Let $$y(x)$$ be the solution of the differential equation

$$x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e},$$

satisfying $$y(1) = 0$$. Then the value of $$2\frac{(y(e))^2}{y(e^2)}$$ is ______.

The given differential equation is

$$x^{2}\,\frac{dy}{dx}+xy = x^{2}+y^{2}, \qquad x \gt \frac1e$$

Divide by $$x^{2}$$ to make the terms dimensionless:

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

The right-hand side contains the ratio $$y/x$$, so set

$$v=\frac{y}{x}\; \Longrightarrow\; y=v\,x$$

Differentiating, $$\dfrac{dy}{dx}=v+x\,\dfrac{dv}{dx}$$.

Substitute this in the differential equation:

$$v+x\,\frac{dv}{dx}=1+v^{2}-v$$

Simplify:

$$x\,\frac{dv}{dx}=v^{2}-2v+1=(v-1)^{2}$$

Separate the variables:

$$\frac{dv}{(v-1)^{2}}=\frac{dx}{x}$$

Integrate both sides:

$$\int\frac{dv}{(v-1)^{2}} = \int\frac{dx}{x}$$
$$-\frac1{v-1}= \ln|x| + C \qquad -(1)$$

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

At $$x=1$$, $$v=\dfrac{y}{x}=0$$. Substitute into $$(1)$$:

$$-\frac1{0-1}=1=\ln|1|+C \Longrightarrow C=1$$

Hence

$$-\frac1{v-1}=1+\ln x$$
$$\frac1{v-1}=-(1+\ln x)$$
$$v-1=-\frac1{1+\ln x}$$
$$v=1-\frac1{1+\ln x}$$

Return to $$y=v\,x$$:

$$y(x)=x\left(1-\frac1{1+\ln x}\right)$$

Now evaluate at the required points.

At $$x=e$$: $$\ln e=1 \;\Rightarrow\; 1+\ln e=2$$
$$y(e)=e\left(1-\frac1{2}\right)=\frac{e}{2}$$

At $$x=e^{2}$$: $$\ln e^{2}=2 \;\Rightarrow\; 1+\ln e^{2}=3$$
$$y(e^{2})=e^{2}\left(1-\frac1{3}\right)=e^{2}\cdot\frac23=\frac{2e^{2}}{3}$$

Finally compute the required expression:

$$2\,\frac{(y(e))^{2}}{y(e^{2})}=2\,\frac{\left(\dfrac{e}{2}\right)^{2}}{\dfrac{2e^{2}}{3}} =2\,\frac{\dfrac{e^{2}}{4}}{\dfrac{2e^{2}}{3}} =\frac{e^{2}}{2}\cdot\frac{3}{2e^{2}} =\frac34=0.75$$

Therefore the desired value is 0.75.