# NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2

Scientific disciplines including Physics, Chemistry, Biology, Mathematics, etc., have been subjects of interest to the human race since the beginning of time. Among the prominent scientific disciplines, Mathematics particularly has a long-drawn history which can be traced back to many early civilizations. The reason is that Mathematics holds great importance in the ordering and smooth functioning of human activities as well as of a majority of human-made or human-controlled operations. Mathematics is thus a critical academic discipline that is taught to students from the beginning of their education.It is also composed of a variety of sub-disciplines, which are vital knowledge systems in themselves. Additionally, Mathematics also has an immeasurable potential for research and experimentation, which makes it ideal for young learners. Within the NCERT academic curriculum, students who wish to pursue Mathematics in greater depth opt for the Mathematics Stream in Classes 11 and 12. Therefore, it goes without saying that Mathematics is an inseparable part of the prescribed NCERT academic curriculum for the students of Class 12. The academic curriculum prescribed by NCERT for Class 12 is aimed at laying a strong foundation in the fundamental sub-disciplines of Mathematics to prepare students for their future endeavours.

This implies the importance of easy access to reliable learning resources like the NCERT Solutions as trustworthy reference materials to facilitate self-learning. The prescribed NCERT academic syllabus for Mathematics in Class 12 is made up of thirteen extensively researched chapters. These chapters have been organised into an ordered sequence for ease of learning. Chapter 9 is titled Differential Equations. Ex 9.2 Class 12 is the second comprehensive assessment, which is a part of this chapter. It has to be objectively comprehended and thoroughly practised to succeed in the examination.

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**NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations (Ex 9.2) Exercise 9.2**

Widespread apprehension and indecisiveness concerning the theme of Differential Equations have been observed among students for a long time. This attitude has been a major cause of concern for students and their instructors alike. Students often find the subject of Differential Equations challenging and as a consequence, may lose interest in it. Differential Equations can be quite an interesting theme if one has the necessary conceptual clarity to complement their efforts to understand the subject. It is also clear that Differential Equations and Exercise 9.2 for Class 12 are critical components of the mathematics curriculum determined for Class 12.As a result, the NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2 can be useful in easing students’ concerns.. The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2 have been designed by Extramarks keeping the general concerns and queries of students and teachers in view. The easy-to-understand format of the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2 can prove crucial in aiding and encouraging self-learning among students.

Differential Equations are a basic and vital part of the wider academic arena of Mathematics as a subject of research. The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2 provide clear-cut and simple solutions to Exercise 9.2 of the chapter on Differential Equations. The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2 have been compiled with careful consideration of the methods of classroom teaching. Therefore, the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2 do not exhibit major discrepancies between classroom lectures and the compiled solutions.

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**Important Topics Covered in NCERT Solution Class 12 Chapter 9 (Exercise 9.2) **

Chapter 9 titled Differential Equations, covers a range of important topics, such as comprehensive questions asking students to verify whether the given function is a solution to the corresponding differential equation. Types of functions, namely implicit and explicit functions have also been covered. The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2 cover all the essential topics which are a part of Exercise 9.2 of Chapter 9. The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2 provide detailed, step-by-step solutions to the problems which comprise Exercise 9.2 Class 12. The formulae used to do the calculations have also been mentioned and covered in detail in order to ensure that students have total clarity about the logical sequence of the calculations. Extramarks is providing the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2 to complement the hard work and perseverance of students.

**Access NCERT Solutions for Class 12 Maths Chapter 9 – Permutations and Combinations**

**NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.2**

Differential Equations have often been perceived and upheld as a dynamic, complex and versatile academic theme within Mathematics. However, within the academic syllabus for Mathematics in Class 12 prescribed by the NCERT, the theme of Differential Equations occupies a prominent place. Differential Equations is indeed a theme comprising conceptually-rich content. This is why it is also regarded as an opening into a wider expanse of academic research in the field of Mathematics. Therefore, it is imperative for students who are looking forward to pursuing a career in Mathematics, to improve their conceptual clarity and comprehension of the theme of Differential Equations.

It is highly recommended that students continue with the regular and consistent practise of questions that are part of the theme of Differential Equations. This would be of great assistance in enhancing the retention abilities of students, which is essential to acquiring and retaining important formulas and their respective derivatives. In accordance with this, students must review and practise the exercises that are part of their textbook with the aid of quality reference material like the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2. Regular practise of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2 will help students achieve excellence in the topic of differential equations.

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**Benefits of NCERT Solutions Class 12 Maths Chapter 9 Exercise 9.2 **

**Benefits of NCERT Solutions Class 12 Maths Chapter 9 Exercise 9.2 **

Differential Equations is a complex topic that contains a variety of information that must be dealt with in various ways and through various learning methods.While at the micro level, it is imperative for students to be familiar with important formulae and their respective derivatives and preferably have them in their memory, conceptual clarity at the holistic level is also mandatory. It is a difficult task to accommodate all of the students’ questions and doubts during classroom teaching sessions and to adequately respond to all of them.These situations may lead to the accumulation of unresolved doubts and queries, which may become a hindrance to attaining the necessary conceptual clarity. This deficiency may pose a significant challenge to students’ sincere efforts to perform well in their exams.

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**Q.1 **

\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{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{\hspace{0.17em}}\text{:}\text{}\text{}\text{y\u201d \u2013 y\u2019 = 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\u2019 = e}}^{\text{x}}\text{}\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\u201d = e}}^{\text{x}}\\ \text{Substituting values of y\u2019 and y\u201d in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y\u201d \u2013 y\u2019 = 0, we get}\\ {\text{y\u201d \u2013 y\u2019 =e}}^{\text{x}}{\text{\u2013 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.2**

\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{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{:}\text{}\text{}\text{y\u2019 \u20132x \u20132 = 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\u2019 = 2x + 2}\\ \text{Substituting values of y\u2019 in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y\u2019 \u20132x \u20132 = 0, we get}\\ \text{y\u2019 \u2013 2x \u20132 = 2x + 2 \u20132x \u20132}\\ \text{}\text{}\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.3**

\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{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{:}\text{}\text{}\text{y\u2019 + 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\u2019 = \u2013sinx}\\ \text{Substituting values of y\u2019 in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation y\u2019 + sinx = 0, we get}\\ \text{y\u2019 + sinx = \u2013sinx + 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.4**

\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{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{:}\text{}\text{}\text{y\u2019 =}\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\u2019 =}\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\u2019 =}\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{}\text{\hspace{0.17em}}\text{=}\frac{\text{x}\left(\sqrt{{\text{1 + x}}^{\text{2}}}\right)}{{\text{1 + x}}^{\text{2}}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\frac{\text{x}}{\sqrt{{\text{1 + x}}^{\text{2}}}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{= y\u2019 = L}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

**Q.5**

\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 = Ax}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{:}\text{}\text{}\text{xy\u2019 = y}\text{}\text{}\left(\text{x}\ne \text{0}\right)\end{array}

**Ans**

\begin{array}{l}\text{y = Ax}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y\u2019 = A}\\ \text{Substituting values of y\u2019 in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation xy\u2019 = y, we get}\\ \text{L}\text{.H}\text{.S}\text{. = xy\u2019}\\ \text{}\text{= x}\left(\text{A}\right)\\ \text{}\text{= Ax = y = R}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

**Q.6**

\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 sinx}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{: xy\u2019 = y + x}\sqrt{{\text{x}}^{\text{2}}{\text{\u2013 y}}^{\text{2}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{x}\ne \text{0}\text{\hspace{0.17em}}\text{and x > y or x < \u2013y}\right)\end{array}

**Ans**

\begin{array}{l}\text{y = xsin x}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{y\u2019 = xcos x + sinx}\\ \text{Substituting values of y\u2019 in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation xy\u2019 = y + x}\sqrt{{\text{x}}^{\text{2}}{\text{\u2013 y}}^{\text{2}}}\text{, we get}\\ \text{L}\text{.H}\text{.S}\text{. = xy\u2019 = x}\left(\text{xcosx + sinx}\right)\\ {\text{= x}}^{\text{2}}\text{cosx + xsinx}\\ \text{R}\text{.H}\text{.S}\text{. = xsinx + x}\sqrt{{\text{x}}^{\text{2}}\text{\u2013}{\left(\text{xsinx}\right)}^{\text{2}}}\\ {\text{= xsinx + x}}^{\text{2}}\sqrt{{\text{1 \u2013 sin}}^{\text{2}}\text{x}}\\ {\text{= xsinx + x}}^{\text{2}}\text{cosx}\\ \text{So, L}\text{.H}\text{.S}\text{. = R}\text{.H}\text{.S}\text{.}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

**Q.7**

\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{xy = logy + C : y\u2019 =}\frac{{\text{y}}^{\text{2}}}{\text{1 \u2013 xy}}\left(\text{xy}\ne \text{0}\right)\end{array}

**Ans**

\begin{array}{l}\text{xy = logy + C}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{xy\u2019 + y =}\frac{\text{1}}{\text{y}}\text{y\u2019 + 0}\\ {\text{xyy\u2019 + y}}^{\text{2}}\text{= y\u2019}\\ {\text{y}}^{\text{2}}\text{= y\u2019 \u2013 xyy\u2019}\\ \text{= y\u2019}\left(\text{1 \u2013 xy}\right)\\ \text{y\u2019 =}\frac{{\text{y}}^{\text{2}}}{\left(\text{1 \u2013 xy}\right)}\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

**Q.8**

\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 \u2013 cosy = x :}\left(\text{ysiny + cosy + x}\right)\text{y\u2019 = y}\end{array}

**Ans**

\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y-\mathrm{cos}y=x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\left(i\right)\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y\u2019}+\mathrm{sin}y.y\u2018=1\\ y\u2018\left(1+\mathrm{sin}y\right)=1\\ \text{}\text{}y\u2018=\frac{1}{1+\mathrm{sin}y}\\ \text{Substituting values of y\u2019 in L}\text{.H}\text{.S}\text{. of given differential}\\ \text{equation}\left(\text{ysiny + cosy + x}\right)\text{y\u2019 = y, we get}\\ \left(y\mathrm{sin}y+\mathrm{cos}y+x\right)y\u2018=\left(y\mathrm{sin}y+\mathrm{cos}y+x\right)\frac{1}{\left(1+\mathrm{sin}y\right)}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(y\mathrm{sin}y+\mathrm{cos}y+y-\mathrm{cos}y\right)\frac{1}{\left(1+\mathrm{sin}y\right)}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\left[\text{From equation}\left(\text{i}\right)\right]\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(y\mathrm{sin}y+y\right)\frac{1}{\left(1+\mathrm{sin}y\right)}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=y\left(\mathrm{sin}y+1\right)\frac{1}{\left(1+\mathrm{sin}y\right)}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=y=R.H.S.\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

**Q.9**

\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}^{\u20131}y\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}:\text{}{y}^{2}y\u2018+{y}^{2}+1=0\end{array}

**Ans**

\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{\hspace{0.17em}}x+y={\mathrm{tan}}^{\u20131}y\text{\hspace{0.17em}}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1+y\u2019}=\left(\frac{1}{1+{y}^{2}}\right)y\u2018\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{1}+\text{\hspace{0.17em}}\text{y\u2019}\right)\left(1+{y}^{2}\right)=y\u2018\\ 1+{y}^{2}+y\u2018+y\u2018{y}^{2}=y\u2018\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1+{y}^{2}+y\u2018{y}^{2}=0\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

**Q.10**

\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}}\text{\hspace{0.17em}}x\in \left(-a,a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}:\text{}x+y\frac{dy}{dx}=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(y\ne 0\right)\end{array}

**Ans**

\begin{array}{l}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=\sqrt{{a}^{2}-{x}^{2}}\text{\hspace{0.17em}}\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{dy}{dx}=\frac{-x}{\sqrt{{a}^{2}-{x}^{2}}}\\ \text{}\text{}\text{}=\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}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=x+y\left(\frac{-x}{y}\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=x-x\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0=R.H.S.\\ \text{Thus,the given function is the solution of the corresponding}\\ \text{differential equation}\text{.}\end{array}

**Q.11 **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.12** 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.

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