# NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2

## Chapter 9 Ex.9.2 Question 1

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

$$y = {e^x} + 1{\rm{ }}:{\rm{ }}y'' - y' = 0$$

### Solution

$$y = {e^x} + 1$$

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{e^x} + 1} \right)\\ &\Rightarrow \; y' = {e^x}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\\frac{d}{{dx}}\left( {y'} \right) &= \frac{d}{{dx}}\left( {{e^x}} \right)\\ &\Rightarrow \; y'' = {e^x}\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From $$\left( 1 \right)$$ and $$\left( 2 \right)$$

\begin{align}y'' - y' &= {e^x} - {e^x}\\ &= 0\end{align}

Thus, the given function is the solution of corresponding differential equation.

## Chapter 9 Ex.9.2 Question 2

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

$$y = {x^2} + 2x + c{\rm{ }}:{\rm{ }}y' - 2x - 2 = 0$$

### Solution

\begin{align}y &= {x^2} + 2x + c\\y' &= \frac{d}{{dx}}\left( {{x^2} + 2x + c} \right)\\ \Rightarrow \; y' &= 2x + 2\end{align}

Therefore,

\begin{align}y' - 2x - 2& = 2x + 2 - 2x - 2\\& = 0\end{align}

Thus, the given function is the solution of the differential equation.

## Chapter 9 Ex.9.2 Question 3

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

$$y = \cos x + C \quad : \quad y' + \sin x = 0$$

### Solution

\begin{align}y &= \cos x + C\\y' &= \frac{d}{{dx}}\left( {\cos x + C} \right)\\ \Rightarrow \; y' &= - \sin x\end{align}

Therefore,

\begin{align}y' + \sin x &= - \sin x + \sin x\\ &= 0\end{align}

Thus, the given function is the solution of the differential equation.

## Chapter 9 Ex.9.2 Question 4

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

$$y = \sqrt {1 + {x^2}} \quad : \quad y' = \frac{{xy}}{{1 + {x^2}}}$$

### Solution

\begin{align}y &= \sqrt {1 + {x^2}} \\y' &= \frac{d}{{dx}}\left( {\sqrt {1 + {x^2}} } \right)\\& = \frac{1}{{2\sqrt {1 + {x^2}} }}.\frac{d}{{dx}}\left( {1 + {x^2}} \right)\\ &= \frac{{2x}}{{2\sqrt {1 + {x^2}} }}\\& = \frac{x}{{2\sqrt {1 + {x^2}} }}\\ &= \frac{x}{{\left( {1 + {x^2}} \right)}} \times \sqrt {1 + {x^2}} \\ &= \frac{x}{{\left( {1 + {x^2}} \right)}}.y\\ &= \frac{{xy}}{{1 + {x^2}}}\end{align}

Thus, the given function is the solution of the differential equation.

## Chapter 9 Ex.9.2 Question 5

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

$$y = Ax \quad:\quad xy' = y\left( {x \ne 0} \right)$$

### Solution

\begin{align}y &= Ax\\y' &= \frac{d}{{dx}}\left( {Ax} \right)\\ &= A\end{align}

Therefore,

\begin{align}xy' &= xA\\& = Ax\\ &= y\end{align}

Thus, the given function is the solution of the differential equation.

## Chapter 9 Ex.9.2 Question 6

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

$$y = x\sin x \quad : \quad xy' = y + x\sqrt {{x^2} - {y^2}} \left( {x \ne 0 \text{ and }x > y} \right)\;or\;x < - y{\rm{ }}$$

### Solution

\begin{align}y &= x\sin x\\y' &= \frac{d}{{dx}}\left( {x\sin x} \right)\\ &= \sin x.\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left( {\sin x} \right)\\ &= \sin x + x\cos x\end{align}

Therefore,

\begin{align}xy' &= x\left( {\sin x + x\cos x} \right)\\ &= x\sin x + {x^2}\cos x\\ &= y + {x^2}.\sqrt {1 - {{\sin }^2}x} \\ &= y + {x^2}\sqrt {1 - {{\left( {\frac{y}{x}} \right)}^2}} \\ &= y + x\sqrt {{x^2} - {y^2}}\end{align}

Thus, the given function is the solution of the differential equation.

## Chapter 9 Ex.9.2 Question 7

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

$$xy = \log y + C \quad : \quad y' = \frac{{{y^2}}}{{1 - xy}}\left( {xy \ne 1} \right)$$

### Solution

\begin{align}&xy = \log y + C\\ &\Rightarrow \; \frac{d}{{dx}}\left( {xy} \right) = \frac{d}{{dx}}\left( {\log y} \right)\\ &\Rightarrow \; y\frac{d}{{dx}}\left( x \right) + x.\frac{{dy}}{{dx}} = \frac{1}{y}\frac{{dy}}{{dx}}\\ &\Rightarrow \; y + xy' = \frac{1}{y}.y'\\ &\Rightarrow \; {y^2} + xyy' = y'\\ &\Rightarrow \; \left( {xy - 1} \right)y' = - {y^2}\\ &\Rightarrow \; y' = \frac{{{y^2}}}{{1 - xy}}\end{align}

Thus, the given function is the solution of the differential equation.

## Chapter 9 Ex.9.2 Question 8

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

$$y - \cos y = x \quad : \quad \left( {y\sin y + \cos y + x} \right)y' = y$$

### Solution

\begin{align}&y - \cos y = x\\& \Rightarrow \; \frac{{dy}}{{dx}} - \frac{d}{{dx}}\left( {\cos y} \right) = \frac{d}{{dx}}(x)\\& \Rightarrow \; y' - \left( { - \sin y} \right).y' = 1\\ &\Rightarrow \; y'\left( {1 + \sin y} \right) = 1\\& \Rightarrow \; y' = \frac{1}{{1 + \sin y}}\end{align}

Therefore,

\begin{align}\left( {y\sin y + \cos y + x} \right)y' &= \left( {y\sin y + \cos y + y - \cos y} \right) \times \frac{1}{{1 + \sin y}}\\ &= y\left( {1 + \sin y} \right).\frac{1}{{1 + \sin y}}\\& = y\end{align}

Thus, the given function is the solution of the differential equation.

## Chapter 9 Ex.9.2 Question 9

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

$$x + y = {\tan ^{ - 1}}y \quad : \quad {y^2}y' + {y^2} + 1 = 0$$

### Solution

\begin{align}&x + y = {\tan ^{ - 1}}y\\& \Rightarrow \; \frac{d}{{dx}}\left( {x + y} \right) = \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}y} \right)\\& \Rightarrow \; 1 + y' = \left[ {\frac{1}{{1 + {y^2}}}} \right]y'\\ &\Rightarrow \; y'\left[ {\frac{1}{{1 + {y^2}}} - 1} \right] = 1\\& \Rightarrow \; y'\left[ {\frac{{1 - \left( {1 + {y^2}} \right)}}{{1 + {y^2}}}} \right] = 1\\& \Rightarrow \; y'\left[ {\frac{{ - {y^2}}}{{1 + {y^2}}}} \right] = 1\\ &\Rightarrow \; y' = \frac{{ - \left( {1 + {y^2}} \right)}}{{{y^2}}}\end{align}

Therefore,

\begin{align}{y^2}y' + {y^2} + 1& = {y^2}\left[ {\frac{{ - (1 + {y^2})}}{{{y^2}}}} \right] + {y^2} + 1\\ &= - 1 - {y^2} + {y^2} + 1\\ &= 0\end{align}

Thus, the given function is the solution of the differential equation.

## Chapter 9 Ex.9.2 Question 10

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

$$y = \sqrt {{a^2} - {x^2}} \;x \in \left( { - a,a} \right) \quad : \quad x + y\frac{{dy}}{{dx}} = 0\left( {y \ne 0} \right)$$

### Solution

\begin{align}&y = \sqrt {{a^2} - {x^2}} \\&\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\sqrt {{a^2} - {x^2}} } \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{a^2} - {x^2}} }}.\frac{d}{{dx}}\left( {{a^2} - {x^2}} \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{a^2} - {x^2}} }}.\left( { - 2x} \right)\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}\end{align}

Therefore,

\begin{align}x + y\frac{{dy}}{{dx}} &= x + \sqrt {{a^2} - {x^2}} \times \frac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}\\ &= x - x\\& = 0\end{align}

Thus, the given function is the solution of the differential equation.

## Chapter 9 Ex.9.2 Question 11

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

(A) $$0$$

(B) $$2$$

(C) $$3$$

(D) $$4$$

### Solution

We know that the number of constants in the general solution of a differential equation of order $$n$$ is equal to its order.

The number of constants in general equation of fourth order differential equation is $$4.$$

Thus, the correct option is D.

## Chapter 9 Ex.9.2 Question 12

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

(A) $$3$$

(B) $$2$$

(C) $$1$$

(D) $$0$$

### Solution

In a particular solution of a differential equation, there are no arbitrary constants.

Thus, the correct option is D.

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