac{\text{x}}{\sqrt{{\text{1 + x}}^{\text{2}}}}\\ \text{Substituting values of y in R}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y’ =}\frac{\text{xy}}{{\text{1 + x}}^{\text{2}}}\text{, we get}\\ \text{R}\text{.H}\text{.S}\text{. =}\frac{\text{xy}}{{\text{1 + x}}^{\text{2}}}\\ \text{=}\frac{\text{x}\left(\sqrt{{\text{1 + x}}^{\text{2}}}\right)}{{\text{1 + x}}^{\text{2}}}\\ \text{=}\frac{\text{x}}{\sqrt{{\text{1 + x}}^{\text{2}}}}\\ \text{= y’ = L}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.17

$MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeyEaiaabccacaqG9aGaaeiiaiaabgea caqG4bGaaGPaVlaaykW7caaMc8UaaCzcaiaaxMaacaaMc8UaaGPaVl aaxMaacaaMc8UaaGPaVlaaykW7caqG6aGaaeiiaiaaxMaacaWLjaGa aeiiaiaabIhacaqG5bGaae4jaiaabccacaqG9aGaaeiiaiaabMhaca WLjaGaaCzcamaabmaabaGaaeiEaiaabccacqGHGjsUcaqGGaGaaeim aaGaayjkaiaawMcaaaaaaa@BCAB@$

Ans.

$MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabMhacaqGGaGaaeypaiaabccacaqGbbGaaeiEaaqaaiaa bseacaqGPbGaaeOzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaae iDaiaabMgacaqGHbGaaeiDaiaabMgacaqGUbGaae4zaiaabccacaqG 3bGaaeOlaiaabkhacaqGUaGaaeiDaiaab6cacaqGGaGaaeiEaiaabY cacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0baabaGa aeyEaiaabEcacaqGGaGaaeypaiaabccacaqGbbaabaGaae4uaiaabw hacaqGIbGaae4CaiaabshacaqGPbGaaeiDaiaabwhacaqG0bGaaeyA aiaab6gacaqGNbGaaeiiaiaabAhacaqGHbGaaeiBaiaabwhacaqGLb Gaae4CaiaabccacaqGVbGaaeOzaiaabccacaqG5bGaae4jaiaabcca caqGPbGaaeOBaiaabccacaqGmbGaaeOlaiaabIeacaqGUaGaae4uai aab6cacaqGGaGaae4BaiaabAgacaqGGaGaae4zaiaabMgacaqG2bGa aeyzaiaab6gacaqGGaGaaeizaiaabMgacaqGMbGaaeOzaiaabwgaca qGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabggacaqGSbGaaeiiaaqa aiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUb GaaeiiaiaabIhacaqG5bGaae4jaiaabccacaqG9aGaaeiiaiaabMha caqGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaa qaaiaabYeacaqGUaGaaeisaiaab6cacaqGtbGaaeOlaiaabccacaqG 9aGaaeiiaiaabIhacaqG5bGaae4jaiaabccaaeaacaWLjaGaaeypai aabccacaqG4bWaaeWaaeaacaqGbbaacaGLOaGaayzkaaaabaGaaCzc aiaab2dacaqGGaGaaeyqaiaabIhacaqGGaGaaeypaiaabccacaqG5b Gaaeiiaiaab2dacaqGGaGaaeOuaiaab6cacaqGibGaaeOlaiaabofa caqGUaaabaGaaeivaiaabIgacaqG1bGaae4CaiaabYcacaqG0bGaae iAaiaabwgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqG GaGaaeOzaiaabwhacaqGUbGaae4yaiaabshacaqGPbGaae4Baiaab6 gacaqGGaGaaeyAaiaabohacaqGGaGaaeiDaiaabIgacaqGLbGaaeii aiaabohacaqGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUb Gaaeiiaiaab+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabcca caqGJbGaae4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Bai aab6gacaqGKbGaaeyAaiaab6gacaqGNbGaaeiiaaqaaiaabsgacaqG PbGaaeOzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabM gacaqGHbGaaeiBaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiD aiaabMgacaqGVbGaaeOBaiaab6caaaaa@0A6F@ $

Q.18

$MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeyEaiaabccacaqG9aGaaeiiaiaabIha caqGGaGaae4CaiaabMgacaqGUbGaaeiEaiaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaeOoaiaabccacaqGGaGaaeiEaiaabMhacaqGNaGa aeiiaiaab2dacaqGGaGaaeyEaiaabccacaqGRaGaaeiiaiaabIhaca qGGaWaaOaaaeaacaqG4bWaaWbaaSqabeaacaqGYaaaaOGaaeiiaiaa bobicaqGGaGaaeyEamaaCaaaleqabaGaiGgGbkdaaaaabeaakiaayk W7caaMc8+aaeWaaeaacaqG4bGaeyiyIKRaaeimaiaaykW7caqGHbGa aeOBaiaabsgacaqGGaGaaeiEaiaabccacaqG+aGaaeiiaiaabMhaca qGGaGaae4BaiaabkhacaqGGaGaaeiEaiaabccacaqG8aGaaeiiaiaa bobicaqG5baacaGLOaGaayzkaaaaaaa@D43D@ $

Ans.

$MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabMhacaqGGaGaaeypaiaabccacaqG4bGaae4CaiaabMga caqGUbGaaeiiaiaabIhaaeaacaqGebGaaeyAaiaabAgacaqGMbGaae yzaiaabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabshacaqG PbGaaeOBaiaabEgacaqGGaGaae4Daiaab6cacaqGYbGaaeOlaiaabs hacaqGUaGaaeiiaiaabIhacaqGSaGaaeiiaiaabEhacaqGLbGaaeii aiaabEgacaqGLbGaaeiDaaqaaiaabMhacaqGNaGaaeiiaiaab2daca qGGaGaaeiEaiaabogacaqGVbGaae4CaiaabccacaqG4bGaaeiiaiaa bUcacaqGGaGaae4CaiaabMgacaqGUbGaaeiEaaqaaiaabofacaqG1b GaaeOyaiaabohacaqG0bGaaeyAaiaabshacaqG1bGaaeiDaiaabMga caqGUbGaae4zaiaabccacaqG2bGaaeyyaiaabYgacaqG1bGaaeyzai aabohacaqGGaGaae4BaiaabAgacaqGGaGaaeyEaiaabEcacaqGGaGa aeyAaiaab6gacaqGGaGaaeitaiaab6cacaqGibGaaeOlaiaabofaca qGUaGaaeiiaiaab+gacaqGMbGaaeiiaiaabEgacaqGPbGaaeODaiaa bwgacaqGUbGaaeiiaiaabsgacaqGPbGaaeOzaiaabAgacaqGLbGaae OCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiBaiaabccaaeaa caqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBai aabccacaqG4bGaaeyEaiaabEcacaqGGaGaaeypaiaabccacaqG5bGa aeiiaiaabUcacaqGGaGaaeiEamaakaaabaGaaeiEamaaCaaaleqaba GaaeOmaaaakiaabccacaqGtaIaaeiiaiaabMhadaahaaWcbeqaaiaa bkdaaaaabeaakiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zai aabwgacaqG0baabaGaaeitaiaab6cacaqGibGaaeOlaiaabofacaqG UaGaaeiiaiaab2dacaqGGaGaaeiEaiaabMhacaqGNaGaaeiiaiaab2 dacaqGGaGaaeiEamaabmaabaGaaeiEaiaabogacaqGVbGaae4Caiaa bIhacaqGGaGaae4kaiaabccacaqGZbGaaeyAaiaab6gacaqG4baaca GLOaGaayzkaaaabaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG GaGaaeiiaiaabccacaqGGaGaaeiiaiaab2dacaqGGaGaaeiEamaaCa aaleqabaGaaeOmaaaakiaabogacaqGVbGaae4CaiaabIhacaqGGaGa ae4kaiaabccacaqG4bGaae4CaiaabMgacaqGUbGaaeiEaaqaaiaabk facaqGUaGaaeisaiaab6cacaqGtbGaaeOlaiaabccacaqG9aGaaeii aiaabIhacaqGZbGaaeyAaiaab6gacaqG4bGaaeiiaiaabUcacaqGGa GaaeiEamaakaaabaGaaeiEamaaCaaaleqabaGaaeOmaaaakiaabcca caqGtaIaaeiiamaabmaabaGaaeiEaiaabohacaqGPbGaaeOBaiaabI haaiaawIcacaGLPaaadaahaaWcbeqaaiaabkdaaaaabeaaaOqaaiaa bccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaae iiaiaabccacaqG9aGaaeiiaiaabIhacaqGZbGaaeyAaiaab6gacaqG 4bGaaeiiaiaabUcacaqGGaGaaeiEamaaCaaaleqabaGaaeOmaaaakm aakaaabaGaaeymaiaabccacaqGtaIaaeiiaiaabohacaqGPbGaaeOB amaaCaaaleqabaGaaeOmaaaakiaabIhaaSqabaaakeaacaqGGaGaae iiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG GaGaaeypaiaabccacaqG4bGaae4CaiaabMgacaqGUbGaaeiEaiaabc cacaqGRaGaaeiiaiaabIhadaahaaWcbeqaaiaabkdaaaGccaqGJbGa ae4BaiaabohacaqG4baabaGaae4uaiaab+gacaqGSaGaaeiiaiaabY eacaqGUaGaaeisaiaab6cacaqGtbGaaeOlaiaabccacaqG9aGaaeii aiaabkfacaqGUaGaaeisaiaab6cacaqGtbGaaeOlaaqaaiaabsfaca qGObGaaeyDaiaabohacaqGSaGaaeiDaiaabIgacaqGLbGaaeiiaiaa bEgacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabAgacaqG1bGaae OBaiaabogacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabMgacaqG ZbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGZbGaae4BaiaabY gacaqG1bGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGVbGaaeOz aiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4yaiaab+gacaqGYb GaaeOCaiaabwgacaqGZbGaaeiCaiaab+gacaqGUbGaaeizaiaabMga caqGUbGaae4zaiaabccaaeaacaqGKbGaaeyAaiaabAgacaqGMbGaae yzaiaabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqG GaGaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6 gacaqGUaaaaaa@850D@ $

Q.19

$MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeiEaiaabMhacaqGGaGaaeypaiaabcca caqGSbGaae4BaiaabEgacaqG5bGaaeiiaiaabUcacaqGGaGaae4qai aabccacaqG6aGaaeiiaiaabMhacaqGNaGaaeiiaiaab2dacaqGGaWa aSaaaeaacaqG5bWaaWbaaSqabeaacaqGYaaaaaGcbaGaaeymaiaabc cacaqGtaIaaeiiaiaabIhacaqG5baaamaabmaabaGaaeiEaiaabMha cqGHGjsUcaqGWaaacaGLOaGaayzkaaaaaaa@B600@ $

Ans.

$MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabIhacaqG5bGaaeiiaiaab2dacaqGGaGaaeiBaiaab+ga caqGNbGaaeyEaiaabccacaqGRaGaaeiiaiaaboeaaeaacaqGebGaae yAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOBaiaabshacaqG PbGaaeyyaiaabshacaqGPbGaaeOBaiaabEgacaqGGaGaae4Daiaab6 cacaqGYbGaaeOlaiaabshacaqGUaGaaeiiaiaabIhacaqGSaGaaeii aiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiaabIhaca qG5bGaae4jaiaabccacaqGRaGaaeiiaiaabMhacaqGGaGaaeypamaa laaabaGaaeymaaqaaiaabMhaaaGaaeyEaiaabEcacaqGGaGaae4kai aabccacaqGWaaabaGaaeiEaiaabMhacaqG5bGaae4jaiaabccacaqG RaGaaeiiaiaabMhadaahaaWcbeqaaiaabkdaaaGccaqGGaGaaeypai aabccacaqG5bGaae4jaaqaaiaabccacaqGGaGaaeiiaiaabccacaqG GaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG5bWaaWbaaSqabe aacaqGYaaaaOGaaeiiaiaab2dacaqGGaGaaeyEaiaabEcacaqGGaGa ae4eGiaabccacaqG4bGaaeyEaiaabMhacaqGNaaabaGaaeiiaiaabc cacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeii aiaabccacaqGGaGaaeiiaiaabccacaqG9aGaaeiiaiaabMhacaqGNa WaaeWaaeaacaqGXaGaaeiiaiaabobicaqGGaGaaeiEaiaabMhaaiaa wIcacaGLPaaaaeaacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabc cacaqGGaGaaeiiaiaabccacaqGGaGaaeyEaiaabEcacaqGGaGaaeyp aiaabccadaWcaaqaaiaabMhadaahaaWcbeqaaiaabkdaaaaakeaada qadaqaaiaabgdacaqGGaGaae4eGiaabccacaqG4bGaaeyEaaGaayjk aiaawMcaaaaaaeaacaqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabs hacaqGObGaaeyzaiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOB aiaabccacaqGMbGaaeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVb GaaeOBaiaabccacaqGPbGaae4CaiaabccacaqG0bGaaeiAaiaabwga caqGGaGaae4Caiaab+gacaqGSbGaaeyDaiaabshacaqGPbGaae4Bai aab6gacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGa aeiiaiaabogacaqGVbGaaeOCaiaabkhacaqGLbGaae4Caiaabchaca qGVbGaaeOBaiaabsgacaqGPbGaaeOBaiaabEgacaqGGaaabaGaaeiz aiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG0b GaaeyAaiaabggacaqGSbGaaeiiaiaabwgacaqGXbGaaeyDaiaabgga caqG0bGaaeyAaiaab+gacaqGUbGaaeOlaaaaaa@F6E3@ $

Q.20

$MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeyEaiaabccacaqGtaIaaeiiaiaaboga caqGVbGaae4CaiaabMhacaqGGaGaaeypaiaabccacaqG4bGaaeiiai aabccacaqGGaGaaeiiaiaabQdacaqGGaGaaeiiaiaabccacaqGGaWa aeWaaeaacaqG5bGaae4CaiaabMgacaqGUbGaaeyEaiaabccacaqGRa GaaeiiaiaabogacaqGVbGaae4CaiaabMhacaqGGaGaae4kaiaabcca caqG4baacaGLOaGaayzkaaGaaeyEaiaabEcacaqGGaGaaeypaiaabc cacaqG5baaaaa@BC7B@ $

Ans.

$MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadMha cqGHsislciGGJbGaai4BaiaacohacaWG5bGaeyypa0JaamiEaiaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaac6cacaGGUaGaaiOl amaabmaabaGaamyAaaGaayjkaiaawMcaaaqaaiaabseacaqGPbGaae OzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqG HbGaaeiDaiaabMgacaqGUbGaae4zaiaabccacaqG3bGaaeOlaiaabk hacaqGUaGaaeiDaiaab6cacaqGGaGaaeiEaiaabYcacaqGGaGaae4D aiaabwgacaqGGaGaae4zaiaabwgacaqG0baabaGaaGPaVlaaykW7ca qG5bGaae4jaiabgUcaRiGacohacaGGPbGaaiOBaiaadMhacaGGUaGa amyEaiaacEcacqGH9aqpcaaIXaaabaGaamyEaiaacEcadaqadaqaai aaigdacqGHRaWkciGGZbGaaiyAaiaac6gacaWG5baacaGLOaGaayzk aaGaeyypa0JaaGymaaqaaiaaxMaacaWLjaGaamyEaiaacEcacqGH9a qpdaWcaaqaaiaaigdaaeaacaaIXaGaey4kaSIaci4CaiaacMgacaGG UbGaamyEaaaaaeaacaqGtbGaaeyDaiaabkgacaqGZbGaaeiDaiaabM gacaqG0bGaaeyDaiaabshacaqGPbGaaeOBaiaabEgacaqGGaGaaeOD aiaabggacaqGSbGaaeyDaiaabwgacaqGZbGaaeiiaiaab+gacaqGMb GaaeiiaiaabMhacaqGNaGaaeiiaiaabMgacaqGUbGaaeiiaiaabYea caqGUaGaaeisaiaab6cacaqGtbGaaeOlaiaabccacaqGVbGaaeOzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGKbGa aeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOBaiaabshaca qGPbGaaeyyaiaabYgacaqGGaaabaGaaeyzaiaabghacaqG1bGaaeyy aiaabshacaqGPbGaae4Baiaab6gacaqGGaWaaeWaaeaacaqG5bGaae 4CaiaabMgacaqGUbGaaeyEaiaabccacaqGRaGaaeiiaiaabogacaqG VbGaae4CaiaabMhacaqGGaGaae4kaiaabccacaqG4baacaGLOaGaay zkaaGaaeyEaiaabEcacaqGGaGaaeypaiaabccacaqG5bGaaeilaiaa bccacaqG3bGaaeyzaiaabccacaqGNbGaaeyzaiaabshaaeaadaqada qaaiaadMhaciGGZbGaaiyAaiaac6gacaWG5bGaey4kaSIaci4yaiaa c+gacaGGZbGaamyEaiabgUcaRiaadIhaaiaawIcacaGLPaaacaWG5b Gaai4jaiabg2da9maabmaabaGaamyEaiGacohacaGGPbGaaiOBaiaa dMhacqGHRaWkciGGJbGaai4BaiaacohacaWG5bGaey4kaSIaamiEaa GaayjkaiaawMcaamaalaaabaGaaGymaaqaamaabmaabaGaaGymaiab gUcaRiGacohacaGGPbGaaiOBaiaadMhaaiaawIcacaGLPaaaaaaaba GaaCzcaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7 cqGH9aqpdaqadaqaaiaadMhaciGGZbGaaiyAaiaac6gacaWG5bGaey 4kaSIaci4yaiaac+gacaGGZbGaamyEaiabgUcaRiaadMhacqGHsisl ciGGJbGaai4BaiaacohacaWG5baacaGLOaGaayzkaaWaaSaaaeaaca aIXaaabaWaaeWaaeaacaaIXaGaey4kaSIaci4CaiaacMgacaGGUbGa amyEaaGaayjkaiaawMcaaaaaaeaacaWLjaGaaCzcaiaaxMaacaWLja GaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaadaWadaqaaiaabAeacaqG YbGaae4Baiaab2gacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabs hacaqGPbGaae4Baiaab6gadaqadaqaaiaabMgaaiaawIcacaGLPaaa aiaawUfacaGLDbaaaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGPaVl aaykW7caaMc8UaaGPaVlabg2da9maabmaabaGaamyEaiGacohacaGG PbGaaiOBaiaadMhacqGHRaWkcaWG5baacaGLOaGaayzkaaWaaSaaae aacaaIXaaabaWaaeWaaeaacaaIXaGaey4kaSIaci4CaiaacMgacaGG UbGaamyEaaGaayjkaiaawMcaaaaaaeaacaWLjaGaaCzcaiaaxMaaca WLjaGaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9iaadMhadaqadaqa aiGacohacaGGPbGaaiOBaiaadMhacqGHRaWkcaaIXaaacaGLOaGaay zkaaWaaSaaaeaacaaIXaaabaWaaeWaaeaacaaIXaGaey4kaSIaci4C aiaacMgacaGGUbGaamyEaaGaayjkaiaawMcaaaaaaeaacaWLjaGaaC zcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9iaa dMhacqGH9aqpcaWGsbGaaiOlaiaadIeacaGGUaGaam4uaiaac6caae aacaqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabshacaqGObGaaeyz aiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMb GaaeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabcca caqGPbGaae4CaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4Cai aab+gacaqGSbGaaeyDaiaabshacaqGPbGaae4Baiaab6gacaqGGaGa ae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabogaca qGVbGaaeOCaiaabkhacaqGLbGaae4CaiaabchacaqGVbGaaeOBaiaa bsgacaqGPbGaaeOBaiaabEgacaqGGaaabaGaaeizaiaabMgacaqGMb GaaeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabgga caqGSbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAai aab+gacaqGUbGaaeOlaaaaaa@D4B7@ $

Q.21

\begin{array}{l}\text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ x+y=ta{n}^{–1}y:y^{2}y‘+y^2+1=0\end{array}

Ans.

\begin{array}{l}x+y={\tan }^{–1}y\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{1+y’}=\left(\frac{1}{1+y^2}\right)y‘\\ \left(\text{1}+\text{y’}\right)\left(1+y^2\right)=y‘\\ 1+y^2+y‘+y‘y^2=y‘\\ 1+y^2+y‘y^2=0\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.22

\begin{array}{l}\text{Verify that the given functions}\left(\text{explicit or implicit}\right)\text{is a solution of the corresponding differential equation:}\\ y=\sqrt{a^2-x^2}x\in \left(-a,a\right):x+y\frac{dy}{dx}=0\left(y\ne 0\right)\end{array}

Ans.

\begin{array}{l}y=\sqrt{a^2-x^2}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{dy}{dx}=\frac{-x}{\sqrt{a^2-x^2}}\\ =\frac{-x}{y}\\ \text{Substituting the value of}\frac{dy}{dx}\text{in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation}x+y\frac{dy}{dx}=0,\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x+y\frac{dy}{dx}\\ =x+y\left(\frac{-x}{y}\right)\\ =x-x\\ =0=R.H.S.\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

Q.23 The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

Ans.

Since, number of constants in a differential equation of order n is equal to its order i.e., n.
Thus, number of arbitrary constants in a differential equation of fourth order are 4.
Thus, option D is correct.

Q.24 The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3 (B) 2 (C) 1 (D) 0

Ans.

In a particular solution, there are no arbitrary constants.

∴ Number of arbitrary constants = 0
Thus, option D is correct.

Q.25 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

\frac{x}{a}+\frac{y}{b}=1

Ans.

\begin{array}{l}\frac{x}{a}+\frac{y}{b}=1\\ \text{Differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{1}{a}+\frac{1}{b}\frac{dy}{dx}=0\\ \text{Againg,}\text{differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{1}{b}\frac{d^{2}y}{d{x}^2}=0\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=0\Rightarrow y‘‘=0\\ \text{Hence, the required differential equation is y”=0}\text{.}\end{array}

Q.26 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y^2=a\left(b^2-x^2\right)

Ans.

$\begin{array}{l}{\mathrm{y}}^2=\mathrm{a}({\mathrm{b}}^2-{\mathrm{x}}^2)\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a}(0-2\mathrm{x})\\ 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=-2\mathrm{ax}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Againg},\mathrm{\hspace{0.17em}}\mathrm{differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ 2\{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\}=-2\mathrm{a}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-\mathrm{\hspace{0.17em}}\mathrm{a}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{ii}\right)\\ \mathrm{Putting}\mathrm{value}\mathrm{of}–\mathrm{a}\mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{to}\mathrm{equation}\left(\mathrm{i}\right),\hspace{0.17em}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=2\{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\}\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+\mathrm{xy}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\\ \mathrm{xyy}‘‘+\mathrm{x}{(\mathrm{y}‘)}^2-\mathrm{yy}‘=0\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{xyy}”+\mathrm{x}{(\mathrm{y}‘)}^{\mathrm{2}}–\mathrm{yy}‘\; =\; 0\mathrm{.}\end{array}$

Q.27 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y=a{e}^{3x}+b{e}^{-2x}

Ans.

\begin{array}{l}y=a{e}^{3x}+b{e}^{-2x}\mathrm{\dots }\left(i\right)\\ \text{Differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ y‘=3a{e}^{3x}-2b{e}^{-2x}\mathrm{\dots }\left(ii\right)\\ \text{Againg,}\text{differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ y‘‘=9a{e}^{3x}+4b{e}^{-2x}\mathrm{\dots }\left(iii\right)\\ \text{Adding equation}\left(\text{ii}\right)\text{and 2×equation}\left(\text{i}\right)\text{,}\text{we get}\\ y‘+2y=3a{e}^{3x}-2b{e}^{-2x}+2\left(a{e}^{3x}+b{e}^{-2x}\right)\\ y‘+2y=3a{e}^{3x}-2b{e}^{-2x}+2a{e}^{3x}+2b{e}^{-2x}\\ y‘+2y=5a{e}^{3x}\Rightarrow a{e}^{3x}=\frac{y‘+2y}{5}\\ \text{Substracting equation}\left(\text{ii}\right)\text{from 3×equation}\left(\text{i}\right)\text{,}\text{we get}\\ 3y-y‘=3a{e}^{3x}+3b{e}^{-2x}-\left(3a{e}^{3x}-2b{e}^{-2x}\right)\\ 3y-y‘=3a{e}^{3x}+3b{e}^{-2x}-3a{e}^{3x}+2b{e}^{-2x}\\ 3y-y‘=5b{e}^{-2x}\\ b{e}^{-2x}=\frac{3y-y‘}{5}\\ {\text{Putting the values of ae}}^{\text{3x}}\text{and}{\text{be}}^{\text{-2x}}\text{in equation}\left(\text{iii}\right)\text{, we get}\\ y‘‘=9\left(\frac{y‘+2y}{5}\right)+4\left(\frac{3y-y‘}{5}\right)\\ \Rightarrow 5y‘‘=9y‘+18y+12y-4y‘\\ \Rightarrow 5y‘‘=5y‘+30y\\ \Rightarrow y‘‘-y‘-6y=0\\ \text{Hence, the required differential equation is y”–y’–6y = 0}\text{.}\end{array}

Q.28 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

{\text{y=e}}^{\text{2x}}\left(\text{a+b}\text{x}\right)

Ans.

\begin{array}{l}y=e^{2x}\left(a+bx\right)\\ \text{Differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ y‘=e^{2x}\left(0+b\right)+2e^{2x}\left(a+bx\right)\\ y‘=b{e}^{2x}+2y\\ \Rightarrow y‘-2y=b{e}^{2x}\mathrm{\dots }\left(i\right)\\ \text{Againg,}\text{differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ y‘‘-2y‘=2b{e}^{2x}\mathrm{\dots }\left(ii\right)\\ \text{Dividing equation}\left(\text{ii}\right)\text{by equation}\left(\text{i}\right)\text{, we get}\\ \frac{y‘‘-2y‘}{y‘-2y}=\frac{2b{e}^{2x}}{b{e}^{2x}}\\ \frac{y‘‘-2y‘}{y‘-2y}=2\\ \Rightarrow y‘‘-2y‘=2\left(y‘-2y\right)\\ \Rightarrow y‘‘-2y‘-2y‘+4y=0\\ \Rightarrow y‘‘-4y‘+4y=0\\ \text{Hence, the required differential equation is y” –4y’ + 4y = 0}\text{.}\end{array}

Q.29 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y=e^x\left(acosx+bsinx\right)

Ans.

\begin{array}{l}y=e^x\left(a\cos x+b\sin x\right)\mathrm{\dots }\left(i\right)\\ \text{Differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ y‘=e^x\left(-a\sin x+b\cos x\right)+e^x\left(a\cos x+b\sin x\right)\\ y‘=e^x\left(-a\sin x+b\cos x\right)+y\left[\text{From}\text{equation}\left(\text{i}\right)\right]\\ \Rightarrow y‘-y=e^x\left(-a\sin x+b\cos x\right)\mathrm{\dots }\left(ii\right)\\ \text{Againg,}\text{differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ y‘‘-y‘=e^x\left(-a\cos x-b\sin x\right)+e^x\left(-a\sin x+b\cos x\right)\\ y‘‘-y‘=-y+y‘-y\left[\text{From equation}\left(ii\right)\right]\\ y‘‘-2y‘+2y=0\\ \text{Hence, the required differential equation is}y‘‘-2y‘+2y=0.\end{array}

Q.30 Form the differential equation of the family of circles touching the y-axis at origin.

Ans.

The centre of the circle touching y-axis at origin lies on x-axis. Let the centre of circle be (a, 0).
Radius of circle will be ‘a’ because circle touches y-axis at origin. So, the equation of circle with centre (a, 0) and radius ‘a’ is as follows
(x – a)² + y² = a²
x² – 2ax + a² + y² = a²
x² + y² = 2ax … (i)

Differentiating equation (i), w.r.t. x, we get

2x + 2yy’ = 2a

x + yy’ = a

Putting value of a in equation (i), we get

x² + y² = 2(x + yy’)x
= 2x² + 2xyy’

y² = x² + 2xyy’

Thus, the required differential equation is 2xyy’ + x² = y².

Q.31 Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Ans.

The equation of parabolas having vertex at origin and axis along positive y-axis is as follows
x² = 4ay …(i)
Differentiating w.r.t. x, we get
2x = 4ay’
or (2x/y’) = 4a
Putting value of 4a in equation (i), we get
x² = (2x/y’)y

x²y’ = 2xy

or xy’ = 2y
or xy’ – 2y = 0

Thus, the required differential equation is
xy’ – 2y = 0.

Q.32 Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Ans.

\begin{array}{l}\text{The}\text{equation of the family of ellipses having foci on}\\ \text{y-axis and centre at origin is as follows}\\ \frac{{\text{x}}^{\text{2}}}{{\text{b}}^{\text{2}}}+\frac{{\text{y}}^{\text{2}}}{{\text{a}}^{\text{2}}}=\text{1}\dots \left(\text{i}\right)\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{\text{2x}}{{\text{b}}^{\text{2}}}+\frac{\text{2yy’}}{{\text{a}}^{\text{2}}}=\text{0}\\ \frac{\text{x}}{{\text{b}}^{\text{2}}}+\frac{\text{yy’}}{{\text{a}}^{\text{2}}}=\text{0}\dots \left(\text{ii}\right)\\ \text{Again,}\text{differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{\text{1}}{{\text{b}}^{\text{2}}}+\frac{\text{1}}{{\text{a}}^{\text{2}}}\left(\text{yy”+y’y’}\right)=\text{0}\\ \frac{\text{1}}{{\text{b}}^{\text{2}}}+\frac{\text{1}}{{\text{a}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}=\text{0}\\ \frac{\text{1}}{{\text{b}}^{\text{2}}}=-\frac{\text{1}}{{\text{a}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}\\ \text{Substituting the value of}\frac{\text{1}}{{\text{b}}^{\text{2}}}\text{in equation}\left(\text{ii}\right)\text{, we get}\\ \text{x}\left[-\frac{\text{1}}{{\text{a}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}\right]+\frac{\text{yy’}}{{\text{a}}^{\text{2}}}=\text{0}\\ \Rightarrow -x\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}+yy‘=0\\ \Rightarrow -xyy‘‘-x{\left(y‘\right)}^2+yy‘=0\\ \Rightarrow xyy‘‘+x{\left(y‘\right)}^2-yy‘=0\\ \text{Thus, the required differential equation is xyy” + x}{\left(\text{y’}\right)}^{\text{2}}\text{–yy’ = 0}\text{.}\end{array}

Q.33 Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Ans.

\begin{array}{l}\text{The differential equation of the family of hyperbolas having foci}\\ \text{on x-axis and centre at}\text{origin}\text{‹}\text{is as follows}\\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\dots \left(\text{i}\right)\\ \text{Differentiating}\text{‹}\text{w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{\text{2x}}{{\text{a}}^{\text{2}}}-\frac{\text{2yy’}}{{\text{b}}^{\text{2}}}=\text{0}\\ \frac{\text{x}}{{\text{a}}^{\text{2}}}-\frac{\text{yy’}}{{\text{b}}^{\text{2}}}=\text{0}\dots \left(\text{ii}\right)\\ \text{Again,}\text{differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{\text{1}}{{\text{a}}^{\text{2}}}-\frac{\text{1}}{{\text{b}}^{\text{2}}}\left(\text{yy”+y’y’}\right)=\text{0}\\ \frac{\text{1}}{{\text{a}}^{\text{2}}}-\frac{\text{1}}{{\text{b}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}=\text{0}\\ \frac{\text{1}}{{\text{a}}^{\text{2}}}=\frac{\text{1}}{{\text{b}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}\\ \text{Substituting the value of}\frac{\text{1}}{{\text{a}}^{\text{2}}}\text{in equation}\left(\text{ii}\right)\text{, we get}\\ \text{x}\left[\frac{\text{1}}{{\text{b}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}\right]-\frac{\text{yy’}}{{\text{b}}^{\text{2}}}=\text{0}\\ \Rightarrow x\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}-yy‘=0\\ \Rightarrow xyy‘‘+x{\left(y‘\right)}^2-yy‘=0\\ \text{Thus, the required differential equation is xyy” + x}{\left(\text{y’}\right)}^{\text{2}}\text{–yy’=0}\text{.}\end{array}

Q.34 Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Ans.

The centre of the circle lies on y-axis. Let the centre of circle be (0, b).
So, the equation of circle with centre (0, b) and radius 3 is as follows
x² + (y – b)² = 3² … (i)

\begin{array}{c}\text{Differentiating equation}\left(\text{i}\right),\text{w}.\text{r}.\text{t}.\text{x},\text{we get}\\ \text{2x}+\text{2}\left(\text{y}–\text{b}\right)\text{y}’=0\\ \text{x}+\left(\text{y}–\text{b}\right)\text{y}’=0\\ \left(\text{y}-\text{b}\right)=\frac{-\text{x}}{\text{y}’}\\ \end{array}

Putting value of a in equation ( i ), we get

\begin{array}{c}{\text{x}}^{\text{2}}+{\left(\frac{-\text{x}}{\text{y}’}\right)}^{\text{2}}=9\\ x^{2}y{‘}^2+x^2=9y{‘}^2\\ {\left(y‘\right)}^2\left(x^2-9\right)+x^2=0\\ This\text{is the required differential equation}\text{.}\end{array}

Q.35

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{differential}\mathrm{equations}\mathrm{has}\\ \mathrm{y}={\mathrm{c}}_{\mathrm{1}}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_{\mathrm{2}}{\mathrm{e}}^{–\mathrm{x}}\mathrm{as}\mathrm{the}\mathrm{general}\mathrm{solution}?\\ \left(\mathrm{A}\right)\mathrm{\hspace{0.17em}}\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{2}}}+\mathrm{y}=\; 0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{B}\right)\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{2}}}-\mathrm{y}=\; 0\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\\ \left(\mathrm{C}\right)\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{2}}}+\; 1\; =\; 0\left(\mathrm{D}\right)\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{2}}}-1=\; 0\end{array}$

Ans.

$\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{x}}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}-{\mathrm{c}}_2\mathrm{\hspace{0.17em}}{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{Again},\mathrm{‹}\mathrm{differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^2}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_2\mathrm{\hspace{0.17em}}{\mathrm{e}}^{-\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{y}\\ \mathrm{or}\mathrm{\hspace{0.17em}}\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^2}-\mathrm{y}=\mathrm{0}\\ \mathrm{Thus},‹‹\mathrm{option}\mathrm{B}\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.36

\begin{array}{l}\text{Which of the following differential equations has y = x as}\\ \text{one of its particular solution?}\\ \left(\text{A}\right)\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=x\left(\text{B}\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+xy=x\\ \left(\text{C}\right)\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0\left(\text{D}\right)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+xy=0\end{array}

Ans.

\begin{array}{l}\text{We have y = x}\text{…}\left(\text{i}\right)\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{dy}{dx}=1\mathrm{\dots }\left(\text{ii}\right)\\ \text{Again,}\text{‹}\text{differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{d^{2}y}{d{x}^2}=0\mathrm{\dots }\left(\text{iii}\right)\\ \text{Putting values of y,}\frac{dy}{dx}and\frac{d^{2}y}{d{x}^2}\text{in each given options,}\\ \left(\text{A}\right)\text{L}\text{.H}\text{.S}.=\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy\\ =0-x^2\left(1\right)+x\left(x\right)\\ =0\ne \text{R}\text{.H}\text{.S}\text{.}\\ \text{Thus, it is not correct option}\text{.}\\ \left(\text{B}\right)\text{L}\text{.H}\text{.S}\text{.}=\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+xy\\ =0-x\left(x\right)+x\left(x\right)\\ =0\ne \text{R}\text{.H}\text{.S}\text{.}\\ \text{Thus, it is not correct option}\text{.}\\ \left(\text{C}\right)\text{L}\text{.H}\text{.S}\text{.}=\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy\\ =0-x^2\left(1\right)+x\left(x\right)\\ =0=\text{R}\text{.H}\text{.S}\text{.}\\ \text{Thus, it is correct option}\text{.}\\ \text{Hence, the correct solution is option C}\text{.}\end{array}

Q.37 For the differential equations, find the general solution:

\frac{dy}{dx}=\frac{1-cosx}{1+cosx}

Ans.

\begin{array}{l}\text{The given differential equation is:}\\ \frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}\\ =\frac{2{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}\left[Cos2x=1-2{\sin }^{2}x=2{\cos }^{2}x-1\right]\\ ={\tan }^2\frac{x}{2}\\ ={\sec }^2\frac{x}{2}-1\\ \text{Separating the variables, we get}\\ \text{dy}=\left({\sec }^2\frac{x}{2}-1\right)dx\\ \text{Integrating both sides of the above equation, we get}\\ {\displaystyle \int dy}={\displaystyle \int \left({\sec }^2\frac{x}{2}-1\right)}dx\\ y=2\tan \frac{x}{2}-x+C\\ \text{This is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}

Q.38 For the differential equations, find the general solution:

\frac{dy}{dx}=\sqrt{4-y^2}\left(-2<y<2\right)

Ans.

\begin{array}{l}\text{The given differential equation is:}\\ \frac{dy}{dx}=\sqrt{4-y^2}\\ \text{Separating}\text{the variables, we get}\\ \frac{dy}{\sqrt{4-y^2}}=dx\\ \text{Integrating both sides of above equation,we get}\\ {\displaystyle \int \frac{dy}{\sqrt{4-y^2}}}={\displaystyle \int dx}\\ {\sin }^{-1}\left(\frac{y}{2}\right)=x+C\\ \frac{y}{2}=\sin \left(x+C\right)\\ y=2\sin \left(x+C\right)\\ \text{This is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}

Q.39 For the differential equations, find the general solution:

\frac{dy}{dx}+y=1\left(y\ne 1\right)

Ans.

\begin{array}{l}\text{The given differential equation is:}\\ \frac{dy}{dx}+y=1\\ \frac{dy}{dx}=1-y\\ Separating\text{the variables, we get}\\ \frac{dy}{1-y}=dx\\ Integrating\text{both sides of the given differential equation, we get}\\ {\displaystyle \int \frac{dy}{1-y}}={\displaystyle \int dx}\\ -\log \left(1-y\right)=x+\log C\\ \Rightarrow -\log \left(1-y\right)-\log C=x\\ \Rightarrow \log C\left(1-y\right)=-x\\ \Rightarrow C\left(1-y\right)=e^{-x}\\ \Rightarrow 1-y=\frac{1}{C}{e}^{-x}\\ \Rightarrow y=1-\frac{1}{C}{e}^{-x}\\ =1-A{e}^{-x}\left[WhereA=\frac{1}{C}\right]\\ This\text{is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}

Q.40 For the differential equations, find the general solution:

se{c}^{2}xtanydx+se{c}^{2}ytanxdy=0

Ans.

\begin{array}{l}\text{The given differential equation is}\\ {\text{sec}}^{\text{2}}\text{x}\text{tan}\text{y}{\text{dx + sec}}^{\text{2}}\text{y}\text{tanx}\text{dy = 0}\\ \text{Dividing by tan x tan y, we get}\\ \frac{{\sec }^{2}x\tan ydx+{\sec }^{2}y\tan xdy}{\text{tan x tan y}}=0\\ \Rightarrow \frac{{\sec }^{2}x}{\tan x}dx+\frac{{\sec }^{2}y}{\tan y}dy=0\\ \Rightarrow \frac{{\sec }^{2}x}{\tan x}dx=-\frac{{\sec }^{2}y}{\tan y}dy\\ \text{Integrating both sides of above equation, we get}\\ {\displaystyle \int \frac{{\sec }^{2}x}{\tan x}}dx=-{\displaystyle \int \frac{{\sec }^{2}y}{\tan y}}dy\\ \Rightarrow \text{log tanx = -log tany + log C}\\ \Rightarrow \text{log tanx + log tany = log C}\\ \Rightarrow \text{log tanx tany = log C}\\ \Rightarrow \text{tanx tany = C}\\ \text{This is the required solution of the given differential equation}\text{.}\end{array}

Q.41 For the differential equations, find the general solution:

\left(e^x+e^{–x}\right)dy-\left(e^x-e^{–x}\right)dx=0

Ans.

\begin{array}{l}The\text{given differential equation is:}\\ \left(e^x+e^{–x}\right)dy-\left(e^x-e^{–x}\right)dx=0\\ \left(e^x-e^{–x}\right)dx=\left(e^x+e^{–x}\right)dy\\ dy=\frac{\left(e^x-e^{–x}\right)}{\left(e^x+e^{–x}\right)}dx\\ Integrating\text{both sides of above equation, we get}\\ {\displaystyle \int dy}={\displaystyle \int \frac{\left(e^x-e^{–x}\right)}{\left(e^x+e^{–x}\right)}}dx\\ y=\log \left(e^x+e^{–x}\right)+C\\ This\text{is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}

Q.42 For the differential equations, find the general solution:

\frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right)

Ans.

\begin{array}{l}\text{The given differential equation is:}\\ \frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right)\\ \Rightarrow \frac{dy}{\left(1+y^2\right)}=\left(1+x^2\right)dx\\ \text{Integrating both sides, we get}\\ {\displaystyle \int \frac{dy}{\left(1+y^2\right)}}={\displaystyle \int \left(1+x^2\right)}dx\\ {\text{tan}}^{-1}y=x+\frac{x^3}{3}+C\\ \text{This is the general solution of given differential equation}\text{.}\end{array}

Q.43 For the differential equations, find the general solution:

ylogydx-xdy=0

Ans.

\begin{array}{l}The\text{given differential equation is:}\\ y\log ydx-xdy=0\\ \Rightarrow y\log ydx=xdy\\ \Rightarrow \frac{dy}{y\log y}=\frac{dx}{x}\\ \text{Integrating both sides, we get}\\ {\displaystyle \int \frac{dy}{y\log y}}={\displaystyle \int \frac{dx}{x}}\\ \Rightarrow \log \log y=\log x+\log C\\ \Rightarrow \log \log y=\log Cx\\ \Rightarrow \log y=Cx\\ \Rightarrow y=e^{Cx}\\ \text{This is the general solution of given differential equation}\text{.}\end{array}

Q.44 For the differential equations, find the general solution:

x^5\frac{dy}{dx}=-y^5

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \mathrm{\hspace{0.17em}}{\mathrm{x}}^5\frac{\mathrm{dy}}{\mathrm{dx}}=-{\mathrm{y}}^5\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{{\mathrm{y}}^5}=-\mathrm{\hspace{0.17em}}\frac{\mathrm{dx}}{{\mathrm{x}}^5}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\int \frac{\mathrm{dy}}{{\mathrm{y}}^5}\mathrm{\hspace{0.17em}}=-\int \frac{\mathrm{dx}}{{\mathrm{x}}^5}\mathrm{\hspace{0.17em}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int {\mathrm{y}}^{-5}\mathrm{dy}\mathrm{\hspace{0.17em}}=-\int {\mathrm{x}}^{-5}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\frac{{\mathrm{y}}^{-4}}{4}+\mathrm{C}‘=\frac{{\mathrm{x}}^{-4}}{4}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{x}}^{-4}+{\mathrm{y}}^{-4}=4\mathrm{C}‘=\mathrm{C}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.45 For the differential equation given below, find the general solution:

\frac{dy}{dx}=si{n}^{–1}x

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}={\sin }^{–1}\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{dy}={\sin }^{–1}\mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int \mathrm{dy}\mathrm{\hspace{0.17em}}=\int {\sin }^{–1}\mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}\mathrm{\hspace{0.17em}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}\mathrm{\hspace{0.17em}}=\int {\sin }^{–1}\mathrm{x}.1\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}={\sin }^{–1}\mathrm{x}\int 1\mathrm{\hspace{0.17em}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}{\sin }^{–1}\mathrm{x}\int 1\mathrm{\hspace{0.17em}}\mathrm{dx})\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\left({\sin }^{–1}\mathrm{x}\right).\mathrm{x}-\int \left(\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{x}\right)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}={\mathrm{xsin}}^{–1}\mathrm{x}-\int \frac{\mathrm{x}}{\sqrt{\mathrm{t}}}\times \mathrm{\hspace{0.17em}}\frac{\mathrm{dt}}{-2\mathrm{x}}+\mathrm{C}\left[\begin{array}{l}\mathrm{Let}\mathrm{}1-{\mathrm{x}}^2=\mathrm{t}\\ \Rightarrow \mathrm{\hspace{0.17em}}-2\mathrm{x}=\frac{\mathrm{dt}}{\mathrm{dx}}\end{array}\right]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}={\mathrm{xsin}}^{–1}\mathrm{x}+\frac{1}{2}\int {\mathrm{t}}^{-\frac{1}{2}}\mathrm{\hspace{0.17em}}\mathrm{dt}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}={\mathrm{xsin}}^{–1}\mathrm{x}+\frac{1}{2}\frac{{\mathrm{t}}^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}={\mathrm{xsin}}^{–1}\mathrm{x}+\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.46 For the differential equation given below, find the general solution:

e^{x}tanydx+\left(1-e^x\right)se{c}^{2}ydy=0

Ans.

$\begin{array}{l}\text{The given differential equation is:}\\ {\text{e}}^{\text{x}}\text{tan y}\text{dx}+(1-e^x){\text{sec}}^{2}ydy=0\\ \frac{e^x}{(1-e^x)}dx+\frac{{\sec }^{2}y}{\tan y}dy=0\left[\text{Dividing both sides by}(1-e^x)\text{tany}\right]\\ \text{Integrating both sides of above equation, we get}\\ \int \frac{e^x}{(1-e^x)}dx+\int \frac{{\sec }^{2}y}{\tan y}dy=C\\ \Rightarrow \int \frac{e^x}{t}\times \frac{dt}{-e^x}+\int \frac{{\sec }^{2}y}{z}\times \frac{dz}{{\sec }^{2}y}=\log C\left[\begin{array}{l}Let\text{t}=\text{1}-{\text{e}}^{\text{x}}andz=\tan y\\ \Rightarrow \frac{dt}{dx}=-e^{x}and\frac{dz}{dy}={\sec }^{2}y\end{array}\right]\\ \Rightarrow -\log t+\log z=\log C\\ \Rightarrow \log \left(\frac{z}{t}\right)=\log C\\ \Rightarrow \frac{\tan y}{1-e^x}=C\\ \Rightarrow \tan y=C(1-e^x)\\ \text{This is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}$

Q.47 For the differential equation given below, find a particular solution satisfying the given condition:

$(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x;\text{‹}y=1whenx=0$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\\ ({\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1)\frac{\mathrm{dy}}{\mathrm{dx}}=2{\mathrm{x}}^2+\mathrm{x}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2{\mathrm{x}}^2+\mathrm{x}}{({\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}+1)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{2{\mathrm{x}}^2+\mathrm{x}}{\{{\mathrm{x}}^2(\mathrm{x}+1)+1(\mathrm{x}+1)\}}\\ \mathrm{\hspace{0.17em}}\mathrm{dy}=\frac{2{\mathrm{x}}^2+\mathrm{x}}{({\mathrm{x}}^2+1)(\mathrm{x}+1)}\mathrm{dx}...\left(\mathrm{i}\right)\\ \mathrm{Let}\mathrm{\hspace{0.17em}}\frac{2{\mathrm{x}}^2+\mathrm{x}}{({\mathrm{x}}^2+1)(\mathrm{x}+1)}=\frac{\mathrm{Ax}+\mathrm{B}}{({\mathrm{x}}^2+1)}+\frac{\mathrm{C}}{(\mathrm{x}+1)}\\ \Rightarrow \mathrm{\hspace{0.17em}}2{\mathrm{x}}^2+\mathrm{x}=(\mathrm{Ax}+\mathrm{B})(\mathrm{x}+1)+\mathrm{C}({\mathrm{x}}^2+1)\\ \Rightarrow \mathrm{\hspace{0.17em}}2{\mathrm{x}}^2+\mathrm{x}={\mathrm{x}}^2(\mathrm{A}+\mathrm{C})+\mathrm{x}(\mathrm{A}+\mathrm{B})+(\mathrm{B}+\mathrm{C})\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{A}+\mathrm{C}=2,\hspace{0.17em}\hspace{0.17em}\mathrm{A}+\mathrm{B}=1\mathrm{and}\mathrm{B}+\mathrm{C}=\mathrm{0}\\ \Rightarrow \mathrm{A}=\frac{3}{2},\mathrm{\hspace{0.17em}}\mathrm{B}=-\frac{1}{2}\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{C}=\frac{1}{2}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{2{\mathrm{x}}^2+\mathrm{x}}{({\mathrm{x}}^2+1)(\mathrm{x}+1)}=\frac{3\mathrm{x}-1}{2({\mathrm{x}}^2+1)}+\frac{1}{2(\mathrm{x}+1)}\\ \mathrm{From}\mathrm{\hspace{0.17em}}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{\hspace{0.17em}}\mathrm{have}\\ \mathrm{\hspace{0.17em}}\mathrm{dy}=\frac{3\mathrm{x}-1}{2({\mathrm{x}}^2+1)}\mathrm{dx}+\frac{1}{2(\mathrm{x}+1)}\mathrm{dx}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\int \mathrm{dy}\mathrm{\hspace{0.17em}}=\int \frac{3\mathrm{x}-1}{2({\mathrm{x}}^2+1)}\mathrm{\hspace{0.17em}}\mathrm{dx}+\int \frac{1}{2(\mathrm{x}+1)}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{y}=\frac{3}{2}\int \frac{\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{\hspace{0.17em}}\mathrm{dx}-\frac{1}{2}\int \frac{1}{{\mathrm{x}}^2+1}\mathrm{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{\mathrm{x}+1}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \mathrm{y}=\frac{3}{2}\times \frac{1}{2}\log ({\mathrm{x}}^2+1)-\frac{1}{2}{\tan }^{-1}\mathrm{x}+\frac{1}{2}\log (\mathrm{x}+1)+\mathrm{C}\\ \mathrm{y}=\frac{1}{4}\{3\log ({\mathrm{x}}^2+1)+2\log (\mathrm{x}+1)\}-\frac{1}{2}{\tan }^{-1}\mathrm{x}+\mathrm{C}\\ \mathrm{y}=\frac{1}{4}\left\{\log {({\mathrm{x}}^2+1)}^3{(\mathrm{x}+1)}^2\right\}-\frac{1}{2}{\tan }^{-1}\mathrm{x}+\mathrm{C}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{Since},\mathrm{\hspace{0.17em}}\mathrm{y}=1\mathrm{when}\mathrm{x}=0\\ 1=\frac{1}{4}\left\{\log {(0+1)}^3{(0+1)}^2\right\}-\frac{1}{2}{\tan }^{-1}0+\mathrm{C}\\ 1=\mathrm{C}\\ \mathrm{Substituting}\mathrm{C}=1\mathrm{in}\mathrm{equation}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}\\ \mathrm{y}=\frac{1}{4}\left\{\log {({\mathrm{x}}^2+1)}^3{(\mathrm{x}+1)}^2\right\}-\frac{1}{2}{\tan }^{-1}\mathrm{x}+1\end{array}$

Q.48 For the differential equation given below, find a particular solution satisfying the given condition:

$\mathbf{\text{x}}{\text{(x}}^2-1)\frac{\mathrm{dy}}{\mathrm{dx}}=1;\mathrm{‹}\mathrm{y}=0\mathrm{when}\mathrm{x}=2$

Ans.

\begin{array}{l}The\text{given differential equation is}\\ x\left(x^2-1\right)\frac{dy}{dx}=1\\ \frac{dy}{dx}=\frac{1}{x\left(x^2-1\right)}\\ \frac{dy}{dx}=\frac{1}{x\left(x-1\right)\left(x+1\right)}\\ dy=\frac{1}{x\left(x-1\right)\left(x+1\right)}dx\mathrm{\dots }\left(i\right)\\ Let\frac{1}{x\left(x-1\right)\left(x+1\right)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\\ \Rightarrow 1=A\left(x^2-1\right)+Bx\left(x+1\right)+Cx\left(x-1\right)\\ Substituting\text{x}=0,1and-1,\text{we get}\end{array} $\begin{array}{l}\mathrm{A}=-1,\hspace{0.17em}\hspace{0.17em}\mathrm{B}=\frac{1}{2}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}\mathrm{C}=\frac{1}{2}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{1}{\mathrm{x}(\mathrm{x}-1)(\mathrm{x}+1)}=\frac{-1}{\mathrm{x}}+\frac{1}{2(\mathrm{x}-1)}+\frac{1}{2(\mathrm{x}+1)}\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{\hspace{0.17em}}\mathrm{have}\\ \mathrm{dy}=\frac{-1}{\mathrm{x}}\mathrm{dx}+\frac{1}{2(\mathrm{x}-1)}\mathrm{dx}+\frac{1}{2(\mathrm{x}+1)}\mathrm{dx}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int \mathrm{dy}=\mathrm{\hspace{0.17em}}-\int \frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{\mathrm{x}-1}\mathrm{\hspace{0.17em}}\mathrm{dx}+\frac{1}{2}\int \frac{1}{\mathrm{x}+1}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=-\mathrm{logx}+\frac{1}{2}\log (\mathrm{x}-1)+\frac{1}{2}\log (\mathrm{x}+1)+\mathrm{logk}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\frac{1}{2}\log \left\{\frac{{\mathrm{k}}^2(\mathrm{x}-1)(\mathrm{x}+1)}{{\mathrm{x}}^2}\right\}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{ii}\right)\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{y}=0\hspace{0.17em}\hspace{0.17em}\mathrm{when}\mathrm{\hspace{0.17em}}\mathrm{x}=2\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0=\frac{1}{2}\log \left\{\frac{{\mathrm{k}}^2(2-1)(2+1)}{2^2}\right\}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0=\log \left\{\frac{{\mathrm{k}}^2\left(1\right)\left(3\right)}{4}\right\}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\log 1=\log \left\{\frac{3{\mathrm{k}}^2}{4}\right\}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1=\frac{3{\mathrm{k}}^2}{4}\Rightarrow {\mathrm{k}}^2=\frac{4}{3}\\ \mathrm{Substituting}{\mathrm{k}}^{\mathrm{2}}=\frac{4}{3}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\frac{1}{2}\log \left\{\frac{\frac{4}{3}(\mathrm{x}-1)(\mathrm{x}+1)}{{\mathrm{x}}^2}\right\}\\ \Rightarrow \mathrm{\hspace{0.17em}}\mathrm{y}=\frac{1}{2}\log \left\{\frac{4({\mathrm{x}}^2-1)}{3{\mathrm{x}}^2}\right\}=\frac{1}{2}\log \left(\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2}\right)-\frac{1}{2}\log \left(\frac{3}{4}\right)\end{array}$

Q.49 For the differential equation given below, find a particular solution satisfying the given condition:

$\text{cos}\text{(}\frac{\mathrm{dy}}{\mathrm{dx}})=\mathrm{a}(\mathrm{a}\in \mathrm{R});\mathrm{‹}\mathrm{y}=1\mathrm{when}x=0$

Ans.

\begin{array}{l}\text{The given differential equation is:}\\ \cos \left(\frac{dy}{dx}\right)=a\\ \frac{dy}{dx}={\cos }^{-1}a\\ dy={\cos }^{-1}adx\\ \text{Integrating both sides, we get}\\ {\displaystyle \int dy}={\cos }^{-1}a{\displaystyle \int dx}\\ y=\left({\cos }^{-1}a\right)x+C\mathrm{\dots }\left(i\right)\\ \text{Now,}\text{y}\text{=}\text{1 when x}\text{=}\text{0}\\ 1=\left({\cos }^{-1}a\right)0+C\\ 1=C\\ \text{Substituting C = 1 in equation}\left(\text{i}\right)\text{,}\text{we get}\\ y=\left({\cos }^{-1}a\right)x+1\\ \frac{y-1}{x}={\cos }^{-1}a\Rightarrow \cos \left(\frac{y-1}{x}\right)=a\end{array}

Q.50 For the differential equation given below, find a particular solution satisfying the given condition:

$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{ytanx};\mathrm{‹}\mathrm{y}=1\mathrm{whenx}=0$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{ytanx}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{\hspace{0.17em}}\mathrm{dy}}{\mathrm{y}}=\mathrm{tanx}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\int \frac{\mathrm{\hspace{0.17em}}\mathrm{dy}}{\mathrm{y}}\mathrm{\hspace{0.17em}}=\int \mathrm{tanx}\mathrm{\hspace{0.17em}}\mathrm{dx}\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\log \mathrm{\hspace{0.17em}}\mathrm{y}=\mathrm{logsec}\mathrm{\hspace{0.17em}}\mathrm{x}+\mathrm{logC}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}=\mathrm{secx}.\mathrm{C}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{y}=1\mathrm{when}\mathrm{x}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1=\mathrm{C}.\sec 0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1=\mathrm{C}\\ \mathrm{Substituting}\mathrm{C}=\; 1\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\hspace{0.17em}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\mathrm{secx}.1\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\mathrm{secx}\end{array}$

Q.51 Find the equation of a curve passing through the point (0, 0) and whose differential equation is y’ = e^x sin x.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{\hspace{0.17em}}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}’={\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\sin \mathrm{x}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\sin \mathrm{x}\end{array}$ $\begin{array}{l}\mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\int \mathrm{dy}\mathrm{\hspace{0.17em}}=\int {\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\sin \mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}\mathrm{\hspace{0.17em}}+\mathrm{C}\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}=\int {\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\sin \mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}\mathrm{\hspace{0.17em}}+\mathrm{C}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{I}=\int {\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\sin \mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{sinx}\int {\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\int {\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx})\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{sinx}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}-\int \left(\mathrm{cosx}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}\right)\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{sinx}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}-\{\mathrm{cosx}\int {\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}\int {\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx})\mathrm{\hspace{0.17em}}\mathrm{dx}\}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{sinx}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}-\{{\mathrm{cosxe}}^{\mathrm{x}}-\int (-\mathrm{sinx})\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{sinx}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}-\int {\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\sin \mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \mathrm{I}\mathrm{\hspace{0.17em}}=\mathrm{sinx}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}-\mathrm{I}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{I}\mathrm{\hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}\mathrm{\hspace{0.17em}}-\mathrm{cosx})\\ \mathrm{I}\mathrm{\hspace{0.17em}}=\frac{{\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}\mathrm{\hspace{0.17em}}-\mathrm{cosx})}{2}\\ \mathrm{From}\mathrm{equation}\mathrm{\hspace{0.17em}}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}\mathrm{\hspace{0.17em}}=\frac{{\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}\mathrm{\hspace{0.17em}}-\mathrm{cosx})}{2}+\mathrm{C}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Since},\mathrm{curve}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}(0,0)\mathrm{.}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0\mathrm{\hspace{0.17em}}=\frac{{\mathrm{e}}^0(\sin 0\mathrm{\hspace{0.17em}}-\cos 0)}{2}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0\mathrm{\hspace{0.17em}}=\frac{1(0\mathrm{\hspace{0.17em}}-1)}{2}+\mathrm{C}\Rightarrow \mathrm{C}=\frac{1}{2}\end{array}$ $\begin{array}{l}\mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\hspace{0.17em}\mathrm{we}\mathrm{\hspace{0.17em}}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}\mathrm{\hspace{0.17em}}=\frac{{\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}\mathrm{\hspace{0.17em}}-\mathrm{cosx})}{2}+\frac{1}{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{y}={\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}\mathrm{\hspace{0.17em}}-\mathrm{cosx})+1\\ 2\mathrm{y}-1={\mathrm{e}}^{\mathrm{x}}(\mathrm{sinx}\mathrm{\hspace{0.17em}}-\mathrm{cosx})\\ \mathrm{Hence},\mathrm{it}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{.}\end{array}$

Q.52

$\begin{array}{l}\mathrm{For}\mathrm{the}\mathrm{differential}\mathrm{equation}\mathrm{xy}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{=}\left(\mathrm{x}+2\right)\left(\mathrm{y}+2\right),\mathrm{find}\mathrm{the}\\ \mathrm{solution}\mathrm{curve}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{point}\left(1,-1\right)\mathrm{.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{\hspace{0.17em}}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{xy}\frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{x}+2)(\mathrm{y}+2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{ydy}}{(\mathrm{y}+2)}=\mathrm{}\frac{(\mathrm{x}+2)}{\mathrm{x}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{(\mathrm{y}+2-2)\mathrm{dy}}{(\mathrm{y}+2)}=\mathrm{}\frac{(\mathrm{x}+2)}{\mathrm{x}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}1\mathrm{dy}-\frac{2}{(\mathrm{y}+2)}\mathrm{dy}=(1+\frac{2}{\mathrm{x}})\mathrm{dx}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \int 1\mathrm{\hspace{0.17em}}\mathrm{dy}-\int \frac{2}{(\mathrm{y}+2)}\mathrm{\hspace{0.17em}}\mathrm{dy}=\int 1\mathrm{\hspace{0.17em}}\mathrm{dx}+2\int \frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}-2\log (\mathrm{y}+2)=\mathrm{x}+2\mathrm{logx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}-\mathrm{x}-\mathrm{C}={\mathrm{logx}}^2+\log {(\mathrm{y}+2)}^2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}-\mathrm{x}-\mathrm{C}={\mathrm{logx}}^2{(\mathrm{y}+2)}^2\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{ii}\right)\\ \mathrm{Since},\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{through}(1,-1).\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-1-1-\mathrm{C}=\log {\left(1\right)}^2{(-1+2)}^2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-2-\mathrm{C}=\log {\left(1\right)}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{C}=-2\\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}-\mathrm{x}+2={\mathrm{logx}}^2{(\mathrm{y}+2)}^2\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{.}\end{array}$

Q.53 Find the equation of the curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Ans.

$\begin{array}{l}\mathrm{According}\mathrm{to}\mathrm{given}\mathrm{condition},\\ \mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}.\mathrm{dy}=\mathrm{x}.\mathrm{dx}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int \mathrm{y}\mathrm{\hspace{0.17em}}\mathrm{dy}=\int \mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{{\mathrm{y}}^2}{2}=\frac{{\mathrm{x}}^2}{2}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{y}}^2={\mathrm{x}}^2+2\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{y}}^2-{\mathrm{x}}^2=2\mathrm{C}...\left(\mathrm{i}\right)\\ \mathrm{The}\mathrm{curve}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}(0-2).\\ {(-2)}^2-0^2=2\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}4=2\mathrm{C}\Rightarrow \mathrm{C}=2\\ \mathrm{Putting}\mathrm{C}=\; 2\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\hspace{0.17em}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{y}}^2-{\mathrm{x}}^2=4\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{.}\end{array}$

Q.54 At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Ans.

$\begin{array}{l}\mathrm{Slope}\left({\mathrm{m}}_{\mathrm{1}}\right)\mathrm{of}\mathrm{curve}\mathrm{at}\mathrm{point}(\mathrm{x},\mathrm{y})=\frac{\mathrm{dy}}{\mathrm{dx}}\\ \mathrm{slope}\left({\mathrm{m}}_{\mathrm{2}}\right)\mathrm{of}\mathrm{line}\mathrm{segment}\mathrm{joining}(\mathrm{x},\mathrm{y})\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{}(-4,-3)=\frac{\mathrm{y}+3}{\mathrm{x}+4}\\ \mathrm{According}\mathrm{to}\mathrm{given}\mathrm{condition}:\\ {\mathrm{m}}_1=2{\mathrm{m}}_2\\ \frac{\mathrm{dy}}{\mathrm{dx}}=2\left(\frac{\mathrm{y}+3}{\mathrm{x}+4}\right)\\ \frac{\mathrm{dy}}{\mathrm{y}+3}=2\left(\frac{\mathrm{dx}}{\mathrm{x}+4}\right)\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int \frac{\mathrm{dy}}{\mathrm{y}+3}\mathrm{\hspace{0.17em}}=2\int \frac{1}{\mathrm{x}+4}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \log (\mathrm{y}+3)=2\log (\mathrm{x}+4)+\mathrm{logC}\\ \log (\mathrm{y}+3)=\log {(\mathrm{x}+4)}^2\mathrm{C}\\ \mathrm{y}+3=\mathrm{C}{(\mathrm{x}+4)}^2\mathrm{\hspace{0.17em}}...\mathrm{\hspace{0.17em}}\left(\mathrm{i}\right)\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{general}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{.}\\ \mathrm{The}\mathrm{curve}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}(-2,1).\\ 1+3=\mathrm{C}{(-2+4)}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{C}=\frac{4}{4}=1\\ \mathrm{Substituting}\mathrm{C}=\; 1\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{y}+3=1{(\mathrm{x}+4)}^2\\ \Rightarrow \mathrm{y}+3={(\mathrm{x}+4)}^2\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{.}\end{array}$

Q.55 The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{radius}\mathrm{of}\mathrm{spherical}\mathrm{balloon}\mathrm{be}\mathrm{r}\mathrm{units}\mathrm{.}\\ \mathrm{Volume}\mathrm{of}\mathrm{balloon}\left(\mathrm{V}\right)=\frac{4}{3}{\mathrm{\pi r}}^3\\ \mathrm{According}\mathrm{to}\mathrm{condition}:\\ \mathrm{\hspace{0.17em}}\frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{k}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{4}{3}{\mathrm{\pi r}}^3\right)=\mathrm{k}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{4}{3}\mathrm{\pi }\times 3{\mathrm{r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k}\\ \hspace{0.17em}\hspace{0.17em}4{\mathrm{\pi r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k}\\ \hspace{0.17em}\hspace{0.17em}4{\mathrm{\pi r}}^2\mathrm{dr}=\mathrm{kdt}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ 4\mathrm{\pi }\int {\mathrm{r}}^2\mathrm{\hspace{0.17em}}\mathrm{dr}=\int \mathrm{k}\mathrm{\hspace{0.17em}}\mathrm{dt}\\ 4\mathrm{\pi }\frac{{\mathrm{r}}^3}{3}=\mathrm{kt}+\mathrm{C}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Now},\mathrm{r}=\; 3\mathrm{at}\mathrm{t}=\; 0\\ 4\mathrm{\pi }\frac{3^3}{3}=\mathrm{k}\left(0\right)+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}36\mathrm{\pi }=\mathrm{C}\\ \mathrm{Again},\mathrm{\hspace{0.17em}}\mathrm{r}=\; 6\mathrm{when}\mathrm{t}=\; 3\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ 4\mathrm{\pi }\frac{6^3}{3}=\mathrm{k}\left(3\right)+36\mathrm{\pi }\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}288\mathrm{\pi }-36\mathrm{\pi }=3\mathrm{k}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{k}=\frac{252\mathrm{\pi }}{3}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=84\mathrm{\pi }\\ \mathrm{Hence},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{have}\\ \frac{4}{3}{\mathrm{\pi r}}^3=84\mathrm{\pi t}+36\mathrm{\pi }\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{r}}^3=63\mathrm{t}+27\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{r}={(63\mathrm{t}+27)}^{\frac{1}{3}}\\ \mathrm{Thus},\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{balloon}\mathrm{after}\mathrm{t}\mathrm{seconds}\mathrm{is}\mathrm{\hspace{0.17em}}{(63\mathrm{t}+27)}^{\frac{1}{3}}.\end{array}$

Q.56 In a bank, principal increases continuously at the rate of r% per year. Find the value of r if ` 100 double itself in 10 years (log_e2= 0.6931).

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{Principal}\mathrm{be}\mathrm{P},\mathrm{time}\mathrm{be}\mathrm{t}\mathrm{years}\mathrm{and}\mathrm{rate}\mathrm{of}\mathrm{interest}\mathrm{be}\mathrm{r}\%\mathrm{p}.\mathrm{a}\mathrm{.}\\ \mathrm{Since},\mathrm{principal}\mathrm{increases}\mathrm{continuously}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{r}\%\mathrm{p}.\mathrm{a}\mathrm{.}\\ \frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{r}\%\hspace{0.17em}\mathrm{of}\mathrm{\hspace{0.17em}}\mathrm{P}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\left(\frac{\mathrm{r}}{100}\right)\mathrm{P}\\ \frac{\mathrm{dP}}{\mathrm{P}}=\left(\frac{\mathrm{r}}{100}\right)\mathrm{dt}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int \frac{\mathrm{dP}}{\mathrm{P}}\mathrm{\hspace{0.17em}}=\int \frac{\mathrm{r}}{100}\mathrm{\hspace{0.17em}}\mathrm{dt}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\log }_{\mathrm{e}}\mathrm{P}=\frac{\mathrm{r}}{100}\mathrm{t}+\mathrm{C}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{Substituting}\mathrm{P}=\; 100\mathrm{and}\mathrm{t}=\; 0\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\log }_{\mathrm{e}}100=\frac{\mathrm{r}}{100}\times 0+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\log }_{\mathrm{e}}100=0+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{C}=\mathrm{\hspace{0.17em}}{\log }_{\mathrm{e}}100\end{array}$ $\begin{array}{l}\mathrm{Again},\mathrm{\hspace{0.17em}}\mathrm{putting}\mathrm{P}=\; 200\mathrm{and}\mathrm{t}=\; 10\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\hspace{0.17em}\mathrm{we}\mathrm{\hspace{0.17em}}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\log }_{\mathrm{e}}200=\frac{\mathrm{r}}{100}\times 10+{\log }_{\mathrm{e}}100\\ {\log }_{\mathrm{e}}200-{\log }_{\mathrm{e}}100=\frac{\mathrm{r}}{10}\\ {\log }_{\mathrm{e}}\frac{200}{100}=\frac{\mathrm{r}}{10}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\log }_{\mathrm{e}}2=\frac{\mathrm{r}}{10}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0.6931=\frac{\mathrm{r}}{10}\Rightarrow \mathrm{r}=6.931\\ \mathrm{Thus},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{interest}\mathrm{is}6.93\%\mathrm{p}.\mathrm{a}\mathrm{.}\end{array}$

Q.57 In a bank, principal increases continuously at the rate of 5% per year. An amount of ₹ 1000 is deposited with this bank, how much will it worth after 10 years (e^0.5 = 1.648).

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{principal}\mathrm{be}\mathrm{P},\mathrm{t}\mathrm{be}\mathrm{time}\mathrm{and}\mathrm{r}\mathrm{be}\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{interest}.\mathrm{Then},\\ \mathrm{rate}\mathrm{of}\mathrm{principal}\mathrm{increasing}\mathrm{per}\mathrm{year}\mathrm{is}:\\ \frac{\mathrm{dP}}{\mathrm{dt}}=\left(\frac{5}{100}\right)\mathrm{P}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{20}\mathrm{P}\\ \Rightarrow \frac{\mathrm{dP}}{\mathrm{P}}=\frac{1}{20}\mathrm{dt}\\ \mathrm{Intergrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int \frac{\mathrm{dP}}{\mathrm{P}}\mathrm{\hspace{0.17em}}=\frac{1}{20}\int \mathrm{dt}\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\log }_{\mathrm{e}}\mathrm{P}=\frac{1}{20}\mathrm{t}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}={\mathrm{e}}^{(\frac{1}{20}\mathrm{t}+\mathrm{C})}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\mathrm{C}}{\mathrm{e}}^{\frac{1}{20}\mathrm{t}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}={\mathrm{Ke}}^{\frac{1}{20}\mathrm{t}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)[\mathrm{Where},\mathrm{‹}\mathrm{K}={\mathrm{e}}^{\mathrm{C}}]\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{t}=\; 0\mathrm{when}\mathrm{P}=\; 1000\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1000={\mathrm{Ke}}^{\frac{1}{20}\left(0\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1000=\mathrm{K}\\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{K}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}=1000{\mathrm{e}}^{\frac{1}{20}\mathrm{t}}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{ii}\right)\\ \mathrm{Now}\mathrm{amount}\mathrm{after}10\mathrm{years},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}=1000{\mathrm{e}}^{\frac{1}{20}\left(10\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1000{\mathrm{e}}^{0.5}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1000\times 1.648\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1648\\ \mathrm{Hence},\mathrm{\hspace{0.17em}}\mathrm{after}10\mathrm{years}\mathrm{the}\mathrm{amount}\mathrm{will}\mathrm{be}₹\hspace{0.17em}1648\mathrm{.}\end{array}$

Q.58 In a culture, the bacteria count is 1,00,000.
The number is increased by 10% in 2 hours.
In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{present}\mathrm{number}\mathrm{of}\mathrm{bacteria}\mathrm{be}\mathrm{N}\mathrm{at}\mathrm{any}\mathrm{instant}\mathrm{t}\mathrm{.}\\ \mathrm{Since},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{growth}\mathrm{of}\mathrm{the}\mathrm{bacteria}\mathrm{is}\mathrm{proportional}\mathrm{to}\mathrm{N}\mathrm{.}\\ \therefore \mathrm{}\frac{\mathrm{dN}}{\mathrm{dt}}\mathrm{\alpha N}\\ \frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{kN}\\ \frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{kdt}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int \frac{\mathrm{dN}}{\mathrm{N}}\mathrm{\hspace{0.17em}}=\int \mathrm{k}\mathrm{\hspace{0.17em}}\mathrm{dt}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{logN}=\mathrm{kt}+\mathrm{C}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Let}\mathrm{number}\mathrm{of}\mathrm{bacteria}\mathrm{at}\mathrm{t}=\; 0\mathrm{be}{\mathrm{N}}_{\mathrm{0}},\mathrm{then}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{logN}}_0=\mathrm{k}\left(0\right)+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{C}={\mathrm{logN}}_0\\ \mathrm{From}\mathrm{equation}(\mathrm{i}),\mathrm{we}\mathrm{have}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{logN}=\mathrm{kt}+{\mathrm{logN}}_0\\ \mathrm{\hspace{0.17em}}\mathrm{logN}-{\mathrm{logN}}_0=\mathrm{kt}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\log {\frac{\mathrm{N}}{\mathrm{N}}}_0=\mathrm{kt}...\left(\mathrm{ii}\right)\\ \mathrm{Since},\mathrm{bacteria}\mathrm{increases}10\%\mathrm{in}2\mathrm{hours},\mathrm{so}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{N}=(1+\frac{10}{100}){\mathrm{N}}_0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\frac{\mathrm{N}}{{\mathrm{N}}_0}=\frac{11}{10}\end{array}$ $\begin{array}{l}\mathrm{Substituting}\mathrm{t}=\; 2\mathrm{and}\frac{\mathrm{N}}{{\mathrm{N}}_0}=\frac{11}{10}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\log \frac{11}{10}=\mathrm{k}\left(2\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{k}=\frac{1}{2}\mathrm{\hspace{0.17em}}\log \left(\frac{11}{10}\right)\\ \mathrm{From}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{have}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\log {\frac{\mathrm{N}}{\mathrm{N}}}_0=\frac{1}{2}\mathrm{\hspace{0.17em}}\log \left(\frac{11}{10}\right)\mathrm{t}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{t}=\frac{\log {\frac{\mathrm{N}}{\mathrm{N}}}_0}{\frac{1}{2}\mathrm{\hspace{0.17em}}\log \left(\frac{11}{10}\right)}\\ =\frac{2\log \frac{\mathrm{N}}{{\mathrm{N}}_0}}{\mathrm{\hspace{0.17em}}\log \left(\frac{11}{10}\right)}\\ \mathrm{So},\mathrm{the}\mathrm{time}\mathrm{taken}\mathrm{to}\mathrm{increase}\mathrm{the}\mathrm{bacteria}\mathrm{from}100000\\ \mathrm{to}200000=\frac{2\log \frac{200000}{100000}}{\mathrm{\hspace{0.17em}}\log \left(\frac{11}{10}\right)}\end{array}$ $\begin{array}{l}\Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{t}=\frac{2\log 2}{\mathrm{\hspace{0.17em}}\log \left(\frac{11}{10}\right)}\\ \mathrm{Hence},\mathrm{\hspace{0.17em}}\mathrm{bacteria}\mathrm{increases}\mathrm{from}100000\mathrm{to}200000\mathrm{in}\frac{2\log 2}{\mathrm{\hspace{0.17em}}\log \left(\frac{11}{10}\right)}\mathrm{\hspace{0.17em}}\mathrm{hrs}\mathrm{.}\end{array}$

Q.59

$\begin{array}{l}\mathrm{The}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{x}+\mathrm{y}}\hspace{0.17em}\hspace{0.17em}\mathrm{is}\\ \left(\mathrm{A}\right){\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{–\mathrm{y}}=\mathrm{C}\left(\mathrm{B}\right){\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{\mathrm{y}}=\mathrm{C}\\ \left(\mathrm{C}\right){\mathrm{e}}^{–\mathrm{x}}+{\mathrm{e}}^{\mathrm{y}}=\mathrm{C}\left(\mathrm{D}\right){\mathrm{e}}^{–\mathrm{x}}+{\mathrm{e}}^{–\mathrm{y}}=\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{differential}\mathrm{equation}\mathrm{is}\\ \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{x}+\mathrm{y}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{y}}\\ \frac{\mathrm{dy}}{{\mathrm{e}}^{\mathrm{y}}}={\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}={\mathrm{e}}^{\mathrm{x}}\mathrm{dx}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \int {\mathrm{e}}^{-\mathrm{y}}\mathrm{\hspace{0.17em}}\mathrm{dy}=\int {\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-{\mathrm{e}}^{-\mathrm{y}}={\mathrm{e}}^{\mathrm{x}}+{\mathrm{C}}_1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{y}}=-{\mathrm{C}}_1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{y}}=\mathrm{C}[\mathrm{Where},\mathrm{C}=-{\mathrm{C}}_1]\\ \mathrm{Thus},\mathrm{option}\mathrm{A}\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.60 Show that the given differential equation is homogeneous and solve it.

(x² + xy) dy = (x² + y²) dx

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}({\mathrm{x}}^2+\mathrm{xy})\mathrm{dy}=({\mathrm{x}}^2+{\mathrm{y}}^2)\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{({\mathrm{x}}^2+{\mathrm{y}}^2)}{({\mathrm{x}}^2+\mathrm{xy})}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{x},\mathrm{y})=\frac{({\mathrm{x}}^2+{\mathrm{y}}^2)}{({\mathrm{x}}^2+\mathrm{xy})}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{f}(\mathrm{\lambda x},\mathrm{\lambda y})=\frac{({\mathrm{\lambda }}^2{\mathrm{x}}^2+{\mathrm{\lambda }}^2{\mathrm{y}}^2)}{({\mathrm{\lambda }}^2{\mathrm{x}}^2+{\mathrm{\lambda }}^2\mathrm{xy})}\\ \mathrm{\hspace{0.17em}}=\frac{{\mathrm{\lambda }}^2({\mathrm{x}}^2+{\mathrm{y}}^2)}{{\mathrm{\lambda }}^2({\mathrm{x}}^2+\mathrm{xy})}={\mathrm{\lambda }}^0\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x},\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\\ \mathrm{Substituting}\mathrm{y}=\mathrm{vx}\mathrm{then}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{({\mathrm{x}}^2+{\mathrm{v}}^{\mathrm{2}}{\mathrm{x}}^2)}{({\mathrm{x}}^2+\mathrm{xvx})}\\ \mathrm{\hspace{0.17em}}=\frac{{\mathrm{x}}^2(1+{\mathrm{v}}^{\mathrm{2}})}{{\mathrm{x}}^2(1+\mathrm{v})}\\ \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{(1+{\mathrm{v}}^{\mathrm{2}})}{(1+\mathrm{v})}-\mathrm{v}\\ \mathrm{\hspace{0.17em}}=\frac{1+{\mathrm{v}}^{\mathrm{2}}-\mathrm{v}-{\mathrm{v}}^2}{(1+\mathrm{v})}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1-\mathrm{v}}{1+\mathrm{v}}\Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{1+\mathrm{v}}{1-\mathrm{v}}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}\\ \Rightarrow \frac{2-(1-\mathrm{v})}{1-\mathrm{v}}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\Rightarrow (\frac{2}{1-\mathrm{v}}-1)\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\int (\frac{2}{1-\mathrm{v}}-1)\mathrm{\hspace{0.17em}}\mathrm{dv}=\int \frac{\mathrm{dx}}{\mathrm{x}}\\ -2\log (1-\mathrm{v})-\mathrm{v}=\mathrm{logx}+\mathrm{logk}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{v}=-2\log (1-\mathrm{v})-\mathrm{logx}-\mathrm{logk}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{v}=-\log \left\{{(1-\mathrm{v})}^2\mathrm{xk}\right\}\\ \frac{\mathrm{y}}{\mathrm{x}}=-\log \left\{{(1-\frac{\mathrm{y}}{\mathrm{x}})}^2\mathrm{xk}\right\}[\mathrm{Put}\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}}]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}{(1-\frac{\mathrm{y}}{\mathrm{x}})}^2\mathrm{xk}={\mathrm{e}}^{-\frac{\mathrm{y}}{\mathrm{x}}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}{\left(\frac{\mathrm{x}-\mathrm{y}}{\mathrm{x}}\right)}^2\mathrm{xk}={\mathrm{e}}^{-\frac{\mathrm{y}}{\mathrm{x}}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}{(\mathrm{x}-\mathrm{y})}^2\frac{\mathrm{k}}{\mathrm{x}}={\mathrm{e}}^{-\frac{\mathrm{y}}{\mathrm{x}}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{(\mathrm{x}-\mathrm{y})}^2=\frac{\mathrm{x}}{\mathrm{k}}{\mathrm{e}}^{-\frac{\mathrm{y}}{\mathrm{x}}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{(\mathrm{x}-\mathrm{y})}^2={\mathrm{Cxe}}^{-\frac{\mathrm{y}}{\mathrm{x}}}[\mathrm{Let}\mathrm{C}=\frac{1}{\mathrm{k}}]\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.61 Show that the given differential equation is homogeneous and solve each of them.

y‘=\frac{x+y}{x}

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}‘=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{x},\mathrm{y})=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=\frac{\mathrm{\lambda x}+\mathrm{\lambda y}}{\mathrm{\lambda x}}\\ =\frac{\mathrm{\lambda }(\mathrm{x}+\mathrm{y})}{\mathrm{\lambda x}}\\ =\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}}={\mathrm{\lambda }}^0\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x},\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\\ \mathrm{For}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{substitute}\mathrm{y}=\mathrm{vx}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}},\\ \mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{x}+\mathrm{vx}}{\mathrm{x}}\\ =\frac{\mathrm{x}(1+\mathrm{v})}{\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=(1+\mathrm{v})\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=(1+\mathrm{v})-\mathrm{v}\end{array}$ $\begin{array}{l}\Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=1\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{v}=\mathrm{logx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{logx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}=\mathrm{xlogx}+\mathrm{Cx}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.62 Show that the given differential equation is homogeneous and solve it.

(x – y)dy – (x + y) dx = 0

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \mathrm{\hspace{0.17em}}(\mathrm{x}-\mathrm{y})\mathrm{dy}-(\mathrm{x}+\mathrm{y})\mathrm{dx}=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}-\mathrm{y}}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Let}\mathrm{F}(\mathrm{x},\mathrm{y})=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}-\mathrm{y}}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=\frac{\mathrm{\lambda x}+\mathrm{\lambda y}}{\mathrm{\lambda x}-\mathrm{\lambda y}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{\lambda }(\mathrm{x}+\mathrm{y})}{\mathrm{\lambda }(\mathrm{x}-\mathrm{y})}={\mathrm{\lambda }}^0\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x},\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\\ \mathrm{For}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{substitute}\mathrm{y}=\mathrm{vx}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{,}\end{array}$ $\begin{array}{l}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{x}+\mathrm{vx}}{\mathrm{x}-\mathrm{vx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{x}(1+\mathrm{v})}{\mathrm{x}(1-\mathrm{v})}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{v}-\mathrm{v}+{\mathrm{v}}^2}{1-\mathrm{v}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1+{\mathrm{v}}^2}{1-\mathrm{v}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{1-\mathrm{v}}{1+{\mathrm{v}}^2}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\frac{1}{1+{\mathrm{v}}^2}-\frac{\mathrm{v}}{1+{\mathrm{v}}^2})\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \int \frac{1}{1+{\mathrm{v}}^2}\mathrm{\hspace{0.17em}}\mathrm{dv}-\int \frac{\mathrm{v}}{1+{\mathrm{v}}^2}\mathrm{\hspace{0.17em}}\mathrm{dv}=\int \frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}{\tan }^{-1}\mathrm{v}-\frac{1}{2}\log (1+{\mathrm{v}}^2)=\mathrm{logx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\frac{1}{2}\log (1+{\frac{\mathrm{y}}{{\mathrm{x}}^2}}^2)+\frac{1}{2}.2\mathrm{logx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\frac{1}{2}\log (\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{{\mathrm{x}}^2}.{\mathrm{x}}^2)+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\frac{1}{2}\log ({\mathrm{x}}^2+{\mathrm{y}}^2)+\mathrm{C}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.63 Show that the given differential equation is homogeneous and solve it.

(x² – y²)dx + 2xy dy = 0

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}({\mathrm{x}}^2-{\mathrm{y}}^2)\mathrm{dx}+2\mathrm{xy}\mathrm{}\mathrm{dy}=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{2\mathrm{xy}}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Let}\mathrm{F}(\mathrm{x},\mathrm{y})=-\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{2\mathrm{xy}}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=-\frac{{\mathrm{\lambda }}^2{\mathrm{x}}^2-{\mathrm{\lambda }}^2{\mathrm{y}}^2}{2{\mathrm{\lambda }}^2\mathrm{xy}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-\frac{{\mathrm{\lambda }}^2({\mathrm{x}}^2-{\mathrm{y}}^2)}{{\mathrm{\lambda }}^2\left(2\mathrm{xy}\right)}={\mathrm{\lambda }}^0\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x},\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}\mathrm{.}\\ \mathrm{So},\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\\ \mathrm{For}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{substitute}\mathrm{y}\hspace{0.17em}=\mathrm{\hspace{0.17em}}\mathrm{vx}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{,}\\ \mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{{\mathrm{x}}^2-{\mathrm{v}}^{\mathrm{2}}{\mathrm{x}}^2}{2\mathrm{x}\left(\mathrm{vx}\right)}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\frac{(1-{\mathrm{v}}^{\mathrm{2}}){\mathrm{x}}^2}{2{\mathrm{x}}^2\mathrm{v}}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\frac{(1-{\mathrm{v}}^{\mathrm{2}})}{2\mathrm{v}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{({\mathrm{v}}^{\mathrm{2}}-1)}{2\mathrm{v}}-\mathrm{v}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{({\mathrm{v}}^{\mathrm{2}}-1)-2{\mathrm{v}}^2}{2\mathrm{v}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-(1+{\mathrm{v}}^2)}{2\mathrm{v}}\\ \frac{2\mathrm{v}}{(1+{\mathrm{v}}^2)}\mathrm{dv}=-\frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\int \frac{2\mathrm{v}}{(1+{\mathrm{v}}^2)}\mathrm{\hspace{0.17em}}\mathrm{dv}=-\int \frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\log (1+{\mathrm{v}}^2)=-\mathrm{logx}+\mathrm{logC}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\log (1+{\mathrm{v}}^2)=\log \left(\frac{\mathrm{C}}{\mathrm{x}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(1+\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2})=\frac{\mathrm{C}}{\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}({\mathrm{x}}^2+{\mathrm{y}}^2)=\mathrm{Cx}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.64 Show that the given differential equation is homogeneous and solve it.

${\text{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^2-2{\mathrm{y}}^2+\mathrm{xy}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^2-2{\mathrm{y}}^2+\mathrm{xy}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^2-2{\mathrm{y}}^2+\mathrm{xy}}{{\mathrm{x}}^2}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\end{array}$ $\begin{array}{l}\text{Let}\text{F}(x,y)=\frac{x^2-2y^2+xy}{x^2}\\ \text{Now,}\text{F}(\lambda x,\lambda y)=\frac{{\lambda }^2{x}^2-2{\lambda }^2{y}^2+{\lambda }^{2}xy}{{\lambda }^2{x}^2}\\ =\frac{{\lambda }^2(x^2-2y^2+xy)}{{\lambda }^2\left(x^2\right)}={\lambda }^0\text{F}(x,y)\\ \text{Therefore, F}\left(x\text{,}y\right)\text{is a homogenous function of degree zero}\text{. So,}\\ \text{the given differential equation is a homogenous differential}\\ \text{equation}\text{.}\\ \text{For solving equation}\left(\text{i}\right)\text{, substitute}y\text{=}vx\text{and}\frac{dy}{dx}=\text{v}+\text{x}\frac{dv}{dx}\text{,}\\ \text{we get}\\ \text{v}+\text{x}\frac{dv}{dx}=\frac{x^2-2{\text{v}}^{\text{2}}{\text{x}}^2+x\text{vx}}{x^2}\\ =\frac{(1-2{\text{v}}^{\text{2}}+v){\text{x}}^2}{x^2}\\ =1-2{\text{v}}^{\text{2}}+v\\ \text{x}\frac{dv}{dx}=1-2{\text{v}}^{\text{2}}+v-v\\ \text{x}\frac{dv}{dx}=1-2{\text{v}}^{\text{2}}\\ \frac{dv}{1-2{\text{v}}^{\text{2}}}=\frac{dx}{x}\\ \text{Integrating both sides, we get}\end{array}$ \begin{array}{l}\frac{\text{1}}{\text{2}}{\displaystyle \int \frac{\text{1}}{{\left(\frac{\text{1}}{\sqrt{\text{2}}}\right)}^{\text{2}}{\text{-v}}^{\text{2}}}}\text{dv=}{\displaystyle \int \frac{\text{dx}}{\text{x}}}\\ \frac{\text{1}}{\text{2}}\text{×}\frac{\text{1}}{\text{2×}\frac{\text{1}}{\sqrt{\text{2}}}}\text{log}\left(\frac{\frac{\text{1}}{\sqrt{\text{2}}}\text{+v}}{\frac{\text{1}}{\sqrt{\text{2}}}\text{-v}}\right)\text{=logx+C}\\ \frac{\text{1}}{\text{2}\sqrt{\text{2}}}\text{log}\left(\frac{\frac{\text{1}}{\sqrt{\text{2}}}\text{+}\frac{\text{y}}{\text{x}}}{\frac{\text{1}}{\sqrt{\text{2}}}\text{–}\frac{\text{y}}{\text{x}}}\right)\text{=logx+C}\\ \frac{\text{1}}{\text{2}\sqrt{\text{2}}}\text{log}\left(\frac{\text{x+}\sqrt{\text{2}}\text{y}}{\text{x-}\sqrt{\text{2}}\text{y}}\right)\text{=logx+C}\\ \text{This is the required solution of the given differential equation}\text{.}\end{array}

Q.65 Show that the given differential equation is homogeneous and solve it.

xdy-ydx=\sqrt{x^2+y^2}dx

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{xdy}-\mathrm{ydx}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{xdy}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{ydx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}+\mathrm{y}}{\mathrm{x}}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Let}\mathrm{F}(\mathrm{x},\mathrm{y})=\frac{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}+\mathrm{y}}{\mathrm{x}}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=\frac{\sqrt{{\mathrm{\lambda }}^2{\mathrm{x}}^2+{\mathrm{\lambda }}^2{\mathrm{y}}^2}+\mathrm{\lambda y}}{\mathrm{\lambda x}}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{\lambda }(\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}+\mathrm{y})}{\mathrm{\lambda x}}={\mathrm{\lambda }}^0\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x},\hspace{0.17em}\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\\ \mathrm{For}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{substitute}\mathrm{y}=\mathrm{vx}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{,}\\ \mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\sqrt{{\mathrm{x}}^2+{\mathrm{v}}^{\mathrm{2}}{\mathrm{x}}^2}+\mathrm{vx}}{\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{(\sqrt{1+{\mathrm{v}}^{\mathrm{2}}}+\mathrm{v})\mathrm{x}}{\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\sqrt{1+{\mathrm{v}}^{\mathrm{2}}}+\mathrm{v}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\sqrt{1+{\mathrm{v}}^{\mathrm{2}}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dv}}{\sqrt{1+{\mathrm{v}}^{\mathrm{2}}}}=\frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\log |\mathrm{v}+\sqrt{1+{\mathrm{v}}^{\mathrm{2}}}|=\log \left|\mathrm{x}\right|+\mathrm{logC}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\log |\frac{\mathrm{y}}{\mathrm{x}}+\sqrt{1+\frac{{\mathrm{y}}^{\mathrm{2}}}{{\mathrm{x}}^2}}|=\mathrm{logC}\left|\mathrm{x}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}|\frac{\mathrm{y}}{\mathrm{x}}+\sqrt{1+\frac{{\mathrm{y}}^{\mathrm{2}}}{{\mathrm{x}}^2}}|=\mathrm{C}\left|\mathrm{x}\right|\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}|\frac{\mathrm{y}}{\mathrm{x}}+\sqrt{1+\frac{{\mathrm{y}}^{\mathrm{2}}}{{\mathrm{x}}^2}}|=\mathrm{C}\left|\mathrm{x}\right|\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}+\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}={\mathrm{Cx}}^2\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.66 Show that the given differential equation is homogeneous and solve each of them.

$\left\{\mathrm{xcos}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right\}\mathrm{ydx}=\left\{\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)–\mathrm{xcos}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right\}\mathrm{xdy}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \{\mathrm{xcos}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\}\mathrm{ydx}=\{\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{xcos}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\}\mathrm{xdy}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\{\mathrm{xcos}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\}\mathrm{y}}{\{\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{xcos}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\}\mathrm{x}}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Let}\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})=\frac{\{\mathrm{xcos}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\}\mathrm{y}}{\{\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{xcos}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\}\mathrm{x}}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=\frac{\{\mathrm{\lambda xcos}\left(\frac{\mathrm{\lambda y}}{\mathrm{\lambda x}}\right)+\mathrm{\lambda ysin}\left(\frac{\mathrm{\lambda y}}{\mathrm{\lambda x}}\right)\}\mathrm{\lambda y}}{\{\mathrm{\lambda ysin}\left(\frac{\mathrm{\lambda y}}{\mathrm{\lambda x}}\right)-\mathrm{\lambda xcos}\left(\frac{\mathrm{\lambda y}}{\mathrm{\lambda x}}\right)\}\mathrm{\lambda x}}\end{array}$ \begin{array}{l}=\frac{{\lambda }^{2}y\left\{x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right)\right\}}{{\lambda }^{2}x\left\{y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right)\right\}}={\lambda }^0\text{F}\left(x,y\right)\\ \text{Therefore, F}\left(x\text{,}y\right)\text{is a homogenous function of degree zero}\text{. So,}\\ \text{the given differential equation is a homogenous differential}\\ \text{equation}\text{.}\\ \text{For solving equation}\left(\text{i}\right)\text{, substitute}y=\text{vx and}\frac{dy}{dx}=\text{v}+\text{x}\frac{dv}{dx}\text{,}\\ \text{we get}\\ \text{v}+\text{x}\frac{dv}{dx}=\frac{\text{vx}\left\{x\cos \left(\frac{\text{vx}}{x}\right)+\text{vx}\sin \left(\frac{\text{vx}}{x}\right)\right\}}{x\left\{\text{vx}\sin \left(\frac{\text{vx}}{x}\right)-x\cos \left(\frac{\text{vx}}{x}\right)\right\}}\\ =\frac{\text{v}\left\{\cos \left(v\right)+\text{v}\sin \left(v\right)\right\}}{\left\{\text{v}\sin \left(v\right)-\cos \left(v\right)\right\}}\\ \text{x}\frac{dv}{dx}=\frac{v\cos v+{\text{v}}^{\text{2}}\sin v}{\text{v}\sin v-\cos v}-v\\ \text{x}\frac{dv}{dx}=\frac{v\cos v+{\text{v}}^{\text{2}}\sin v-{\text{v}}^{\text{2}}\sin v+v\cos v}{\text{v}\sin v-\cos v}\\ \text{x}\frac{dv}{dx}=\frac{2v\cos v}{\text{v}\sin v-\cos v}\\ \left(\text{v}\sin v-\cos v\right)\frac{dv}{v\cos v}=\frac{2dx}{x}\\ \text{Integrating both sides, we get}\\ {\displaystyle \int \frac{\text{v}\sin v}{v\cos v}}dv-{\displaystyle \int \frac{\cos v}{v\cos v}}dv=2{\displaystyle \int \frac{1}{x}}dx\end{array} \begin{array}{l}\end{array} \begin{array}{l}\log \sec v-\log \left|v\right|=2\log \left|x\right|+\log C\\ \Rightarrow \log \left(\frac{\sec v}{v}\right)=\log C{x}^2\\ \Rightarrow \frac{\sec \frac{y}{x}}{\frac{y}{x}}=C{x}^2\\ \Rightarrow x\sec \frac{y}{x}=C{x}^{2}y\\ \Rightarrow xy\cos \left(\frac{y}{x}\right)=\frac{1}{C}\\ =k\left[\text{Let}k=\frac{1}{C}\right]\\ \Rightarrow xy\cos \left(\frac{y}{x}\right)=k\\ \text{This is the required solution of the given differential equation}\text{.}\end{array}

Q.67 Show that the given differential equation is homogeneous and solve each of them.

$\text{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}+\mathrm{xsin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=0$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}+\mathrm{xsin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}-\mathrm{xsin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}{\mathrm{x}}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\end{array}$ \begin{array}{l}\text{Let}\text{F}\left(x,y\right)=\frac{y-x\sin \left(\frac{y}{x}\right)}{x}\\ \text{Now,}\text{F}\left(\lambda x,\lambda y\right)=\frac{\lambda y-\lambda x\sin \left(\frac{\lambda y}{\lambda x}\right)}{\lambda x}\\ \text{=}\frac{\text{λ}\left\{\text{y-xsin}\left(\frac{\text{y}}{\text{x}}\right)\right\}}{\text{λx}}{\text{=λ}}^{\text{0}}\text{F}\left(\text{x,y}\right)\\ \text{Therefore, F}\left(\text{x,y}\right)\text{is a homogenous function of degree zero}\text{. So,}\\ \text{the given differential equation is a homogenous differential}\\ \text{equation}\text{.}\\ \text{For solving equation}\left(\text{i}\right)\text{, substitute}y\text{=}vx\text{and}\frac{dy}{dx}=\text{v}+\text{x}\frac{dv}{dx}\text{,}\\ \text{we get}\\ \text{v}+\text{x}\frac{dv}{dx}=\frac{\text{vx}-x\sin \left(\frac{\text{vx}}{x}\right)}{x}\\ =\frac{\left\{\text{v}-\sin v\right\}x}{x}\\ \Rightarrow \text{x}\frac{dv}{dx}=\text{v}-\sin v-v\\ \Rightarrow \frac{dv}{\sin v}=-\frac{dx}{x}\\ \Rightarrow \mathrm{cosec}vdv=-\frac{dx}{x}\\ \text{Integrating both sides, we get}\end{array} \begin{array}{l}{\displaystyle \int \mathrm{cosec}v}dv=-{\displaystyle \int \frac{1}{x}}dx\\ \Rightarrow \log \left|\mathrm{cosec}v-\cot v\right|=-\log \left|x\right|+\log C\\ \Rightarrow \log \left|\mathrm{cosec}v-\cot v\right|=\log \frac{C}{\left|x\right|}\\ \Rightarrow \mathrm{cosec}\frac{y}{x}-\cot \frac{y}{x}=\frac{C}{x}\\ \Rightarrow \frac{1}{\sin \frac{y}{x}}-\frac{\cos \frac{y}{x}}{\sin \frac{y}{x}}=\frac{C}{x}\\ \Rightarrow x\left\{1-\cos \left(\frac{y}{x}\right)\right\}=Csin\left(\frac{y}{x}\right)\\ \text{This is the required solution of the given differential equation}\text{.}\end{array}

Q.68 Show that the given differential equation is homogeneous and solve it.

$\text{y}\mathrm{dx}+\mathrm{x}\log \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\mathrm{dy}-2\mathrm{x}\mathrm{dy}=0$

Ans.

${\displaystyle \begin{array}{l}\text{The given differential equation is:}\\ \mathrm{y}\mathrm{d}\mathrm{x}+\mathrm{x}\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\mathrm{d}\mathrm{y}-<!--\; -\; -->2\mathrm{x}\mathrm{d}\mathrm{y}=0\\ \mathrm{y}\mathrm{d}\mathrm{x}+\left\{\mathrm{x}\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-<!--\; -\; -->2\mathrm{x}\right\}\mathrm{d}\mathrm{y}=0\\ \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{y}}{\left\{2\mathrm{x}-<!--\; -\; -->\mathrm{x}\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right\}}& ...\left(\mathrm{i}\right)\end{array}}$ ${\displaystyle \begin{array}{l}\mathrm{L}\mathrm{e}\mathrm{t}\mathrm{F}\left(\mathrm{x},\mathrm{y}\right)=\frac{\mathrm{y}}{\left\{2\mathrm{x}-<!--\; -\; -->\mathrm{x}\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right\}}\\ \mathrm{N}\mathrm{o}\mathrm{w},\mathrm{F}\left(\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{x},\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{y}\right)=\frac{\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{y}}{\left\{2\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{x}-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{x}\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{y}}{\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{x}}\right)\right\}}\\ =\frac{\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{y}}{\mathrm{\lambda <!--\; \lambda \; -->}\left\{2\mathrm{x}-<!--\; -\; -->\mathrm{x}\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right\}}={\mathrm{\lambda <!--\; \lambda \; -->}}^0\mathrm{F}\left(\mathrm{x},\mathrm{y}\right)\\ \text{Therefore, F}\left(\mathrm{x},\mathrm{y}\right)\text{is a homogenous function of degree zero}.\mathrm{S}\mathrm{o},\\ \text{the given differential equation is a homogenous differential}\\ \mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}.\\ \text{For solving equation}\left(\mathrm{i}\right),\mathrm{s}\mathrm{u}\mathrm{b}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{t}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{y}=\mathrm{v}\mathrm{x}\mathrm{a}\mathrm{n}\mathrm{d}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}},\\ \mathrm{w}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{t}\\ \mathrm{v}+\mathrm{x}\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{v}\mathrm{x}}{\left\{2\mathrm{x}-<!--\; -\; -->\mathrm{x}\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{\mathrm{v}\mathrm{x}}{\mathrm{x}}\right)\right\}}\\ \mathrm{x}\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{v}}{2-<!--\; -\; -->\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}}-<!--\; -\; -->\mathrm{v}\\ \mathrm{x}\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{v}-<!--\; -\; -->2\mathrm{v}+\mathrm{v}\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}}{2-<!--\; -\; -->\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}}\\ \mathrm{x}\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{v}\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}-<!--\; -\; -->\mathrm{v}}{2-<!--\; -\; -->\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}}\\ \left\{\frac{2-<!--\; -\; -->\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}}{\mathrm{v}\left(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}-<!--\; -\; -->1\right)}\right\}\mathrm{d}\mathrm{v}=\frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}\end{array}}$ ${\displaystyle \begin{array}{l}\left\{\frac{1+\left(1-<!--\; -\; -->\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}\right)}{\mathrm{v}\left(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}-<!--\; -\; -->1\right)}\right\}\mathrm{d}\mathrm{v}=\frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}\\ \left\{\frac{1}{\mathrm{v}\left(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}-<!--\; -\; -->1\right)}-<!--\; -\; -->\frac{1}{\mathrm{v}}\right\}\mathrm{d}\mathrm{v}=\frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}\\ \mathrm{I}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{b}\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{s},\mathrm{w}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{t}\\ \int <!--\; \int \; -->\frac{1}{\mathrm{v}\left(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}-<!--\; -\; -->1\right)}\mathrm{d}\mathrm{v}-<!--\; -\; -->\int <!--\; \int \; -->\frac{1}{\mathrm{v}}\mathrm{d}\mathrm{v}=\int <!--\; \int \; -->\frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}\\ \mathrm{l}\mathrm{o}\mathrm{g}\left|\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}-<!--\; -\; -->1\right|-<!--\; -\; -->\mathrm{l}\mathrm{o}\mathrm{g}\left|\mathrm{v}\right|=\mathrm{l}\mathrm{o}\mathrm{g}\left|\mathrm{x}\right|+\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{C}\\ \Rightarrow <!--\; \Rightarrow \; -->\mathrm{l}\mathrm{o}\mathrm{g}\frac{\left(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}-<!--\; -\; -->1\right)}{\mathrm{v}}=\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{C}\mathrm{x}\\ \Rightarrow <!--\; \Rightarrow \; -->\frac{\left(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{v}-<!--\; -\; -->1\right)}{\mathrm{v}}=\mathrm{C}\mathrm{x}\\ \Rightarrow <!--\; \Rightarrow \; -->\frac{\mathrm{x}}{\mathrm{y}}\left(\mathrm{l}\mathrm{o}\mathrm{g}\frac{\mathrm{y}}{\mathrm{x}}-<!--\; -\; -->1\right)=\mathrm{C}\mathrm{x}\\ \mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-<!--\; -\; -->1=\mathrm{C}\mathrm{y}\\ \mathrm{T}\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}.\end{array}}$

Q.69 Show that the given differential equation is homogeneous and solve each of them.

${\text{(1+e}}^{\frac{\mathrm{x}}{\mathrm{y}}})\mathrm{dx}+{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}(1-\frac{\mathrm{x}}{\mathrm{y}})\mathrm{dy}=0$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ (1+{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}})\mathrm{dx}+{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}(1-\frac{\mathrm{x}}{\mathrm{y}})\mathrm{dy}=0\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{(1+{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}})}{{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}(1-\frac{\mathrm{x}}{\mathrm{y}})}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dx}}{\mathrm{dy}}=-\frac{{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}(1-\frac{\mathrm{x}}{\mathrm{y}})}{(1+{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}})}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Let}\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})=-\frac{{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}(1-\frac{\mathrm{x}}{\mathrm{y}})}{(1+{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}})}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=-\frac{{\mathrm{e}}^{\frac{\mathrm{\lambda x}}{\mathrm{\lambda y}}}(1-\frac{\mathrm{\lambda x}}{\mathrm{\lambda y}})}{(1+{\mathrm{e}}^{\frac{\mathrm{\lambda x}}{\mathrm{\lambda y}}})}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\frac{{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}(1-\frac{\mathrm{x}}{\mathrm{y}})}{(1+{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}})}={\mathrm{\lambda }}^0\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x},\hspace{0.17em}\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\end{array}$ \begin{array}{l}\end{array} \begin{array}{l}\text{For solving equation}\left(\text{i}\right)\text{, substitute}x\text{=}vy\text{and}\frac{dx}{dy}=\text{v}+\text{y}\frac{dv}{dy}\text{,}\\ \text{we get}\\ \text{v}+\text{y}\frac{dv}{dy}=-\frac{e^{\frac{\text{vy}}{\text{y}}}\left(1-\frac{\text{vy}}{\text{y}}\right)}{\left(1+e^{\frac{\text{vy}}{\text{y}}}\right)}\\ =-\frac{e^v\left(1-v\right)}{\left(1+e^v\right)}\\ \text{y}\frac{dv}{dy}=-\frac{e^v\left(1-v\right)}{\left(1+e^v\right)}-v\\ =\frac{v{e}^v-e^v-v-v{e}^v}{\left(1+e^v\right)}\\ =-\frac{\left(e^v+v\right)}{\left(1+e^v\right)}\\ \frac{\left(1+e^v\right)}{\left(e^v+v\right)}dv=-\frac{dy}{y}\\ \text{Integrating both sides, we get}\\ {\displaystyle \int \frac{\left(1+e^v\right)}{\left(e^v+v\right)}}dv=-{\displaystyle \int \frac{1}{y}}dy\\ \Rightarrow \log \left|v+e^v\right|=-\log y+\log C\end{array} \begin{array}{l}\Rightarrow \log \left|\frac{x}{y}+e^{\frac{x}{y}}\right|=\log \frac{C}{y}\\ \Rightarrow \frac{x}{y}+e^{\frac{x}{y}}=\frac{C}{y}\\ \Rightarrow x+y{e}^{\frac{x}{y}}=C\\ \text{This is the required solution of the given differential equation}\text{.}\end{array}

Q.70 For the differential equation, find the particular situation satisfying the given condition:

(x + y) dy + (x – y) dx = 0; y = 1 when x = 1

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ (\mathrm{x}+\mathrm{y})\mathrm{dy}+(\mathrm{x}-\mathrm{y})\mathrm{dx}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(\mathrm{x}+\mathrm{y})\mathrm{dy}=-(\mathrm{x}-\mathrm{y})\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(\mathrm{y}-\mathrm{x})}{\mathrm{x}+\mathrm{y}}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{Let}\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})=\frac{(\mathrm{y}-\mathrm{x})}{\mathrm{x}+\mathrm{y}}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=\frac{(\mathrm{\lambda y}-\mathrm{\lambda x})}{\mathrm{\lambda x}+\mathrm{\lambda y}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{\lambda }(\mathrm{y}-\mathrm{x})}{\mathrm{\lambda }(\mathrm{x}+\mathrm{y})}={\mathrm{\lambda }}^0\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x}\mathrm{,}\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\end{array}$ $\begin{array}{l}\mathrm{equation}\mathrm{.}\\ \mathrm{For}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{substitute}\mathrm{y}\hspace{0.17em}=\mathrm{\hspace{0.17em}}\mathrm{vx}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{,}\\ \mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{vx}-\mathrm{x}}{\mathrm{x}+\mathrm{vx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{(\mathrm{v}-1)\mathrm{x}}{(1+\mathrm{v})\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{(\mathrm{v}-1)}{(1+\mathrm{v})}-\mathrm{v}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{\mathrm{v}-1-\mathrm{v}-{\mathrm{v}}^2}{(1+\mathrm{v})}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-(1+{\mathrm{v}}^2)}{(1+\mathrm{v})}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{(1+\mathrm{v})}{(1+{\mathrm{v}}^2)}\mathrm{dv}=-\frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\int \frac{1}{1+{\mathrm{v}}^2}\mathrm{\hspace{0.17em}}\mathrm{dv}+\int \frac{\mathrm{v}}{1+{\mathrm{v}}^2}\mathrm{\hspace{0.17em}}\mathrm{dv}=-\int \frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\tan }^{-1}\mathrm{v}+\frac{1}{2}\log (1+{\mathrm{v}}^2)=-\mathrm{logx}+\mathrm{k}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2{\tan }^{-1}\mathrm{v}+\log (1+{\mathrm{v}}^2)=-2\mathrm{logx}+2\mathrm{k}\\ 2{\tan }^{-1}\mathrm{v}+\log (1+{\mathrm{v}}^2)+2\mathrm{logx}=2\mathrm{k}\end{array}$ \begin{array}{l}\end{array} \begin{array}{l}2{\tan }^{-1}\frac{y}{x}+\log \left(1+{\frac{y}{x^2}}^2\right)x^2=2k\\ 2{\tan }^{-1}\frac{y}{x}+\log \left(x^2+y^2\right)=2k\mathrm{\dots }\left(ii\right)\\ \text{Now, putting y}=1atx=1.\\ 2{\tan }^{-1}\frac{1}{1}+\log \left(1^2+1^2\right)=2k\\ 2{\tan }^{-1}1+\log \left(2\right)=2k\\ 2\left(\frac{\pi }{4}\right)+\log 2=2k\\ \Rightarrow \left(\frac{\pi }{2}\right)+\log 2=2k\\ \text{Substituting the value of 2k in equation}\left(\text{ii}\right)\text{, we get}\\ 2{\tan }^{-1}\frac{y}{x}+\log \left(x^2+y^2\right)=\left(\frac{\pi }{2}\right)+\log 2\\ \text{This is the required solution of the given differential equation}\text{.}\end{array}

Q.71 For the differential equation, find the particular situation satisfying the given condition:

x² dy + (xy + y²) dx = 0; y = 1 when x = 1

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{x}}^2\mathrm{dy}+(\mathrm{xy}+{\mathrm{y}}^2)\mathrm{dx}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{x}}^2\mathrm{dy}=-(\mathrm{xy}+{\mathrm{y}}^2)\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{(\mathrm{xy}+{\mathrm{y}}^2)}{{\mathrm{x}}^2}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{x},\mathrm{y})=-\frac{(\mathrm{xy}+{\mathrm{y}}^2)}{{\mathrm{x}}^2}\end{array}$ $\begin{array}{l}\mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=-\frac{({\mathrm{\lambda }}^2\mathrm{xy}+{\mathrm{\lambda }}^2{\mathrm{y}}^2)}{{\mathrm{\lambda }}^2{\mathrm{x}}^2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\frac{{\mathrm{\lambda }}^2(\mathrm{xy}+{\mathrm{y}}^2)}{{\mathrm{\lambda }}^2{\mathrm{x}}^2}={\mathrm{\lambda }}^0\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x},\hspace{0.17em}\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\\ \mathrm{For}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{substitute}\mathrm{y}\hspace{0.17em}=\mathrm{\hspace{0.17em}}\mathrm{vx}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{,}\\ \mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{(\mathrm{xvx}+{\mathrm{v}}^2{\mathrm{x}}^2)}{{\mathrm{x}}^2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\frac{(\mathrm{v}+{\mathrm{v}}^2){\mathrm{x}}^2}{{\mathrm{x}}^2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=-\mathrm{v}-{\mathrm{v}}^2-\mathrm{v}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-{\mathrm{v}}^2-2\mathrm{v}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dv}}{\mathrm{v}(\mathrm{v}+2)}=-\frac{\mathrm{dx}}{\mathrm{x}}\\ \frac{1}{2}\mathrm{\hspace{0.17em}}\frac{\{(\mathrm{v}+2)-\mathrm{v}\}\mathrm{dv}}{\mathrm{v}(\mathrm{v}+2)}=-\frac{\mathrm{dx}}{\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\frac{1}{2}(\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{v}+2})\mathrm{dv}=-\frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \frac{1}{2}\int \frac{1}{\mathrm{v}}\mathrm{\hspace{0.17em}}\mathrm{dv}-\frac{1}{2}\int \frac{1}{\mathrm{v}+2}\mathrm{\hspace{0.17em}}\mathrm{dv}=-\int \frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\end{array}$ $\begin{array}{l}\mathrm{\hspace{0.17em}}\frac{1}{2}\log \left|\mathrm{v}\right|-\frac{1}{2}\log |\mathrm{v}+2|=-\log \left|\mathrm{x}\right|+\mathrm{logC}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\log \left(\frac{\mathrm{v}}{\mathrm{v}+2}\right)=2\log \left(\frac{\mathrm{C}}{\mathrm{x}}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\log \left(\frac{\frac{\mathrm{y}}{\mathrm{x}}}{\frac{\mathrm{y}}{\mathrm{x}}+2}\right)=\log {\left(\frac{\mathrm{C}}{\mathrm{x}}\right)}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\log \left(\frac{\mathrm{y}}{\mathrm{y}+2\mathrm{x}}\right)=\log {\left(\frac{\mathrm{C}}{\mathrm{x}}\right)}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{y}}{\mathrm{y}+2\mathrm{x}}={\left(\frac{\mathrm{C}}{\mathrm{x}}\right)}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{{\mathrm{x}}^2\mathrm{y}}{\mathrm{y}+2\mathrm{x}}={\mathrm{C}}^2...\left(\mathrm{ii}\right)\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{y}=1\mathrm{when}\mathrm{x}=\mathrm{1}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{1^2\left(1\right)}{1+2\left(1\right)}={\mathrm{C}}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{3}={\mathrm{C}}^2\\ \mathrm{Substituting}\mathrm{value}\mathrm{of}{\mathrm{C}}^{\mathrm{2}}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{{\mathrm{x}}^2\mathrm{y}}{\mathrm{y}+2\mathrm{x}}=\frac{1}{3}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3{\mathrm{x}}^2\mathrm{y}=\mathrm{y}+2\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}+2\mathrm{x}=3{\mathrm{x}}^2\mathrm{y}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.72 For the differential equations, find the particular situation satisfying the given condition:

${\text{[xsin}}^2\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{y}]\mathrm{dx}+\mathrm{xdy}=0;\mathrm{y}=\frac{\mathrm{\pi }}{}\mathrm{}\mathrm{when}\mathrm{x}=1$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}[{\mathrm{xsin}}^2\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{y}]\mathrm{dx}+\mathrm{xdy}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{xdy}=-[{\mathrm{xsin}}^2\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{y}]\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{[{\mathrm{xsin}}^2\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{y}]}{\mathrm{x}}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{x},\mathrm{y})=-\frac{[{\mathrm{xsin}}^2\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{y}]}{\mathrm{x}}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=-\frac{[{\mathrm{\lambda xsin}}^2\left(\frac{\mathrm{\lambda y}}{\mathrm{\lambda x}}\right)-\mathrm{\lambda y}]}{\mathrm{\lambda x}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-\frac{\mathrm{\lambda }[{\mathrm{xsin}}^2\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{y}]}{\mathrm{\lambda x}}={\mathrm{\lambda }}^0\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x},\hspace{0.17em}\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\end{array}$ $\begin{array}{l}\mathrm{For}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{substitute}\mathrm{y}=\mathrm{vx}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{,}\\ \mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{[{\mathrm{xsin}}^2\left(\frac{\mathrm{vx}}{\mathrm{x}}\right)-\mathrm{vx}]}{\mathrm{x}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=-{\sin }^2\mathrm{v}+\mathrm{v}-\mathrm{v}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{cosec}}^{\mathrm{2}}\mathrm{v}\mathrm{\hspace{0.17em}}\mathrm{dv}=-\frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\int {\mathrm{cosec}}^{\mathrm{2}}\mathrm{v}\mathrm{\hspace{0.17em}}\mathrm{dv}=-\int \frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\mathrm{cotv}=-\log \left|\mathrm{x}\right|+\mathrm{logC}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\cot \frac{\mathrm{y}}{\mathrm{x}}=\log \left|\mathrm{Cx}\right|...\left(\mathrm{ii}\right)\\ \mathrm{Now},\mathrm{y}=\frac{\mathrm{\pi }}{4}\mathrm{at}\mathrm{x}=1\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\cot \frac{\left(\frac{\mathrm{\pi }}{4}\right)}{1}=\log |\mathrm{C}.1|\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1=\log \left|\mathrm{C}\right|\Rightarrow \mathrm{C}=\mathrm{e}\\ \mathrm{Substituting}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\cot \frac{\mathrm{y}}{\mathrm{x}}=\log \left|\mathrm{ex}\right|\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.73 For the differential equations, find the particular situation satisfying the given condition:

$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}+\mathrm{cosec}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=0;\mathrm{}\mathrm{y}=0\mathrm{when}\mathrm{x}=1$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}+\mathrm{cosec}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}-\mathrm{cosec}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{x},\mathrm{y})=\frac{\mathrm{y}}{\mathrm{x}}-\mathrm{cosec}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=\frac{\mathrm{\lambda y}}{\mathrm{\lambda x}}-\mathrm{cosec}\left(\frac{\mathrm{\lambda y}}{\mathrm{\lambda x}}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{\mathrm{y}}{\mathrm{x}}-\mathrm{cosec}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)={\mathrm{\lambda }}^0\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x},\hspace{0.17em}\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\\ \mathrm{For}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{substitute}\mathrm{y}\hspace{0.17em}=\mathrm{\hspace{0.17em}}\mathrm{vx}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{,}\\ \mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{vx}}{\mathrm{x}}-\mathrm{cosec}\left(\frac{\mathrm{vx}}{\mathrm{x}}\right)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}-\mathrm{cosecv}-\mathrm{v}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dv}}{\mathrm{cosecv}}=-\frac{\mathrm{dx}}{\mathrm{x}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}-\int \mathrm{sinv}\mathrm{\hspace{0.17em}}\mathrm{dv}=\int \frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{cosv}=\log \left|\mathrm{x}\right|+\mathrm{logC}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\log \left|\mathrm{Cx}\right|\mathrm{\hspace{0.17em}}...\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}\mathrm{Now},\mathrm{y}=0\mathrm{at}\mathrm{x}=\mathrm{1}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\cos \left(\frac{0}{1}\right)=\log |\mathrm{C}.1|\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\cos 0=\mathrm{logC}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1=\mathrm{logC}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{C}=\mathrm{e}\\ \mathrm{Putting}\mathrm{C}=\mathrm{e}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\log \left|\mathrm{ex}\right|\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.74 For the differential equation given below, find a particular situation satisfying the given condition:

${\text{2xy+y}}^2-2{\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}=0;\mathrm{‹}\mathrm{y}=2\mathrm{whenx}=1$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}2\mathrm{xy}+{\mathrm{y}}^2-2{\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ \hspace{0.17em}\hspace{0.17em}-2{\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}=-(2\mathrm{xy}+{\mathrm{y}}^2)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(2\mathrm{xy}+{\mathrm{y}}^2)}{2{\mathrm{x}}^2}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{x},\mathrm{y})=\frac{(2\mathrm{xy}+{\mathrm{y}}^2)}{2{\mathrm{x}}^2}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=\frac{({\mathrm{\lambda }}^{2}2\mathrm{xy}+{\mathrm{\lambda }}^2{\mathrm{y}}^2)}{2{\mathrm{\lambda }}^2{\mathrm{x}}^2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{{\mathrm{\lambda }}^2(2\mathrm{xy}+{\mathrm{y}}^2)}{2{\mathrm{\lambda }}^2{\mathrm{x}}^2}={\mathrm{\lambda }}^0\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})\end{array}$ $\begin{array}{l}\mathrm{Therefore},\mathrm{F}(\mathrm{x},\hspace{0.17em}\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\\ \mathrm{For}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{substitute}\mathrm{y}\hspace{0.17em}=\mathrm{\hspace{0.17em}}\mathrm{vx}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{,}\\ \mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{(2\mathrm{xvx}+{\mathrm{v}}^2{\mathrm{x}}^2)}{2{\mathrm{x}}^2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{(2\mathrm{v}+{\mathrm{v}}^2){\mathrm{x}}^2}{2{\mathrm{x}}^2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{2\mathrm{v}+{\mathrm{v}}^2}{2}-\mathrm{v}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{{\mathrm{v}}^2}{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dv}}{{\mathrm{v}}^2}=\frac{\mathrm{dx}}{2\mathrm{x}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int 2{\mathrm{v}}^{-2}\mathrm{\hspace{0.17em}}\mathrm{dv}=\int \frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{2{\mathrm{v}}^{-1}}{-1}=\log \left|\mathrm{x}\right|+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{2}{\mathrm{v}}=\log \left|\mathrm{x}\right|+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{2}{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}=\log \left|\mathrm{x}\right|+\mathrm{C}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{2\mathrm{x}}{\mathrm{y}}=\log \left|\mathrm{x}\right|+\mathrm{C}...\left(\mathrm{ii}\right)\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\mathrm{y}=2\mathrm{at}\mathrm{x}=1.\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{2\left(1\right)}{2}=\log \left|1\right|+\mathrm{C}\Rightarrow \mathrm{C}=-1\\ \mathrm{Substituting}\mathrm{C}=\; –1\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{2\mathrm{x}}{\mathrm{y}}=\log \left|\mathrm{x}\right|-1\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\frac{2\mathrm{x}}{1-\log \left|\mathrm{x}\right|}(\mathrm{x}\ne 0,\mathrm{x}\ne \mathrm{e})\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.75

$\begin{array}{l}\mathrm{A}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{h}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{by}\mathrm{making}\mathrm{the}\mathrm{substitution}\mathrm{.}\\ \left(\mathrm{A}\right)\mathrm{\hspace{0.17em}}\mathrm{y}=\mathrm{vx}\left(\mathrm{B}\right)\mathrm{\hspace{0.17em}}\mathrm{v}=\mathrm{yx}\left(\mathrm{C}\right)\mathrm{\hspace{0.17em}}\mathrm{x}=\mathrm{vy}\left(\mathrm{D}\right)\mathrm{\hspace{0.17em}}\mathrm{x}=\mathrm{v}\end{array}$

Ans.

$\begin{array}{l}\mathrm{A}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{h}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{by}\mathrm{making}\mathrm{the}\mathrm{substitution}\\ \mathrm{as}\mathrm{x}=\mathrm{vy}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{C}\mathrm{.}\end{array}$

Q.76 Which of the following is a homogeneous differential equation?
(A) (4x + 6y + 5)dy – (3y + 2x + 4) dx = 0

(B) (xy) dx – (x³ + y³) dy = 0

(C) (x³ + 2y²) dx + 2xy dy = 0

(D) y² dx + (x²– xy – y²) dy = 0

Ans.

A function F(x, y) is said to be homogeneous function of degree n if F(λx, λy)= λⁿ F(x, y) for any nonzero constant λ.

$\begin{array}{l}\mathrm{Let}\mathrm{us}\mathrm{check}\mathrm{option}\mathrm{D},\\ {\mathrm{y}}^2\mathrm{dx}+({\mathrm{x}}^2-\mathrm{xy}-{\mathrm{y}}^2)\mathrm{dy}=0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}({\mathrm{x}}^2-\mathrm{xy}-{\mathrm{y}}^2)\mathrm{dy}=-{\mathrm{y}}^2\mathrm{dx}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{{\mathrm{y}}^2}{({\mathrm{x}}^2-\mathrm{xy}-{\mathrm{y}}^2)}\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{F}(\mathrm{x},\mathrm{y})=-\frac{{\mathrm{y}}^2}{({\mathrm{x}}^2-\mathrm{xy}-{\mathrm{y}}^2)}\\ \mathrm{then}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{\lambda x},\mathrm{\hspace{0.17em}}\mathrm{\lambda y})=-\frac{{\mathrm{\lambda }}^2{\mathrm{y}}^2}{({\mathrm{\lambda }}^2{\mathrm{x}}^2-{\mathrm{\lambda }}^2\mathrm{xy}-{\mathrm{\lambda }}^2{\mathrm{y}}^2)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-{\mathrm{\lambda }}^2\frac{{\mathrm{\lambda }}^2{\mathrm{y}}^2}{({\mathrm{x}}^2-\mathrm{xy}-{\mathrm{y}}^2)}={\mathrm{\lambda }}^0\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{in}\mathrm{option}\mathrm{D}\mathrm{is}\\ \mathrm{homogenous}\mathrm{equation}\mathrm{.}\end{array}$

Q.77 Find the general solution of the differential equation given below.

$\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y}=\mathrm{sinx}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y}=\mathrm{sinx}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=2\mathrm{andQ}=\mathrm{sinx}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int 2\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{2\mathrm{x}}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{relation},\\ \mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{2\mathrm{x}}=\int (\mathrm{sinx}\times {\mathrm{e}}^{2\mathrm{x}})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\hspace{0.17em}\hspace{0.17em}...\mathrm{\hspace{0.17em}}\left(\mathrm{i}\right)\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{I}=\int {\mathrm{sinxe}}^{2\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}=\mathrm{sinx}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx})\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}=\mathrm{sinx}\left[\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right]-\int \mathrm{cosx}.\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}=\frac{1}{2}\mathrm{sinx}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{2\mathrm{x}}-\frac{1}{2}\int \mathrm{cosx}.\mathrm{\hspace{0.17em}}{\mathrm{e}}^{2\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\\ \hspace{0.17em}\hspace{0.17em}=\frac{1}{2}\mathrm{sinx}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{2\mathrm{x}}-\frac{1}{2}\{\mathrm{cosx}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx})\mathrm{\hspace{0.17em}}\mathrm{dx}\}\\ \hspace{0.17em}\hspace{0.17em}=\frac{1}{2}\mathrm{sinx}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{2\mathrm{x}}-\frac{1}{2}\{\mathrm{cosx}\left[\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right]-\int (-\mathrm{sinx}\frac{{\mathrm{e}}^{2\mathrm{x}}}{2})\mathrm{\hspace{0.17em}}\mathrm{dx}\}\end{array}$ \begin{array}{l}I=\frac{1}{2}sinx{e}^{2x}-\frac{1}{4}\left(cosx{e}^{2x}+I\right)\\ I=\frac{1}{2}sinx{e}^{2x}-\frac{1}{4}cosx{e}^{2x}-\frac{1}{4}I\\ \frac{5}{4}I=\frac{1}{4}{e}^{2x}\left(2sinx-cosx\right)\\ I=\frac{1}{5}{e}^{2x}\left(2sinx-cosx\right)\\ \text{So, from equation}\left(\text{i}\right)\text{, we have}\\ y{e}^{2x}=\frac{1}{5}{e}^{2x}\left(2sinx-cosx\right)+C\\ \Rightarrow y=\frac{1}{5}\left(2sinx-cosx\right)+C{e}^{–2x}\\ \text{This is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}

Q.78 Find the general solution of the differential equation given below.

$\frac{\mathrm{dy}}{\mathrm{dx}}+3\mathrm{y}={\mathrm{e}}^{-2\mathrm{x}}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\frac{\mathrm{dy}}{\mathrm{dx}}+3\mathrm{y}={\mathrm{e}}^{-2\mathrm{x}}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=3\mathrm{and}\mathrm{Q}={\mathrm{e}}^{-2\mathrm{x}}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int 3\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{3\mathrm{x}}\end{array}$ \begin{array}{l}\text{The solution of given differential equation is given by the relation,}\\ \text{y}\left(I.F.\right)={\displaystyle \int \left(Q\times I.F.\right)}dx+C\\ y{e}^{3x}={\displaystyle \int \left(e^{-2x}\times e^{3x}\right)}dx+C\\ ={\displaystyle \int e^x}dx+C\\ y{e}^{3x}=e^x+C\\ y=e^{-2x}+C{e}^{-3x}\\ \text{This is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}

Q.79 Find the general solution of the differential equation given below.

\frac{dy}{dx}+\frac{y}{x}=x^2

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}={\mathrm{x}}^2\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=\frac{1}{\mathrm{x}}\mathrm{and}\mathrm{Q}={\mathrm{x}}^2\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int \frac{1}{\mathrm{x}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\mathrm{logx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{x}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{relation},\\ \mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{yx}=\int ({\mathrm{x}}^2\times \mathrm{x})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\end{array}$ \begin{array}{l}={\displaystyle \int x^3}dx+C\\ xy=\frac{x^4}{4}+C\\ \text{This is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}

Q.80 Find the general solution of the differential equation given below.

$\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\mathrm{secx}\right)\mathrm{y}=\mathrm{tanx}(\mathbf{0}\le \mathbf{x}<\frac{\mathrm{\pi }}{2})$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\mathrm{secx}\right)\mathrm{y}=\mathrm{tanx}(0\le \mathrm{x}<\frac{\mathrm{\pi }}{2})\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=\mathrm{secx}\mathrm{and}\mathrm{Q}=\mathrm{tanx}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int \mathrm{secx}\mathrm{\hspace{0.17em}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\log (\mathrm{secx}+\mathrm{tanx})}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{secx}+\mathrm{tanx}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{relation},\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}(\mathrm{secx}+\mathrm{tanx})=\int (\mathrm{secx}+\mathrm{tanx})\mathrm{tanx}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}(\mathrm{secx}+\mathrm{tanx})=\int \mathrm{secxtanx}\mathrm{\hspace{0.17em}}\mathrm{dx}+\int {\tan }^2\mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{secx}+\int ({\sec }^2\mathrm{x}-1)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{secx}+\mathrm{tanx}-\mathrm{x}+\mathrm{C}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\end{array}$

Q.81 Find the general solution of the differential equation given below.

${\text{cos}}^2\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=\tan \mathrm{\hspace{0.17em}}\mathrm{x}(\mathbf{0}\le \mathbf{x}<\frac{\mathrm{\pi }}{2})$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}{\cos }^2\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=\tan \mathrm{\hspace{0.17em}}\mathrm{x}(0\le \mathrm{x}<\frac{\mathrm{\pi }}{2})\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\left({\sec }^2\mathrm{x}\right)\mathrm{y}={\sec }^2\mathrm{xtan}\mathrm{\hspace{0.17em}}\mathrm{x}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}={\sec }^2\mathrm{x}\mathrm{and}\mathrm{Q}={\sec }^2\mathrm{xtan}\mathrm{\hspace{0.17em}}\mathrm{x}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int {\sec }^2\mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\mathrm{tanx}}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{relation},\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{\mathrm{tanx}}=\int ({\sec }^2\mathrm{x}\hspace{0.17em}\hspace{0.17em}\mathrm{tanx})\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{tanx}}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{\mathrm{tanx}}=\int {\mathrm{te}}^{\mathrm{t}}{\sec }^2\mathrm{x}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dt}}{{\sec }^2\mathrm{x}}+\mathrm{C}\left[\begin{array}{l}\mathrm{Let}\mathrm{\hspace{0.17em}}\mathrm{t}=\mathrm{tanx}\\ \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}={\sec }^2\mathrm{x}\end{array}\right]\\ \hspace{0.17em}\hspace{0.17em}{\mathrm{ye}}^{\mathrm{tanx}}=\int {\mathrm{te}}^{\mathrm{t}}\mathrm{\hspace{0.17em}}\mathrm{dt}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{t}\int {\mathrm{e}}^{\mathrm{t}}\mathrm{\hspace{0.17em}}\mathrm{dt}-\int (\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{t}\mathrm{\hspace{0.17em}}\int {\mathrm{e}}^{\mathrm{t}}\mathrm{\hspace{0.17em}}\mathrm{dt})\mathrm{\hspace{0.17em}}\mathrm{dt}+\mathrm{C}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\mathrm{te}}^{\mathrm{t}}-\int 1.{\mathrm{e}}^{\mathrm{t}}\mathrm{\hspace{0.17em}}\mathrm{dt}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\mathrm{te}}^{\mathrm{t}}-{\mathrm{e}}^{\mathrm{t}}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\mathrm{e}}^{\mathrm{t}}(\mathrm{t}-1)+\mathrm{C}\\ \mathrm{\hspace{0.17em}}{\mathrm{ye}}^{\mathrm{tanx}}={\mathrm{e}}^{\mathrm{tanx}}(\mathrm{tanx}-1)+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}=(\mathrm{tanx}-1)+{\mathrm{Ce}}^{-\mathrm{tanx}}\end{array}$ \begin{array}{l}\text{This is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}

Q.82 Find the general solution of the differential equation given below.

$\text{x}\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y}={\mathrm{x}}^2\mathrm{\hspace{0.17em}}\mathrm{logx}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y}={\mathrm{x}}^2\mathrm{\hspace{0.17em}}\mathrm{logx}\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+2\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{logx}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=\frac{2}{\mathrm{x}}\mathrm{and}\mathrm{Q}=\mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{logx}\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int \frac{2}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={{\mathrm{e}}^2}^{\mathrm{logx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{{}^{{\mathrm{logx}}^2}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{x}}^2\end{array}$ $\begin{array}{l}\mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{relation},\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{yx}}^2=\int \mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{logx}.\mathrm{\hspace{0.17em}}{\mathrm{x}}^2\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{yx}}^2=\int {\mathrm{x}}^3\mathrm{\hspace{0.17em}}\mathrm{logx}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{logx}\int {\mathrm{x}}^3\mathrm{\hspace{0.17em}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\int {\mathrm{x}}^3\mathrm{\hspace{0.17em}}\mathrm{dx})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{logx}.\left[\frac{{\mathrm{x}}^4}{4}\right]-\int \left(\frac{1}{\mathrm{x}}\left[\frac{{\mathrm{x}}^4}{4}\right]\right)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{logx}.\left[\frac{{\mathrm{x}}^4}{4}\right]-\frac{1}{4}\int {\mathrm{x}}^3\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{logx}.\left[\frac{{\mathrm{x}}^4}{4}\right]-\frac{1}{4}\left[\frac{{\mathrm{x}}^4}{4}\right]+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{yx}}^2=\frac{{\mathrm{x}}^4}{16}(4\mathrm{logx}-1)+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\frac{{\mathrm{x}}^2}{16}(4\mathrm{logx}-1)+{\mathrm{Cx}}^{-2}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\end{array}$

Q.83 Find the general solution of the differential equation given below.

$\text{xlogx}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=\frac{}{\mathrm{x}}\mathrm{logx}$

Ans.

$\text{The given differential equation is xlogx}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=\frac{2}{\mathrm{x}}\mathrm{logx}$ $\begin{array}{l}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{xlogx}}=\frac{2}{{\mathrm{x}}^2}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=\frac{1}{\mathrm{xlogx}}\mathrm{and}\mathrm{Q}=\frac{2}{{\mathrm{x}}^2}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int \frac{1}{\mathrm{xlogx}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\log \left(\mathrm{logx}\right)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{logx}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\\ \mathrm{relation},\\ \mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \mathrm{ylogx}=\int (\frac{2}{{\mathrm{x}}^2}\times \mathrm{logx})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \mathrm{ylogx}=\mathrm{logx}\int \frac{2}{{\mathrm{x}}^2}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}\int \frac{2}{{\mathrm{x}}^2}\mathrm{\hspace{0.17em}}\mathrm{dx})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \mathrm{ylogx}=2\{\mathrm{logx}(-\frac{1}{\mathrm{x}})-\int \frac{1}{\mathrm{x}}\times -\frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\}+\mathrm{C}\\ \mathrm{ylogx}=2\{-\frac{1}{\mathrm{x}}\mathrm{logx}+\int {\mathrm{x}}^{-2}\mathrm{\hspace{0.17em}}\mathrm{dx}\}+\mathrm{C}\\ \mathrm{ylogx}=2(-\frac{1}{\mathrm{x}}\mathrm{logx}+\frac{{\mathrm{x}}^{-1}}{-1})+\mathrm{C}\end{array}$ $\begin{array}{l}\mathrm{ylogx}=-\frac{2}{\mathrm{x}}\mathrm{logx}-\frac{2}{\mathrm{x}}+\mathrm{C}\\ \mathrm{ylogx}=-\frac{2}{\mathrm{x}}(\mathrm{logx}+1)+\mathrm{C}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\end{array}$

Q.84 Find the general solution of the differential equation given below.

${\text{(1+x}}^2)\mathrm{dy}+2\mathrm{xydx}=\mathrm{cotx}\mathrm{dx}(\mathrm{x}\ne 0)$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(1+{\mathrm{x}}^2)\mathrm{dy}+2\mathrm{xydx}=\mathrm{cotxdx}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{x}\ne 0)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2\mathrm{x}}{(1+{\mathrm{x}}^2)}\mathrm{y}=\frac{\mathrm{cotx}}{(1+{\mathrm{x}}^2)}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=\frac{2\mathrm{x}}{(1+{\mathrm{x}}^2)}\mathrm{and}\mathrm{Q}=\frac{\mathrm{cotx}}{(1+{\mathrm{x}}^2)}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int \frac{2\mathrm{x}}{(1+{\mathrm{x}}^2)}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\log (1+{\mathrm{x}}^2)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(1+{\mathrm{x}}^2)\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\\ \end{array}$

relation,

\begin{array}{l}\text{y}\left(I.F.\right)={\displaystyle \int \left(Q\times I.F.\right)}dx+C\\ y\left(1+x^2\right)={\displaystyle \int \left\{\frac{\cot x}{\left(1+x^2\right)}\times \left(1+x^2\right)\right\}}dx+C\\ y\left(1+x^2\right)={\displaystyle \int \cot x}dx+C\\ y\left(1+x^2\right)=\log \left|\sin x\right|+C\\ y={\left(1+x^2\right)}^{-1}\log \left|\sin x\right|+C{\left(1+x^2\right)}^{-1}\\ \text{This is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}

Q.85 Find the general solution of the differential equation given below.

$\text{x}\frac{dy}{dx}+y-x+xycotx=0$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\\ \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}-\mathrm{x}+\mathrm{xycotx}=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{(1+\mathrm{xcotx})}{\mathrm{x}}\mathrm{y}=1\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=\frac{(1+\mathrm{xcotx})}{\mathrm{x}}\mathrm{}\mathrm{and}\mathrm{}\mathrm{Q}=1\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int \frac{(1+\mathrm{xcotx})}{\mathrm{x}}\mathrm{dx}}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int (\frac{1}{\mathrm{x}}+\mathrm{cotx})\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\log \left|\mathrm{x}\right|+\log \left|\mathrm{sinx}\right|}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={{\mathrm{e}}^{\log }}^{\left|\mathrm{xsinx}\right|}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{xsinx}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\\ \mathrm{relation},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \mathrm{y}\left(\mathrm{xsinx}\right)=\int \{1\times \left(\mathrm{xsinx}\right)\}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \mathrm{y}\left(\mathrm{xsinx}\right)=\int \mathrm{xsinx}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \mathrm{y}\left(\mathrm{xsinx}\right)=\mathrm{x}\int \mathrm{sinx}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\int \mathrm{sinx}\mathrm{\hspace{0.17em}}\mathrm{dx})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \mathrm{y}\left(\mathrm{xsinx}\right)=-\mathrm{xcosx}-\int (-\mathrm{cosx})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \mathrm{y}\left(\mathrm{xsinx}\right)=-\mathrm{xcosx}+\mathrm{sinx}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\frac{1}{\mathrm{x}}-\mathrm{cotx}+\frac{\mathrm{C}}{\mathrm{x}}\mathrm{cosecx}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\end{array}$

Q.86 Find the general solution of the differential equation given below.

$\text{(x+y)}\frac{\mathrm{dy}}{\mathrm{dx}}=1$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}(\mathrm{x}+\mathrm{y})\frac{\mathrm{dy}}{\mathrm{dx}}=1\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{(\mathrm{x}+\mathrm{y})}\end{array}$ $\begin{array}{l}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=(\mathrm{x}+\mathrm{y})\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dx}}{\mathrm{dy}}-\mathrm{x}=\mathrm{y}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=-1\mathrm{}\mathrm{and}\mathrm{}\mathrm{Q}=\mathrm{y}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdy}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int -1\mathrm{dy}}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\mathrm{e}}^{-\mathrm{y}}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\\ \mathrm{relation},\\ \mathrm{x}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{xe}}^{-\mathrm{y}}=\int \{\mathrm{y}\times {\mathrm{e}}^{-\mathrm{y}}\}\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}{\mathrm{xe}}^{-\mathrm{y}}=\int {\mathrm{ye}}^{-\mathrm{y}}\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ {\mathrm{xe}}^{-\mathrm{y}}=\mathrm{y}\int {\mathrm{e}}^{-\mathrm{y}}\mathrm{\hspace{0.17em}}\mathrm{dy}-\int (\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{y}\int {\mathrm{e}}^{-\mathrm{y}}\mathrm{\hspace{0.17em}}\mathrm{dy})\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ =-{\mathrm{ye}}^{-\mathrm{y}}+\int (1.{\mathrm{e}}^{-\mathrm{y}})\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-{\mathrm{ye}}^{-\mathrm{y}}-{\mathrm{e}}^{-\mathrm{y}}+\mathrm{C}\\ \Rightarrow \mathrm{x}=-\mathrm{y}-1+{\mathrm{Ce}}^{\mathrm{y}}(\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}{\mathrm{e}}^{–\mathrm{y}})\\ \Rightarrow \mathrm{x}+\mathrm{y}+1={\mathrm{Ce}}^{\mathrm{y}}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\end{array}$

Q.87 Find the general solution of the differential equation given below.

$\text{ydx+}{\text{(x−y}}^2)\mathrm{dy}=0$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{ydx}+(\mathrm{x}-{\mathrm{y}}^2)\mathrm{dy}=0\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{(\mathrm{x}-{\mathrm{y}}^2)}\\ \Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=-\frac{(\mathrm{x}-{\mathrm{y}}^2)}{\mathrm{y}}\\ \Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=-(\frac{\mathrm{x}}{\mathrm{y}}-\mathrm{y})\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{y}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=\frac{1}{\mathrm{y}}\mathrm{}\mathrm{and}\mathrm{}\mathrm{Q}=\mathrm{y}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdy}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int \frac{1}{\mathrm{y}}\mathrm{dy}}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\mathrm{logy}}=\mathrm{y}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\\ \mathrm{relation},\\ \mathrm{x}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{xy}=\int \{\mathrm{y}\times \mathrm{y}\}\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{xy}=\int {\mathrm{y}}^2\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\end{array}$ \begin{array}{l}xy=\frac{y^3}{3}+C\\ x=\frac{y^2}{3}+\frac{C}{y}\\ \text{This is the required general solution of the given differential}\\ \text{equation}\text{.}\end{array}

Q.88 Find the general solution of the differential equation given below.

${\text{(x+3y}}^2)\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}(\mathrm{y}>0)$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}(\mathrm{x}+3{\mathrm{y}}^2)\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}(\mathrm{y}>0)\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{(\mathrm{x}+3{\mathrm{y}}^2)}\\ \Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{(\mathrm{x}+3{\mathrm{y}}^2)}{\mathrm{y}}\\ \Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=(\frac{\mathrm{x}}{\mathrm{y}}+3\mathrm{y})\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dx}}{\mathrm{dy}}-\frac{\mathrm{x}}{\mathrm{y}}=3\mathrm{y}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=-\frac{1}{\mathrm{y}}\mathrm{}\mathrm{and}\mathrm{}\mathrm{Q}=3\mathrm{y}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdy}}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int -\frac{1}{\mathrm{y}}\mathrm{dy}}\hspace{0.17em}\hspace{0.17em}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{-\mathrm{logy}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{\mathrm{y}}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\\ \mathrm{relation},\\ \mathrm{x}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}\frac{1}{\mathrm{y}}=\int \{3\mathrm{y}\times \frac{1}{\mathrm{y}}\}\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\frac{1}{\mathrm{y}}=\int 3\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}=3\mathrm{y}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}=3{\mathrm{y}}^2+\mathrm{Cy}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\end{array}$

Q.89 For the differential equations, find a particular solution satisfying the given condition:

$\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{ytanx}=\mathrm{sinx};\mathrm{y}=0\mathrm{whenx}=\frac{\mathrm{\pi }}{}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{ytanx}=\mathrm{sinx};\\ \mathrm{y}=0\mathrm{when}\mathrm{x}=\frac{\mathrm{\pi }}{3}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\end{array}$ $\begin{array}{l}\mathrm{P}=2\mathrm{tanx}\mathrm{and}\mathrm{Q}=\mathrm{sinx}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int 2\mathrm{tanx}\mathrm{\hspace{0.17em}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={{\mathrm{e}}^2}^{\mathrm{logsecx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{{}^{{\mathrm{logsec}}^2\mathrm{x}}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\sec }^2\mathrm{x}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{relation},\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ {\mathrm{ysec}}^2\mathrm{x}=\int (\mathrm{sinx}\times {\sec }^2\mathrm{x})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ {\mathrm{ysec}}^2\mathrm{x}=\int \mathrm{tanx}.\mathrm{secx}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\mathrm{secx}+\mathrm{C}\mathrm{‹}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\mathrm{cosx}+{\mathrm{Ccos}}^2\mathrm{x}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{This}\mathrm{‹}\mathrm{is}\mathrm{the}\mathrm{general}\mathrm{equation}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{y}=0\hspace{0.17em}\hspace{0.17em}\mathrm{when}\mathrm{\hspace{0.17em}}\mathrm{x}=\frac{\mathrm{\pi }}{3},\mathrm{\hspace{0.17em}}\mathrm{so}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0=\cos \frac{\mathrm{\pi }}{3}+{\mathrm{Ccos}}^2\frac{\mathrm{\pi }}{3}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}0=\frac{1}{2}+\mathrm{C}\frac{1}{4}\Rightarrow \mathrm{C}=-2\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\mathrm{cosx}-2{\cos }^2\mathrm{x}\\ \mathrm{This}\mathrm{‹}\mathrm{is}\mathrm{the}\mathrm{particular}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.90 For the differential equation given below, find a particular solution satisfying the given condition:

${\text{(1+x}}^2)\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{xy}=\frac{1}{1+{\mathrm{x}}^2};\mathrm{y}=0\mathrm{when}x=1$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}(1+{\mathrm{x}}^2)\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{xy}=\frac{1}{1+{\mathrm{x}}^2};\\ \mathrm{y}=0\mathrm{whenx}=1\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2\mathrm{x}}{(1+{\mathrm{x}}^2)}\mathrm{y}=\frac{1}{{(1+{\mathrm{x}}^2)}^2}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=\frac{2\mathrm{x}}{(1+{\mathrm{x}}^2)}\mathrm{and}\mathrm{Q}=\frac{1}{{(1+{\mathrm{x}}^2)}^2}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdy}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int \frac{2\mathrm{x}}{(1+{\mathrm{x}}^2)}\mathrm{dy}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\log (1+{\mathrm{x}}^2)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=(1+{\mathrm{x}}^2)\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\\ \mathrm{relation},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \mathrm{y}(1+{\mathrm{x}}^2)=\int \{\frac{1}{{(1+{\mathrm{x}}^2)}^2}\times (1+{\mathrm{x}}^2)\}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ =\int \frac{1}{(1+{\mathrm{x}}^2)}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\end{array}$ $\begin{array}{l}\mathrm{y}(1+{\mathrm{x}}^2)={\tan }^{-1}\mathrm{x}+\mathrm{C}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{y}=0\hspace{0.17em}\hspace{0.17em}\mathrm{at}\mathrm{x}=1\\ 0(1+1^2)={\tan }^{-1}1+\mathrm{C}\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0=\frac{\mathrm{\pi }}{4}+\mathrm{C}\Rightarrow \mathrm{C}=-\frac{\mathrm{\pi }}{4}\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right),\hspace{0.17em}\mathrm{we}\mathrm{have}\end{array}$ $\begin{array}{l}\mathrm{y}(1+{\mathrm{x}}^2)={\tan }^{-1}\mathrm{x}-\frac{\mathrm{\pi }}{4}\\ \mathrm{This}\mathrm{‹}\mathrm{is}\mathrm{the}\mathrm{particular}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.91 For the differential equation given below, find a particular solution satisfying the given condition:

$\frac{\mathrm{dy}}{\mathrm{dx}}-3\mathrm{ycotx}=\sin 2\mathrm{x};\mathrm{y}=2\mathrm{whenx}=\frac{\mathrm{\pi }}{}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\frac{\mathrm{dy}}{\mathrm{dx}}-3\mathrm{ycotx}=\sin 2\mathrm{x};\\ \mathrm{y}=2\mathrm{whenx}=\frac{\mathrm{\pi }}{2}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=-3\mathrm{cotx}\mathrm{and}\mathrm{Q}=\sin 2\mathrm{x}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int -3\mathrm{cotxdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{-3\mathrm{logsinx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{{\sin }^3\mathrm{x}}={\mathrm{cosec}}^3\mathrm{x}\end{array}$ $\begin{array}{l}\mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\\ \mathrm{relation},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ {\mathrm{ycosec}}^3\mathrm{x}=\int \{\sin 2\mathrm{x}\times {\mathrm{cosec}}^3\mathrm{x}\}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ =\int 2\frac{\mathrm{cosx}}{{\sin }^2\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ =\int 2\mathrm{cotx}.\mathrm{cosecx}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ {\mathrm{ycosec}}^3\mathrm{x}=-2\mathrm{cosecx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}=-2{\sin }^2\mathrm{x}+{\mathrm{Csin}}^3\mathrm{x}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\mathrm{y}=2\hspace{0.17em}\hspace{0.17em}\mathrm{at}\mathrm{x}=\frac{\mathrm{\pi }}{2}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2=-2{\sin }^2\frac{\mathrm{\pi }}{2}+{\mathrm{Csin}}^3\frac{\mathrm{\pi }}{2}\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2=-2+\mathrm{C}\Rightarrow \mathrm{C}=4\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right),\hspace{0.17em}\mathrm{we}\mathrm{have}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}=-2{\sin }^2\mathrm{x}+4{\sin }^3\mathrm{x}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{particular}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\end{array}$

Q.92 Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{slope}\mathrm{of}\mathrm{a}\mathrm{curve}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{at}\mathrm{point}(\mathrm{x},\mathrm{y})\mathrm{be}\frac{\mathrm{dy}}{\mathrm{dx}}.\\ \mathrm{According}\mathrm{to}\mathrm{question},\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}+\mathrm{y}\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=\mathrm{x}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=-1\mathrm{}\mathrm{and}\mathrm{}\mathrm{Q}=\mathrm{x}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int -1\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\\ \mathrm{relation},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=\int (\mathrm{x}\times {\mathrm{e}}^{-\mathrm{x}})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=\mathrm{x}\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{x}\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=\mathrm{x}(-{\mathrm{e}}^{-\mathrm{x}})-\int 1.(-{\mathrm{e}}^{-\mathrm{x}})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=-{\mathrm{xe}}^{-\mathrm{x}}+\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=-{\mathrm{xe}}^{-\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=-\mathrm{x}-1+{\mathrm{Ce}}^{\mathrm{x}}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{x}+\mathrm{\hspace{0.17em}}\mathrm{y}+1={\mathrm{Ce}}^{\mathrm{x}}...\left(\mathrm{i}\right)\\ \mathrm{Since},\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{origin},\mathrm{then}\\ \hspace{0.17em}\hspace{0.17em}0+\mathrm{\hspace{0.17em}}0+1={\mathrm{Ce}}^0\Rightarrow \mathrm{C}=1\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right),\hspace{0.17em}\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{x}+\mathrm{\hspace{0.17em}}\mathrm{y}+1={\mathrm{e}}^{\mathrm{x}}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{passing}\mathrm{through}\mathrm{origin}\mathrm{is}\mathrm{\hspace{0.17em}}\\ \mathrm{x}+\mathrm{\hspace{0.17em}}\mathrm{y}+1={\mathrm{e}}^{\mathrm{x}}.\end{array}$

Q.93 Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{slope}\mathrm{of}\mathrm{a}\mathrm{curve}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{at}\mathrm{point}(\mathrm{x},\mathrm{y})\mathrm{be}\frac{\mathrm{dy}}{\mathrm{dx}}.\\ \mathrm{According}\mathrm{to}\mathrm{question},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}+5=\mathrm{x}+\mathrm{y}\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=\mathrm{x}-5\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=-1\mathrm{}\mathrm{and}\mathrm{}\mathrm{Q}=\mathrm{x}-5\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int -1\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\\ \mathrm{relation},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=\int (\mathrm{x}-5)\times {\mathrm{e}}^{-\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=\int (\mathrm{x}-5){\mathrm{e}}^{-\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=\{(\mathrm{x}-5)\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int (\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-5)\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx})\mathrm{\hspace{0.17em}}\mathrm{dx}\}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=\{-(\mathrm{x}-5){\mathrm{e}}^{-\mathrm{x}}+\int {\mathrm{e}}^{-\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}\}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=-(\mathrm{x}-5){\mathrm{e}}^{-\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{-\mathrm{x}}=(4-\mathrm{x}){\mathrm{e}}^{-\mathrm{x}}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}=(4-\mathrm{x})+{\mathrm{Ce}}^{\mathrm{x}}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{i}\right)\\ \mathrm{The}\mathrm{curve}\mathrm{passes}\mathrm{through}(0,2),\hspace{0.17em}\mathrm{then}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2=(4-0)+{\mathrm{Ce}}^0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2=4+\mathrm{C}\Rightarrow \mathrm{C}=-2\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}=(4-\mathrm{x})-2{\mathrm{e}}^{\mathrm{x}}\\ \mathrm{Thus},\mathrm{this}\mathrm{is}\mathrm{the}\mathrm{required}\hspace{0.17em}\hspace{0.17em}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{.}\end{array}$

Q.94

$\begin{array}{l}\mathrm{The}\mathrm{}\mathrm{integrating}\mathrm{}\mathrm{factor}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{differential}\mathrm{}\mathrm{equation}\\ \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=2{\mathrm{x}}^2\hspace{0.17em}\hspace{0.17em}\mathrm{is}\end{array}$ ${\text{(A)e}}^{–\mathrm{x}}\left(\mathrm{B}\right){\mathrm{e}}^{–\mathrm{y}}\left(\mathrm{C}\right)\frac{1}{\mathrm{x}}\left(\mathrm{D}\right)\mathrm{x}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=2{\mathrm{x}}^2\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=2\mathrm{x}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=-\frac{1}{\mathrm{x}}\mathrm{}\mathrm{and}\mathrm{}\mathrm{Q}=2\mathrm{x}\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int -\frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}}\mathrm{\hspace{0.17em}}={\mathrm{e}}^{-\mathrm{logx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{1}{\mathrm{x}}\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{C}\mathrm{.}\end{array}$

Q.95

$\begin{array}{l}\mathrm{The}\mathrm{Integrating}\mathrm{Factor}\mathrm{of}\mathrm{the}\mathrm{differential}equation\\ \left(1-{\mathrm{y}}^2\right)\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{yx}=\mathrm{ay}\left(-1<\mathrm{y}<1\right)\mathrm{is}\\ \left(\mathrm{A}\right)\frac{1}{{\mathrm{y}}^2-1}\left(\mathrm{B}\right)\frac{1}{\sqrt{{\mathrm{y}}^2-1}}\left(\mathrm{C}\right)\frac{1}{1-{\mathrm{y}}^2}\left(\mathrm{D}\right)\frac{1}{\sqrt{1-{\mathrm{y}}^2}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}(1-{\mathrm{y}}^2)\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{yx}=\mathrm{ay}(-1<\mathrm{y}<1)\\ \Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{y}}{(1-{\mathrm{y}}^2)}\mathrm{x}=\mathrm{a}\frac{\mathrm{y}}{(1-{\mathrm{y}}^2)}\\ \mathrm{Comparing}\mathrm{it}\mathrm{with}\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \mathrm{P}=\frac{\mathrm{y}}{(1-{\mathrm{y}}^2)}\mathrm{and}\mathrm{Q}=\mathrm{a}\frac{\mathrm{y}}{(1-{\mathrm{y}}^2)}\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{Pdy}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\mathrm{e}}^{\int \frac{\mathrm{y}}{(1-{\mathrm{y}}^2)}\mathrm{\hspace{0.17em}}\mathrm{dy}}\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\mathrm{e}}^{-\frac{1}{2}\log (1-{\mathrm{y}}^2)}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={{\mathrm{e}}^{\log (1-{\mathrm{y}}^2)}}^{{}^{-\frac{1}{2}}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={(1-{\mathrm{y}}^2)}^{-\frac{1}{2}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{\sqrt{1-{\mathrm{y}}^2}}\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.96 For the differential equations given below, indicate its order and degree (if defined).

$\begin{array}{l}\left(\mathrm{i}\right)\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+5\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2-6\mathrm{y}=\mathrm{logx}\left(\mathrm{ii}\right){\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^3-4{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+7\mathrm{y}=\mathrm{sinx}\\ \left(\mathrm{iii}\right)\frac{{\mathrm{d}}^4\mathrm{y}}{{\mathrm{dx}}^4}-\sin \left(\frac{{\mathrm{d}}^3\mathrm{y}}{{\mathrm{dx}}^3}\right)=0\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+5\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2-6\mathrm{y}=\mathrm{logx}\\ \Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+5\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2-6\mathrm{y}-\mathrm{logx}=0\\ \mathrm{The}\mathrm{highest}\mathrm{order}\mathrm{derivative}\mathrm{present}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{is}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}.\mathrm{So},\mathrm{its}\mathrm{order}\mathrm{is}2\mathrm{and}\mathrm{highest}\mathrm{power}\mathrm{raised}\mathrm{by}\\ \mathrm{}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\mathrm{is}\mathrm{one}.\mathrm{Hence}\mathrm{its}\mathrm{degree}\mathrm{is}\mathrm{one}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^3-4{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+7\mathrm{y}=\mathrm{sinx}\\ \Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^3-4{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+7\mathrm{y}-\mathrm{sinx}=0\\ \mathrm{The}\mathrm{highest}\mathrm{order}\mathrm{derivative}\mathrm{present}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{is}\frac{\mathrm{dy}}{\mathrm{dx}}.\mathrm{So},\mathrm{its}\mathrm{order}\mathrm{is}1\mathrm{and}\mathrm{highest}\mathrm{power}\mathrm{raised}\mathrm{by}\\ \mathrm{}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{is}\mathrm{three}.\mathrm{Hence}\mathrm{its}\mathrm{degree}\mathrm{is}3\mathrm{.}\end{array}$ $\begin{array}{l}\left(\mathrm{iii}\right)\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{{\mathrm{d}}^4\mathrm{y}}{{\mathrm{dx}}^4}-\sin \left(\frac{{\mathrm{d}}^3\mathrm{y}}{{\mathrm{dx}}^3}\right)=0\\ \mathrm{The}\mathrm{highest}\mathrm{order}\mathrm{derivative}\mathrm{present}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{is}\frac{{\mathrm{d}}^4\mathrm{y}}{{\mathrm{dx}}^4}.\mathrm{So},\mathrm{its}\mathrm{order}\mathrm{is}4.\\ \mathrm{Since}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{polynomial}\mathrm{equation}\mathrm{.}\\ \mathrm{Hence},\mathrm{its}\mathrm{degree}\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{.}\end{array}$

Q.97 For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{y}={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{–\mathrm{x}}+{\mathrm{x}}^2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}:\mathrm{x}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+2\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{xy}+{\mathrm{x}}^2-2=0\\ \left(\mathrm{ii}\right)\mathrm{y}={\mathrm{e}}^{\mathrm{x}}(\mathrm{acosx}+\mathrm{bsinx})\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}:\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-2\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y}=0\\ \left(\mathrm{iii}\right)\mathrm{y}=\mathrm{xsin}3\mathrm{x}\hspace{0.17em}\hspace{0.17em}:\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+9\mathrm{y}-6\cos 3\mathrm{x}=0\\ \left(\mathrm{iv}\right){\mathrm{x}}^2=2{\mathrm{y}}^2\mathrm{logy}:({\mathrm{x}}^2+{\mathrm{y}}^2)\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{xy}=0\end{array}$

Ans.

$\begin{array}{l}\left(i\right)\text{The given equation is:}\\ y=a{e}^x+b{e}^{–x}+x^2\mathrm{\dots }\left(i\right)\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{dy}{dx}=a{e}^x-b{e}^{-x}+2x\\ \end{array}$

Again, differentiating w.r.t. x, we get

\begin{array}{l}\frac{d^{2}y}{d{x}^2}=a{e}^x+b{e}^{-x}+2\\ \text{Substituting the value of}\frac{d^{2}y}{d{x}^2}\text{and}\frac{dy}{dx}\text{in given differential}\\ \text{equation, we get:}\\ \text{L}\text{.H}\text{.S}\text{.}=x\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}-xy+x^2-2\\ =x\left(a{e}^x+b{e}^{-x}+2\right)+2\left(a{e}^x-b{e}^{-x}+2x\right)-x\left(a{e}^x+b{e}^{–x}+x^2\right)\\ +x^2-2\\ =\bcancel{ax{e}^x}+\bcancel{bx{e}^{-x}}+2x+2a{e}^x-2b{e}^{-x}+4x-\bcancel{ax{e}^x}-\bcancel{bx{e}^{–x}}-x^3\\ +x^2-2\\ =2a{e}^x-2b{e}^{-x}-x^3+x^2+6x-2\ne 0\\ \text{L}\text{.H}\text{.S}\text{.}\ne \text{R}\text{.H}\text{.S}\text{.}\\ \text{Thus, the given function is not a solution of corresponding}\\ \text{diffrential equation}\text{.}\end{array} \begin{array}{l}\left(ii\right)\text{The given equation is:}\\ y=e^x\left(a\cos x+b\sin x\right)\mathrm{\dots }\left(i\right)\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{dy}{dx}=e^x\frac{d}{dx}\left(a\cos x+b\sin x\right)+\left(a\cos x+b\sin x\right)\frac{d}{dx}{e}^x\\ =e^x\left(-a\sin x+b\cos x\right)+\left(a\cos x+b\sin x\right)e^x\\ =e^x\left(-a\sin x+b\cos x+a\cos x+b\sin x\right)\\ \text{Again, differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{d^{2}y}{d{x}^2}=e^x\frac{d}{dx}\left(-a\sin x+b\cos x+a\cos x+b\sin x\right)\end{array} \begin{array}{l}+\left(-a\sin x+b\cos x+a\cos x+b\sin x\right)\frac{d}{dx}{e}^x\\ =e^x\left(-a\cos x-bsinx-a\sin x+b\cos x\right)\\ +\left(-a\sin x+b\cos x+a\cos x+b\sin x\right)e^x\\ =e^x\left(\begin{array}{l}-\bcancel{a\cos x}-\bcancel{bsinx}-a\sin x+b\cos x-a\sin x+b\cos x\\ +\bcancel{a\cos x}+\bcancel{b\sin x}\end{array}\right)\\ =e^x\left(-2a\sin x+2b\cos x\right)\\ \text{Substituting the value of}\frac{d^{2}y}{d{x}^2}\text{and}\frac{dy}{dx}\text{in given differential}\\ \text{equation, we get:}\\ L.H.S.=\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+2y\\ =e^x\left(-2a\sin x+2b\cos x\right)-2e^x\left(-a\sin x+b\cos x+a\cos x+b\sin x\right)\\ +2e^x\left(a\cos x+b\sin x\right)\\ =e^x\left(\begin{array}{l}-2a\sin x+2b\cos x+2a\sin x-2b\cos x-2a\cos x-2b\sin x\\ +2a\cos x+2b\sin x\end{array}\right)\\ =e^x\left(0\right)=0\\ \text{L}\text{.H}\text{.S}\text{.=R}\text{.H}\text{.S}\text{.}\\ \text{Thus, the given function is a solution of corresponding}\\ \text{diffrential equation}\text{.}\\ \left(iii\right)\text{The given equation is:}\\ y=x\sin 3x\mathrm{\dots }\left(i\right)\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{dy}{dx}=x\frac{d}{dx}\sin 3x+\sin 3x\frac{d}{dx}x\end{array} \begin{array}{l}=x.3\cos 3x+\sin 3x.1\\ =3x\cos 3x+\sin 3x\\ \text{Again, differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{d^{2}y}{d{x}^2}=3x\frac{d}{dx}\cos 3x+\cos 3x\frac{d}{dx}3x+\frac{d}{dx}sin3x\\ =-9x\sin 3x+3\cos 3x+3\cos 3x\\ =-9x\sin 3x+6\cos 3x\\ \text{Substituting the value of}\frac{d^{2}y}{d{x}^2}\text{and}\frac{dy}{dx}\text{in given differential}\\ \text{equation, we get:}\\ \text{L}\text{.H}\text{.S}\text{.}=\frac{d^{2}y}{d{x}^2}+9y-6\cos 3x\\ =-9x\sin 3x+6\cos 3x+9x\sin 3x-6\cos 3x\\ =0\\ \text{L}\text{.H}\text{.S}\text{.=R}\text{.H}\text{.S}\text{.}\\ \text{Thus, the given function is a solution of corresponding}\\ \text{diffrential equation}\text{.}\\ \left(iv\right)\text{The given equation is:}\\ x^2=2y^2\log y\mathrm{\dots }\left(i\right)\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ 2x=2y^2\frac{d}{dx}\log y+2\log y\frac{d}{dx}{y}^2\\ =2y^2\frac{1}{y}\frac{dy}{dx}+2\log y\left(2y\frac{dy}{dx}\right)\\ =\left(2y+4y\log y\right)\frac{dy}{dx}\end{array} \begin{array}{l}x=\left(y+2y\log y\right)\frac{dy}{dx}\\ \frac{dy}{dx}=\frac{x}{y\left(1+2\log y\right)}\\ \text{Substituting the value of}\frac{dy}{dx}\text{in given differential}\\ \text{equation, we get:}\\ \text{L}\text{.H}\text{.S}\text{.=}\left(x^2+y^2\right)\frac{dy}{dx}-xy\\ =\left(2y^2\log y+y^2\right)\frac{x}{y\left(1+2\log y\right)}-xy\\ =y^2\left(2\log y+1\right)\frac{x}{y\left(1+2\log y\right)}-xy\\ =xy-xy\\ =0=\text{R}\text{.H}\text{.S}\text{.}\\ \text{Thus, the given function is a solution of corresponding}\\ \text{diffrential equation}\text{.}\end{array}

Q.98 Form the differential equation representing the family of curves given by (x – a)² +2y² = a², where a is an arbitrary constant.

Ans.

\begin{array}{l}\text{The given equation is:}\\ {\left(\text{x}-\text{a}\right)}^{\text{2}}+{\text{2y}}^{\text{2}}={\text{a}}^{\text{2}}\\ x^2-2ax+a^2+2y^2=a^2\\ x^2-2ax+2y^2=0\\ \end{array}

Differentiating w.r.t. x, we get

\begin{array}{l}\text{2}x-2a+4y\frac{dy}{dx}=0\\ \frac{dy}{dx}=\frac{2a-2x}{4y}\\ =\frac{a-x}{2y}\mathrm{\dots }\left(ii\right)\\ \text{From equation}\left(i\right),a=\frac{x^2+2y^2}{2x}\\ \text{Putting value of a in equation}\left(\text{ii}\right)\text{,}\text{we get}\\ \frac{dy}{dx}=\frac{\left(\frac{x^2+2y^2}{2x}\right)-x}{2y}\\ =\frac{x^2+2y^2-2x^2}{4xy}\\ =\frac{2y^2-x^2}{4xy}\\ \text{Which is the differential equation of the family of curves}\text{.}\end{array}

Q.99 Prove that x² – y² = c (x² + y²)² is the general solution of differential equation (x³ – 3xy²) dx = (y³ – 3x²y) dy, where c is a parameter.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ ({\mathrm{x}}^{\mathrm{3}}-3{\mathrm{xy}}^{\mathrm{2}})\mathrm{dx}=({\mathrm{y}}^{\mathrm{3}}-3{\mathrm{x}}^{\mathrm{2}}\mathrm{y})\mathrm{dy}\end{array}$ $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{({\mathrm{x}}^{\mathrm{3}}-3{\mathrm{xy}}^{\mathrm{2}})}{({\mathrm{y}}^{\mathrm{3}}-3{\mathrm{x}}^{\mathrm{2}}\mathrm{y})}\\ \mathrm{Let}\mathrm{\hspace{0.17em}}\mathrm{F}(\mathrm{x},\mathrm{y})=\frac{({\mathrm{x}}^{\mathrm{3}}-3{\mathrm{xy}}^{\mathrm{2}})}{({\mathrm{y}}^{\mathrm{3}}-3{\mathrm{x}}^{\mathrm{2}}\mathrm{y})},\mathrm{\hspace{0.17em}}\mathrm{then}\\ \mathrm{F}(\mathrm{\lambda x},\mathrm{\lambda y})=\frac{({\mathrm{\lambda }}^3{\mathrm{x}}^{\mathrm{3}}-\mathrm{3}{\mathrm{\lambda }}^3{\mathrm{xy}}^{\mathrm{2}})}{({\mathrm{\lambda }}^3{\mathrm{y}}^{\mathrm{3}}-\mathrm{3}{\mathrm{\lambda }}^3{\mathrm{x}}^{\mathrm{2}}\mathrm{y})}\\ =\frac{{\mathrm{\lambda }}^3({\mathrm{x}}^{\mathrm{3}}-3{\mathrm{xy}}^{\mathrm{2}})}{{\mathrm{\lambda }}^3({\mathrm{y}}^{\mathrm{3}}-3{\mathrm{x}}^{\mathrm{2}}\mathrm{y})}={\mathrm{\lambda }}^0\mathrm{F}(\mathrm{x},\mathrm{y})\\ \mathrm{Therefore},\mathrm{F}(\mathrm{x},\mathrm{y})\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{function}\mathrm{of}\mathrm{degree}\mathrm{zero}.\mathrm{So},\\ \mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\\ \mathrm{For}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{substitute}\mathrm{y}=\mathrm{vx}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{,}\\ \mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{({\mathrm{x}}^{\mathrm{3}}-3{\mathrm{v}}^{\mathrm{2}}{\mathrm{x}}^{\mathrm{3}})}{({\mathrm{v}}^{\mathrm{3}}{\mathrm{x}}^{\mathrm{3}}-3{\mathrm{vx}}^{\mathrm{3}})}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{(\mathrm{1}-3{\mathrm{v}}^{\mathrm{2}})}{({\mathrm{v}}^{\mathrm{3}}-3\mathrm{v})}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{(\mathrm{1}-3{\mathrm{v}}^{\mathrm{2}})}{({\mathrm{v}}^{\mathrm{3}}-3\mathrm{v})}-\mathrm{v}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\mathrm{1}-3{\mathrm{v}}^{\mathrm{2}}-{\mathrm{v}}^4+3{\mathrm{v}}^2}{({\mathrm{v}}^{\mathrm{3}}-3\mathrm{v})}\end{array}$ \begin{array}{l}=\frac{\left(1-v^4\right)}{\left({\text{v}}^{\text{3}}-\text{3v}\right)}\\ \frac{\left({\text{v}}^{\text{3}}-\text{3v}\right)dv}{\left(1-v^4\right)}=\frac{dx}{x}\\ \text{Integrating both sides, we get}\\ {\displaystyle \int \frac{\left({\text{v}}^{\text{3}}-\text{3v}\right)dv}{\left(1-v^4\right)}}={\displaystyle \int \frac{1}{x}}dx\mathrm{\dots }\left(\text{i}\right)\\ \text{Let}I={\displaystyle \int \frac{\left({\text{v}}^{\text{3}}-\text{3v}\right)dv}{\left(1-v^4\right)}}\\ ={\displaystyle \int \frac{{\text{v}}^{\text{3}}dv}{\left(1-v^4\right)}}-3{\displaystyle \int \frac{\text{v}dv}{\left(1-v^4\right)}}\\ =-\frac{1}{4}\log \left(1-v^4\right)-\frac{3}{4}\log \left|\frac{1-v^2}{1+v^2}\right|\\ \text{From eqeuation}\left(\text{i}\right)\text{,}\text{we}\text{get}\\ -\frac{1}{4}\log \left(1-v^4\right)-\frac{3}{4}\log \left|\frac{1-v^2}{1+v^2}\right|=\log \left|x\right|+\log C\\ -\frac{1}{4}\log \left\{\left(1-v^4\right){\left|\frac{1-v^2}{1+v^2}\right|}^3\right\}=\log C\left|x\right|\\ \log \left\{\left(1-v^4\right){\left|\frac{1-v^2}{1+v^2}\right|}^3\right\}=-4\log C\left|x\right|\end{array} $\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\log \left\{(1-{\mathrm{v}}^4){\left(\frac{1-{\mathrm{v}}^2}{1+{\mathrm{v}}^2}\right)}^3\right\}=\log {\left(\mathrm{C}\left|\mathrm{x}\right|\right)}^{-4}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(1-{\mathrm{v}}^4){\left(\frac{1+{\mathrm{v}}^2}{1-{\mathrm{v}}^2}\right)}^3={\left(\mathrm{C}\left|\mathrm{x}\right|\right)}^{-4}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(1-\frac{{\mathrm{y}}^4}{{\mathrm{x}}^4}){\left(\frac{1+\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}}{1-\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}}\right)}^3={\left(\mathrm{C}\left|\mathrm{x}\right|\right)}^{-4}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\frac{{\mathrm{x}}^4-{\mathrm{y}}^4}{{\mathrm{x}}^4}\right){\left(\frac{\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{{\mathrm{x}}^2}}{\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{{\mathrm{x}}^2}}\right)}^3={(\mathrm{C}‘\left|\mathrm{x}\right|)}^{-4}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{({\mathrm{x}}^2-{\mathrm{y}}^2)({\mathrm{x}}^2+{\mathrm{y}}^2)}{{\mathrm{x}}^4}{\left(\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{{\mathrm{x}}^2-{\mathrm{y}}^2}\right)}^3=\frac{1}{{(\mathrm{C}‘\left|\mathrm{x}\right|)}^4}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^4}{{\mathrm{x}}^4{({\mathrm{x}}^2-{\mathrm{y}}^2)}^2}=\frac{1}{{(\mathrm{C}‘)}^4{\mathrm{x}}^4}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{(\mathrm{C}‘)}^4{({\mathrm{x}}^2+{\mathrm{y}}^2)}^4={({\mathrm{x}}^2-{\mathrm{y}}^2)}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}({\mathrm{x}}^2-{\mathrm{y}}^2)={(\mathrm{C}‘)}^2{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}({\mathrm{x}}^2-{\mathrm{y}}^2)=\mathrm{C}{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2[\mathrm{Let}\hspace{0.17em}\hspace{0.17em}\mathrm{C}={(\mathrm{C}‘)}^2]\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.100 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{circle}\mathrm{with}\mathrm{centre}(\mathrm{a},\mathrm{a})\mathrm{and}\mathrm{radius}\mathrm{a}\mathrm{which}\\ \mathrm{touches}\mathrm{the}\mathrm{coordinate}\mathrm{aes}\mathrm{is}:\\ {(\mathrm{x}-\mathrm{a})}^2+{(\mathrm{y}-\mathrm{a})}^2={\mathrm{a}}^2...\left(\mathrm{i}\right)\\ \mathrm{Differentiating}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{2}(\mathrm{x}-\mathrm{a})+2(\mathrm{y}-\mathrm{a})\mathrm{y}‘=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(\mathrm{x}-\mathrm{a})+(\mathrm{y}-\mathrm{a})\mathrm{y}‘=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}-\mathrm{a}+\mathrm{yy}‘-\mathrm{ay}‘=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}+\mathrm{yy}‘-\mathrm{a}(1+\mathrm{y}‘)=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{a}=\frac{\mathrm{x}+\mathrm{yy}‘}{(1+\mathrm{y}‘)}\\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\{\mathrm{x}-\frac{\mathrm{x}+\mathrm{yy}‘}{(1+\mathrm{y}‘)}\}}^2+{\{\mathrm{y}-\frac{\mathrm{x}+\mathrm{yy}‘}{(1+\mathrm{y}‘)}\}}^2={\left\{\frac{\mathrm{x}+\mathrm{yy}‘}{(1+\mathrm{y}‘)}\right\}}^2\\ \Rightarrow {\left\{\frac{\mathrm{x}(1+\mathrm{y}‘)-\mathrm{x}-\mathrm{yy}‘}{(1+\mathrm{y}‘)}\right\}}^2+{\left\{\frac{\mathrm{y}(1+\mathrm{y}‘)-\mathrm{x}-\mathrm{yy}‘}{(1+\mathrm{y}‘)}\right\}}^2={\left\{\frac{\mathrm{x}+\mathrm{yy}‘}{(1+\mathrm{y}‘)}\right\}}^2\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\left\{\frac{\mathrm{x}+\mathrm{xy}‘-\mathrm{x}-\mathrm{yy}‘}{(1+\mathrm{y}‘)}\right\}}^2+{\left\{\frac{\mathrm{y}+\mathrm{yy}‘-\mathrm{x}-\mathrm{yy}‘}{(1+\mathrm{y}‘)}\right\}}^2={\left\{\frac{\mathrm{x}+\mathrm{yy}‘}{(1+\mathrm{y}‘)}\right\}}^2\end{array}$ \begin{array}{l}\Rightarrow {\left\{\frac{xy‘-yy‘}{\left(1+y‘\right)}\right\}}^2+{\left\{\frac{y-x}{\left(1+y‘\right)}\right\}}^2={\left\{\frac{x+yy‘}{\left(1+y‘\right)}\right\}}^2\\ \Rightarrow {\left(x-y\right)}^{2}y{‘}^2+{\left(x-y\right)}^2={\left(x+yy‘\right)}^2\\ \Rightarrow {\left(x-y\right)}^2\left(y{‘}^2+1\right)={\left(x+yy‘\right)}^2\\ \Rightarrow {\left(x+yy‘\right)}^2={\left(x-y\right)}^2\left(y{‘}^2+1\right)\\ \text{Which is the required differential equation}\text{.}\end{array}

Q.101

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}\\ \frac{\mathrm{dy}}{\mathrm{dx}}+\sqrt{\frac{1-{\mathrm{y}}^2}{1-{\mathrm{x}}^2}}=0.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}+\sqrt{\frac{1-{\mathrm{y}}^2}{1-{\mathrm{x}}^2}}=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=-\sqrt{\frac{1-{\mathrm{y}}^2}{1-{\mathrm{x}}^2}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\sqrt{1-{\mathrm{y}}^2}}=-\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\sin }^{–1}\mathrm{y}=-{\sin }^{–1}\mathrm{x}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\sin }^{–1}\mathrm{x}+{\sin }^{–1}\mathrm{y}=\mathrm{C}\end{array}$ \begin{array}{l}\text{Thus, the general solution of the given differential equation is:}\\ {\text{sin}}^{\text{-1}}\text{x}+{\text{sin}}^{\text{-1}}\text{y}=C\end{array}

Q.102

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}\\ \mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{{\mathrm{y}}^2+\mathrm{y}+1}{{\mathrm{x}}^2+\mathrm{x}+1}=0\mathrm{is}\mathrm{given}\mathrm{by}(\mathrm{x}+\mathrm{y}+1)=\mathrm{A}(1-\mathrm{x}-\mathrm{y}-2\mathrm{xy}),\\ \mathrm{where}\mathrm{A}\mathrm{is}\mathrm{parameter}\mathrm{.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{{\mathrm{y}}^2+\mathrm{y}+1}{{\mathrm{x}}^2+\mathrm{x}+1}=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{{\mathrm{y}}^2+\mathrm{y}+1}{{\mathrm{x}}^2+\mathrm{x}+1}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{({\mathrm{y}}^2+\mathrm{y}+1)}=-\frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{x}+1}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dy}}{({\mathrm{y}}^2+\mathrm{y}+1)}+\frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{x}+1}=0\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\int \frac{\mathrm{dy}}{({\mathrm{y}}^2+\mathrm{y}+1)}\mathrm{\hspace{0.17em}}+\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{x}+1}=\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int \frac{\mathrm{dy}}{{(\mathrm{y}+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\mathrm{\hspace{0.17em}}+\int \frac{\mathrm{dx}}{{(\mathrm{x}+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\mathrm{C}\\ \Rightarrow \frac{1}{\left(\frac{\sqrt{3}}{2}\right)}{\tan }^{-1}\left(\frac{2\mathrm{y}+1}{\sqrt{3}}\right)+\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}\mathrm{\hspace{0.17em}}{\tan }^{-1}\left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)=\mathrm{C}\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{\hspace{0.17em}}{\tan }^{-1}\left(\frac{2\mathrm{y}+1}{\sqrt{3}}\right)+\mathrm{\hspace{0.17em}}{\tan }^{-1}\left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)=\frac{\sqrt{3}}{2}\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\tan }^{-1}\left\{\frac{\left(\frac{2\mathrm{y}+1}{\sqrt{3}}\right)+\left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)}{1-\left(\frac{2\mathrm{y}+1}{\sqrt{3}}\right)\left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)}\right\}=\frac{\sqrt{3}}{2}\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\tan }^{-1}\left\{\frac{\sqrt{3}(2\mathrm{y}+1+2\mathrm{x}+1)}{3-(2\mathrm{y}+1)(2\mathrm{x}+1)}\right\}=\frac{\sqrt{3}}{2}\mathrm{C}\\ \Rightarrow \mathrm{\hspace{0.17em}}{\tan }^{-1}\left\{\frac{\sqrt{3}(2\mathrm{y}+2\mathrm{x}+2)}{3-(4\mathrm{xy}+2\mathrm{x}+2\mathrm{y}+1)}\right\}=\frac{\sqrt{3}}{2}\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\tan }^{-1}\left\{\frac{2\sqrt{3}(\mathrm{y}+\mathrm{x}+1)}{2(1-2\mathrm{xy}-\mathrm{x}-\mathrm{y})}\right\}=\frac{\sqrt{3}}{2}\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left\{\frac{\sqrt{3}(\mathrm{y}+\mathrm{x}+1)}{(1-2\mathrm{xy}-\mathrm{x}-\mathrm{y})}\right\}=\tan \left(\frac{\sqrt{3}}{2}\mathrm{C}\right)\\ \Rightarrow (\mathrm{y}+\mathrm{x}+1)=\frac{1}{\sqrt{3}}\tan \left(\frac{\sqrt{3}}{2}\mathrm{C}\right)(1-2\mathrm{xy}-\mathrm{x}-\mathrm{y})\\ \Rightarrow (\mathrm{y}+\mathrm{x}+1)=\mathrm{A}(1-2\mathrm{xy}-\mathrm{x}-\mathrm{y})\mathrm{\hspace{0.17em}}\\ [\mathrm{Where}\mathrm{\hspace{0.17em}}\mathrm{A}=\frac{1}{\sqrt{3}}\tan \left(\frac{\sqrt{3}}{2}\mathrm{C}\right)]\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$ \begin{array}{l}\end{array}

Q.103

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{point}\\ \mathrm{\hspace{0.17em}}(\mathrm{0},\frac{\mathrm{\pi }}{\mathrm{4}})\hspace{0.17em}\hspace{0.17em}\mathrm{whose}\mathrm{differential}\mathrm{equation}\mathrm{is}\\ \mathrm{sinxcosy}\mathrm{dx}+\mathrm{cosxsiny}\mathrm{dy}=0.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\\ \mathrm{sinxcosydx}+\mathrm{cosxsinydy}=0\\ \mathrm{Dividing}\mathrm{equation}\mathrm{by}\cos \mathrm{x}\cos \mathrm{y},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{sinxcosydx}+\mathrm{cosxsinydy}}{\cos \mathrm{x}\cos \mathrm{y}}=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{tanxdx}+\mathrm{tanydy}=0\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\int \mathrm{tanx}\mathrm{\hspace{0.17em}}\mathrm{dx}+\int \mathrm{tany}\mathrm{\hspace{0.17em}}\mathrm{dy}=\mathrm{logC}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{logsecx}+\mathrm{logsecy}=\mathrm{logC}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{logsecxsecy}=\mathrm{logC}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{secxsecy}=\mathrm{C}...\left(\mathrm{i}\right)\\ \mathrm{The}\mathrm{curve}\mathrm{passes}\mathrm{through}(0,\frac{\mathrm{\pi }}{4}),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sec 0\sec \frac{\mathrm{\pi }}{4}=\mathrm{C}\\ \Rightarrow 1.\sqrt{2}=\mathrm{C}\Rightarrow \mathrm{C}=\sqrt{2}\\ \mathrm{Putting}\mathrm{C}=\sqrt{2}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{secxsecy}=\sqrt{2}\Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{cosy}=\frac{\mathrm{secx}}{\sqrt{2}}\\ \mathrm{Which}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{.}\end{array}$

Q.104 Find the particular solution of the differential equation (1 + e^2x) dy + ( 1 + y²) e^x dx = 0, given that y = 1 when x = 0.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{1}+{\mathrm{e}}^{2\mathrm{x}})\mathrm{dy}+(\mathrm{1}+{\mathrm{y}}^{\mathrm{2}}){\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=0\\ \Rightarrow \frac{\mathrm{dy}}{(\mathrm{1}+{\mathrm{y}}^{\mathrm{2}})}+\frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}}{(\mathrm{1}+{\mathrm{e}}^{2\mathrm{x}})}=0\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \int \frac{\mathrm{dy}}{(\mathrm{1}+{\mathrm{y}}^{\mathrm{2}})}\mathrm{\hspace{0.17em}}+\int \frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}}{\{\mathrm{1}+{\left({\mathrm{e}}^{\mathrm{x}}\right)}^2\}}\mathrm{\hspace{0.17em}}=0\\ {\tan }^{-1}\mathrm{y}+{\tan }^{-1}{\mathrm{e}}^{\mathrm{x}}=\mathrm{C}...\left(\mathrm{i}\right)\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\mathrm{y}=1\mathrm{at}\mathrm{x}=0\\ {\tan }^{-1}1+{\tan }^{-1}{\mathrm{e}}^0=\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\frac{\mathrm{\pi }}{4}+{\tan }^{-1}1=\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{4}=\mathrm{C}\Rightarrow \mathrm{C}=\frac{\mathrm{\pi }}{2}\\ \mathrm{Putting}\hspace{0.17em}\hspace{0.17em}\mathrm{C}=\mathrm{\hspace{0.17em}}\frac{\mathrm{\pi }}{2}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{\hspace{0.17em}}{\tan }^{-1}\mathrm{y}+{\tan }^{-1}{\mathrm{e}}^{\mathrm{x}}=\frac{\mathrm{\pi }}{2}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{particular}\mathrm{question}\mathrm{of}\mathrm{given}\mathrm{differential}\\ \mathrm{equation}\mathrm{.}\end{array}$

Q.105

${\text{Solve the differential equation y e}}^{\frac{\mathrm{x}}{\mathrm{y}}}\mathrm{dx}=({\mathrm{xe}}^{\frac{\mathrm{x}}{\mathrm{y}}}+{\mathrm{y}}^2)\mathrm{dy}\hspace{0.17em}\hspace{0.17em}(\mathrm{y}\ne 0).$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{ye}}^{\frac{\mathrm{x}}{\mathrm{y}}}\mathrm{dx}=({\mathrm{xe}}^{\frac{\mathrm{x}}{\mathrm{y}}}+{\mathrm{y}}^2)\mathrm{dy}\hspace{0.17em}\hspace{0.17em}(\mathrm{y}\ne 0)\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{ye}}^{\frac{\mathrm{x}}{\mathrm{y}}}\frac{\mathrm{dx}}{\mathrm{dy}}=({\mathrm{xe}}^{\frac{\mathrm{x}}{\mathrm{y}}}+{\mathrm{y}}^2)\\ \Rightarrow {\mathrm{ye}}^{\frac{\mathrm{x}}{\mathrm{y}}}\frac{\mathrm{dx}}{\mathrm{dy}}-{\mathrm{xe}}^{\frac{\mathrm{x}}{\mathrm{y}}}={\mathrm{y}}^2\\ \Rightarrow (\mathrm{y}\frac{\mathrm{dx}}{\mathrm{dy}}-\mathrm{x}){\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}={\mathrm{y}}^2\\ \Rightarrow \frac{(\mathrm{y}\frac{\mathrm{dx}}{\mathrm{dy}}-\mathrm{x})}{{\mathrm{y}}^2}{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}=1\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Let}\hspace{0.17em}\hspace{0.17em}{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}=\mathrm{t}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{y},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}=\frac{\mathrm{dt}}{\mathrm{dx}}\Rightarrow {\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}\left(\frac{\mathrm{y}\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{x}-\mathrm{x}\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{y}}{{\mathrm{y}}^2}\right)=\frac{\mathrm{dt}}{\mathrm{dx}}\end{array}$ $\begin{array}{l}\Rightarrow {\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}\left(\frac{\mathrm{y}\frac{\mathrm{dx}}{\mathrm{dy}}-\mathrm{x}}{{\mathrm{y}}^2}\right)=\frac{\mathrm{dt}}{\mathrm{dx}}...\left(\mathrm{ii}\right)\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\mathrm{equation}\left(\mathrm{ii}\right),\hspace{0.17em}\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dt}}{\mathrm{dx}}=1\Rightarrow \mathrm{dt}=\mathrm{dx}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \int \mathrm{dt}=\int \mathrm{dx}\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{t}=\mathrm{x}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{e}}^{\frac{\mathrm{x}}{\mathrm{y}}}=\mathrm{x}+\mathrm{C}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\end{array}$

Q.106 Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = – 1, when x = 0.

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{differential}\mathrm{\hspace{0.17em}}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(\mathrm{x}-\mathrm{y})(\mathrm{dx}+\mathrm{dy})=\mathrm{dx}-\mathrm{dy}\\ (\mathrm{x}-\mathrm{y})\mathrm{dx}+(\mathrm{x}-\mathrm{y})\mathrm{dy}=\mathrm{dx}-\mathrm{dy}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{x}-\mathrm{y}+\mathrm{1})\mathrm{dy}=(\mathrm{1}-\mathrm{x}+\mathrm{y})\mathrm{dx}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(\mathrm{1}-\mathrm{x}+\mathrm{y})}{(\mathrm{x}-\mathrm{y}+\mathrm{1})}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-(\mathrm{x}-\mathrm{y})}{1+(\mathrm{x}-\mathrm{y})}...\left(\mathrm{i}\right)\end{array}$ \begin{array}{l}Letv=x-y\Rightarrow \frac{dv}{dx}=1-\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=1-\frac{dv}{dx}\\ \text{Substituting value of}\left(\text{x}-\text{y}\right)\text{and}\frac{dy}{dx}\text{in equation}\left(\text{i}\right)\text{, we get}\\ 1-\frac{dv}{dx}=\frac{1-v}{1+v}\\ \Rightarrow \frac{dv}{dx}=1-\frac{1-v}{1+v}\\ =\frac{1+v-1+v}{1+v}\\ \Rightarrow \frac{dv}{dx}=\frac{2v}{1+v}\\ \Rightarrow \frac{1+v}{v}dv=2dx\\ \text{Integrating both sides, we get}\\ {\displaystyle \int \left(\frac{1}{v}+1\right)}dv=2{\displaystyle \int dx}\\ \Rightarrow \log \left|v\right|+v=2x+C\\ \Rightarrow \log \left|x-y\right|+x-y=2x+C\\ \Rightarrow \log \left|x-y\right|=x+y+C\mathrm{\dots }\left(ii\right)\\ Now,\text{y}=-1\text{at x}=\text{0}\\ \Rightarrow \log \left|1\right|=0-1+C\Rightarrow C=1\\ \text{Substituting C}=\text{1 in equation}\left(\text{ii}\right)\text{, we get}\\ \log \left|x-y\right|=x+y+1\\ \text{This is the required particular solution of the given}\\ \text{differential equation}\text{.}\end{array}

Q.107

$\text{Solve the differential equation}\text{[}\frac{{\mathrm{e}}^{-2\sqrt{\mathrm{x}}}}{\sqrt{\mathrm{x}}}-\frac{\mathrm{y}}{\sqrt{\mathrm{x}}}]\frac{\mathrm{dx}}{\mathrm{dy}}=1(\mathrm{x}\ne 0)$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{differential}\mathrm{equation}\mathrm{\hspace{0.17em}}\mathrm{is}:\\ [\frac{{\mathrm{e}}^{-2\sqrt{\mathrm{x}}}}{\sqrt{\mathrm{x}}}-\frac{\mathrm{y}}{\sqrt{\mathrm{x}}}]\frac{\mathrm{dx}}{\mathrm{dy}}=1(\mathrm{x}\ne 0)\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=[\frac{{\mathrm{e}}^{-2\sqrt{\mathrm{x}}}}{\sqrt{\mathrm{x}}}-\frac{\mathrm{y}}{\sqrt{\mathrm{x}}}]\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\sqrt{\mathrm{x}}}=\frac{{\mathrm{e}}^{-2\sqrt{\mathrm{x}}}}{\sqrt{\mathrm{x}}}\\ \mathrm{Comparing}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{with}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{P}=\frac{1}{\sqrt{\mathrm{x}}}\mathrm{and}\mathrm{Q}=\frac{{\mathrm{e}}^{-2\sqrt{\mathrm{x}}}}{\sqrt{\mathrm{x}}}\\ \mathrm{Now},\mathrm{I}.\mathrm{F}\mathrm{.}={\mathrm{e}}^{\int \mathrm{P}\mathrm{\hspace{0.17em}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int \frac{1}{\sqrt{\mathrm{x}}}\mathrm{\hspace{0.17em}}\mathrm{dx}}={\mathrm{e}}^{2\sqrt{\mathrm{x}}}\\ \mathrm{The}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}(\mathrm{I}.\mathrm{F}.)=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}\left({\mathrm{e}}^{2\sqrt{\mathrm{x}}}\right)=\int (\frac{{\mathrm{e}}^{-2\sqrt{\mathrm{x}}}}{\sqrt{\mathrm{x}}}\times {\mathrm{e}}^{2\sqrt{\mathrm{x}}})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\int \left(\frac{1}{\sqrt{\mathrm{x}}}\right)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\end{array}$ \begin{array}{l}y\left(e^{2\sqrt{x}}\right)=2\sqrt{x}+C\\ \text{This is the general solution of the given equation}\text{.}\end{array}

Q.108

$\begin{array}{l}\mathrm{Find}\mathrm{a}\mathrm{particular}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}\\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ycotx}=4\mathrm{xcosecx}\hspace{0.17em}\hspace{0.17em}(\mathrm{x}\ne 0),\mathrm{given}\mathrm{thaty}=0\\ \mathrm{when}\mathrm{x}=\frac{\mathrm{\pi }}{2}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ycotx}=4\mathrm{xcosecx}\hspace{0.17em}\hspace{0.17em}(\mathrm{x}\ne 0)\\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{\hspace{0.17em}}\mathrm{where}\hspace{0.17em}\hspace{0.17em}\mathrm{P}=\mathrm{cotx}\mathrm{and}\mathrm{Q}=4\mathrm{xcosecx}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{P}\mathrm{\hspace{0.17em}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int \mathrm{cotx}\mathrm{\hspace{0.17em}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\mathrm{logsinx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\mathrm{sinx}\\ \mathrm{The}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{y}(\mathrm{I}.\mathrm{F}\mathrm{.})=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{ysinx}=\int (4\mathrm{xcosecx}\times \mathrm{sinx})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\int 4\mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\end{array}$ \begin{array}{l}=4\frac{x^2}{2}+C\\ \Rightarrow \text{y}\sin x=2x^2+C\mathrm{\dots }\left(i\right)\\ \text{Which is general solution of the given differential equation}\text{.}\\ \text{Now, y}=0\text{when}x=\frac{\pi }{2}\\ \text{From equation}\left(\text{i}\right)\text{, we have}\\ \left(0\right)\sin \frac{\pi }{2}=2{\left(\frac{\pi }{2}\right)}^2+C\Rightarrow C=-\frac{{\pi }^2}{2}\\ \text{Putting value of C in equation}\left(\text{i}\right)\text{, we get}\\ \text{y}\sin x=2x^2-\frac{{\pi }^2}{2}\\ \text{This is the particular solution of the given differential equation}\text{.}\end{array}

Q.109

$\begin{array}{l}\mathrm{Find}\mathrm{a}\mathrm{particular}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}\\ (\mathrm{x}+1)\frac{\mathrm{dy}}{\mathrm{dx}}=2{\mathrm{e}}^{–\mathrm{y}}-1,\mathrm{given}\mathrm{that}\mathrm{y}=0\mathrm{when}\mathrm{x}=0.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ (\mathrm{x}+1)\frac{\mathrm{dy}}{\mathrm{dx}}=2{\mathrm{e}}^{–\mathrm{y}}-1\\ \Rightarrow \frac{\mathrm{dy}}{(2{\mathrm{e}}^{–\mathrm{y}}-1)}=\frac{\mathrm{dx}}{(\mathrm{x}+1)}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\int \frac{\mathrm{dy}}{(2{\mathrm{e}}^{–\mathrm{y}}-1)}\mathrm{\hspace{0.17em}}=\int \frac{1}{(\mathrm{x}+1)}\mathrm{\hspace{0.17em}}\mathrm{dx}\end{array}$ \begin{array}{l}\Rightarrow {\displaystyle \int \frac{e^{y}dy}{\left(2-e^y\right)}}={\displaystyle \int \frac{1}{\left(x+1\right)}}dx\\ \Rightarrow -\log \left(2-e^y\right)=\log \left|x+1\right|+\log C\\ \Rightarrow \log {\left(2-e^y\right)}^{-1}=\log \left|C\left(x+1\right)\right|\\ \Rightarrow {\left(2-e^y\right)}^{-1}=C\left(x+1\right)\\ \Rightarrow \left(2-e^y\right)=\frac{1}{C\left(x+1\right)}\mathrm{\dots }\left(i\right)\\ \text{Now},\text{y}=0\text{when}x=\text{0}\\ \left(2-e^0\right)=\frac{1}{C\left(0+1\right)}\\ \Rightarrow \left(2-1\right)=\frac{1}{C}\\ \Rightarrow C=1\\ \text{Putting value C in equation}\left(\text{i}\right)\text{, we get}\\ \left(2-e^y\right)=\frac{1}{1\left(x+1\right)}\\ \Rightarrow e^y=2-\frac{1}{x+1}\\ =\frac{2x+2-1}{x+1}\\ y=\log \left|\frac{2x+1}{x+1}\right|,x\ne -1\\ \text{This is the particular solution of the given differential equation}\text{.}\end{array}

Q.110 The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time.
If the population of the village was 20,000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{number}\mathrm{of}\mathrm{people}\mathrm{in}\mathrm{a}\mathrm{village}\mathrm{at}\mathrm{any}\mathrm{instant}\left(\mathrm{t}\right)\mathrm{be}\mathrm{P}\mathrm{.}\\ \mathrm{Then},\mathrm{according}\mathrm{to}\mathrm{question}:\\ \frac{\mathrm{dP}}{\mathrm{dt}}\mathrm{\alpha }\mathrm{\hspace{0.17em}}\mathrm{P}\\ \frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{kP}[\mathrm{Where}\mathrm{k}\mathrm{is}\mathrm{a}\mathrm{proportionality}\mathrm{constant}\mathrm{.}]\\ \frac{\mathrm{dP}}{\mathrm{P}}=\mathrm{kdt}\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\int \frac{\mathrm{dP}}{\mathrm{P}}\mathrm{\hspace{0.17em}}=\mathrm{k}\int \mathrm{dt}\mathrm{\hspace{0.17em}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{logP}=\mathrm{kt}+\mathrm{C}...\left(\mathrm{i}\right)\\ \mathrm{In}\mathrm{the}\mathrm{year}1999,\mathrm{t}=\; 0\mathrm{and}\mathrm{P}=\; 20,000\\ \therefore \mathrm{\hspace{0.17em}}\log 20,000=\mathrm{k}\left(0\right)+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{C}=\log 20,000\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{logP}=\mathrm{kt}+\log 20,000\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}...\left(\mathrm{ii}\right)\\ \mathrm{In}\mathrm{the}\mathrm{year}2004,\mathrm{t}=\; 5\mathrm{\hspace{0.17em}}\mathrm{years}\mathrm{and}\mathrm{P}=\; 25,000,\mathrm{then}\mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right),\\ \mathrm{we}\mathrm{have}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\log 25,000=\mathrm{k}\left(5\right)+\log 20,000\end{array}$ \begin{array}{l}\log \frac{\text{25,000}}{\text{20,000}}=5k\\ k=\frac{1}{5}\log \left(\frac{5}{4}\right)\\ \text{Putting value of k in equation}\left(\text{ii}\right)\text{, we have}\\ \log P=\frac{1}{5}\log \left(\frac{5}{4}\right)t+\log \text{20,000}\\ \text{In the year 2009, t}=\text{10 years}\\ \log P=\text{10}\times \frac{1}{5}\log \left(\frac{5}{4}\right)+\log \text{20,000}\\ \Rightarrow \log P=2\log \left(\frac{5}{4}\right)+\log \text{20,000}\\ \Rightarrow \log P=\log {\left(\frac{5}{4}\right)}^2+\log \text{20,000}\\ =\log \left\{{\left(\frac{5}{4}\right)}^2\times 20,000\right\}\\ P=\frac{5}{4}\times \frac{5}{4}\times 20,000=31250\\ \text{Therefore, the population of the village in year 2009 is 31250}\text{.}\end{array}

Q.111

$\begin{array}{l}\mathrm{The}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}\\ \frac{\mathrm{ydx}–\mathrm{xdy}}{\mathrm{y}}=\; 0\; \hspace{0.17em}\mathrm{is}\\ \left(\mathrm{A}\right)\mathrm{xy}=\mathrm{C}\left(\mathrm{B}\right)\mathrm{x}={\mathrm{Cy}}^{\mathrm{2}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{C}\right)\mathrm{}\mathrm{y}=\mathrm{Cx}\left(\mathrm{D}\right)\mathrm{y}={\mathrm{Cx}}^{\mathrm{2}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \frac{\mathrm{ydx}-\mathrm{xdy}}{\mathrm{y}}=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{dx}}{\mathrm{x}}-\frac{\mathrm{dy}}{\mathrm{y}}=0\\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ \mathrm{\hspace{0.17em}}\int \frac{1}{\mathrm{x}}\mathrm{\hspace{0.17em}}\mathrm{dx}-\int \frac{1}{\mathrm{y}}\mathrm{\hspace{0.17em}}\mathrm{dy}=\mathrm{logC}‘\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{logx}-\mathrm{logy}=\mathrm{logC}‘\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\log \left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\mathrm{logC}‘\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{C}‘\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\frac{1}{\mathrm{C}‘}\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{y}=\mathrm{Cxwhere},\mathrm{C}=\frac{1}{\mathrm{C}‘}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{C}\mathrm{.}\end{array}$

Q.112

$\begin{array}{l}\mathrm{The}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{a}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{type}\\ \frac{\mathrm{dy}}{\mathrm{dx}}+{\mathrm{P}}_1\mathrm{x}={\mathrm{Q}}_1\mathrm{\hspace{0.17em}}\mathrm{is}\\ \left(\mathrm{A}\right){\mathrm{ye}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dy}}=\int \left({\mathrm{Q}}_1{\mathrm{e}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dy}}\right)\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ \left(\mathrm{B}\right)\mathrm{y}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dX}}=\int \left({\mathrm{Q}}_1{\mathrm{e}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dX}}\right)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \left(\mathrm{C}\right)\mathrm{x}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dy}}=\int \left({\mathrm{Q}}_1{\mathrm{e}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dy}}\right)\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ \left(\mathrm{D}\right){\mathrm{xe}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dx}}=\int \left({\mathrm{Q}}_1{\mathrm{e}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dx}}\right)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{\hspace{0.17em}}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{type}\\ \frac{\mathrm{dx}}{\mathrm{dy}}+{\mathrm{P}}_1\mathrm{x}={\mathrm{Q}}_1\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{.}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{I}.\mathrm{F}\mathrm{.}={\mathrm{e}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dy}}\\ \mathrm{The}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \mathrm{x}(\mathrm{I}.\mathrm{F}\mathrm{.})=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ \mathrm{x}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dy}}=\int ({\mathrm{Q}}_1\times {\mathrm{e}}^{\int {\mathrm{P}}_1\mathrm{\hspace{0.17em}}\mathrm{dy}})\mathrm{\hspace{0.17em}}\mathrm{dy}+\mathrm{C}\\ \mathrm{Therefore},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{C}\mathrm{.}\end{array}$

Q.113

$\begin{array}{l}\mathrm{The}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}\\ {\mathrm{e}}^{\mathrm{x}}\mathrm{dy}+({\mathrm{ye}}^{\mathrm{x}}+2\mathrm{x})\mathrm{dx}=0\mathrm{is}\\ \left(\mathrm{A}\right){\mathrm{xe}}^{\mathrm{y}}+{\mathrm{x}}^2=\mathrm{C}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{B}\right)\mathrm{x}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{\mathrm{y}}+{\mathrm{y}}^2=\mathrm{C}\\ \left(\mathrm{C}\right){\mathrm{ye}}^{\mathrm{x}}+{\mathrm{x}}^2=\mathrm{C}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left(\mathrm{D}\right){\mathrm{ye}}^{\mathrm{y}}+{\mathrm{x}}^2=\mathrm{C}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{differential}\mathrm{equation}\mathrm{is}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{e}}^{\mathrm{x}}\mathrm{dy}+({\mathrm{ye}}^{\mathrm{x}}+2\mathrm{x})\mathrm{dx}=0\\ \Rightarrow {\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}+{\mathrm{ye}}^{\mathrm{x}}=-2\mathrm{x}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=-2{\mathrm{xe}}^{-\mathrm{x}}\\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\mathrm{\hspace{0.17em}}\mathrm{where}\hspace{0.17em}\hspace{0.17em}\mathrm{P}=1\mathrm{and}\mathrm{Q}=-2{\mathrm{xe}}^{-\mathrm{x}}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int \mathrm{P}\mathrm{\hspace{0.17em}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\int 1\mathrm{\hspace{0.17em}}\mathrm{dx}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\mathrm{e}}^{\mathrm{x}}\\ \mathrm{The}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{is}:\\ \hspace{0.17em}\hspace{0.17em}\mathrm{y}(\mathrm{I}.\mathrm{F}\mathrm{.})=\int (\mathrm{Q}\times \mathrm{I}.\mathrm{F}.)\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{ye}}^{\mathrm{x}}=\int (-2{\mathrm{xe}}^{-\mathrm{x}}\times {\mathrm{e}}^{\mathrm{x}})\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-2\int \mathrm{x}\mathrm{\hspace{0.17em}}\mathrm{dx}+\mathrm{C}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=-2\frac{{\mathrm{x}}^2}{2}+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{ye}}^{\mathrm{x}}=-{\mathrm{x}}^2+\mathrm{C}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{ye}}^{\mathrm{x}}+{\mathrm{x}}^2=\mathrm{C}\\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{C}\mathrm{.}\end{array}$