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

Exercise 9.2

1. The function is $\mathbf{y = {e^x} + 1}$ . Show that the function is a solution of the differential equation, $\mathbf{y'' - y' = 0}$.

Ans: The given equation is $y = {e^x} + 1$.

Differentiating equation on both the sides of with respect to $x$, the equation becomes,

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{e^x} + 1} \right)$

The derivative of ${e^x}$ is ${e^x}$ and the derivative of a constant, $\dfrac{d}{{dx}}\left( a \right) = 1$, where $a$ is constant.

Thus, 

$ \Rightarrow \dfrac{{dy}}{{dx}} = {e^x} \\$

$ \Rightarrow y' = {e^x} \\  $

Now, differentiate the obtained equation again with respect to $x$,

$ \Rightarrow \dfrac{d}{{dx}}\left( {y'} \right) = \dfrac{d}{{dx}}\left( {{e^x}} \right) \\$

$ \Rightarrow y'' = {e^x} \\ $

Therefore, consider the given equation to be considered for evaluation, $y'' - y' = 0$.

Now, the left-hand side of the given equation is $y'' - y'$.

Substitute ${e^x}$ for $y''$ and ${e^x}$ for $y'$ in the left-hand side of the equation,

$ \Rightarrow y'' - y' = {e^x} - {e^x} \\$

$ \Rightarrow y'' - y' = 0 \\ $

Therefore, the LHS of the equation is found to be equivalent to the right hand side of the equation.

As a result, the provided function is the differential equation's solution.

2. The function is $\mathbf{y = {x^2} + 2x + C}$. Show that the function is a solution of the differential equation, $\mathbf{y' - 2x - 2 = 0}$.

Ans: The given equation is $y = {x^2} + 2x + C$.

Differentiating equation on both the sides of with respect to $x$, the equation becomes,

$y' = \dfrac{d}{{dx}}\left( {{x^2} + 2x + C} \right)$

The derivative of ${x^n}$ is $n{x^{n - 1}}$ and the derivative of $x$, $\dfrac{d}{{dx}}\left( x \right) = 1$, whereas the derivative $\dfrac{d}{{dx}}\left( a \right) = 1$, where $a$ is constant.

Thus, 

$ \Rightarrow y' = \dfrac{d}{{dx}}\left( {{x^2} + 2x + C} \right) \\ $

$ \Rightarrow y' = \dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( {2x} \right) + \dfrac{d}{{dx}}\left( C \right) \\ $

 $ \Rightarrow y' = 2x + 2 \\ $

Now, consider the LHS of the equation that is to be taken for evaluation, $y' - 2x - 2 = 0$.

Substitute the obtained value for $y'$ in the LHS of the above equation.

$ \Rightarrow y' - 2x - 2 = \left( {2x - 2} \right) - 2x - 2 \\ $

$ \Rightarrow y' - 2x - 2 = 2x - 2 - 2x + 2 \\ $

$ \Rightarrow y' - 2x - 2 = 0 \\ $

Therefore, the LHS of the equation is found to be equivalent to the right hand side of the equation.

As a result, the provided function is the differential equation's solution.

3. The function is  $\mathbf{y = \cos x + C}$. Show that the function is a solution of the differential equation, $\mathbf{y' + \sin x = 0}$.

Ans: The given equation is $y = \cos x + C$.

Differentiating equation on both the sides of with respect to $x$, the equation becomes,

$y' = \dfrac{d}{{dx}}\left( {\cos x + C} \right)$

The derivative of$\cos x$ is $ - \sin x$, whereas the derivative $\dfrac{d}{{dx}}\left( a \right) = 1$, where $a$ is constant.

Thus, 

$\Rightarrow y' = \dfrac{d}{{dx}}\left( {\cos x + C} \right)  \\ $

 $ \Rightarrow y' = \dfrac{d}{{dx}}\left( {\cos x} \right) + \dfrac{d}{{dx}}\left( C \right) \\ $

$ \Rightarrow y' =  - \sin x \\ $

Now, consider the LHS of the equation that is to be considered for evaluation,$y' + \sin x = 0$.

Substitute the obtained value for $y'$ in the LHS of the above equation.

$ \Rightarrow y' + \sin x = \left( { - \sin x} \right) + \sin x \\$

 $ \Rightarrow y' + \sin x = 0 \\ $

Therefore, the LHS of the equation is found to be equivalent to the right hand side of the equation.

As a result, the provided function is the differential equation's solution.

4. The function is  $\mathbf{y = \sqrt {1 + {x^2}} }$. Show that the function a solution of the differential equation, $\mathbf{y' = \dfrac{{xy}}{{1 + {x^2}}}}$.

Ans: The given equation is $y = \sqrt {1 + {x^2}} $.

Differentiating equation on both the sides of with respect to $x$, the equation becomes,

$y' = \dfrac{d}{{dx}}\left( {\sqrt {1 + {x^2}} } \right)$

The derivative of $\sqrt x $ is $\dfrac{1}{{2\sqrt x }}$ and the derivative of ${x^n}$, $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, whereas the derivative $\dfrac{d}{{dx}}\left( a \right) = 1$, where $a$ is constant.

Thus, 

$ \Rightarrow y' = \dfrac{d}{{dx}}\left( {\sqrt {1 + {x^2}} } \right) \\ $

$ \Rightarrow y' = \dfrac{1}{{2\sqrt {1 + {x^2}} }} \cdot \dfrac{d}{{dx}}\left( {1 + {x^2}} \right)\\ $

$ \Rightarrow y' = \dfrac{{2x}}{{2\sqrt {1 + {x^2}} }}  \\ $

$ \Rightarrow y' = \dfrac{x}{{\sqrt {1 + {x^2}} }}  \\  $

Now, multiply and divide the numerator and denominator of the right hand side of the above equation by $\sqrt {1 + {x^2}} $. 

$ \Rightarrow y' = \dfrac{x}{{\sqrt {1 + {x^2}} }}\left( {\dfrac{{\sqrt {1 + {x^2}} }}{{\sqrt {1 + {x^2}} }}} \right) \\ $

$ \Rightarrow y' = \dfrac{x}{{1 + {x^2}}} \cdot \sqrt {1 + {x^2}} \\ $

Substitute $y$ for $\sqrt {1 + {x^2}} $ in the above equation,

$ \Rightarrow y' = \dfrac{x}{{1 + {x^2}}} \cdot y \\ $

 $ \Rightarrow y' = \dfrac{{xy}}{{1 + {x^2}}} \\ $

Therefore, the LHS of the equation is found to be equivalent to the right hand side of the equation.

As a result, the provided function is the differential equation's solution.

5. The function is  $\mathbf{y = Ax}$. Show that the function is a solution of the differential equation, $\mathbf{xy' = y\left( {x \ne 0} \right)}$.

Ans: The given equation is $y = Ax$.

Differentiating equation on both the sides of with respect to $x$, the equation becomes,

$y' = \dfrac{d}{{dx}}\left( {Ax} \right)$

The derivative of$x$ is equal to 1.

Thus, 

$ \Rightarrow y' = \dfrac{d}{{dx}}\left( {Ax} \right) \\$

$ \Rightarrow y' = A\dfrac{d}{{dx}}\left( x \right) \\$

$ \Rightarrow y' = A\left( 1 \right) \\ $

$ \Rightarrow y' = A \\ $

Now, consider the LHS of the equation to be evaluated, that is, $xy' = y\left( {x \ne 0} \right)$.

Substitute the obtained value for $y'$ in the LHS of the above equation.

$ \Rightarrow xy' = xA \\ $

$ \Rightarrow xy' = Ax  \\ $

Thus, $y = Ax$.

Therefore, the LHS of the equation is found to be equivalent to the right hand side of the equation.

As a result, the provided function is the differential equation's solution.

6. The function is  $\mathbf{y = x\sin x}$. Show that the function is a solution of the differential equation,$\mathbf{xy' = y + x\sqrt {{x^2} - {y^2}} {\text{  }}\left( {x \ne 0{\text{ and }}x > y{\text{ or }}x <  - y} \right)}$.

Ans: The given equation is $y = x\sin x$.

Differentiating equation on both the sides of with respect to $x$, the equation becomes,

$y' = \dfrac{d}{{dx}}\left( {x\sin x} \right)$

The product rule of differential equations says that, 

$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}\left( v \right) + v\dfrac{d}{{dx}}\left( u \right)$

The derivative of $\sin x$ is $\cos x$ and the derivative of $x$ is 1.

$ \Rightarrow y' = \dfrac{d}{{dx}}\left( {x\sin x} \right) \\ $

$ \Rightarrow y' = x\dfrac{d}{{dx}}\left( {\sin x} \right) + \sin x\dfrac{d}{{dx}}\left( x \right) \\ $

$ \Rightarrow y' = x\cos x + \sin x  \\ $

Now, consider the LHS of the equation $xy' = y + x\sqrt {{x^2} - {y^2}} {\text{ }}$.

Substitute the obtained value for $y'$ in the LHS of the above equation.

$ \Rightarrow xy' = x\left( {x\cos x + \sin x} \right) \\ $

$ \Rightarrow xy' = {x^2}\cos x + x\sin x \\ $

It is known that ${\sin ^2}x + {\cos ^2}x = 1$, thus, $\cos x = \sqrt {1 - {{\sin }^2}x} $ and $\sin x = \dfrac{y}{x}$ as $y = x\sin x$.

Thus,

$ \Rightarrow xy' = {x^2}\sqrt {1 - {{\sin }^2}x}  + y \\ $

$ \Rightarrow xy' = {x^2}\sqrt {1 - {{\left( {\dfrac{y}{x}} \right)}^2}}  + y \\ $

$ \Rightarrow xy' = \dfrac{{{x^2}}}{x}\sqrt {{x^2} - {y^2}}  + y  \\ $

$ \Rightarrow xy' = x\sqrt {{x^2} - {y^2}}  + y \\ $

Thus, $xy' = y + x\sqrt {{x^2} - {y^2}} $

Therefore, the LHS of the equation is found to be equivalent to the right hand side of the equation.

As a result, the provided function is the differential equation's solution.

7. The function is  $\mathbf{xy = \log y + C}$. Show that the function is a solution of the differential equation, $\mathbf{y' = \dfrac{{{y^2}}}{{1 - xy}}\left( {xy \ne 1} \right)}$.

Ans: The given equation is $xy = \log y + C$.

Differentiating equation on both the sides of with respect to $x$, the equation becomes,

$\dfrac{d}{{dx}}\left( {xy} \right) = \dfrac{d}{{dx}}\left( {\log y} \right)$

The product rule of differential equations says that, 

$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}\left( v \right) + v\dfrac{d}{{dx}}\left( u \right)$

Also, the derivative of $x$ is 1, whereas the derivative of $\log x$ is $\dfrac{1}{x}$.

Thus, 

$ \Rightarrow \dfrac{d}{{dx}}\left( {xy} \right) = \dfrac{d}{{dx}}\left( {\log y} \right) \\ $

$ \Rightarrow y\dfrac{d}{{dx}}\left( x \right) + x\dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {\log y} \right) \\ $

$ \Rightarrow y\left( 1 \right) + xy' = \dfrac{1}{y}y' \\ $

$ \Rightarrow {y^2} + xyy' = y' \\ $

Now, consider the LHS of the equation $y' = \dfrac{{{y^2}}}{{1 - xy}}\left( {xy \ne 1} \right)$.

Substitute the obtained value for $y'$ in the LHS of the above equation.

$ \Rightarrow {y^2} + xyy' = y' \\ $

$ \Rightarrow y' - xyy' = {y^2}  \\ $

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

$ \Rightarrow y' = \dfrac{{{y^2}}}{{\left( {1 - xy} \right)}} \\ $

Therefore, the LHS of the equation is found to be equivalent to the right hand side of the equation.

As a result, the provided function is the differential equation's solution.

8. The function is  $\mathbf{y - \cos y = x}$. Show that the function is a solution of the differential equation, $\mathbf{\left( {y\sin y + \cos y + x} \right)y' = 1}$.

Ans: The given equation is $y - \cos y = x$.

Differentiating equation on both the sides of with respect to $x$, the equation becomes,

$\dfrac{d}{{dx}}y - \dfrac{d}{{dx}}\cos y = \dfrac{d}{{dx}}x$

The derivative of$\cos x$ is $ - \sin x$, whereas the derivative $\dfrac{d}{{dx}}\left( a \right) = 1$, where $a$ is constant and the derivative of $x$ is 1.

Thus, 

$ \Rightarrow \dfrac{d}{{dx}}y - \dfrac{d}{{dx}}\cos y = \dfrac{d}{{dx}}x \\ $

$ \Rightarrow y'\sin y \cdot y' = 1 \\ $

$ \Rightarrow y'\left( {1 + \sin y} \right) = 1 \\ $

$ \Rightarrow y' = \dfrac{1}{{1 + \sin y}}  \\ $

Now, consider the LHS of the equation $\left( {y\sin y + \cos y + x} \right)y' = 1$.

Substitute the obtained value for $y'$ in the LHS of the above equation.

$ \Rightarrow \left( {y\sin y + \cos y + x} \right)y' = \left( {y\sin y + \cos y + y - \cos y} \right) \times \dfrac{1}{{1 + \sin y}} \\ $

$ \Rightarrow y\left( {1 + \sin y} \right) \cdot \dfrac{1}{{\sin y}} = y \\ $

Therefore, the LHS of the equation is found to be equivalent to the right hand side of the equation.

As a result, the provided function is the differential equation's solution.

9. The function is  $\mathbf{x + y = {\tan ^{ - 1}}y}$. Show that the function is a solution of the differential equation, $\mathbf{{y^2}y' + {y^2} + 1 = 0}$.

Ans: The given equation is $x + y = {\tan ^{ - 1}}y$.

Differentiating equation on both the sides of with respect to $x$, the equation becomes,

$\dfrac{d}{{dx}}\left( {x + y} \right) = \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}y} \right)$

The derivative of ${\tan ^{ - 1}}x$ is $\dfrac{1}{{1 + {x^2}}}$ and the derivative of $x$ is 1.

$ \Rightarrow \dfrac{d}{{dx}}\left( {x + y} \right) = \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}y} \right) \\ $

$ \Rightarrow 1 + y' = \left[ {\dfrac{1}{{1 + {y^2}}}} \right]y' \\ $

$ \Rightarrow y'\left[ {\dfrac{1}{{1 + {y^2}}} - 1} \right] = 1 \\ $

$ \Rightarrow y'\left[ {\dfrac{{1 - \left( {1 + {y^2}} \right)}}{{1 + {y^2}}}} \right] = 1 \\$

$ \Rightarrow y'\left[ {\dfrac{{ - {y^2}}}{{1 + {y^2}}}} \right] = 1 \\$

$ \Rightarrow y' = \dfrac{{\left( { - 1 + {y^2}} \right)}}{{{y^2}}} \\ $

Now, consider the LHS of the equation ${y^2}y' + {y^2} + 1 = 0$.

Substitute the obtained value for $y'$ in the LHS of the above equation.

$ \Rightarrow {y^2}\left[ {\dfrac{{ - \left( {1 + {y^2}} \right)}}{{{y^2}}}} \right] + {y^2} + 1 = 0  \\ $

$ \Rightarrow  - 1 - {y^2} + {y^2} + 1 = 0  \\ $ 

Thus, ${y^2}y' + {y^2} + 1 = 0$

Therefore, the LHS of the equation is found to be equivalent to the right hand side of the equation.

As a result, the provided function is the differential equation's solution.

10. The function is  $\mathbf{y = \sqrt {{a^2} - {x^2}} x \in \left( { - a,a} \right)}$. Show that the function is a solution of the differential equation, $\mathbf{x + y\dfrac{{dy}}{{dx}} = 0\left( {y \ne 0} \right)}$.

Ans: The given equation is $y = \sqrt {{a^2} - {x^2}} $.

Differentiating equation on both the sides of with respect to $x$, the equation becomes,

$y' = \dfrac{d}{{dx}}\left( {\sqrt {{a^2} - {x^2}} } \right)$

The derivative of $\sqrt x $ is $\dfrac{1}{{2\sqrt x }}$ and the derivative of ${x^n}$, $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, whereas the derivative $\dfrac{d}{{dx}}\left( a \right) = 1$, where $a$ is constant.

Thus, 

$ \Rightarrow y' = \dfrac{d}{{dx}}\left( {\sqrt {{a^2} - {x^2}} } \right)  \\ $

 $ \Rightarrow y' = \dfrac{1}{{2\sqrt {{a^2} - {x^2}} }} \cdot \dfrac{d}{{dx}}\left( {{a^2} - {x^2}} \right)  \\ $

$ \Rightarrow y' = \dfrac{{ - 2x}}{{2\sqrt {{a^2} - {x^2}} }} \\ $

$ \Rightarrow y' = \dfrac{{ - x}}{{\sqrt {{a^2} - {x^2}} }} \\ $

Now, consider the left-hand side of the equation to be considered for evaluation, that is,$x + y\dfrac{{dy}}{{dx}} = 0$.

Substitute $y$ for $\dfrac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}$  and $\sqrt {{a^2} - {x^2}} $ for $y$ in the LHS of the above equation,

$ \Rightarrow x + y\dfrac{{dy}}{{dx}} = x + \sqrt {{a^2} - {x^2}}  \cdot \dfrac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}  \\ $

 $ \Rightarrow y' = x - x \\ $

 $ \Rightarrow y' = 0 \\ $

Therefore, the LHS of the equation is found to be equivalent to the right hand side of the equation.

As a result, the provided function is the differential equation's solution.

11. The numbers of arbitrary constants in the general solution of a differential equation of fourth order are:

1. 0

2. 2

3. 3

4. 4

Ans: A differential equation solution is an expression for the dependent variable in terms of one or more independent variables that fulfils the relation. All potential solutions are included in the general solution, which usually includes arbitrary constants or arbitrary functions.

It is known that the number of constants in the general solution of a differential equation of order $n$ is equal to the order of the differential equation.

As a result, the general equation of the fourth order differential equation has four constants. As a result, D is the right response.

12. The numbers of arbitrary constants in the particular solution of a differential equation of third order are:

1. 3

2. 2

3. 1

4. 0

Ans:  The general solution becomes the particular solution of the problem when the arbitrary constant of the general solution obtains a unique value.

A differential equation’s particular solution is derived by applying the boundary conditions.

Also, it is known that in a particular solution of a differential equation, there are no arbitrary constants found. Thus, the correct response for is (D) which is 0.

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

NCERT Solution Class 12 Maths of Chapter 9 All Exercises

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

Opting for the NCERT solutions for Ex 9.2 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.2 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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