Solving ODE $yy'=2x^3$ with $y(1)=3$

Question

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?

Answer 1

$$yy'=2x^3$$ $$\frac{1}{2} (y^2)'=2x^3$$ Simply integrate $$\frac{1}{2} y^2=\frac 12x^4+K$$ $$ y^2=x^4+K$$ $$y(1)^2=x^4+K \implies K=9-1=8$$ $$ y^2=x^4+8$$

Answer 2

\begin{eqnarray*} y dy = 2 x^3 dx. \end{eqnarray*} \begin{eqnarray*} y^2 = x^4+K. \end{eqnarray*} $y(1)=3$ gives $K=8$.

Answer 3

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 $$ y=\sqrt {x^4+8}$$