NCERT Solutions Class 12 Maths Chapter 9 Differential Equations: Download Free PDFs, Formulas & Weightage

Class 12 Math Chapter 9 Differential Equations carry 7–9 marks in the CBSE Class 12 Board Exam, which includes a 2-mark question (VSA) related to Definitions, order, and degree, and 4-6 Marks question (LA) related to solving differential equations and application problems. Class 12 Differential Equations also hold good weight in JEE Main, which typically carries 4 –12 marks questions related to various concepts mentioned in the chapter. Students can check key topics and important formulae below;

Class 12 Differential Equations Key Topics

• Basic Concepts such as Definition, order, degree, and general form of differential equations.
• Formation of Differential Equations & Elimination of arbitrary constants from equations
• Types: Ordinary and Partial Differential Equations
• Solutions of Differential Equations: General solution and particular solution.
• Methods of Solving Differential Equations: Variable Separation Method and more
• Homogeneous and Linear Differential Equations
• Applications: growth and decay-related problems based on mostly first and second-order derivative differential equations.

Check out Dropped Topics of Differential Equations – 9.4 Formation of Differential Equations Whose General Solution is Given, Example 25, Ques. 3, 5 and 15 (Miscellaneous Exercise), Point Six of the Summary.

Differential Equations Important Formulae for CBSE and Competitive Exams

Students can check the important formulae for CBSE and other engineering entrance exams such as JEE Main, CUSAT CAT, and more... below;

• General Solution:

  $$y = \int f(x)dx + C$$
  

• Variable Separation:

  $$\int \frac{1}{g(y)}dy = \int f(x)dx$$
  

• Linear Differential Equations:

  • General form: $\frac{dy}{dx} + Py = Q$
    
  • Solution form: $$y\cdot e^{\int Pdx}=\int Q\cdot e^{\int Pdx}dx+C$$

• Homogeneous Equations:

  • Substitution: $y = vx$or $x = vy$
    
  • Simplified separable form: $$\frac{dy}{dx} = f\left(\frac{y}{x}\right)$$
    

• Exact Differential Equations:

  • General form: $M(x, y)dx + N(x, y)dy = 0$
    
  • Solution: $$\int Mdx + \int \left(N - \frac{\partial}{\partial y} \left( \int Mdx \right)\right)dy = C$$
• Exponential Growth/Decay Differential Equations: $$\frac{dy}{dt}=ky\text{Solution: }y=y_0{e}^{kt}$$

Shiksha has prepared solutions for the complete differential equations chapter in one solution PDF. Students can download the Class 12 Math Chapter 9 Differential Equations solution PDF for free. The Class 12 Maths Differential Equations NCERT Solution PDF can be very useful for all student boards as well as JEE Main, and other entrance exams. Download the Differential Equation Solution PDF for free;

Class 12 Math Chapter 9 Differential Equations Solution: Free PDF Download

Class 12 Differential Equations is an important chapter, which consists of essential concepts of Calculus. Various exercises of Differential Equations deal with key concepts such as order and degree of a differential equation, formation of differential equations by eliminating arbitrary constants, and methods to solve them, including variable separation, homogeneous equations, and linear differential equations.  Shiksha provides exercise-wise solutions with step-by-step explanations, Students can take the help of solutions to master these concepts. Students can check the Exercise-wise Chapter 9 Differential Equation Class 12 math solutions here.

Class 12 Differential Equations Exercise 9.1 focuses on Fundamental concepts such as understanding the order and degree of a differential equation, identifying differential equations from given relationships, and forming differential equations by eliminating arbitrary constants from given functions. Differential Equation Exercise 9.1 Solutions consists of 12 Questions. Students can check the complete Exercise 9.1 Solutions below;

Class 12 Chapter 9 Differential Equations Exercise 9.1 NCERT Solution

Determine order and degree (if defined) of differential equations given in Questions 1 to 10:

Q1.    $\frac{d^{4}y}{d{x}^4}+sin\left(y\text{'}\right)=0$

A.1. The highest order derivation present in the differential equation (D.E.) is  $\frac{d^{4}y}{d{x}^4}$ , so its order is 4.

As, the given D.E.is not a polynomial equation in its derivative ,its degree is not defined.

Q2. y' + 5y = 0

A.2. The highest order derivation present in the D.E. is y , so its order is 1.

As the given D.E. is a polynomial equation in its derivative its degree is 1.

Q3.  ${\left(\frac{ds}{dt}\right)}^4+3s\frac{d^{2}s}{d{t}^2}=0$

A.3. The highest order derivation present in the D.E. is  $\frac{d^{2}s}{d{t}^2}$ so its order is 2 .

As the given D.E. is a polynomial equation in its derivative its degree is 1.

Q4.  $\frac{d^{2}y}{{\left(d{x}^2\right)}^2}+cos\left(\frac{dy}{dx}\right)=0$

A.4.  $\frac{d^{4}y}{d{x}^4}$ As the given D.E. is not a polynomial equation in its derivative, its degree is not defined.

Q5.    $\frac{d^{2}y}{d{x}^2}=cos3x+sin3x$

A.5.  $\frac{d^{4}y}{d{x}^4}$ As the given D.E. is a polynomial equation in its derivative, its degree is 1.

Q6. (y′′′) + (y′′) + (y′)⁴ + y⁵ =0

A.6. The highest order derivative present in the D.E. is  $y^{\left|\right||}$ so its order is 3.

As the given D.E. is a polynomial equation in its derivation , its degree is 2.

Q7. y''' + 2y'' + y' = 0

A.7. The highest order present in the D.E. is  $y^{\left|\right||}$ so its order is 3.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

Q8. y′ + y = e^x

A.8. The given order derivative present in the D.E. is  $y^{|}$ so its order is 1.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

Q9. y'′ + (y')²  + 2y = 0

A.9. The highest order derivative present in the D.E. is  $y^{\left|\right|}$ so its order is 2.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

Q10.  y'′ + 2y' + sin y = 0

A.10. The highest order derivative present in the D.E. is  $y^{\left|\right|}$ so its order is 2.

As the given D.E. is polynomial equation in its derivative, its degree is 1.

Q11. The degree of the differential equation   ${\left(\frac{d^{2}y}{d{x}^2}\right)}^3+{\left(\frac{dy}{dx}\right)}^2+sin\left(\frac{dy}{dx}\right)+1=0$  is:

(A) 3 

(B) 2 

(C) 1 

(D) Not defined

A.11. In the given D.E,

$sin\frac{dy}{dx}$ is a trigonometric function of derivative  $\frac{dy}{dx}$ . So it is not a polynomial equation so its derivative is not defined.

Hence, Degree of the given D.E. is not defined.

$\therefore$ Option (D) is correct.

Q12. The order of the differential equation   $2x^2\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+y=0$  is:

(A) 2 

(B) 1 

(C) 0 

(D) Not defined

A.12. The highest order derivative present in the given D.E. is  $\frac{d^{2}y}{d{x}^2}$ and its order is 2.

$\therefore$ Option (A) is correct.

Differential Equations Exercise 9.2 of Class 12 Mathematics focuses on problems related to solving differential equations using the variable separation method. Students need to rewrite a given differential equation so that the variables x and $y$ are separated, in the form $\frac{dy}{dx} = f(x)g(y)$, and followed by integration to find the solution. Differential Equations Exercise 9.2  Solutions will consist of explanations of 12 Questions. Students can check the complete Exercise 9.2 Solutions below;

Class 12 Chapter 9 Differential Equations Exercise 9.2 NCERT Solution

Note: In each of the Questions, 1 to 6 verifies that the given functions (explicit) is a solution of the corresponding differential equation:

Q1.  y = e^x   + 1  :  y″ – y′ = 0

A.1. Given  $f{x}^n$ is  $y=e^x+1$

Differentiating with  $x$ we get,

$y^{|}=\frac{dy}{dx}=e^x$

Again,

$y^{\left|\right|}=\frac{d^{2}y}{d{x}^2}=e^x$

Substituting value of  $y^{\left|\right|}$ and  $y^{|}$ in the given D.E. we get

$L.H.S.=y^{\left|\right|}-y^{|}=e^x-e^x=0=R.H.S$

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E.

Q2. y = x²  + 2x + C  :  y′ – 2x – 2 = 0

A.2. Given ,  $f{x}^n$ is  $y=x^2+2x+c$

So,  $y^{|}=2x+2$

Substituting value of  $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}-2x-2=2x+2-2x-2=0=R.H.S$

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E.

Q3. y = cos x + C : y′ + sin x = 0

A.3. Given,  $f{x}^n$ is  $y=cosx+c$

So,  $y^{|}=-sinx$

Putting the value of  $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}+sinx=-sinx+sinx=0=R.H.S$

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E.

Q4.  y = √1 + x² ; y/= $\frac{xy}{1+x^2}$

Q5. y = Ax   :  xy′ = y (x ≠ 0)

A.5. Given,  $y=Ax:$

So,  $y^{|}=\frac{Adx}{dx}=A$

Putting value of  $y^{|}$ in L.H.S. of the given D.E.

L.H.S=  $x{y}^{|}=xA=Ax=y$ =R.H.S

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E.

A.6. Given,  $y=xsinx$

So,  $y^{|}=x\frac{d}{dx}sinx+sinx\frac{dx}{dx}=xcosx+sinx$

Now, L.H.S of the given D.E  $=x{y}^{|}$

$=x\left(xcosx+sinx\right)$

$=x^{2}cosx+xsinx$

 

Q7. xy = log y + C :       

$y\text{'}=\frac{y^2}{1-xy}\left(xy\ne 1\right)$

A.7. Given,  $xy=logy+c$

Differentiate w.r.t. x we have

$\begin{array}{l}x\frac{dy}{dx}+y\frac{dx}{dx}=\frac{d}{dx}logy+\frac{d}{dx}C\\ \Rightarrow x\frac{dy}{dx}+y=\frac{1}{y}\frac{dy}{dx}+0\\ \Rightarrow x\frac{dy}{dx}-\frac{1}{y}\frac{dy}{dx}=-y\\ \Rightarrow \frac{dy}{dx}\left[x-\frac{1}{y}\right]=-y\\ \Rightarrow \frac{dy}{dx}\left[\frac{xy-1}{y}\right]=-y\\ \Rightarrow \frac{dy}{dx}=\frac{-y^2}{xy-1}=\frac{\left(-1\right)\times -y^2}{\left(-1\right)\times \left(xy-1\right)}=\frac{y^2}{1-xy}\\ \therefore y^{|}=\frac{y^2}{1-xy}\end{array}$

Hence, y is a Solution of the given D.E

Q8. y – cos y = x :  (y sin y + cos y + x) y′ = y

A.8. Given  $y-cosy=x$

Differentiate w.r.t ‘x’ we get

$\begin{array}{l}\frac{dy}{dx}-\left(-siny\right)\frac{dy}{dx}=\frac{dx}{dx}\\ \Rightarrow \frac{dy}{dx}\left[1+siny\right]=1\\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+siny}=y^{|}\end{array}$

So, L.H.S of given D.E  $=\left(ysiny+cosy+x\right)y^{|}$

$\begin{array}{l}=\left(ysiny+cosy+y-cosx\right)\left[\frac{1}{1+siny}\right]\\ =\frac{y\left(1+siny\right)}{\left(1+siny\right)}=y=R.H.S\end{array}$

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E.

Q9. x + y = tan^-1 y   :   y²y' + y² + 1 = 0

A.9. Given,  $x+y=ta{n}^{-1}y$

Differentiate with ‘x’ we get

$\begin{array}{l}1+\frac{dy}{dx}=\frac{1}{1+y^2}\frac{dy}{dx}\\ =1+y^{|}=\frac{1}{1+y^2}{y}^{|}\\ =\left(1+y^{|}\right)\left(1+y^2\right)=y^{|}\\ =1+y^2{y}^{|}+y^{|}+y^2=y^{|}\\ =y^2{y}^{|}+y^2+1=0\end{array}$

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E

Q11. 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

A.11. The number of arbitrary constant is general solution of D.E of 4^th order is four.

$\therefore$ Option (D) is correct.

Q12. 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

A.12. In a particular solution, there are no arbitrary constant.

Hence, option (D) is correct.

Class 12 Math Differential Equation Exercise 9.3 focuses on solving homogeneous differential equations. Homogeneous differential equations can be expressed in terms of a single variable by substituting $y = vx$or $x = vy$, and then solving it using integration. Differential Equations Exercise 9.3 Solutions are comprised of solutions of 12 Questions. Students can check the complete Exercise 9.3 Solutions below;

Class 12 Chapter 9 Differential Equations Exercise 9.3 NCERT Solution

Note: In each of the questions 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b

Q1.  $\frac{x}{a}+\frac{y}{b}=1$

A.1. Given: Equation of the family of curves   $\frac{x}{a}+\frac{y}{b}=1..........(i)$

Differentiating both sides of the given equation with respect to x, we get:

$\begin{array}{l}\frac{1}{a}+\frac{1}{b}\frac{dy}{dx}=0\\ \Rightarrow \frac{1}{a}+\frac{1}{b}{y}^{\text{'}}=0\end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l}0+\frac{1}{b}{y}^{"}=0\\ \Rightarrow \frac{1}{b}{y}^{"}=0\\ \Rightarrow y^{"}=0\end{array}$

Hence, the required differential equation of the given curve is  $y^{"}=0$ .

Q2.  $y^2=a\left(b^2-x^2\right)$

A.2. Given: Equation of the family of curves   $y^2=a\left(b^2-x^2\right)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}2y\frac{dy}{dx}=a\left(-2x\right)\\ \Rightarrow 2y{y}^{\text{'}}=-2ax\\ \Rightarrow y{y}^{\text{'}}=-ax..........(1)\end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l}{y}^{\text{'}}.y^{\text{'}}+y{y}^{"}=-a\\ \Rightarrow {\left(y^{\text{'}}\right)}^2+y{y}^{"}=-a..........(2)\end{array}$

Dividing equation (2) by equation (1), we get:

$\begin{array}{l}\frac{{\left(y^{\text{'}}\right)}^2+y{y}^{"}}{y{y}^{\text{'}}}=\frac{-a}{-ax}\\ \Rightarrow xy{y}^{"}+x{\left(y^{\text{'}}\right)}^2-y{y}^{"}=0\end{array}$

This is the required differential equation of the given curve.

Q3.  $y=a{e}^{3x}+b{e}^{-2x}$

A.3. Given: Equation of the family of curves   $y=a{e}^{3x}+b{e}^{-2x}..........(i)$

Differentiating both sides with respect to x, we get:

$y^{\text{'}}=3a{e}^{3x}-2b{e}^{-2x}..........(ii)$

Again, differentiating both sides with respect to x, we get:

$y^{"}=9a{e}^{3x}-4b{e}^{-2x}..........(iii)$

Multiplying equation (i) with (ii) and then adding it to equation (ii), we get:

$\begin{array}{l}\left(2a{e}^{3x}+2b{e}^{-2x}\right)+\left(3a{e}^{3x}-2b{e}^{-2x}\right)=2y+y^{\text{'}}\\ \Rightarrow 5a{e}^{3x}=2y+y^{\text{'}}\\ \Rightarrow a{e}^{3x}=\frac{2y+y^{\text{'}}}{5}\end{array}$

Now, multiplying equation (i) with (iii) and subtracting equation (ii) from it, we get:

$\begin{array}{l}\left(3a{e}^{3x}+2b{e}^{-2x}\right)-\left(3a{e}^{3x}-2b{e}^{-2x}\right)=3y-y^{\text{'}}\\ \Rightarrow 5b{e}^{-2x}=3y-y^{\text{'}}\\ \Rightarrow b{e}^{-2x}=\frac{3y-y^{\text{'}}}{5}\end{array}$

Substituting the values of  $a{e}^{3x}$ and  $b{e}^{-2x}$ in equation (iii), we get:

$\begin{array}{l}{y}^{"}=9.\frac{\left(2y-y^{\text{'}}\right)}{5}+4\frac{\left(3y-y^{\text{'}}\right)}{5}\\ \Rightarrow y^{"}=\frac{18y+9y^{\text{'}}}{5}+\frac{12y-4y^{\text{'}}}{5}\\ \Rightarrow y^{"}=\frac{30y+5y^{\text{'}}}{5}\\ \Rightarrow y^{"}=6y+y^{\text{'}}\\ \Rightarrow y^{"}-y^{\text{'}}-6y=0\end{array}$

This is the required differential equation of the given curve.

Q4.  $y=e^{2x}\left(a+bx\right)$

A.4. Given: Equation of the family of curves   $y=e^{2x}\left(a+bx\right)..........(1)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}{y}^{\text{'}}=2e^{2x}\left(a+bx\right)+e^{2x}.b\\ \Rightarrow y^{\text{'}}=e^{2x}\left(2a+2bx+b\right)..........(2)\end{array}$

Multiplying equation (1) with (2) and then subtracting it from equation (2), we get:

$\begin{array}{l}{y}^{\text{'}}-2y=e^{2x}\left(2a+2bx+b\right)-e^{2x}\left(2a+2bx\right)\\ \Rightarrow y^{\text{'}}-2y=b{e}^{2x}..........(3)\end{array}$

Differentiating both sides with respect to x, we get:

$y^{"}-2y^{\text{'}}=2b{e}^{2x}..........(4)$

Dividing equation (4) by equation (3), we get:

$\begin{array}{l}\frac{y^{"}-2y^{\text{'}}}{y^{\text{'}}-2y}=2\\ \Rightarrow y^{"}-2y^{\text{'}}=2y^{\text{'}}-4y\\ \Rightarrow y^{"}-4y^{\text{'}}+4y=0\end{array}$

This is the required differential equation of the given curve.

Q5.  $y=e^x\left(acosx+bsinx\right)$

A.5. Given: Equation of the family of curves  $y=e^x\left(acosx+bsinx\right)..........(i)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}{y}^{\text{'}}=e^x\left(acosx+bsinx\right)+e^x\left(-acosx+bsinx\right)\\ \Rightarrow y^{\text{'}}=e^x\left[\left(a+b\right)cosx-\left(a-b\right)sinx\right]..........(2)\end{array}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}{y}^{"}=e^x\left[\left(a+b\right)cosx-\left(a-b\right)sinx\right]+e^x\left[-\left(a+b\right)sinx-\left(a-b\right)cosx\right]\\ y^{"}=e^x\left[2bcosx-2asinx\right]\\ y^{"}=2e^x\left(bcosx-asinx\right)\\ \Rightarrow \frac{y^{"}}{2}=e^x\left(bcosx-asinx\right)..........(3)\end{array}$

Adding equations (1) and (3), we get:

$\begin{array}{l}y+\frac{y^{"}}{2}=e^x\left[\left(a+b\right)cosx-\left(a-b\right)sinx\right]\\ \Rightarrow y+\frac{y^{"}}{2}=y^{\text{'}}\\ \Rightarrow 2y+y^{"}=2y^{\text{'}}\\ \Rightarrow y^{"}-2y^{\text{'}}+2y=0\end{array}$

This is the required differential equation of the given curve.

Q6. Form the differential equation of the family of circles touching the y-axis at the origin.

A.6. The centre of the circle touching the y-axis at origin lies on the x-axis.

Let (a, 0) be the centre of the circle.

Since it touches the y-axis at origin, its radius is a.

Now, the equation of the circle with centre (a, 0) and radius (a) is

$\begin{array}{l}{\left(x-a\right)}^2+y^2=a^2.\\ \Rightarrow x^2+y^2=2ax..........(1)\end{array}$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2x+2y{y}^{\text{'}}=2a\\ \Rightarrow x+y{y}^{\text{'}}=a\end{array}$

Now, on substituting the value of a in equation (1), we get:

$\begin{array}{l}{x}^2+y^2=2\left(x+y{y}^{\text{'}}\right)x\\ \Rightarrow x^2+y^2=2x^2+2xy{y}^{\text{'}}\\ \Rightarrow 2xy{y}^{\text{'}}+x^2=y^2\end{array}$

This is the required differential equation.

Q7. Find the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

A.7. The equation of the parabola having the vertex at origin and the axis along the positive y-axis is:

$x^2 =4ay$

Differentiating equation (1) with respect to x, we get:

$2x=4ay\text{'}$

Dividing equation (2) by equation (1), we get:

$\begin{array}{l}\frac{2x}{x^2}=\frac{4a{y}^{\text{'}}}{4ay}\\ \Rightarrow \frac{2}{x}=\frac{y^{\text{'}}}{y}\\ \Rightarrow x{y}^{\text{'}}=2y\\ \Rightarrow x{y}^{\text{'}}-2y=0\end{array}$

This is the required differential equation.

Q8. Form the differential equation of family of ellipse having foci on y-axis and centre at the origin.

A.8. The equation of the family of ellipses having foci on the y-axis and the centre at origin is as follows:

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1..........(1)$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}\frac{2x}{b^2}+\frac{2y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow \frac{x}{b^2}+\frac{y{y}^{\text{'}}}{a^2}..........(2)\end{array}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}\Rightarrow \frac{1}{b^2}+\frac{y^{\text{'}}.y^{\text{'}}+y.y^{"}}{a^2}=0\\ \Rightarrow \frac{1}{b^2}+\frac{1}{a^2}\left({y^{\text{'}}}^2+y{y}^{"}\right)=0\\ \Rightarrow \frac{1}{b^2}=-\frac{1}{a^2}\left({y^{\text{'}}}^2+y{y}^{"}\right)\end{array}$

Substituting this value in equation (2), we get:

$\begin{array}{l}x\left[-\frac{1}{a^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)\right]+\frac{y{y}^{"}}{a^2}=0\\ \Rightarrow -x{\left(y^{\text{'}}\right)}^2-xy{y}^{"}+y{y}^{\text{'}}=0\\ \Rightarrow xy{y}^{"}+x{\left(y^{\text{'}}\right)}^2=0\end{array}$

This is the required differential equation

Q9. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.

A.9. The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1..........(1)$

Differentiating both sides of equation (1) with respect to x, we get:

$\begin{array}{l}\frac{2x}{a^2}-\frac{2y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow \frac{x}{a^2}-\frac{y{y}^{\text{'}}}{b^2}=0..........(2)\end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l}\frac{1}{a^2}-\frac{y^{\text{'}}.y^{\text{'}}+y.y^{"}}{b^2}=0\\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)=0\end{array}$

Substituting the value of  $\frac{1}{a^2}$ in equation (2), we get:

$\begin{array}{l}\frac{x}{b^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)-\frac{y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow x{\left(y^{\text{'}}\right)}^2+xy{y}^{"}-y{y}^{\text{'}}=0\\ \Rightarrow xy{y}^{"}+x{\left(y^{\text{'}}\right)}^2-y{y}^{\text{'}}=0\end{array}$

This is the required differential equation.

Q10. Form the differential equation of the family of circles having centres on y-axis and radius 3 units.

A.10. Let the centre of the circle on y-axis be (0, b).

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:

$\begin{array}{l}{x}^2+{\left(y-b\right)}^2=3^2\\ \Rightarrow x^2+{\left(y-b\right)}^2=9..........(1)\end{array}$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2x+2(y-b).y^{\text{'}}=0\\ \Rightarrow (y-b).y^{\text{'}}=-x\\ \Rightarrow y-b=\frac{-x}{y^{\text{'}}}\end{array}$

Substituting the value of  $(y-b)$ in equation (1), we get:

$\begin{array}{l}{x}^2+{\left(\frac{-x}{y^{\text{'}}}\right)}^2=9\\ \Rightarrow x^2\left[1+\frac{1}{{\left(y^{\text{'}}\right)}^2}\right]=9\\ \Rightarrow x^2\left({\left(y^{\text{'}}\right)}^2+1\right)=9{\left(y^{\text{'}}\right)}^2\\ \Rightarrow (x^2-9){\left(y^{\text{'}}\right)}^2+x^2=0\end{array}$

This is the required differential equation.

Q11. Which of the following differential equation has   $y=c_1{e}^x+c_2{e}^{-x}$  as the general solution:

$\begin{array}{l}(A)\frac{d^{2}y}{d{x}^2}+y=0\\ (B)\frac{d^{2}y}{d{x}^2}-y=0\\ (C)\frac{d^{2}y}{d{x}^2}+1=0\\ (D)\frac{d^{2}y}{d{x}^2}-1=0\end{array}$

A.11. Given:  $y=c_1{e}^x+c_2{e}^{-x}.........(1)$

Differentiating with respect to x, we get:

$\frac{dy}{dx}=c_1{e}^x-c_2{e}^{-x}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=c_1{e}^x+c_2{e}^{-x}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=y\\ \Rightarrow \frac{d^{2}y}{d{x}^2}-y=0\end{array}$

This is the required differential equation of the given equation of curve.

Hence, the correct answer is B.

Q12. Which of the following differential equations has y = x as one of its particular solutions:

$\begin{array}{l}(A)\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=x\\ (B)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+xy=x\\ (C)\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0\\ (D)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+xy=0\end{array}$

A12. The given equation of curve is   $y = x$ ..

Differentiating with respect to x, we get:

$\frac{dy}{dx}=1 ..........(1)$

Again, differentiating with respect to x, we get:

$\frac{d^{2}y}{d{x}^2}=0..........(2)$

Now, on substituting the values of y,  $\frac{d^{2}y}{d{x}^2}$ and  $\frac{dy}{dx}$   from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0-x^2.1+x.x\\ =-x^2+x^2\\ =0\end{array}$

Therefore, option (C) is correct.

Differential Equations Exercise 9.4 deals with solving linear differential equations of various forms and different solution methods. Linear differential equations are of the form $\frac{dy}{dx} + Py = Q$, where $P$and $Q$are functions of $x$. Exercise 9.4 Solutions of Class 12 Differential Equation consists of a detailed solution of 23 Questions. Students can check the complete Exercise 9.4 Solutions below;

Class 12 Chapter 9 Differential Equations Exercise 9.4 NCERT Solution

For each of the following differential equations in Exercise 1 to 10, find the general solution:

Q1.  $\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$

A.1. The given D.E is

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

By separable of variable,

$\begin{array}{l}\Rightarrow dy=\frac{1-cosx}{1+cosx}dx\left\{\begin{array}{l}cos2x=1-2si{n}^{2}x\\ =2si{n}^{2}x=1-cos2x\\ =2si{n}^2\frac{x}{2}=1-cosx\\ cos2x=2co{s}^{2}x-1\end{array}\right\}\\ \Rightarrow dy=\frac{2si{n}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}dx\\ \Rightarrow dy=ta{n}^2\frac{x}{2}dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int ta{n}^2\frac{x}{2}dx\left\{se{c}^{2}x=1+ta{n}^2\right\}}}\\ \Rightarrow y={\displaystyle \int \left(se{c}^2\frac{x}{2}-1\right)}dx\end{array}$

$\Rightarrow y=\frac{tan\frac{x}{2}}{\frac{1}{2}}-x+c$ c = constant

$\Rightarrow y=2tan\frac{x}{2}-x+c$ is the general solution.

Q3.  $\frac{dy}{dx}+y=1\left(y\ne 1\right)$

A3. Given,  $\frac{dy}{dx}+y=1$

$\Rightarrow \frac{dy}{dx}=1-y=-\left(y-1\right)$

By separable of variable,

$\frac{dy}{\left(y-1\right)}=-dx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{\left(y-1\right)}=-{\displaystyle \int dx}}\\ \Rightarrow log\left|y-1\right|=-x+c\\ \Rightarrow \left|y-1\right|=e^{-x+c}\\ \Rightarrow y-1=\pm e^{-x}.e^c\end{array}$

$\Rightarrow y=1+Ac$ where  $A=\pm e^c$

Is the general solution.

Q4.  $se{c}^{2}xtanydx+se{c}^{2}ytanxdy=0$

A.4. Given,  $se{c}^{2}xtanydx+se{c}^{2}ytanxdy=0$

Dividing throughout by ‘  $tanxtany$ ’ we get,

$\begin{array}{l}\frac{se{c}^{2}xtany}{tanxtany}dx+\frac{se{c}^{2}ytanx}{tanxtany}dy=0\\ \Rightarrow \frac{se{c}^{2}x}{tanx}dx+\frac{se{c}^{2}y}{tany}dy=0\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \frac{se{c}^{2}x}{tanx}dx+{\displaystyle \int \frac{se{c}^{2}y}{tany}dy=logc}}\\ \Rightarrow log\left|tanx\right|+log\left|tany\right|=logc\left\{{\displaystyle \int \frac{f^{|}\left(x\right)}{f\left(x\right)}}dx-log\left|f\left(x\right)\right|\right\}\\ \Rightarrow log\left|\left(tanx+tany\right)\right|=logc\end{array}$

$\Rightarrow tanxtany=\pm c$ is the required general solution.

Q5.  $\left(e^x+e^{-x}\right)dy-\left(e^x-e^{-x}\right)dx=0$

A.5. Given,  $\left(e^x+e^{-x}\right)dy-\left(e^x-e^{-x}\right)dx=0$

$\begin{array}{l}\Rightarrow \left(e^x+e^{-x}\right)dy=\left(e^x-e^{-x}\right)dx\\ \Rightarrow dy=\frac{e^x-e^{-x}}{e^x+e^{-x}}dx\end{array}$

Integrating both sides

$\Rightarrow {\displaystyle \int dy=}\frac{e^x-e^{-x}}{e^x+e^{-x}}dx\left\{\because {\displaystyle \int \frac{f^{|}\left(x\right)}{f\left(x\right)}}dx=log\left|x\right|\right\}$

$\Rightarrow y=log\left|e^x+e^{-x}\right|+c$ is the required general solution.

Q6.  $\frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right)$

A.6. Given,  $\frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right)$

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

Integrating both sides

$\begin{array}{l}{\displaystyle \int \frac{dy}{\left(1+y^2\right)}dy={\displaystyle \int \left(x^2+1\right)dx}}\\ \Rightarrow ta{n}^{-1}\frac{y}{1}=\frac{x^3}{3}+x+c\end{array}$

$\Rightarrow ta{n}^{-1}y=\frac{x^3}{3}+x+c$ is the general solution.

Q7.  $ylogydx-xdy=0$

A.7. Given,  $ylogydx-xdy=0$

$\begin{array}{l}\Rightarrow ylogydx=xdy\\ \Rightarrow \frac{dy}{ylogy}=\frac{dx}{x}\end{array}$

Integration both sides,

${\displaystyle \int \frac{dy}{ylogy}={\displaystyle \int \frac{dx}{x}}}$

Put log  $y=t\Rightarrow \frac{1}{y}=\frac{dt}{dy}\Rightarrow \frac{dy}{y}=dt$

Hence,  ${\displaystyle \int \frac{dt}{t}={\displaystyle \int \frac{dx}{x}}}$

$\begin{array}{l}\Rightarrow log\left|t\right|=log\left|x\right|+log\left|c\right|=log\left|xc\right|\\ \Rightarrow t=\pm xc\end{array}$

$\Rightarrow logy=ax$ where  $a=\pm c$

$\Rightarrow y=e^{ax}$ is the general solution.

Q8.  $x^5\frac{dy}{dx}=-y^5$

A.8. Given,  $x^5\frac{dy}{dx}=-y^5$

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

Integrating both sides

$\begin{array}{l}{\displaystyle \int \frac{dy}{y^5}=-{\displaystyle \int \frac{dx}{x^5}}}\\ \Rightarrow {\displaystyle \int y^{-5}dy=-{\displaystyle \int x^{-5}dx}}\end{array}$