finding the solution of a separable differential equation at a certain point

Question

problem:

$y' = x$ exp $(y-x^2)$: when $x = 0, y = 0$

at first I wasn't really sure what exp meant but I assumed that it meant $e^{f(x)}$. so my solution goes like this:

$\frac{dy}{dx}=xe^{y-x^2}$

$dy=xe^{y-x^2}dx$

$ln(dy) = ln(xdx) +y-x^2$

$ln(dy)-y = ln(xdx) - x^2$

applying $e$ to both sides : $dy -e^y = xdx - e^{x^2}$

then I get stuck; but to be honest I don't really know if I started this right so sorry if your eyes hurt from seeing this.

Answer 1

The problem is that $\ln(dy)$ as no clear signification here.

Try to write $x e^{y-x^2}=x e^{-x^2} e^y$ and: $$e^{-y} dy = x e^{-x^2} dx$$ and then integrate using the initial conditions.