If $x^{3}dy+xy·dx=x^{2}dy+2ydx; y\left(2\right)=e$ and $x>1$, then $y\left(4\right)$ is equal to :

Given $x^{3}dy+xy·dx=x^{2}dy+2ydx$

$\Rightarrow \left(x^3-x^2\right)dy=\left(2-x\right)ydx$

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

Integrating both sides with respect to $x$, we get

$\int \frac{dy}{y}=\int \frac{\left(2-x\right)}{x^2\left(x-1\right)}dx+k ............\left(i\right)$

Let $\frac{\left(2-x\right)}{x^2\left(x-1\right)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{\left(x-1\right)}$

$\Rightarrow \left(2-x\right)=Ax\left(x-1\right)+B\left(x-1\right)+C{x}^2$

Putting $x=0\Rightarrow 2=-B\Rightarrow B=-2$

Putting $x=1\Rightarrow 2-1=C\Rightarrow C=1$

Putting $x=2\Rightarrow 2-2=A\left(2\right)\left(1\right)+B\left(1\right)+C\left(2^2\right)\Rightarrow 2A+2=0\Rightarrow A=-1$

From equation $\left(i\right)$, we get

$\int \frac{dy}{y}=\int \left(\frac{-1}{x}+\frac{-2}{x^2}+\frac{1}{x-1}\right)dx+k$

$\Rightarrow \ln y=-\ln x+\frac{2}{x}+\ln \left|x-1\right|+k ..........\left(ii\right)$

Given $y\left(2\right)=e$ i.e. at $x=2, y=e$

$\Rightarrow \ln e=-\ln 2+\frac{2}{2}+\ln \left|2-1\right|+k$

$\Rightarrow 1=-\ln 2+1+0+k$

$\Rightarrow k=\ln 2$

Putting in equation $\left(ii\right)$, we get

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

Now putting $x=4$, we get

$\ln y=-\ln 4+\frac{2}{4}+\ln \left|4-1\right|+\ln 2$

$\Rightarrow \ln y=-2\ln 2+\frac{1}{2}+\ln 3+\ln 2$

$\Rightarrow \ln y=-\ln 2+\frac{1}{2}+\ln 3$

$\Rightarrow \ln y=\frac{1}{2}+\ln \frac{3}{2}$

$\Rightarrow \ln \frac{2y}{3}=\frac{1}{2}$

$\Rightarrow \frac{2y}{3}=e^{\frac{1}{2}}$

$\Rightarrow y=\frac{3}{2}\sqrt{e}$

$\Rightarrow y\left(4\right)=\frac{3}{2}\sqrt{e}$