1. Rewrite the differential equation

Given: $2x^2\,dy+(e^y-2x)\,dx=0, \qquad x>0$

Divide by $dx$: $2x^2\frac{dy}{dx}+e^y-2x=0$

So, $2x^2\frac{dy}{dx}=2x-e^y$ $\frac{dy}{dx}=\frac{1}{x}-\frac{e^y}{2x^2}$

2. Use a substitution

Let $u=e^{-y}$ Then $\frac{du}{dx}=-e^{-y}\frac{dy}{dx}=-u\frac{dy}{dx}$

From $\frac{dy}{dx}=\frac{1}{x}-\frac{e^y}{2x^2}$ we get $\frac{du}{dx}=-u\left(\frac{1}{x}-\frac{e^y}{2x^2}\right)$ Since $u=e^{-y}$, we have $ue^y=1$. Hence $\frac{du}{dx}=-\frac{u}{x}+\frac{1}{2x^2}$

Therefore the linear differential equation in $u$ is: $\frac{du}{dx}+\frac{1}{x}u=\frac{1}{2x^2}$

3. Solve the linear equation

Integrating factor: $\text{I.F.}=e^{\int \frac{1}{x}dx}=e^{\ln x}=x$ (since $x>0$)

Multiply throughout by $x$: $x\frac{du}{dx}+u=\frac{1}{2x}$ So, $\frac{d}{dx}(xu)=\frac{1}{2x}$

Integrate: $xu=\frac{1}{2}\ln x+C$

Substitute back $u=e^{-y}$: $xe^{-y}=\frac{1}{2}\ln x+C$

4. Use the condition $y(e)=1$

At $x=e$, $y=1$: $e\cdot e^{-1}=\frac{1}{2}\ln e+C$ $1=\frac{1}{2}+C$ $C=\frac{1}{2}$

Hence, $xe^{-y}=\frac{1}{2}(\ln x+1)$

5. Find $y(1)$

At $x=1$: $1\cdot e^{-y(1)}=\frac{1}{2}(\ln 1+1)=\frac{1}{2}$ So, $e^{-y(1)}=\frac{1}{2}$ $-y(1)=\ln\left(\frac{1}{2}\right)=-\ln 2$ $y(1)=\ln 2$

6. Match with options

$y(1)=\ln 2=\log_e 2$ So the correct option is C.