Problem 07 | Separation of Variables

$y' = x \exp (y - x^2)$

$\dfrac{dy}{dx} = x e^{y - x^2}$

$\dfrac{dy}{dx} = x e^y e^{-x^2}$

$\dfrac{dy}{e^y} = x e^{-x^2} \, dx$

$e^{-y} \, dy = e^{-x^2}(x \, dx)$

$\displaystyle -\int e^{-y} \, (-dy) = -\frac{1}{2} \int e^{-x^2}(-2x \, dx)$

$-e^{-y} = -\frac{1}{2} e^{-x^2} - c$

$e^{-y} = \frac{1}{2} e^{-x^2} + c$
 

when x = 0, y = 0
$e^{-0} = \frac{1}{2} e^{-0^2} + c$

$1 = \frac{1}{2} + c$

$c = \frac{1}{2}$
 

thus,
$e^{-y} = \frac{1}{2} e^{-x^2} + \frac{1}{2}$

$e^{-y} = \frac{1}{2} e^{-x^2} + \frac{1}{2}$

$\dfrac{1}{e^y} = \dfrac{e^{-x^2} + 1}{2}$

$\dfrac{2}{e^{-x^2} + 1} = e^y$

$\ln \dfrac{2}{e^{-x^2} + 1} = \ln e^y$

$\ln \dfrac{2}{e^{-x^2} + 1} = y \ln e$

$y = \ln \dfrac{2}{e^{-x^2} + 1}$

$y = \ln \dfrac{2}{1 + \exp (-x^2)}$           answer