Let y(x) be the solution of the differential equation x^2 (d y)/(d...

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