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(D) The given differential equation is $2 y e^{x / y^{2}} d x+\left(y^{2}-4 x e^{x / y^{2}}\right) d y=0$.Rearranging the terms,we get $2 e^{x / y^{2}

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$-e^{2} \log _{e}(2)$

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(D) The given differential equation is $2 y e^{x / y^{2}} d x+\left(y^{2}-4 x e^{x / y^{2}}\right) d y=0$.Rearranging the terms,we get $2 e^{x / y^{2}}[y d x-2 x d y]+y^{2} d y=0$.Dividing by $y^{3}$,we obtain $2 e^{x / y^{2}}\left[\frac{y^{2} d x-2 x y d y}{y^{4}}\right]+\frac{1}{y} d y=0$.This can be written as $2 e^{x / y^{2}} d\left(\frac{x}{y^{2}}\right)+\frac{1}{y} d y=0$.Integrating both sides,we get $\int 2 e^{x / y^{2}} d\left(\frac{x}{y^{2}}\right)+\int \frac{1}{y} d y=C$.This yields $2 e^{x / y^{2}}+\ln |y|=C$.Given $x(1)=0$,substituting $x=0$ and $y=1$ gives $2 e^{0}+\ln(1)=C$,so $C=2$.The equation of the curve is $2 e^{x / y^{2}}+\ln |y|=2$.To find $x(e)$,substitute $y=e$ into the equation: $2 e^{x / e^{2}}+\ln(e)=2$.$2 e^{x / e^{2}}+1=2 \Rightarrow 2 e^{x / e^{2}}=1$.$e^{x / e^{2}}=\frac{1}{2} \Rightarrow \frac{x}{e^{2}}=\ln\left(\frac{1}{2}\right) = -\ln(2)$.Thus,$x(e)=-e^{2} \log _{e}(2)$.

Let $x=x(y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x+\left(y^{2}-4 x e^{x / y^{2}}\right) d y=0$ such that $x(1)=0$. Then,$x(e)$ is equal to

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