Answer
$y=e^{\frac{x^2}{2}-1}$
Work Step by Step
We are given that $ \dfrac{dy}{dx}=x(y+1)$
We will separate the variables to obtain:
$\dfrac{dy}{y+1}=x dx$
Integrate to obtain:
$\int \dfrac{dy}{y+1}=\int x \ dx$
This implies that $\ln |y+1|=\dfrac{x^2}{2}+C$
After applying the initial conditions, $y=0$ when $x=0$, we get $C=0$
Therefore, we have: $\ln |y+1|=\dfrac{x^2}{2}\implies e^{\ln |y+1|}=e^{\frac{x^2}{2}}$
or, $y=e^{\frac{x^2}{2}-1}$
