NCERT Solutions Class 12 Maths Chapter 9 Differential Equation Exercise 9.4

Q.1 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.2 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.3 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.4 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.5 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.6 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.7 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.8 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.9 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.10 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.11

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

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.12

\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}