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(C) Given the differential equation: $2 x^{2} dy + (e^{y} - 2x) dx = 0$.Divide by $2 x^{2} dx$: $\frac{dy}{dx} + \frac{e^{y} - 2x}{2 x^{2}} = 0 \Right

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$\log_{e} 2$

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(C) Given the differential equation: $2 x^{2} dy + (e^{y} - 2x) dx = 0$.Divide by $2 x^{2} dx$: $\frac{dy}{dx} + \frac{e^{y} - 2x}{2 x^{2}} = 0 \Rightarrow \frac{dy}{dx} + \frac{e^{y}}{2 x^{2}} - \frac{1}{x} = 0$.Rearrange: $\frac{dy}{dx} - \frac{1}{x} = -\frac{e^{y}}{2 x^{2}} \Rightarrow e^{-y} \frac{dy}{dx} - \frac{e^{-y}}{x} = -\frac{1}{2 x^{2}}$.Let $z = e^{-y}$,then $\frac{dz}{dx} = -e^{-y} \frac{dy}{dx}$.Substituting this into the equation: $-\frac{dz}{dx} - \frac{z}{x} = -\frac{1}{2 x^{2}} \Rightarrow \frac{dz}{dx} + \frac{z}{x} = \frac{1}{2 x^{2}}$.This is a linear differential equation with integrating factor $IF = e^{\int \frac{1}{x} dx} = e^{\log_{e} x} = x$.The solution is $z \cdot x = \int x \cdot \frac{1}{2 x^{2}} dx + C = \int \frac{1}{2x} dx + C = \frac{1}{2} \log_{e} x + C$.Since $z = e^{-y}$,we have $x e^{-y} = \frac{1}{2} \log_{e} x + C$.Given $y(e) = 1$,substitute $x = e$ and $y = 1$: $e \cdot e^{-1} = \frac{1}{2} \log_{e} e + C \Rightarrow 1 = \frac{1}{2} + C \Rightarrow C = \frac{1}{2}$.Thus,$x e^{-y} = \frac{1}{2} \log_{e} x + \frac{1}{2} = \frac{1}{2} \log_{e} (ex)$.To find $y(1)$,substitute $x = 1$: $1 \cdot e^{-y(1)} = \frac{1}{2} \log_{e} (e \cdot 1) = \frac{1}{2} \log_{e} e = \frac{1}{2}$.$e^{-y(1)} = \frac{1}{2} \Rightarrow e^{y(1)} = 2 \Rightarrow y(1) = \log_{e} 2$.

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

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