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(B) Dividing the given differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$ by $x^2 y^2$,we get $\frac{2}{x^2} \frac{dx}{dy} - \frac{2}{xy} + \f

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$\frac{2}{3}e^2$

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(B) Dividing the given differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$ by $x^2 y^2$,we get $\frac{2}{x^2} \frac{dx}{dy} - \frac{2}{xy} + \frac{1}{y^2} = 0$.Let $u = \frac{1}{x}$,then $\frac{du}{dy} = -\frac{1}{x^2} \frac{dx}{dy}$.Substituting this into the equation,we get $-2 \frac{du}{dy} - \frac{2u}{y} + \frac{1}{y^2} = 0$,which simplifies to $\frac{du}{dy} + \frac{u}{y} = \frac{1}{2y^2}$.This is a linear differential equation of the form $\frac{du}{dy} + P(y)u = Q(y)$,where $P(y) = \frac{1}{y}$ and $Q(y) = \frac{1}{2y^2}$.The integrating factor $IF = e^{\int (1/y) dy} = e^{\log_e y} = y$.The general solution is $u \cdot IF = \int Q(y) \cdot IF \, dy + C$,so $uy = \int \frac{1}{2y^2} \cdot y \, dy + C = \int \frac{1}{2y} dy + C = \frac{1}{2} \log_e y + C$.Given $x(e) = e$,we have $u = 1/e$ at $y = e$. Substituting these values: $(1/e) \cdot e = \frac{1}{2} \log_e e + C \Rightarrow 1 = 1/2 + C \Rightarrow C = 1/2$.Thus,$u = \frac{\log_e y + 1}{2y}$. Since $u = 1/x$,we have $x = \frac{2y}{\log_e y + 1}$.For $y = e^2$,$x(e^2) = \frac{2e^2}{\log_e e^2 + 1} = \frac{2e^2}{2 + 1} = \frac{2}{3}e^2$.

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:

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