If y=y(x) solves e^y-x(dydx) (schematic: dy-dxe^y-x=xdx), with y(0)=0, then y(1) equals:

Question

If $y=y(x)$ solves $e^{y-x}\left(\dfrac{dy}{dx}ight)\ (	ext{schematic: }\dfrac{dy-dx}{e^{y-x}}=xdx)$, with $y(0)=0$, then $y(1)$ equals:

Accepted Answer

$-e^{x-y}=\dfrac{x^2}{2}+c$; using $(0,0)$: $c=-1$, so $e^{x-y}=\dfrac{2-x^2}{2}$, i.e. $y=x+\ln(x+1)$ (per source key). At $x=1$: $y=1+\ln2$.

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