Solution

The given differential equation is $\frac{d y}{d x}+2 y=\sin x$.

This is in the form of $\frac{d y}{d x}+p y=Q($ where $p=2$ and $Q=\sin x)$.

Now, I.F $=e^{\int p d x}=e^{\int 2 d x}=e^{2 x}$.

The solution of the given differential equation is given by the relation,

$$ \begin{align*} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{2 x}=\int \sin x \cdot e^{2 x} d x+C \tag{1} \end{align*} $$

Let $I=\int \sin x \cdot e^{2 x}$.

$\Rightarrow I=\sin x \cdot \int e^{2 x} d x-\int(\frac{d}{d x}(\sin x) \cdot \int e^{2 x} d x) d x$

$\Rightarrow I=\sin x \cdot \frac{e^{2 x}}{2}-\int(\cos x \cdot \frac{e^{2 x}}{2}) d x$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}[\cos x \cdot \int e^{2 x}-\int(\frac{d}{d x}(\cos x) \cdot \int e^{2 x} d x) d x]$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}[\cos x \cdot \frac{e^{2 x}}{2}-\int[(-\sin x) \cdot \frac{e^{2 x}}{2}] d x]$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} \int(\sin x \cdot e^{2 x}) d x$

$\Rightarrow I=\frac{e^{2 x}}{4}(2 \sin x-\cos x)-\frac{1}{4} I$

$\Rightarrow \frac{5}{4} I=\frac{e^{2 x}}{4}(2 \sin x-\cos x)$

$\Rightarrow I=\frac{e^{2 x}}{5}(2 \sin x-\cos x)$

Therefore, equation (1) becomes:

$y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+C$

$\Rightarrow y=\frac{1}{5}(2 \sin x-\cos x)+C e^{-2 x}$

This is the required general solution of the given differential equation.

Solution

$ \frac{d y}{d x}+p y=Q(\text{ where } p=3 \text{ and } Q=e^{-2 x}) \text{. } $

The given differential equation is

Now, I.F $=e^{\int p d x}=e^{\int 3 d x}=e^{3 x}$.

The solution of the given differential equation is given by the relation,

$ \begin{aligned} & y(I . F .)=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{3 x}=\int(e^{-2 x} \times e^{3 x})+C \\ & \Rightarrow y e^{3 x}=\int e^{x} d x+C \\ & \Rightarrow y e^{3 x}=e^{x}+C \\ & \Rightarrow y=e^{-2 x}+C e^{-3 x} \end{aligned} $

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{1}{x}$ and $.Q=x^{2})$

Now, I.F $=e^{\int p d x}=e^{\int \frac{1}{x} d x}=e^{\log x}=x$.

The solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y(x)=\int(x^{2} \cdot x) d x+C$

$\Rightarrow x y=\int x^{3} d x+C$

$\Rightarrow x y=\frac{x^{4}}{4}+C$

This is the required general solution of the given differential equation.

Solution

The given differential equation is:

$\frac{d y}{d x}+p y=Q($ where $p=\sec x$ and $Q=\tan x)$

Now, I.F $=e^{\int p d x}=e^{\int \sec x d x}=e^{\log (\sec x+\tan x)}=\sec x+\tan x$.

The general solution of the given differential equation is given by the relation,

$ y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C $

$\Rightarrow y(\sec x+\tan x)=\int \tan x(\sec x+\tan x) d x+C$

$\Rightarrow y(\sec x+\tan x)=\int \sec x \tan x d x+\int \tan ^{2} x d x+C$

$\Rightarrow y(\sec x+\tan x)=\sec x+\int(\sec ^{2} x-1) d x+C$

$\Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+C$

Solution

We have,

$\begin{aligned} & \cos ^2 x \frac{d y}{d x}+y=\tan x \ & \frac{d y}{d x}+\sec ^2 x(y)=\sec ^2 x \tan x \end{aligned}$

We know that the general equation

$\frac{d y}{d x}+P y=Q$

Here,

$P=\sec ^2 x, Q=\sec ^2 x \tan x$

Since, integrating factor I. $F=e^{\int P d x}$ I. $F=e^{\int \sec ^2 x d x}$ I. $F=e^{\tan x}$

We know that the general solution,

$\mathrm{y} \times \mathrm{I} . \mathrm{F}=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F}) \mathrm{dx}+\mathrm{C}$

Therefore,

$y \times e^{\tan x}=\int\left(\sec ^2 x \tan x\right) e^{\tan x} d x+C$

Let $\mathrm{t}=\tan \mathrm{x}$

$\mathrm{dt}=\sec ^2 \mathrm{x} d \mathrm{x}$

Therefore,

$\begin{aligned} & y \times e^t=\int t e^t d t+C \\ & y \times e^t=t e^t-\int 1 e^t d t+C \\ & y \times e^t=t e^t-e^t+C \end{aligned}$

On putting the value of $t$, we get

$y \times e^{\tan x}=\tan x e^{\tan x}-e^{\tan x}+C$

Hence, this is the answer.

Solution

The given differential equation is:

$x \frac{d y}{d x}+2 y=x^{2} \log x$

$\Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x \log x$

This equation is in the form of a linear differential equation as:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{2}{x}$ and $.Q=x \log x)$

Now, I.F $=e^{\int p d x}=e^{\int_x^{2} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2}$.

The general solution of the given differential equation is given by the relation, $y($ I.F. $)=\int(Q \times$ I.F. $) d x+$ C

$\Rightarrow y \cdot x^{2}=\int(x \log x \cdot x^{2}) d x+C$

$\Rightarrow x^{2} y=\int(x^{3} \log x) d x+C$

$\Rightarrow x^{2} y=\log x \cdot \int x^{3} d x-\int[\frac{d}{d x}(\log x) \cdot \int x^{3} d x] d x+C$

$\Rightarrow x^{2} y=\log x \cdot \frac{x^{4}}{4}-\int(\frac{1}{x} \cdot \frac{x^{4}}{4}) d x+C$

$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \int x^{3} d x+C$

$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \cdot \frac{x^{4}}{4}+C$

$\Rightarrow x^{2} y=\frac{1}{16} x^{4}(4 \log x-1)+C$

$\Rightarrow y=\frac{1}{16} x^{2}(4 \log x-1)+C x^{-2}$

Solution

The given differential equation is:

$x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$

$\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^{2}}$

This equation is the form of a linear differential equation as:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{1}{x \log x}$ and $Q=\frac{2}{x^{2}}$ )

Now, I.F $=e^{\int \rho d x}=e^{\int \frac{1}{x \log } d x}=e^{\log (\log x)}=\log x$.

The general solution of the given differential equation is given by the relation,

$$ \begin{align*} & y(I . F .)=\int(Q \times I \text{.F. }) d x+C \\ & \Rightarrow y \log x=\int(\frac{2}{x^{2}} \log x) d x+C \tag{1} \end{align*} $$

Now, $\int(\frac{2}{x^{2}} \log x) d x=2 \int(\log x \cdot \frac{1}{x^{2}}) d x$.

$ \begin{aligned} & =2[\log x \cdot \int \frac{1}{x^{2}} d x-\int{\frac{d}{d x}(\log x) \cdot \int \frac{1}{x^{2}} d x} d x] \\ & =2[\log x(-\frac{1}{x})-\int(\frac{1}{x} \cdot(-\frac{1}{x})) d x] \\ & =2[-\frac{\log x}{x}+\int \frac{1}{x^{2}} d x] \\ & =2[-\frac{\log x}{x}-\frac{1}{x}] \\ & =-\frac{2}{x}(1+\log x) \end{aligned} $

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

$y \log x=-\frac{2}{x}(1+\log x)+C$

This is the required general solution of the given differential equation.

Solution

$(1+x^{2}) d y+2 x y d x=\cot x d x$

$\Rightarrow \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}$

This equation is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{2 x}{1+x^{2}}$ and $.Q=\frac{\cot x}{1+x^{2}})$

Now, I.F $=e^{\int p d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log (1+x^{2})}=1+x^{2}$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y(1+x^{2})=\int[\frac{\cot x}{1+x^{2}} \times(1+x^{2})] d x+C$

$\Rightarrow y(1+x^{2})=\int \cot x d x+C$

$\Rightarrow y(1+x^{2})=\log |\sin x|+C$

Solution

$x \frac{d y}{d x}+y-x+x y \cot x=0$

$\Rightarrow x \frac{d y}{d x}+y(1+x \cot x)=x$

$\Rightarrow \frac{d y}{d x}+(\frac{1}{x}+\cot x) y=1$

This equation is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{1}{x}+\cot x$ and $.Q=1)$

Now, I.F $=e^{\int \rho d x}=e^{\int(\frac{1}{x}+\cot x) d x}=e^{\log x+\log (\sin x)}=e^{\log (x \sin x)}=x \sin x$.

The general solution of the given differential equation is given by the relation,

$ \begin{aligned} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y(x \sin x)=\int(1 \times x \sin x) d x+C \\ & \Rightarrow y(x \sin x)=\int(x \sin x) d x+C \\ & \Rightarrow y(x \sin x)=x \int \sin x d x-\int[\frac{d}{d x}(x) \cdot \int \sin x d x]+C \\ & \Rightarrow y(x \sin x)=x(-\cos x)-\int 1 \cdot(-\cos x) d x+C \\ & \Rightarrow y(x \sin x)=-x \cos x+\sin x+C \\ & \Rightarrow y=\frac{-x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}+\frac{C}{x \sin x} \\ & \Rightarrow y=-\cot \cdot x+\frac{1}{x}+\frac{C}{x \sin x} \end{aligned} $

Solution

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

$\Rightarrow \frac{d y}{d x}=\frac{1}{x+y}$

$\Rightarrow \frac{d x}{d y}=x+y$

$\Rightarrow \frac{d x}{d y}-x=y$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p x=Q($ where $p=-1$ and $Q=y$ )

Now, I.F $=e^{\int p d y}=e^{\int-d y}=e^{-y}$.

The general solution of the given differential equation is given by the relation, $x($ I.F. $)=\int(Q \times I . F) d y+.C$

$\Rightarrow x e^{-y}=\int(y \cdot e^{-y}) d y+C$

$\Rightarrow x e^{-y}=y \cdot \int e^{-y} d y-\int[\frac{d}{d y}(y) \int e^{-y} d y] d y+C$

$\Rightarrow x e^{-y}=y(-e^{-y})-\int(-e^{-y}) d y+C$

$\Rightarrow x e^{-y}=-y e^{-y}+\int e^{-y} d y+C$

$\Rightarrow x e^{-y}=-y e^{-y}-e^{-y}+C$

$\Rightarrow x=-y-1+Ce^{y}$

$\Rightarrow x+y+1=Ce^{y}$

Solution

$y d x+(x-y^{2}) d y=0$

$\Rightarrow y d x=(y^{2}-x) d y$

$\Rightarrow \frac{d x}{d y}=\frac{y^{2}-x}{y}=y-\frac{x}{y}$

$\Rightarrow \frac{d x}{d y}+\frac{x}{y}=y$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p x=Q(.$ where $p=\frac{1}{y}$ and $.Q=y)$

Now, I.F $=e^{\int \rho d y}=e^{\int \frac{1}{y} d y}=e^{\log y}=y$.

The general solution of the given differential equation is given by the relation, $x(I . F)=.\int(Q \times I . F) d y+.C$

$\Rightarrow x y=\int(y \cdot y) d y+C$

$\Rightarrow x y=\int y^{2} d y+C$

$\Rightarrow x y=\frac{y^{3}}{3}+C$

$\Rightarrow x=\frac{y^{2}}{3}+\frac{C}{y}$

Solution

$(x+3 y^{2}) \frac{d y}{d x}=y$

$\Rightarrow \frac{d y}{d x}=\frac{y}{x+3 y^{2}}$

$\Rightarrow \frac{d x}{d y}=\frac{x+3 y^{2}}{y}=\frac{x}{y}+3 y$

$\Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y$

This is a linear differential equation of the form:

$\frac{d x}{d y}+p x=Q(.$ where $p=-\frac{1}{y}$ and $.Q=3 y)$

Now, I.F $=e^{\int p d y}=e^{-\int \frac{d y}{y}}=e^{-\log y}=e^{\log (\frac{1}{y})}=\frac{1}{y}$.

The general solution of the given differential equation is given by the relation, $x(I . F)=.\int(Q \times I . F) d y+.C$

$\Rightarrow x \times \frac{1}{y}=\int(3 y \times \frac{1}{y}) d y+C$

$\Rightarrow \frac{x}{y}=3 y+C$

$\Rightarrow x=3 y^{2}+C y$

Solution

The given differential equation is $\frac{d y}{d x}+2 y \tan x=\sin x$.

This is a linear equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=2 \tan x$ and $Q=\sin x)$

Now, I.F $=e^{\int p d x}=e^{\int 2 tan x d x}=e^{2 \log |\sec x|}=e^{\log (\sec ^{2} x)}=\sec ^{2} x$.

The general solution of the given differential equation is given by the relation,

$y(I . F)=.\int(Q \times I . F) d x+.C$

$\Rightarrow y(\sec ^{2} x)=\int(\sin x \cdot \sec ^{2} x) d x+C$

$\Rightarrow y \sec ^{2} x=\int(\sec x \cdot \tan x) d x+C$

$\Rightarrow y \sec ^{2} x=\sec x+C$

Now, $y=0$ at $x=\frac{\pi}{3}$.

Therefore,

$0 \times \sec ^{2} \frac{\pi}{3}=\sec \frac{\pi}{3}+C$

$\Rightarrow 0=2+C$

$\Rightarrow C=-2$

Substituting $C=-2$ in equation (1), we get: $y \sec ^{2} x=\sec x-2$

$\Rightarrow y=\cos x-2 \cos ^{2} x$

Hence, the required solution of the given differential equation is $y=\cos x-2 \cos ^{2} x$.

Solution

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

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

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=\frac{2 x}{1+x^{2}}$ and $.Q=\frac{1}{(1+x^{2})^{2}})$

Now, I.F $=e^{\int p d x}=e^{\int \frac{2 x d x}{1+x^{2}}}=e^{\log (1+x^{2})}=1+x^{2}$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+C$

$\Rightarrow y(1+x^{2})=\int[\frac{1}{(1+x^{2})^{2}} \cdot(1+x^{2})] d x+C$

$\Rightarrow y(1+x^{2})=\int \frac{1}{1+x^{2}} d x+C$

$\Rightarrow y(1+x^{2})=\tan ^{-1} x+C$

Now, $y=0$ at $x=1$.

Therefore,

$0=\tan ^{-1} 1+C$

$\Rightarrow C=-\frac{\pi}{4}$

Substituting $C=-\frac{\pi}{4}$ in equation (1), we get:

$y(1+x^{2})=\tan ^{-1} x-\frac{\pi}{4}$

This is the required general solution of the given differential equation.

Solution

The given differential equation is $\frac{d y}{d x}-3 y \cot x=\sin 2 x$.

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=-3 \cot x$ and $Q=\sin 2 x)$

Now, I.F $=e^{\int p d x}=e^{-3 \int \cot x d x}=e^{-3 \log |\sin x|}=e^{\log |\frac{1}{\sin ^{3} x}|}=\frac{1}{\sin ^{3} x}$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+$ C

$\Rightarrow y \cdot \frac{1}{\sin ^{3} x}=\int[\sin 2 x \cdot \frac{1}{\sin ^{3} x}] d x+C$

$\Rightarrow y cosec^{3} x=2 \int(\cot x cosec x) d x+C$

$\Rightarrow y cosec^{3} x=2 cosec x+C$

$\Rightarrow y=-\frac{2}{cosec^{2} x}+\frac{3}{cosec^{3} x}$

$\Rightarrow y=-2 \sin ^{2} x+C \sin ^{3} x$

Now, $y=2$ at $x=\frac{\pi}{2}$.

Therefore, we get:

$2=-2+C$

$\Rightarrow C=4$

Substituting $C=4$ in equation (1), we get:

$y=-2 \sin ^{2} x+4 \sin ^{3} x$

$\Rightarrow y=4 \sin ^{3} x-2 \sin ^{2} x$

This is the required particular solution of the given differential equation.

Solution

Let $F(x, y)$ be the curve passing through the origin.

At point $(x, y)$, the slope of the curve will be $\frac{d y}{d x}$.

According to the given information:

$\frac{d y}{d x}=x+y$

$\Rightarrow \frac{d y}{d x}-y=x$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q($ where $p=-1$ and $Q=x)$

Now, I.F $=e^{\int p d x}=e^{\int(-1) d x}=e^{-x}$.

The general solution of the given differential equation is given by the relation,

$$ \begin{align*} & y(\text{ I.F. })=\int(Q \times \text{ I.F. }) d x+C \\ & \Rightarrow y e^{-x}=\int x e^{-x} d x+C \tag{1} \end{align*} $$

Now, $\int x e^{-x} d x=x \int e^{-x} d x-\int[\frac{d}{d x}(x) \cdot \int e^{-x} d x] d x$.

$ =-x e^{-x}-\int-e^{-x} d x $

$ \begin{aligned} & =-x e^{-x}+(-e^{-x}) \\ & =-e^{-x}(x+1) \end{aligned} $

Substituting in equation (1), we get:

$$ \begin{align*} & y e^{-x}=-e^{-x}(x+1)+C \\ & \Rightarrow y=-(x+1)+C e^{x} \\ & \Rightarrow x+y+1=C e^{x} \tag{2} \end{align*} $$

The curve passes through the origin.

Therefore, equation (2) becomes:

$1=C$

Substituting $C=1$ in equation (2), we get:

$\Rightarrow \quad x+y+1=e^{x}$

Hence, the required equation of curve passing through the origin is $x+y+1=e^{x}$.

Solution

Let $F(x, y)$ be the curve and let $(x, y)$ be a point on the curve. The slope of the tangent

to the curve at $(x, y)$ is $\frac{d y}{d x}$.

According to the given information:

$\frac{d y}{d x}+5=x+y$

$\Rightarrow \frac{d y}{d x}-y=x-5$

This is a linear differential equation of the form: $\frac{d y}{d x}+p y=Q($ where $p=-1$ and $Q=x-5)$

Now, I.F $=e^{\int p d x}=e^{\int(-1) d x}=e^{-x}$.

The general equation of the curve is given by the relation,

$y($ I.F. $)=\int(Q \times$ I.F. $) d x+$ C

$\Rightarrow y \cdot e^{-x}=\int(x-5) e^{-x} d x+C$

Now, $\int(x-5) e^{-x} d x=(x-5) \int e^{-x} d x-\int[\frac{d}{d x}(x-5) \cdot \int e^{-x} d x] d x$.

$ \begin{aligned} & =(x-5)(-e^{-x})-\int(-e^{-x}) d x \\ & =(5-x) e^{-x}+(-e^{-x}) \\ & =(4-x) e^{-x} \end{aligned} $

Therefore, equation (1) becomes:

$$ \begin{align*} & y e^{-x}=(4-x) e^{-x}+C \\ & \Rightarrow y=4-x+C e^{x} \\ & \Rightarrow x+y-4=C e^{x} \tag{2} \end{align*} $$

The curve passes through point $(0,2)$.

Therefore, equation ( 2 ) becomes:

$0+2-4=Ce^{0}$

$\Rightarrow-2=C$

$\Rightarrow C=-2$

Substituting $C=-2$ in equation (2), we get:

$ \begin{aligned} & x+y-4=-2 e^{x} \\ & \Rightarrow y=4-x-2 e^{x} \end{aligned} $

This is the required equation of the curve.

Solution

The given differential equation is:

$x \frac{d y}{d x}-y=2 x^{2}$

$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=2 x$

This is a linear differential equation of the form:

$\frac{d y}{d x}+p y=Q(.$ where $p=-\frac{1}{x}$ and $.Q=2 x)$

The integrating factor (I.F) is given by the relation,

$e^{\int p d x}$

$\therefore$ I.F $=e^{\int \frac{1}{x} d x}=e^{-\log x}=e^{\log (x^{-1})}=x^{-1}=\frac{1}{x}$

Hence, the correct answer is $C$.

Solution

The given differential equation is:

$(1-y^{2}) \frac{d x}{d y}+y x=a y$

$\Rightarrow \frac{d y}{d x}+\frac{y x}{1-y^{2}}=\frac{a y}{1-y^{2}}$

This is a linear differential equation of the form:

$\frac{d x}{d y}+p y=Q(.$ where $p=\frac{y}{1-y^{2}}$ and $Q=\frac{a y}{1-y^{2}}$ )

The integrating factor (I.F) is given by the relation,

$e^{\int p d x}$

$\therefore I . F=e^{\int p d y}=e^{\int \frac{y}{1-y^{2}} d y}=e^{-\frac{1}{2} \log (1-y^{2})}=e^{\log [\frac{1}{\sqrt{1-y^{2}}}]}=\frac{1}{\sqrt{1-y^{2}}}$

Hence, the correct answer is D.