Answer
$\displaystyle y = -\ln(1-\frac{x^2}{2})$
Work Step by Step
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$\displaystyle \frac{dy}{dx} = xe^{-y}$
--------------------------------Separation of Variables:
$\displaystyle \int e^{-y}dy=\int xdx$
$\displaystyle -e^{-y} = \frac{x^2}{2}+C$
--------------------------------Find $C$, the constant of integration
$y(0) = 0$
$\displaystyle -e^0 = \frac{0^2}{2}+C$
$-1 = C$
--------------------------------Plug $C$ back in and isolate $y$:
$\displaystyle -e^{-y} = \frac{x^2}{2}-1$
$\displaystyle e^{-y} = 1-\frac{x^2}{2}$
$\displaystyle -y = \ln|1-\frac{x^2}{2}|$
$\displaystyle y = -\ln(1-\frac{x^2}{2}) = \ln(\frac{2}{2-x^2})$
