The general Solution of $\frac{{dy}}{{dx}} = 2x{e^{{x^2} - y}}$ is

Given that, $\frac{{dy}}{{dx}} = 2x{e^{{x^2} - y}} = 2x{e^{{x^2}}} \cdot {e^{ - y}}$

$\Rightarrow$ ${e^y}\frac{{dy}}{{dx}} = 2x{e^{{x^2}}}$

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

On integrating both sides,

we get
$\int {{e^y}} dy = 2\int x {e^{{x^2}}}dx$

Put ${x^2} = t$ in RHS integral,

we get
$2xdx = dt$

$\int {{e^y}} dy = \int {{e^t}} dt$

$\Rightarrow$ ${e^y} = {e^t} + C$

$\Rightarrow$ ${e^y} = {e^{{x^2}}} + C$