Let $$x = x(y)$$ be the solution of the differential equation $$2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$$, $$y > 1$$, $$x(e) = e$$. Then $$x(e^2)$$ is equal to :

The given differential equation is $$2y^{2}\frac{dx}{dy}-2xy+x^{2}=0,\; y\gt 1$$.

First isolate $$\dfrac{dx}{dy}$$:
$$2y^{2}\frac{dx}{dy}=2xy-x^{2}$$
$$\frac{dx}{dy}= \frac{2xy-x^{2}}{2y^{2}}=\frac{x}{y}-\frac{x^{2}}{2y^{2}}$$ $$-(1)$$

Equation $$(1)$$ contains the terms $$x$$ and $$x^{2}$$, so put $$u=\frac{1}{x}$$. Then $$u=\dfrac{1}{x}\;\Rightarrow\;\dfrac{du}{dy}=-\dfrac{1}{x^{2}}\dfrac{dx}{dy}$$.

Using $$(1)$$:
$$\frac{du}{dy}= -\frac{1}{x^{2}}\left(\frac{x}{y}-\frac{x^{2}}{2y^{2}}\right)= -\frac{1}{xy}+\frac{1}{2y^{2}}$$
But $$u=\dfrac{1}{x}$$, hence $$-\dfrac{1}{xy}=-\dfrac{u}{y}$$. Therefore
$$\frac{du}{dy}+\frac{u}{y}=\frac{1}{2y^{2}}$$ $$-(2)$$

Equation $$(2)$$ is linear in $$u(y)$$ with integrating factor $$\mu(y)=e^{\int (1/y)\,dy}=y$$.

Multiply by the integrating factor:
$$y\frac{du}{dy}+u=\frac{1}{2y}$$
Left side is the derivative of $$yu$$, so
$$\frac{d}{dy}(yu)=\frac{1}{2y}$$.

Integrate with respect to $$y$$:
$$yu=\int\frac{1}{2y}\,dy=\frac{1}{2}\ln y+C$$, where $$C$$ is the constant of integration.

Solve for $$u$$ and then for $$x$$:
$$u=\frac{1}{2y}\ln y+\frac{C}{y}$$
$$\frac{1}{x}=\frac{1}{2y}\ln y+\frac{C}{y}$$
$$x(y)=\frac{1}{\dfrac{1}{2y}\ln y+\dfrac{C}{y}}=\frac{y}{\dfrac{1}{2}\ln y+C}$$ $$-(3)$$

Use the initial condition $$x(e)=e$$:
Put $$y=e$$ in $$(3)$$:
$$e=\frac{e}{\dfrac{1}{2}\ln e+C}=\frac{e}{\dfrac{1}{2}(1)+C}$$
Hence $$\dfrac{1}{2}+C=1 \;\Rightarrow\; C=\frac{1}{2}$$.

Substitute the value of $$C$$ back into $$(3)$$:
$$x(y)=\frac{y}{\dfrac{1}{2}\ln y+\dfrac{1}{2}}=\frac{y}{\dfrac{1}{2}(\ln y+1)}=\frac{2y}{\ln y+1}$$ $$-(4)$$

Finally, evaluate at $$y=e^{2}$$:
For $$y=e^{2}$$, $$\ln y=\ln(e^{2})=2$$.
Using $$(4)$$:
$$x(e^{2})=\frac{2e^{2}}{2+1}=\frac{2e^{2}}{3}$$.

Hence $$x(e^{2})=\dfrac{2e^{2}}{3}$$.

Option B which is: $$\dfrac{2e^{2}}{3}$$