Solution of Differential Equation and Value Evaluation

To find the solution $x = x(y)$ of the given differential equation: $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0, \quad y > 1$

We can rewrite the equation by dividing all terms by $y^2$: $2\frac{dx}{dy} - \frac{2x}{y} + \frac{x^2}{y^2} = 0$

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

This is a homogeneous differential equation as well as a Bernoulli's differential equation. Let us solve it using the substitution method for a homogeneous equation.

Let $x = vy$, where $v$ is a function of $y$. Differentiating with respect to $y$, we get: $\frac{dx}{dy} = v + y\frac{dv}{dy}$

Substituting $x = vy$ and $\frac{dx}{dy} = v + y\frac{dv}{dy}$ into the differential equation: $v + y\frac{dv}{dy} - v = -\frac{v^2 y^2}{2y^2}$

$y\frac{dv}{dy} = -\frac{v^2}{2}$

Separating the variables $v$ and $y$: $\frac{dv}{v^2} = -\frac{dy}{2y}$

Integrating both sides: $\int v^{-2} \, dv = -\frac{1}{2} \int \frac{1}{y} \, dy$

$-\frac{1}{v} = -\frac{1}{2} \ln y + C_1$

$\frac{1}{v} = \frac{1}{2} \ln y + C$

Since $v = \frac{x}{y}$, we have $\frac{1}{v} = \frac{y}{x}$: $\frac{y}{x} = \frac{1}{2} \ln y + C$

We are given the initial condition $x(e) = e$, which means $x = e$ when $y = e$: $\frac{e}{e} = \frac{1}{2} \ln(e) + C$

$1 = \frac{1}{2}(1) + C \implies C = \frac{1}{2}$

Thus, the implicit solution is: $\frac{y}{x} = \frac{1}{2} \ln y + \frac{1}{2} = \frac{1}{2}(\ln y + 1)$

Now, to find $x(e^2)$, we substitute $y = e^2$ into the equation: $\frac{e^2}{x(e^2)} = \frac{1}{2} (\ln(e^2) + 1)$

$\frac{e^2}{x(e^2)} = \frac{1}{2} (2 + 1) = \frac{3}{2}$

$x(e^2) = \frac{2}{3} e^2$

Hence, the correct option is B.