Answer
$ \ln |y|=\dfrac{e^{x^2}}{2} +c$
Work Step by Step
In order to to solve the given differential equation we will have to separate the variables and then integrate.
Here, we have
$y'=xye^{x^2}$ or, $\int \dfrac{dy}{y}=\int xe^{x^2} dx$
or,$\int \dfrac{dy}{y}=(\dfrac{1}{2})\int (2x) e^{x^2} dx$
Plug $x^2=p \implies 2x dx=dp$
Then, we have $ \ln |y|=\dfrac{e^{p}}{2} +c$
Hence, $ \ln |y|=\dfrac{e^{x^2}}{2} +c$
