Calculus 8th Edition

$\displaystyle y = -\ln(1-\frac{x^2}{2})$
---------------------------------------- $\displaystyle \frac{dy}{dx} = xe^{-y}$ --------------------------------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 $y$: $\displaystyle -e^{-y} = \frac{x^2}{2}-1$ $\displaystyle e^{-y} = 1-\frac{x^2}{2}$ $\displaystyle -y = \ln|1-\frac{x^2}{2}|$ $\displaystyle y = -\ln(1-\frac{x^2}{2}) = \ln(\frac{2}{2-x^2})$