NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

Given function is
$2x^2\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+y=0$
We can rewrite it as
$2x.y^{{}^{″}}-3y^{{}^{\prime }}+y=0$
Now, it is clear from the above that the highest order derivative present in differential equation is $y^{{}^{″}}$

Therefore, order of given differential equation $2x^2\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+y=0$ is 2

Therefore, the answer is (A)

Question: Verify that the given functions (explicit or implicit) are a solution of the corresponding differential equation:

1. $y=e^x+1:y^{″}-y^{\prime }=0$

2. $y=x^2+2x+C:y^{\prime }-2x-2=0$

3. $y=\cos x+C:y^{\prime }+\sin x=0$

4. $y=\sqrt{1+x^2}:y^{\prime }=\frac{xy}{1+x^2}$

5. $y=Ax:x{y}^{\prime }=y(x\ne 0)$

6. $y=x\sin x:x{y}^{\prime }=y+x\sqrt{x^2-y^2}(x\ne 0\text{and}x>yorx<-y)$

7. $xy=\log y+C:y^{\prime }=\frac{y^2}{1-xy}(xy\ne 1)$

8. $y-cosy=x:(y\sin y+\cos y+x)y^{\prime }=y$

9. $x+y={\tan }^{-1}y:y^2{y}^{\prime }+y^2+1=0$

10. $y=\sqrt{a^2-x^2}x\in (-a,a):x+y\frac{dy}{dx}=0(y\ne 0)$

Answer(1):

Given,

$y=e^x+1$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}{e}^x}{\mathrm{d}x}=e^x$

Again, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}{y}^{\prime }}{\mathrm{d}x}=\frac{\mathrm{d}{e}^x}{\mathrm{d}x}=e^x$

$⟹y^{″}=e^x$

Substituting the values of y’ and y'' in the given differential equations,

y'' - y' = e ^x - e ^x = 0 = RHS.

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

Answer(2):

Given,

$y=x^2+2x+C$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(x^2+2x+C)=2x+2$

Substituting the values of y’ in the given differential equations,

$y^{\prime }-2x-2=2x+2-2x-2=0=RHS$ .

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

Answer(3):

Given,

$y=\cos x+C$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(cost+C)=-sinx$

Substituting the values of y’ in the given differential equations,

$y^{\prime }-\sin x=-sinx-sinx=-2sinx\ne RHS$ .

Therefore, the given function is not the solution of the corresponding differential equation.

Answer(4):

Given,

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

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(\sqrt{1+x^2})=\frac{2x}{2\sqrt{1+x^2}}=\frac{x}{\sqrt{1+x^2}}$

Substituting the values of y in the RHS.

$\frac{x\sqrt{1+x^2}}{1+x^2}=\frac{x}{\sqrt{1+x^2}}=LHS$ .

Therefore, the given function is a solution of the corresponding differential equation.

Answer(5):

Given,

$y=Ax$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(Ax)=A$

Substituting the values of y' in LHS,

$x{y}^{\prime }=x(A)=Ax=y=RHS$ .

Therefore, the given function is a solution of the corresponding differential equation.

Answer(6):

Given,

$y=x\sin x$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(xix)=six+xcosx$

Substituting the values of y' in LHS,

$x{y}^{\prime }=x(ssix+xcosx)$

Substituting the values of y in RHS.

$xsinx+x\sqrt{x^2-x^{2}si{n}^{2}x}=xix+x^2\sqrt{1-sin{x}^2}=x(sinx+xcosx)=LHS$

Therefore, the given function is a solution of the corresponding differential equation.

Answer(7):

Given,

$xy=\log y+C$

Now, differentiating both sides w.r.t. x,

$$\begin{gathered} y+x\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(logy)=\frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x} \\ ⟹y^2+xy{y}^{\prime }=y^{\prime } \\ ⟹y^2=y^{\prime }(1-xy) \\ ⟹y^{\prime }=\frac{y^2}{1-xy} \end{gathered}$$

Substituting the values of y' in LHS,

$y^{\prime }=\frac{y^2}{1-xy}=RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Answer(8):

Given,

$y-cosy=x$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y}{\mathrm{d}x}+siny\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(x)=1$

$⟹$ y' + siny.y' = 1

$⟹$ y'(1 + siny) = 1

$⟹y^{\prime }=\frac{1}{1+siny}$

Substituting the values of y and y' in LHS,

$((x+cosy)\sin y+\cos y+x)(\frac{1}{1+siny})$

$=[x(1+siny)+cosy(1+siny)]\frac{1}{1+siny}$

= (x + cosy) = y = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Answer(9):

Given,

$x+y={\tan }^{-1}y$

Now, differentiating both sides w.r.t. x,

$$\begin{gathered} 1+\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{1+y^2}\frac{\mathrm{d}y}{\mathrm{d}x} \\ ⟹1+y^2=y^{\prime }(1-(1+y^2))=-y^2{y}^{\prime } \\ ⟹y^{\prime }=-\frac{1+y^2}{y^2} \end{gathered}$$

Substituting the values of y' in LHS,

$y^2(-\frac{1+y^2}{y^2})+y^2+1=-1-y^2+y^2+1=0=RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Answer(10):

Given,

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

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(\sqrt{a^2-x^2})=\frac{-2x}{2\sqrt{a^2-x^2}}=\frac{-x}{\sqrt{a^2-x^2}}$

Substituting the values of y and y' in LHS,

$x+y\frac{dy}{dx}=x+(\sqrt{a^2-x^2})(\frac{-x}{\sqrt{a^2-x^2}})=0=RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Question:11 : he number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4

Answer:

(D) 4

The number of constants in the general solution of a differential equation of order n is equal to its order.

Question: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

Answer:

(D) 0

In a particular solution of a differential equation, there is no arbitrary constant.

Question:1 Find the general solution: $\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

Answer:

Given,

$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

$$\begin{gathered} ⟹\frac{dy}{dx}=\frac{2si{n}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}=ta{n}^2\frac{x}{2} \\ ⟹dy=(se{c}^2\frac{x}{2}-1)dx \end{gathered}$$

$$\begin{gathered} ⟹\int dy=\int se{c}^2\frac{x}{2}dx-\int dx \\ ⟹y=2ta{n}^{-1}\frac{x}{2}-x+C \end{gathered}$$

Question:2 Find the general solution: $\frac{dy}{dx}=\sqrt{4-y^2}(-2<y<2)$

Answer:

Given,the question

$\frac{dy}{dx}=\sqrt{4-y^2}$

$$\begin{gathered} ⟹\frac{dy}{\sqrt{4-y^2}}=dx \\ ⟹\int \frac{dy}{\sqrt{4-y^2}}=\int dx \end{gathered}$$

$(\int \frac{dy}{\sqrt{a^2-y^2}}=si{n}^{-1}\frac{y}{a})$

The required general solution:

$⟹si{n}^{-1}\frac{y}{2}=x+C$

Question:3 Find the general solution: $\frac{dy}{dx}+y=1(y\ne 1)$

Answer:

Given,the question

$\frac{dy}{dx}+y=1$

$$\begin{gathered} ⟹\frac{dy}{dx}=1-y \\ ⟹\int \frac{dy}{1-y}=\int dx \end{gathered}$$

$(\int \frac{dx}{x}=lnx)$

$$\begin{gathered} ⟹-log(1-y)=x+C(WecanwriteC=logk) \\ ⟹logk(1-y)=-x \\ ⟹1-y=\frac{1}{k}{e}^{-x} \end{gathered}$$

The required general equation

$⟹y=1-\frac{1}{k}{e}^{-x}$

Question:4 Find the general solution: ${\sec }^{2}x\tan ydx+{\sec }^{2}y\tan xdy=0$

Answer:

Given,

${\sec }^{2}x\tan ydx+{\sec }^{2}y\tan xdy=0$

$$\begin{gathered} ⟹\frac{se{c}^{2}y}{tany}dy=-\frac{se{c}^{2}x}{tanx}dx \\ ⟹\int \frac{se{c}^{2}y}{tany}dy=-\int \frac{se{c}^{2}x}{tanx}dx \end{gathered}$$

Now, let tany = t and tax = u

$se{c}^{2}ydy=dtandse{c}^{2}xdx=du$

$$\begin{gathered} ⟹\int \frac{dt}{t}=-\int \frac{du}{u} \\ ⟹logt=-logu+logk \\ ⟹t=\frac{1}{ku} \\ ⟹tany=\frac{1}{ktanx} \end{gathered}$$

Question:5 Find the general solution:

$(e^x+e^{-x})dy-(e^x-e^{-x})dx=0$

Answer:

Given, the question

$(e^x+e^{-x})dy-(e^x-e^{-x})dx=0$

$⟹dy=\frac{(e^x-e^{-x})}{(e^x+e^{-x})}dx$

Let,

$$\begin{gathered} (e^x+e^{-x})=t \\ ⟹(e^x-e^{-x})dx=dt \end{gathered}$$

$$\begin{gathered} ⟹\int dy=\int \frac{dt}{t} \\ ⟹y=logt+C \\ ⟹y=log(e^x+e^{-x})+C \end{gathered}$$

This is the general solution

Question:6 Find the general solution: $\frac{dy}{dx}=(1+x^2)(1+y^2)$

Answer:

Given, the question

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

$⟹\int \frac{dy}{(1+y^2)}=\int (1+x^2)dx$

$(\int \frac{dx}{(1+x^2)}=ta{n}^{-1}x+c)$

$⟹ta{n}^{-1}y=x+\frac{x^3}{3}+C$

Question:7 Find the general solution: $y\log ydx-xdy=0$

Answer:

Given,

$y\log ydx-xdy=0$

$⟹\frac{1}{ylogy}dy=\frac{1}{x}dx$

let logy = t

=> 1/ydy = dt

$$\begin{gathered} ⟹\int \frac{dt}{t}=\int \frac{1}{x}dx \\ ⟹\log t=\log x+\log k \\ ⟹t=kx \\ ⟹\log y=kx \end{gathered}$$

This is the general solution

Question:8 Find the general solution: $x^5\frac{dy}{dx}=-y^5$

Answer:

Given, the question

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

$$\begin{gathered} ⟹\int \frac{dy}{y^5}=-\int \frac{dx}{x^5} \\ ⟹\frac{y^{-4}}{-4}=-\frac{x^{-4}}{-4}+C \\ ⟹\frac{1}{y^4}+\frac{1}{x^4}=C \end{gathered}$$

This is the required general equation.

Question:9 Find the general solution: $\frac{dy}{dx}={\sin }^{-1}x$

Answer:

Given, the question

$\frac{dy}{dx}={\sin }^{-1}x$

$⟹\int dy=\int {\sin }^{-1}xdx$

Now,