Answer
Given $$\frac{dy}{dx}= xe^y,\ \ \ \ y(0)= 0$$
Since
\begin{align*}
e^{-y}dy&=xdx\\
\int e^{-y}dy&=\int xdx\\
-e^{-y}&=\frac{1}{2}x^2+c
\end{align*}
At $x=0, \ y=0 $, then $c= -1$ and
$$-e^{-y} =\frac{1}{2}x^2 -1$$
Hence
$$y= -\ln \left(1-\frac{1}{2}x^2\right)$$
Work Step by Step
Given $$\frac{dy}{dx}= xe^y,\ \ \ \ y(0)= 0$$
Since
\begin{align*}
e^{-y}dy&=xdx\\
\int e^{-y}dy&=\int xdx\\
-e^{-y}&=\frac{1}{2}x^2+c
\end{align*}
At $x=0, \ y=0 $, then $c= -1$ and
$$-e^{-y} =\frac{1}{2}x^2 -1$$
Hence
$$y= \ln \left(1-\frac{1}{2}x^2\right)^{-1}$$
