Solving initial value problem Im solving initial value problem
$$ \frac {dy}{dx} + xy = xy^2; y(0)=3$$
After applying Bernoulli's equation method i obtained
$$ \frac {dv}{dx} -xv = -x$$
So, $$ p(x) = -x, q(x) = -x $$
For finding integrating factor 
$$u(x)=e^ {-\int xdx}=e^ {-\frac {x^2}{2}}$$
$$ y= \frac {\int u(x)q(x)dx+C}{u(x)}={\int u(x)q(x)dx}$$
So,
$$-\int xe^ {-\frac {x^2}{2}}dx $$
Please help further or guide me if i did something wrong.

This seems fine. the only point is $v=\frac1y$, so $y=\frac1v$ and 
$$u(x)v(x)-1\cdot\frac13=u(x)v(x)-u(0)v(0)=\int_0^x (u(t)v(t))'=\int_0^x u(t)q(t)dt=-\int_0^x te^{-\frac{t^2}{2}}dt$$
So 
$$v(x)=e^{\frac{x^2}{2}}\left(\frac13-\int_0^x te^{-\frac{t^2}{2}}dt\right),\hspace{10pt} y(x)=\frac{1}{v(x)}$$
