Solving ODE $yy'=2x^3$ with $y(1)=3$ I'm stuck solving $yy'=2x^3$ with $y(1)=3$
I know that I'm looking to get the equation into the form of: $y' + P(x)y = Q(x)$ and then find the integrating factor $e^{\int(P(x)dx}$, but then how do I find my $P(x)$ in this case? And what do I do after with the initial condition?

The differential equation 
$$yy'=2x^3$$ is separable. 
Integrate both sides and you get $$ \frac {y^2}{2} = \frac {x^4}{2}+c $$
Solve for $y,$ and with initial value $ y(1)=3,$ we get $$ y=\sqrt {x^4+8}$$
