Let \(y(x)\) be the solution of the differential equation
\(x^2 \frac{d y}{d x}+x y=x^2+y^2, x>\frac{1}{e}\)
satisfying \(y(1)=0\). Then the value of \(2 \frac{(y(e))^2}{y\left(e^2\right)}\) is _____ .

Put \(\mathrm{y}=\mathrm{vx} \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)
D.E.
\(\begin{aligned}
& \mathrm{x}^2\left(v+\mathrm{x} \frac{\mathrm{~d} v}{\mathrm{dx}}\right)+\mathrm{x}^2 v=\mathrm{x}^2\left(1+v^2\right) \\
& \Rightarrow v+x \frac{d v}{d x}+v=1+v^2 \\
& \Rightarrow x \frac{d v}{d x}=1+v^2-2 v \\
& \Rightarrow \int \frac{d v}{(v-1)^2}=\int \frac{d x}{x} \\
& \Rightarrow-\frac{1}{v-1}=\ln |x|+\mathrm{C}
\end{aligned}\)
\(\Rightarrow \frac{\mathrm{x}}{\mathrm{x}-\mathrm{y}}=\ln |x|+\mathrm{C}=\ln \mathrm{x}+\mathrm{C} \quad\left(\right.\) Since \(\left.\mathrm{x}>\frac{1}{\mathrm{e}}\right)\)
Given \(y(1)=0\)
\(\Rightarrow C=1\)
So \(\frac{x}{x-y}=\ln (e x)\)
Now \(y(e)=\frac{e}{2}\) and \(y\left(e^2\right)=\frac{2 e^2}{3}\)
\(\therefore \frac{2(y(e))^2}{y\left(e^2\right)}=\frac{2 \cdot \frac{e^2}{4}}{\frac{2 e^2}{3}}=\frac{3}{4}=0.75\)