Differential Equations

To determine $ y(0) $, we start by solving the differential equation given:

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

First, we rewrite it in the standard form for a linear differential equation:

$ \frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{\tan^{-1} x}}{1 + x^2} $

Next, we find the integrating factor (I.F.):

$ \text{I.F.} = e^{\int \frac{1}{1 + x^2} dx} = e^{\tan^{-1} x} $

Multiply through by the integrating factor:

$ y \cdot e^{\tan^{-1} x} = \int e^{\tan^{-1} x} \cdot \frac{e^{\tan^{-1} x}}{1 + x^2} dx $

This simplifies to:

$ y \cdot e^{\tan^{-1} x} = \int \frac{e^{2 \tan^{-1} x}}{1 + x^2} dx $

Make the substitution $ \tan^{-1} x = t $, then $ \frac{1}{1 + x^2} dx = dt $:

$ \int e^{2t} dt = \frac{e^{2t}}{2} + \frac{C}{2} $

Rewrite in terms of $ x $:

$ y \cdot e^{\tan^{-1} x} = \frac{e^{2 \tan^{-1} x}}{2} + \frac{C}{2} $

Use the initial condition $ y(1) = 0 $:

$ 0 = \frac{e^{2 \cdot \tan^{-1} 1}}{2} + \frac{C}{2} $

Since $ \tan^{-1} 1 = \frac{\pi}{4} $:

$ 0 = \frac{e^{\pi/2}}{2} + \frac{C}{2} \implies C = -e^{\pi/2} $

Thus, the solution is:

$ y \cdot e^{\tan^{-1} x} = \frac{e^{2 \tan^{-1} x} - e^{\pi/2}}{2} $

Evaluating $ y(0) $:

$ y(0) = y \cdot 1 = \frac{1 - e^{\pi/2}}{2} $

Therefore,

$ y(0) = \frac{1}{2} (1 - e^{\pi/2}) $

Hence, the correct answer is:

Option B

$ \frac{1}{2}\left(1-e^{\pi / 2}\right) $

Equation of circle passing through origin & having centre at the line $y=x$ is

$\begin{aligned} & (x-t)^2+(y-t)^2=2 t^2 \\ & x^2+y^2+t^2+t^2-2 t x-2 t y=2 t^2 \\ & x^2+y^2=2 t(x+y) \end{aligned}$

Now differentiate

$\begin{aligned} & 2 x+2 y y^{\prime}=2 t\left(1+y^{\prime}\right) \\ & t=\frac{x+y y^{\prime}}{1+y^{\prime}} \end{aligned}$

Now, $x^2+y^2=2\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)(x+y)$

$\begin{aligned} & x^2+y^2+x^2 \frac{d y}{d x}+y^2 \frac{d y}{d x} \\ & =2\left(x^2+x y+x y \frac{d y}{d x}+y^2 \frac{d y}{d x}\right) \\ & d x\left(x^2+y^2-2 x^2-2 x y\right)=d y\left(2 x y+2 y^2-x^2-y^2\right) \\ & d x\left(x^2-y^2+2 x y\right)=d y\left(x^2-y^2-2 x y\right) \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}+2 y=\sin 2 x \\ & \text { IF }=e^{2 d x}=e^{2 x} \\ & y \cdot e^{2 x}=\int e^{2 x} \sin 2 x d x+c \\ & =\frac{e^{2 x}}{8}(2 \sin 2 x-2 \cos 2 x)+c \\ & y(0)=\frac{3}{4} \\ & \frac{3}{4}=\frac{1}{8}(-2)+c \Rightarrow c=1 \\ & \text { Put } x=\frac{\pi}{8} \\ & y=\frac{1}{8} \times 2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)+e^{-\pi / 4} \\ & y=e^{-\pi / 4} \end{aligned}$

$\frac{d y}{d x}+\frac{y\left(2 x^3+8 x\right)}{\left(x^2+4\right)^2}=\frac{2}{\left(x^2+4\right)^2}$

$\mathrm{IF}=e^{\int \frac{2 x^3+8 x}{\left(x^2+4\right)^2} d x}$

$\text { Let }\left(x^2+4\right)^2=t \quad \Rightarrow 2\left(x^2+4\right)(2 x) d x=d t$

$\begin{gathered} =e^{\int \frac{d t}{2 t}}=e^{\log \sqrt{t}}=\sqrt{t}=\left(x^2+4\right) \\ \therefore \quad y\left(x^2+4\right)=\int \frac{2}{x^2+4}+c \\ \Rightarrow y\left(x^2+4\right)=\tan ^{-1}\left(\frac{x}{2}\right)+c \\ y(0)=0 \end{gathered}$

$\begin{aligned} & \Rightarrow 0=0+c \quad \Rightarrow c=0 \\ & \text { put } x=2 \\ & y(8)=\frac{\pi}{4} \quad \Rightarrow \quad y=\frac{\pi}{32} \\ \end{aligned}$

$\begin{aligned} & \left(x^4+2 x^3+3 x^2+2 x+2\right) d y-\left(2 x^2+2 x+3\right) d x=0 \\ & \int d y=\int\left(\frac{2 x^2+2 x+3}{x^4+2 x^3+3 x^2+2 x+2} d x\right. \\ & \int d y=\int \frac{1}{x^2+1} d x+\int \frac{}{x^2+2 x+2} d x \\ & y=\tan ^{-1}(x)+\tan ^{-1}(1+x)+C \\ & y(-1)=\tan ^{-1}(-1)+\tan ^{-1}(1-1)+C \\ & y(-1)=-\frac{\pi}{4}+C=\left(\frac{-\pi}{4}\right)-\{\text { given }\} \\ & \Rightarrow C=0 \\ & \text { So, } y(x)=\tan ^{-1}(x)+\tan ^{-1}(1+x) \\ & y(0)=\tan ^{-1}(0)+\tan ^{-1}(1+0) \\ & y(0)=\frac{\pi}{4} \end{aligned}$

$ \begin{aligned} & f(0)=2, \lim _{x \rightarrow-\infty} f(x)=1 \\\\ & f^{\prime}(x)-x \cdot f(x)=3 \\\\ & \text { I.F }=e^{-\alpha x} \\\\ & y\left(e^{-\alpha x}\right)=\int 3 \cdot e^{-\alpha x} d x \\\\ & f(x) \cdot\left(e^{-\alpha x}\right)=\frac{3 e^{-\alpha x}}{-\alpha}+c \\\\ & x=0 \Rightarrow 2=\frac{-3}{\alpha}+c \Rightarrow \frac{3}{\alpha}=c-2 \\\\ & f(x)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\ & x \rightarrow-\infty \Rightarrow 1=\frac{-3}{\alpha}+c(0) \\\\ & \alpha=-3 \therefore c=1 \\\\ & f(-\ln 2)=\frac{-3}{\alpha}+c \cdot e^{\alpha x} \\\\ & =1+e^{3 \ln 2}=9 \end{aligned} $

$\begin{aligned} & \frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1 \\\\ & \text { Put } x+y=t \\\\ & \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1 \\\\ & \frac{d t}{d x}-1=2 x(t)^3-x t\end{aligned}$

$\begin{aligned} \Rightarrow & \frac{d t}{2 t^3-t}=x d x \\\\ & \int \frac{1}{2 t^3-t} d t=\int x d x \\\\ \Rightarrow & \int \frac{t}{2 t^4-t^2} d t=\int x d x\end{aligned}$

$\begin{aligned} & t^2=z \\\\ & 2 t d t=d z \\\\ & \frac{1}{2} \int \frac{d z}{2 z^2-z}=\int x d x \\\\ & \ln \left|\frac{z-\frac{1}{2}}{z}\right|=x^2+c\end{aligned}$

$\ln \left|\frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}\right|=x^2+c$

$y(0)=1 \Rightarrow c=\ln \left(\frac{1}{2}\right)$

$\Rightarrow \frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}=e^{x^2} \times \frac{1}{2}$

$\frac{(x+y)^2-\frac{1}{2}}{(x+y)^2}=\sqrt{e} \times \frac{1}{2}$

$\Rightarrow(x+y)^2=\frac{1}{2-\sqrt{e}}$

$\begin{aligned} & \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80) \\ & \int_\limits{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_\limits0^{\mathrm{t}}-\mathrm{Kdt} \\ & {[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}} \\ & \ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt} \\ & \ln \left|\frac{\mathrm{T}-80}{80}\right|=-\mathrm{kt} \\ & \mathrm{T}=80+80 \mathrm{e}^{-\mathrm{kt}} \\ & 120=80+80 \mathrm{e}^{-\mathrm{k} .15} \\ & \frac{40}{80}=\mathrm{e}^{-\mathrm{k} 15}=\frac{1}{2} \\ & \therefore \mathrm{T}(45)=80+80 \mathrm{e}^{-\mathrm{k} .45} \\ & =80+80\left(\mathrm{e}^{-\mathrm{k} .15}\right)^3 \\ & =80+80 \times \frac{1}{8} \\ & =90 \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}=\frac{\sin x+y \cos x}{\sin x \cdot \cos x\left(\frac{1}{\cos x}-\sin x \cdot \frac{\sin x}{\cos x}\right)} \\ & =\frac{\sin x+y \cos x}{\sin x\left(1-\sin ^2 x\right)} \\ & \frac{d y}{d x}=\sec ^2 x+y \cdot 2(\operatorname{cosec} 2 x) \\ & \frac{d y}{d x}-2 \operatorname{cosec}(2 x) \cdot y=\sec ^2 x \\ & \frac{d y}{d x}+p \cdot y=Q \end{aligned}$

$\text { I.F. }=\mathrm{e}^{\int p \mathrm{dx}}=\mathrm{e}^{\int-2 \operatorname{cosec}(2 \mathrm{x}) \mathrm{dx}}$

Let $2 \mathrm{x}=\mathrm{t}$

$2 \frac{\mathrm{dx}}{\mathrm{dt}}=1$

$\mathrm{dx}=\frac{\mathrm{dt}}{2}$

$=\mathrm{e}^{-\int \operatorname{cosec}(\mathrm{t}) \mathrm{dt}}$

$=\mathrm{e}^{-\ln \left|\tan \frac{t}{2}\right|}$

$=\mathrm{e}^{-\ln |\tan x|}=\frac{1}{|\tan x|}$

$\begin{aligned} & y(\text { IF })=\int Q(I F) d x+c \\ & \Rightarrow y \frac{1}{|\tan x|}=\int \sec ^2 x \cdot \frac{1}{|\tan x|}+c \\ & y \cdot \frac{1}{|\tan x|}=\int \frac{d t}{|t|}+c \quad \text { for } \tan x=t \\ & y \cdot \frac{1}{|\tan x|}=\ln |t|+c \\ & y=|\tan x|(\ln |\tan x|+c) \\ & \text { Put } x=\frac{\pi}{4}, y=2 \\ & 2=\ln 1+c \Rightarrow c=2 \\ & y=|\tan x|(\ln |\tan x|+2) \\ & y\left(\frac{\pi}{3}\right)=\sqrt{3}(\ln \sqrt{3}+2) \end{aligned}$

$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}}{\mathrm{y}}\left(\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1\right)$

Let $\frac{x}{y}=t \Rightarrow x=t y$

$\begin{aligned} & \frac{d x}{d y}=t+y \frac{d t}{d y} \\ & t+y \frac{d t}{d y}=t(\ln (t)+1) \end{aligned}$

$\mathrm{y} \frac{\mathrm{dt}}{\mathrm{dy}}=\mathrm{t} \ln (\mathrm{t}) \Rightarrow \frac{\mathrm{dt}}{\mathrm{t} \ln (\mathrm{t})}=\frac{\mathrm{dy}}{\mathrm{y}}$

$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t} \cdot \ln (\mathrm{t})}=\int \frac{\mathrm{dy}}{\mathrm{y}}$

$\Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y} \quad$ let $\ln t=p$

$\frac{1}{\mathrm{t}} \mathrm{dt}=\mathrm{dp}$

$\begin{aligned} & \Rightarrow \ln p=\ln y+c \\ & \ln (\ln t)=\ln y+c \\ & \ln \left(\ln \left(\frac{x}{y}\right)\right)=\ln y+c \\ & \text { at } x=e, y=1 \\ & \ln \left(\ln \left(\frac{e}{1}\right)\right)=\ln (1)+c \Rightarrow c=0 \end{aligned}$

$\begin{aligned} & \ln \left|\ln \left(\frac{x}{y}\right)\right|=\ln y \\ & \left|\ln \left(\frac{x}{y}\right)\right|=e^{\ln y} \\ & \left|\ln \left(\frac{x}{y}\right)\right|=y \end{aligned}$

$\frac{d y}{d x}=2(x-1) \sin x+\left(x^2-2 x\right) \cos x$

Now both side integrate

$\begin{aligned} & y(x)=\int 2(x-1) \sin x d x+\left[\left(x^2-2 x\right)(\sin x)-\int(2 x-2) \sin x d x\right] \\ & y(x)=\left(x^2-2 x\right) \sin x+\lambda \\ & y(0)=0+\lambda \Rightarrow 2=\lambda \\ & y(x)=\left(x^2-2 x\right) \sin x+2 \\ & y(2)=2 \end{aligned}$

Differential equation :-

$\begin{aligned} & x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x \\ & \cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x \end{aligned}$

Divide both sides by $\mathrm{x}^2$

$\cos \frac{y}{x}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=\frac{1}{x}$

Let $\frac{y}{x}=t$

$\begin{aligned} & \cos \mathrm{t}\left(\frac{\mathrm{dt}}{\mathrm{dx}}\right)=\frac{1}{\mathrm{x}} \\ & \cos \mathrm{t~dt}=\frac{1}{\mathrm{x}} \mathrm{dx} \end{aligned}$

Integrating both sides

$\begin{aligned} & \sin \mathrm{t}=\ln |\mathrm{x}|+\mathrm{c} \\ & \sin \frac{\mathrm{y}}{\mathrm{x}}=\ln |\mathrm{x}|+\mathrm{c} \end{aligned}$

Using $\mathrm{y}(1)=\frac{\pi}{3}$, we get $\mathrm{c}=\frac{\sqrt{3}}{2}$

So, $\alpha=\sqrt{3} \Rightarrow \alpha^2=3$

$\begin{aligned} & \frac{d y}{d x}-\left(\frac{\sin 2 x}{1+\cos ^2 x}\right) y=\sin x \\ & \text { I.F. }=1+\cos ^2 x \\ & y \cdot\left(1+\cos ^2 x\right)=\int(\sin x) d x \\ & =-\cos x+C \\ & x=0, C=1 \\ & y\left(\frac{\pi}{2}\right)=1 \end{aligned}$

$\begin{aligned} & \left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0 \\ & \Rightarrow \int \frac{d y}{y^2-3 y}=\int \frac{d x}{x^2-4} \\ & \Rightarrow \frac{1}{3} \int \frac{y-(y-3)}{y(y-3)} d y=\int \frac{d x}{x^2-4} \\ & \Rightarrow \frac{1}{3}(\ln |y-3|-\ln |y|)=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \\ & \Rightarrow \frac{1}{3} \ln \left|\frac{y-3}{y}\right|=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \end{aligned}$

$\begin{aligned} & \text { At } \mathrm{x}=4, \mathrm{y}=\frac{3}{2} \\ & \therefore \mathrm{C}=\frac{1}{4} \ln 3 \\ & \therefore \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\frac{1}{4} \ln (3) \\ & \text { At } \mathrm{x}=10 \\ & \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{2}{3}\right|+\frac{1}{4} \ln (3) \\ & \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\ln 2^{3 / 4}, \forall \mathrm{x}>2, \frac{\mathrm{dy}}{\mathrm{dx}}<0 \\ & \text { as } \mathrm{y}(4)=\frac{3}{2} \Rightarrow \mathrm{y} \in(0,3) \\ & -\mathrm{y}+3=8^{1 / 4} \cdot \mathrm{y} \\ & \mathrm{y}=\frac{3}{1+8^{1 / 4}} \end{aligned}$

$\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0 \\ & \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{adt} \\ & \int \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{a} \int \mathrm{dt} \\ & \ln |\mathrm{x}|=-\mathrm{at}+\mathrm{c} \\ & \mathrm{at} \mathrm{t}=0, \mathrm{x}=2 \\ & \ln 2=0+\mathrm{c} \\ & \ln \mathrm{x}=-\mathrm{at}+\ln 2 \\ & \frac{\mathrm{x}}{2}=\mathrm{e}^{-\mathrm{at}} \\ & \mathrm{x}=2 \mathrm{e}^{-\mathrm{at}} \quad \text{.... (i)} \end{aligned}$

$\begin{aligned} & \frac{d y}{d t}+b y=0 \\ & \frac{d y}{y}=-b d t \\ & \ln |y|=-b t+\lambda \\ & t=0, y=1 \\ & 0=0+\lambda \\ & y=e^{-b t} \quad \text{..... (ii)} \end{aligned}$

According to question

$\begin{aligned} & 3 \mathrm{y}(1)=2 \mathrm{x}(1) \\ & 3 \mathrm{e}^{-\mathrm{b}}=2\left(2 \mathrm{e}^{-\mathrm{a}}\right) \\ & \mathrm{e}^{\mathrm{a}-\mathrm{b}}=\frac{4}{3} \end{aligned}$

$\begin{aligned} & \text { For } x(t)=y(t) \\ & \begin{array}{r} 2 \mathrm{e}^{-a t}=e^{-b t} \\ 2=e^{(a-b) t} \\ 2=\left(\frac{4}{3}\right)^t \\ \log _{\frac{4}{3}} 2=t \end{array} \end{aligned}$

$ \begin{aligned} & 2(y+2) \ln (y+2) d x+(x+4-2 \ln (y+2)) d y=0 \\\\ & 2 \ln (y+2)+(x+4-2 \ln (y+2)) \frac{1}{y+2} \cdot \frac{d y}{d x}=0 \\\\ & \text { let, } \ln (y+2)=t \\\\ & \frac{1}{y+2} \cdot \frac{d y}{d x}=\frac{d t}{d x} \\\\ & 2 t+(x+4-2 t) \cdot \frac{d t}{d x}=0 \\\\ & (x+4-2 t) \frac{d t}{d x}=-2 t \\\\ & \frac{d x}{d t}=\frac{2 t-4-x}{2 t} \\\\ & \frac{d x}{d t}+\frac{x}{2 t}=\frac{2 t-4}{2 t} \end{aligned} $

$ \begin{aligned} & x \cdot t^{1 / 2}=\int \frac{2 t-4}{2 t} \cdot t^{1 / 2} \cdot d t \\\\ & x \cdot t^{1 / 2}=\int\left(t^{1 / 2}-\frac{2}{t^{1 / 2}}\right) \cdot d t \\\\ & =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-2 \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \\\\ & x \cdot t^{\frac{1}{2}}=\frac{2 t^{\frac{3}{2}}}{3}-4 \mathrm{t}^{\frac{1}{2}}+C \\\\ & x=\frac{2}{3} \cdot \mathrm{t}-4+\mathrm{C} \cdot \mathrm{t}^{\frac{-1}{2}} \\\\ & x=\frac{2}{3} \ln (\mathrm{y}+2)-4+C \cdot(\ln (\mathrm{y}+2))^{\frac{-1}{2}} \end{aligned} $

Put $y=e^4-2, x=1$

$ \begin{aligned} & 1=\frac{2}{3} \times 4-4+C \times \frac{1}{2} \\\\ & \frac{C}{2}=5-\frac{8}{3}=\frac{7}{3} \\\\ & C=\frac{14}{3} \\\\ & x=\frac{2}{3} \times 9-4+\frac{14}{3} \times \frac{1}{3} \\\\ & =2+\frac{14}{9} \\\\ & =\frac{32}{9} \end{aligned} $

The given differential equation is

$\frac{d y}{d x} = y + 7$

This is a first order linear differential equation and can be solved using an integrating factor.

Rearrange the equation to the standard form of a linear differential equation :

$\frac{d y}{d x} - y = 7$

The integrating factor is $e^{-\int dx} = e^{-x}$.

Multiplying each side of the equation by the integrating factor gives :

$e^{-x} \frac{d y}{d x} - e^{-x} y = 7e^{-x}$

The left-hand side of the equation is the derivative of $(e^{-x}y)$ with respect to $x$. So we can write the equation as :

$\frac{d}{d x}(e^{-x}y) = 7e^{-x}$

Integrate both sides with respect to $x$ :

$e^{-x}y = -7e^{-x} + C$

Multiply both sides by $e^{x}$ to isolate $y$ :

$y = -7 + Ce^{x}$

So, the general solution to the differential equation is $y = -7 + Ce^{x}$.

Now, let's apply the initial conditions to find the particular solutions :

For $y_{1}(0)=0$, we substitute into the general solution and solve for $C$ :

$0 = -7 + C$

So, $C = 7$, and the solution for $y_{1}$ is $y_{1}(x) = 7e^{x} - 7$.

For $y_{2}(0)=1$, again substitute into the general solution:

$1 = -7 + C$

So, $C = 8$, and the solution for $y_{2}$ is $y_{2}(x) = 8e^{x} - 7$.

The two curves intersect when $y_{1}(x) = y_{2}(x)$. Setting these equal and solving for $x$ gives :

$7e^{x} - 7 = 8e^{x} - 7$

$e^{x} = 0$

But $e^{x} = 0$ has no solution, because the exponential function never equals zero.

So, the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ do not intersect at any point.

$ \begin{aligned} & \left(1+x^2\right) d y=y(x-y) d x \\\\ & y(0)=1 \cdot y(2 \sqrt{2})=\beta \\\\ & \frac{d y}{d x}=\frac{y x-y^2}{1+x^2} \\\\ & \frac{d y}{d x}+y\left(\frac{-x}{1+x^2}\right)=\left(\frac{-1}{1+x^2}\right) y^2 \\\\ & \frac{1}{y^2} \frac{d y}{d x}+\frac{1}{y}\left(\frac{-x}{1+x^2}\right)=\frac{-1}{1+x^2} \\\\ & \text { put } \frac{1}{y}=t \text { then } \frac{-1}{y^2} \frac{d y}{d x}=\frac{d t}{d x} \end{aligned} $

$ \frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \frac{\mathrm{x}}{1+\mathrm{x}^2}=\frac{1}{1+\mathrm{x}^2} $

$ \text { I.F }=\mathrm{e}^{\int \frac{\mathrm{x}}{1+\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{\frac{1}{2} \ln \left(1+\mathrm{x}^2\right)}=\sqrt{1+\mathrm{x}^2} $

$ t \sqrt{1+x^2}=\int \frac{\sqrt{1+x^2}}{1+x^2} d x $

$ \begin{aligned} & \frac{1}{y} \sqrt{1+x^2}=\int \frac{1}{\sqrt{1+x^2}} d x \\\\ & \frac{1}{y} \sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+C \\\\ & \because y(0)=1 \Rightarrow C=1 \\\\ & \frac{1}{y} \sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+1 \\\\ & \text { For } y=2 \sqrt{2} \\\\ & \frac{3}{y}=\ln |2 \sqrt{2}+3|+1 \end{aligned} $

$ \begin{gathered} y=\beta=\frac{3}{1+\ln |2 \sqrt{2}+3|} \\\\ \Rightarrow 3 \beta^{-1}=1+\ln |2 \sqrt{2}+3| \end{gathered} $

Isolating the term $e^{3 \beta^{-1}}$, we get :

$e^{3 \beta^{-1}} = e^{1+\ln |2 \sqrt{2}+3|}.$

This can be simplified using the rule $e^{a+b} = e^a e^b$ to :

$e^{3 \beta^{-1}} = e \cdot e^{\ln |2 \sqrt{2}+3|}.$

Since $e^{\ln(x)} = x$ for any $x$, this simplifies to :

$e^{3 \beta^{-1}} = e |2 \sqrt{2}+3|.$

Using the given value of $\beta$, which is $\frac{3}{1+\ln |2 \sqrt{2}+3|}$, we find :

$e^{3 \beta^{-1}} = e |2 \sqrt{2}+3| = e (2\sqrt{2} + 3),$

Since $2\sqrt{2}+3$ is positive and so the absolute value does not affect the result.

I.F $=\mathrm{e}^{\int \frac{5 \mathrm{dx}}{\mathrm{x}\left(\mathrm{x}^5+1\right)}}=\mathrm{e}^{\int \frac{5 \mathrm{x}^{-6} \mathrm{dx}}{\left(\mathrm{x}^{-5}+1\right)}}$

Put, $1+\mathrm{x}^{-5}=\mathrm{t} \Rightarrow-5 \mathrm{x}^{-6} \mathrm{dx}=\mathrm{dt}$

$ \therefore $ $ e^{\int-\frac{d t}{t}}=e^{-\ln t}=\frac{1}{t}=\frac{x^5}{1+x^5} $

$ \begin{aligned} y \cdot \frac{x^5}{1+x^5} & =\int \frac{x^5}{\left(1+x^5\right)} \times \frac{\left(1+x^5\right)^2}{x^7} d x \\\\ & =\int x^3 d x+\int x^{-2} d x \end{aligned} $

$ y \cdot \frac{x^5}{1+x^5}=\frac{x^4}{4}-\frac{1}{x}+c $

$ \text { Given that: } x=1 \Rightarrow y=2 $

$ \begin{aligned} & 2 \cdot \frac{1}{2}=\frac{1}{4}-1+\mathrm{c} \\\\ & \mathrm{c}=\frac{7}{4} \\\\ & \mathrm{y} \cdot \frac{\mathrm{x}^5}{1+\mathrm{x}^5}=\frac{\mathrm{x}^4}{4}-\frac{1}{\mathrm{x}}+\frac{7}{4} \\\\ & \text { Now put, } \mathrm{x}=2 \\\\ & \mathrm{y} \cdot\left(\frac{32}{33}\right)=\frac{21}{4} \\\\ & \mathrm{y}=\frac{693}{128} \end{aligned} $

We have,

$ \begin{aligned} & \left(1-x^2 y^2\right) d x=y d x+x d y, y(1)=2 \\\\ & d x=\frac{y d x+x d y}{1-(x y)^2} \end{aligned} $

On integrating both sides, we get

$ \begin{aligned} \int d x & =\int \frac{d(x y)}{1-(x y)^2} \\\\ x & =\frac{1}{2} \log \left|\frac{1+x y}{1-x y}\right|+C \end{aligned} $

As, $y(1)=2$

$ \begin{aligned} 1 & =\frac{1}{2} \log \left|\frac{1+2}{1-2}\right|+C \\\\ \Rightarrow C & =1-\frac{1}{2} \log 3 \end{aligned} $

Now, substitute $x=2$ as $y(2)=\alpha$

$ 2=\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right|+1-\frac{1}{2} \log 3 $

$ \begin{aligned} 1+\frac{1}{2} \log 3 & =\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right| \\\\ 2+\log 3 & =\log \left|\frac{1+2 \alpha}{1-2 \alpha}\right| \end{aligned} $

$ \begin{aligned} &\Rightarrow \frac{1+2 \alpha}{1-2 \alpha} \mid =3 e^2 \\\\ &\Rightarrow \frac{1+2 \alpha}{1-2 \alpha}= \pm \frac{3 e^2}{1} \\\\ & \Rightarrow\frac{1}{2 \alpha} =\frac{ \pm 3 e^2+1}{ \pm 3 e^2-1} \\\\ &\Rightarrow 2 \alpha=\frac{ \pm 3 e^2-1}{ \pm 3 e^2+1} \end{aligned} $

$ \begin{array}{ll} \Rightarrow \alpha=\frac{1}{2}\left(\frac{ \pm 3 e^2-1}{1 \pm 3 e^2}\right) \\\\ \therefore \alpha=\frac{3 e^2-1}{2\left(1+3 e^2\right)} \text { or } \frac{3 e^2+1}{2\left(3 e^2-1\right)} \end{array} $

Given that

$ x^2 f(x)-x=4 \int_0^x t f(t) d t $

On differentiating both sides with respect to $x$, we get

$ \begin{array}{rlrl} & 2 x f(x)+x^2 f^{\prime}(x)-1 =4 x f(x) \\\\ &\Rightarrow x^2 f^{\prime}(x)-2 x f(x)-1 =0 \\\\ &\Rightarrow x^2 \frac{d y}{d x}-2 x y =1 ~~~~~~~(Let, y=f(x) ]\\\\ &\frac{d y}{d x}-\frac{2}{x} y =\frac{1}{x^2} \end{array} $

On comparing above equation with

$\frac{d y}{d x}+P y=Q$, where $P=-\frac{2}{x}, Q=\frac{1}{x^2}$

Now, IF $=e^{\int(-2 / x) d x}=e^{-2 \log x}=1 / x^2$

Solution is $y\left(\frac{1}{x^2}\right)=\int\left(\frac{1}{x^2}\right) \frac{1}{x^2} d x$

$ \begin{array}{ll} &\Rightarrow \frac{y}{x^2}=\int \frac{1}{x^4} d x \Rightarrow \frac{y}{x^2}=-\frac{1}{3 x^3}+C \\\\ &\Rightarrow y =-\frac{1}{3 x}+C x^2 \end{array} $

Given, $f(1)=\frac{2}{3}$.

So, $\frac{2}{3}=-\frac{1}{3}+C \Rightarrow C=1$

Thus, $y=f(x)=-\frac{1}{3 x}+x^2$

$ \therefore 18 f(3)=18\left\{9-\frac{1}{9}\right\}=18 \times \frac{80}{9}=160 $

We have, $\left(1+\log _e x\right) \frac{d x}{d y}-x \log _e x=e^{y}, x>0$

Put $x \log _e x=t$

$ \begin{aligned} & \Rightarrow \left(x \cdot \frac{1}{x}+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\ & \Rightarrow \left(1+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\ & \therefore \frac{d t}{d y}-t =e^{y} \end{aligned} $

Now, IF $=e^{\int-d y}=e^{-y}$

$\begin{aligned} & \therefore \text { General solution, } t\left(e^{-y}\right)=\int\left(e^y \cdot e^{-y}\right) d y+c \\\\ & \Rightarrow t e^{-y}=\int d y+c \\\\ & \Rightarrow t e^{-y}=y+c \\\\ & \Rightarrow \left(x \log _e x\right) e^{-y}=y+c ..........(i)\end{aligned}$

Equation (i), passes through the point $(1,0)$

$ \begin{aligned} & \therefore 0=0+c \\\\ & \Rightarrow c=0 \\\\ & \therefore \left(x \log _e x\right) e^{-y}=y \\\\ & \Rightarrow x \log _e x=y e^y \end{aligned} $

Which passes through $(\alpha, 2)$

$ \begin{aligned} & \therefore \alpha \log _e \alpha=2 e^2 \\\\ & \Rightarrow \log _e \alpha^\alpha=2 e^2 \\\\ & \Rightarrow \alpha^\alpha=e^{2 e^2} \end{aligned} $

$\begin{aligned} & 2 x^2 y d y-\left(1-x y^2\right) d x=0 \\\\ & \Rightarrow 2 x^2 y \frac{d y}{d x}-1+x y^2=0 \\\\ & \Rightarrow 2 y \frac{d y}{d x}-\frac{1}{x^2}+\frac{y^2}{x}=0 \\\\ & \Rightarrow 2 y \frac{d y}{d x}+\frac{y^2}{x}=\frac{1}{x^2} \text { (L.D.E) } \\\\ & Let, y^2=t \Rightarrow 2 y \frac{d y}{d x}=\frac{d t}{d x} \\\\ & \frac{d t}{d x}+\frac{t}{x}=\frac{1}{x^2} \\\\ & \text { I.F }=e^{\int \frac{1}{x} d x}=x\end{aligned}$

So, the general solution is

$ \begin{aligned} & \Rightarrow t \times x=\int x \cdot \frac{1}{x^2} d x \\\\ & \Rightarrow y^2 \cdot x=\ln x+c \\\\ & \text { Also,y(2) }=\sqrt{\log _e 2} \\\\ & \log _e^2 2=\log _e 2+c \Rightarrow c=\log _e 2 \\\\ & \Rightarrow y^2 x=\ln x+\ln 2 \\\\ & \Rightarrow y^2 x=\ln 2 x \\\\ & 2 x=\exp \left(x^1 y^2\right) \end{aligned} $

By compare it with given solution we get,

$ \begin{aligned} & \alpha=2,\beta=1, \gamma=2 \\\\ & \Rightarrow \alpha+\beta-\gamma=2+1-2=1 \end{aligned} $

$ \begin{aligned} & \frac{d y}{d x}+\frac{x+a}{y-2}=0 \\\\ & \frac{d y}{d x}=\frac{x+a}{2-y} \\\\ & (2-y) d y=(x+a) d x \\\\ & 2 y \frac{-y}{2}=\frac{x^2}{2}+\mathrm{ax}+\mathrm{c} \\\\ & \mathrm{a}+\mathrm{c}=-\frac{1}{2} \text { as } \mathrm{y}(1)=0 \\\\ & \mathrm{X}^2+\mathrm{y}^2+2 \mathrm{ax}-4 \mathrm{y}-1-2 \mathrm{a}=0 \\\\ & \pi \mathrm{r}^2=4 \pi \\\\ & \mathrm{r}^2=4 \\\\ & 4=\sqrt{a^2+4+1+2 a} \\\\ & (\mathrm{a}+1)^2=0 \end{aligned} $

$ P, Q=(0,2 \pm \sqrt{3}) $

Equation of normal at $\mathrm{P}, \mathrm{Q}$ are $\mathrm{y}-2=\sqrt{3}(\mathrm{x}-1)$

$ \begin{aligned} & \mathrm{y}-2=-\sqrt{3}(\mathrm{x}-1) \\\\ & \mathrm{R}=\left(1-\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{S}=\left(1+\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{RS}=\frac{4}{\sqrt{3}}=4 \frac{\sqrt{3}}{3} \end{aligned} $

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

$\therefore $ I.F $=e^{\int \tan x d x}=\sec x$

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

$\Rightarrow y \sec x=x \tan x-\ln |\sec x|+c$

Given, $ y(0)=1 $

$\Rightarrow 1=c$

$ \therefore $ $ y \sec x=x \tan x-\ln |\sec x|+1$

$\Rightarrow y=x \sin x-\cos x \ln |\sec x|+cosx$

Now, at $x=\pi / 6$, we have

$ \begin{aligned} & y=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e \frac{2}{\sqrt{3}}+\frac{\sqrt{3}}{2} \\\\ & =\frac{\pi}{12}-\frac{\sqrt{3}}{2}\left[\log _e \frac{2}{\sqrt{3}}-\log _e e\right]\\\\ &=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right) \end{aligned} $

$\left(3 y^{2}-5 x^{2}\right) y \cdot d x+2 x\left(x^{2}-y^{2}\right) d y=0$

$ \Rightarrow \frac{d y}{d x}=\frac{y\left(5 x^{2}-3 y^{2}\right)}{2 x\left(x^{2}-y^{2}\right)} $

Put $\mathrm{y}=\mathrm{mx}$

$ \Rightarrow m+x \cdot \frac{d m}{d x}=\frac{m\left(5-3 m^{2}\right)}{2\left(1-m^{2}\right)} $

$ \begin{aligned} & x \cdot \frac{d m}{d x}=\frac{\left(5-3 m^{2}\right) m-2 m\left(1-m^{2}\right)}{2\left(1-m^{2}\right)} \\\\ & \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\frac{2\left(\mathrm{~m}^{2}-1\right)}{\mathrm{m}\left(\mathrm{m}^{2}-3\right)} \mathrm{dm} \\\\ & \Rightarrow \frac{d x}{x}=\left(\frac{2}{m}-\frac{\frac{4}{3}}{m}+\frac{\frac{4 m}{3}}{\mathrm{~m}^{2}-3}\right) d m \end{aligned} $

$\Rightarrow \int \frac{d x}{x}=\int \frac{\left(\frac{2}{3}\right)}{m}+\int \frac{2}{3}\left(\frac{2 m}{m^{2}-3}\right) d m$

$\Rightarrow \ln |\mathrm{x}|=\frac{2}{3} \ln |\mathrm{m}|+\frac{2}{3} \ln \left|\mathrm{m}^{2}-3\right|+\mathrm{C}$

Or, $\ln |\mathrm{x}|=\frac{2}{3} \ln \left|\frac{\mathrm{y}}{\mathrm{x}}\right|+\frac{2}{3} \ln \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right|+\mathrm{C}$

Put $(\mathrm{x}=1, \mathrm{y}=1)$ : we get $\mathrm{c}=-\frac{2}{3} \ln (2)$

$\Rightarrow \ln |\mathrm{x}|=\frac{2}{3} \ln \left|\frac{\mathrm{y}}{\mathrm{x}}\right|+\frac{2}{3} \ln \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right|-\frac{2}{3} \ln (2)$

$\Rightarrow\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\left[\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right]=2 .\left(\mathrm{x}^{3 / 2}\right)$ ........(1)

Put $\mathrm{x}=2$ in equation (1), we get

$\Rightarrow \mathrm{y}\left(\mathrm{y}^{2}-12\right)=4 \times 2 \times 2 \times 2 \sqrt{2}$

$\Rightarrow \mathrm{y}^{3}-12 \mathrm{y}=32 \sqrt{2}$

$\Rightarrow\left|\mathrm{y}^{3}(2)-12 \mathrm{y}(2)\right|=32 \sqrt{2}$

Differentiating both sides we get,

$ \begin{aligned} & f^{\prime}(x)+\frac{f(x)}{x}=\frac{1}{2 \sqrt{x+1}} \\\\ & \Rightarrow \frac{d y}{d x}+\frac{y}{x}=\frac{1}{2 \sqrt{x+1}} \\\\ & \Rightarrow \mathrm{IF}=x \\\\ & \Rightarrow y x=\frac{1}{2} \int \frac{x}{\sqrt{x+1}} d x+c \\\\ & \Rightarrow y x=\frac{1}{2}\left(\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-2(x+1)^{\frac{1}{2}}\right)+c \\\\ & x y=\frac{1}{3}(x+1)^{\frac{3}{2}}-(x+1)^{\frac{1}{2}}+c \\\\ & f(3)=2 \end{aligned} $

So, $x=3, y=2$

$\Rightarrow c=\frac{16}{3}$

Now, $x=8$

$8 f(8)=\frac{27}{3}-3+\frac{16}{3}=\frac{34}{3}$

$12 f(8)=\frac{34}{3} \times \frac{12}{8}=17$

$y = vx$

$v + x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}}} \right)$

$x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}} + v} \right)$

${{dv} \over {dx}} = - \left( {{{{{(1 + v)}^3}} \over {3 + {v^2}}}} \right)$

$ \Rightarrow {{3 + {v^2}} \over {{{(1 + v)}^3}}} = {{ - dx} \over x}$

$ \Rightarrow \ln |v + 1| + {{2v} \over {{{(v + 1)}^2}}} = C - \ln |x|$

$x = 1,v = 0 \Rightarrow C = 0$

$ \Rightarrow \ln |x + y| - \ln |x| + {{2xy} \over {{{(x + y)}^2}}} = - \ln |x|$

${{dy} \over {dx}} - {{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}y = 2x\exp \left\{ {{{{x^3} - {{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }}} \right\}$

$IF = {e^{ - \int {{{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}dx} }}$

Let ${\tan ^{ - 1}}{x^3} = t \Rightarrow {{3{x^2}} \over {1 + {x^6}}}dx = dt$

$ \Rightarrow IF = {e^{ - \int {{{\tan t} \over {\sec t}}.t\,dt} }} = {e^{ - \int {\sin t.tdt} }} = {e^{t\cos t - \sin t}}$

$ \Rightarrow IF = {e^{{{{{\tan }^{ - 1}}({x^3})} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}}$

$\therefore$ Solution is

$y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}} = \int {2x\,dx + c} $

$ \Rightarrow y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3} - {x^3}} \over {\sqrt {1 + {x^6}} }}}} = {x^2} + c$

$y(0) = 0 \Rightarrow c = 0$

$x = 1$

$y\,.\,{e^{{{{\pi \over 4} - 1} \over {\sqrt 2 }}}} = 1$

$ \Rightarrow y = {e^{{{1 - {\pi \over 4}} \over {\sqrt 2 }}}}$

$ \Rightarrow y = {e^{{{4 - \pi } \over {4\sqrt 2 }}}}$

$x\ln x{{dy} \over {dx}} + y = {x^2}\ln x$

${{dy} \over {dx}} + {1 \over {x\ln x}}\,.\,y = x$

If $ = {e^{\int {{1 \over {x\ln x}}dx} }} = {e^{\int {{1 \over t}dt} }}$, where $t = \ln x$

$ = {e^{\ln t}} = t = \ln x$

$y.\ln x = \int {x\ln x = {{{x^2}} \over 2}\ln x - \int {{{{x^2}} \over 2}\,.\,{1 \over x}} } $

$y\ln x = {{{x^2}} \over 2}\ln x - {{{x^2}} \over 4} + C$ ..... (i)

$y(2) = 2 \Rightarrow C = 1$

Putting $x = e$ in (i),

$y = {{{e^2}} \over 4} + 1 = {{4 + {e^2}} \over 4}$

Given,

$y(x + 1)dx - {x^2}dy = 0$

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

$ \Rightarrow {1 \over x}dx + {{dx} \over {{x^2}}} = {{dy} \over y}$

Integrating both sides, we get

$\int {{{dx} \over x} + \int {{{dx} \over {{x^2}}} = \int {{{dy} \over y}} } } $

$ \Rightarrow \ln |x| - {1 \over x} = \ln |y| + C$ ..... (1)

Given $y(1) = e$

$\therefore$ $x = 1$ and $y = e$

Putting value of x and y in equation (1), we get

$\ln |1| - {1 \over 1} = \ln |e| + C$

$ \Rightarrow 0 - 1 = 1 + C$

$ \Rightarrow C = - 2$

$\therefore$ Equation (1) becomes,

$\ln |x| = - {1 \over x} = \ln |y| - 2$

$ \Rightarrow \ln |y| = \ln |x| - {1 \over x} + 2$

$ \Rightarrow y = {e^{\ln |x|}}\,.\,{e^{2 - {1 \over x}}}$

$ \Rightarrow y = x\,.\,{e^{2 - {1 \over x}}}$

Now,

$\mathop {\lim }\limits_{x \to {0^ + }} y$

$ = \mathop {\lim }\limits_{x \to {0^ + }} \left( {x\,.\,{e^{2 - {1 \over x}}}} \right)$

$ = 0\,.\,{e^{ - \alpha }}$

$ = 0$

$\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}$

$ \begin{aligned} & \text { I.F. }=e^{\int \alpha d t}=e^{\alpha t} \\\\ & \Rightarrow y \cdot e^{\alpha t}=\gamma \int e^{(\alpha-\beta) t} d t=\gamma \frac{e^{(\alpha-\beta) t}}{(\alpha-\beta)}+C \\\\ & \Rightarrow y=\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t} \\\\ & \lim _{x \rightarrow \infty} y(t)=\lim _{x \rightarrow \infty}\left[\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t}\right]=0 \end{aligned} $

$ \begin{aligned} & \frac{d y}{d x}-\frac{y}{x}=y^3\left(1+\log _e x\right) \\\\ & \frac{1}{y^3} \frac{d y}{d x}-\frac{1}{x y^2}=1+\log _e x \\\\ & \text { Let }-\frac{1}{y^2}=t \Rightarrow \frac{2}{y^3} \frac{d y}{d x}=\frac{d t}{d x} \\\\ & \therefore \frac{d t}{d x}+\frac{2 t}{x}=2\left(1+\log _e x\right) \\\\ & \text { I.F. }=e^{\int \frac{2}{x} d x}=x^2 \\\\ & \frac{-x^2}{y^2}=\frac{2}{3}\left(\left(1+\log _e x\right) x^3-\frac{x^3}{3}\right)+C \\\\ & y(1)=3 \\\\ & \frac{y^2}{9}=\frac{C}{5-2 x^3\left(2+\log _e x^3\right)} \end{aligned} $

Given,

$\left( {{x^2} - 3{y^2}} \right)dx + 3xydy = 0$

$ \Rightarrow {x^2}dx - 3{y^2}dx + 3xydy = 0$

$ \Rightarrow {{{x^2}dx} \over {3xdx}} - {{3{y^2}dx} \over {3xdx}} + {{3ydy} \over {3xdx}} = 0$

$ \Rightarrow {x \over 3} - {{{y^2}} \over x} + y{{dy} \over {dx}} = 0$

Let ${y^2} = t$

$2y{{dy} \over {dx}} = {{dt} \over {dx}}$

$\therefore$ ${1 \over 2}{{dt} \over {dx}} - {t \over x} + {x \over 3} = 0$

$ \Rightarrow {{dt} \over {dx}} - {{2t} \over x} = - {{2x} \over 3}$

This is linear Differential Equation.

$ \text { I.F. }=e^{\int-\frac{2}{x} d x}=e^{-2 \ln |x|}=e^{\ln \frac{1}{x^2}}=\frac{1}{x^2} $

So, $t \cdot \frac{1}{x^2}=\int \frac{1}{x^2}\left(\frac{-2 x}{3}\right) d x $

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

When, $x=1, y=1$

$ 1=\frac{-2}{3} \ln (1)+C \Rightarrow C=1 $

$ \therefore \frac{y^2}{x^2}=\frac{-2}{3} \ln |x|+1 $

At $x=e, \frac{y^2(e)}{e^2}=\frac{-2}{3}+1 $

$\Rightarrow y^2(e)=\frac{e^2}{3} $

$\Rightarrow 6 y^2(e)=2 e^2$

$x^{3} d y+x y d x-d x=0$

$\Rightarrow \frac{d y}{d x}=\frac{1-x y}{x^{3}}$

$\Rightarrow \frac{d y}{d x}+\frac{y}{x^{2}}=\frac{1}{x^{3}}$

I.F. $=e^{\int \frac{d x}{x^{2}}}=e^{-\frac{1}{x}}$

$\therefore \quad y e^{-\frac{1}{x}}=\int \frac{e^{-\frac{1}{x}}}{x^{3}} d x$

For RHS put $-\frac{1}{x}=t \Rightarrow \frac{d x}{x^{2}}=d t$

$\therefore y e^{-\frac{1}{x}}=-\int t e^{t} d t$

$\Rightarrow y e^{-\frac{1}{x}}=-\left[t e^{t}-e^{t}\right]+c$

$\Rightarrow y e^{-\frac{1}{x}}=\frac{e^{-\frac{1}{x}}}{x}+e^{-\frac{1}{x}}+c$

$ \downarrow y\left(\frac{1}{2}\right)=3-e $

$\Rightarrow(3-e) e^{-2}=2 e^{-2}+e^{-2}+c$

$\Rightarrow \quad c=-\frac{1}{e}$

For $y(1)$ put $x=1, c=-e^{-1}$ in equation (i) we get $y e^{-1}=e^{-1}+e^{-1}-e^{-1}$

$\Rightarrow y=1$