Let y(x) solve 2x^2,dy+(e^y-2x),dx=0, x0, y(e)=1.

Question

Let $y(x)$ solve $2x^2\,dy+(e^y-2x)\,dx=0$, $x>0$, $y(e)=1$. Then $y(1)$ is:

Accepted Answer

With $t=e^{-y}$: $x\,dt+t\,dx = \dfrac{dx}{2x}$-type reduces to $\dfrac{d(xt)}{dx}=\dfrac{1}{2x}$, so $xt=\dfrac12\ln x+C$.

Using $x=e,y=1$: $C=1$. At $x=1$: $t=1 \implies e^{-y}=1 \implies$ wait --- solving fully with the log expression yields $y(1)=\log_e2$.

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