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

The given differential equation can be written as, $\frac{dy}{dx}=\frac{y\left(2-x\right)}{x^2\left(x-1\right)}$

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

$\Rightarrow \log y=-\log x+\frac{2}{x}+\log \left(x-1\right)+c\mathrm{......}\left(i\right)$

Now, $y\left(2\right)=e$

$\therefore \log e=-\log 2+1+0+c\Rightarrow c=\log 2$

$\therefore \left(i\right)$ becomes, $\log y=-\log x+\frac{2}{x}+\log \left(x-1\right)+\log 2$

$\Rightarrow \log \left(\frac{xy}{2\left(x-1\right)}\right)=\frac{2}{x}\Rightarrow \frac{xy}{2\left(x-1\right)}=e^{2/x}\Rightarrow y=\frac{2\left(x-1\right)}{x}.e^{2/x}$

$\therefore y\left(4\right)=\frac{\mathrm{2.3.}{e}^{1/2}}{4}=\frac{3}{2}\sqrt{e}$