Solve First Order Differential Equation and Evaluate Expression

To solve the given differential equation $x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}$

we can rewrite it by dividing the entire equation by $x^2$: $\frac{dy}{dx} + \frac{y}{x} = 1 + \left(\frac{y}{x}\right)^2$

This is a homogeneous first-order differential equation. Let $y = vx$, where $v$ is a function of $x$. Differentiating with respect to $x$, we get: $\frac{dy}{dx} = v + x \frac{dv}{dx}$

Substituting $y = vx$ and $\frac{dy}{dx}$ into the differential equation gives: $v + x \frac{dv}{dx} + v = 1 + v^2$ $x \frac{dv}{dx} = v^2 - 2v + 1$ $x \frac{dv}{dx} = (v - 1)^2$

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

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

Since $v = \frac{y}{x}$, this becomes: $-\frac{1}{\frac{y}{x} - 1} = \ln x + C$ $\frac{x}{x - y} = \ln x + C$

We are given the initial condition $y(1) = 0$. Substituting $x = 1$ and $y = 0$: $\frac{1}{1 - 0} = \ln 1 + C \implies C = 1$

Thus, the particular solution is: $\frac{x}{x - y} = 1 + \ln x$

Rearranging for $x - y$: $x - y = \frac{x}{1 + \ln x}$ $y(x) = x - \frac{x}{1 + \ln x} = \frac{x \ln x}{1 + \ln x}$

Now, we evaluate $y(e)$ and $y(e^2)$:

1. For $x = e$: $y(e) = \frac{e \ln e}{1 + \ln e} = \frac{e \cdot 1}{1 + 1} = \frac{e}{2}$

2. For $x = e^2$: $y(e^2) = \frac{e^2 \ln(e^2)}{1 + \ln(e^2)} = \frac{e^2 \cdot 2}{1 + 2} = \frac{2e^2}{3}$

Finally, we evaluate the required expression $2 \frac{(y(e))^2}{y(e^2)}$: $2 \frac{(y(e))^2}{y(e^2)} = 2 \cdot \frac{\left(\frac{e}{2}\right)^2}{\frac{2e^2}{3}} = 2 \cdot \frac{\frac{e^2}{4}}{\frac{2e^2}{3}} = 2 \cdot \left(\frac{e^2}{4} \times \frac{3}{2e^2}\right) = 2 \cdot \frac{3}{8} = \frac{3}{4} = 0.75$

The value of the expression is $0.75$.