$\text{Let}y\left(x\right)\text{be the solution of the differential equation}$
$x^2\frac{dy}{dx}+xy=x^2+y^2,x>\frac{1}{e},$
$\text{satisfying}y\left(1\right)=0.$

$\text{Then the value of}\mathrm{ 2}\frac{{y\left(e\right)}^2}{y\left(e^{-2}\right)}\text{is}\text{\_\_\_\_\_\_}$

The correct answer is 0.75.

Step-by-step Derivation:

We are given the first-order differential equation:
$x^2\frac{dy}{dx}+xy=x^2+y^2$
for $x>\frac{1}{e}$, with the initial condition $y\left(1\right)=0$.

First, we divide the entire differential equation by $x^2$ to write it in homogeneous form:
$\frac{dy}{dx}+\frac{y}{x}=1+{\left(\frac{y}{x}\right)}^2$

Let us use the substitution $y=vx$, which gives:
$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Substituting these into the differential equation:
$v+x\frac{dv}{dx}+v=1+v^2$
$x\frac{dv}{dx}=v^2-2v+1$
$x\frac{dv}{dx}={(v-1)}^2$

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

Integrating both sides:
$\int \frac{dv}{{(v-1)}^2}=\int \frac{dx}{x}$
$-\frac{1}{v-1}=\ln \left|x\right|+C$

Substituting back $v=\frac{y}{x}$:
$-\frac{1}{\frac{y}{x}-1}=\ln \left|x\right|+C$
$\frac{x}{x-y}=\ln \left|x\right|+C$

Applying the initial condition $y\left(1\right)=0$:
$\frac{1}{1-0}=\ln \left(1\right)+C\Rightarrow C=1$

For $x>\frac{1}{e}$, we obtain:
$\frac{x}{x-y}=\ln x+1$
$\frac{x-y}{x}=\frac{1}{\ln x+1}$
$1-\frac{y}{x}=\frac{1}{\ln x+1}$
$\frac{y}{x}=1-\frac{1}{\ln x+1}=\frac{\ln x}{\ln x+1}$
$y(x)=\frac{x\ln x}{\ln x+1}$

Note that the domain of the solution is restricted to $x>\frac{1}{e}$. Thus, $e^{-2}=\frac{1}{e^2}$ lies outside this domain. Evaluating the term in the denominator as $y\left(e^2\right)$ (which is the correct representation matching the domain constraint):

1. Find $y\left(e\right)$:
$y\left(e\right)=\frac{e\ln e}{}$
ln e + 1 = e 2

2. Find $y\left(e^2\right)$:
$y\left(e^2\right)=\frac{e^2\ln \left(e^2\right)}{\ln \left(e^2\right)+1}=\frac{2e^2}{2+1}=\frac{2e^2}{3}$

3. Calculate the value of the expression:
$2\cdot \frac{{\left[y\right(e\left)\right]}^2}{y\left(e^2\right)}=2\cdot \frac{{\left(\frac{e}{2}\right)}^2}{\frac{2e^2}{3}}=2\cdot \frac{e^2/4}{2e^2/3}=2\cdot \frac{3}{8}=\frac{3}{4}=0.75$