Hi, the question is

$\displaystyle y'=\frac{y}{x}+2x^2$

and y(1)=0

I put it to this form

$\displaystyle y' -y(x^{-1})=2x^2$

so $\displaystyle f(x)=x^{-1}$ and $\displaystyle h(x)=2x^2$

is that right?

Integrating Factor $\displaystyle I(x)=exp(\int f(x)dx)$

$\displaystyle =exp(\int x^{-1}dx)$

which means I(x) = e

So i mulptiply both sides by e?

$\displaystyle \frac{d}{dx}ey=2ex^2$

and then i'm lost and confused, any help will be marvellous