dy/dx - 2y = 4x^2
with y(0) = -1
any help would be appreciated
This is first order linear, so it is solved using the Integrating Factor.
The integrating factor is
$\displaystyle e^{\int{-2\,dx}} = e^{-2x}$.
Multiply both sides by the integrating factor:
$\displaystyle e^{-2x}\frac{dy}{dx} - 2e^{-2x}y = 4x^2e^{-2x}$
$\displaystyle \frac{d}{dx}\left(e^{-2x}y\right) = 4x^2e^{-2x}$
$\displaystyle e^{-2x}y = \int{4x^2e^{-2x}\,dx}$
Now evaluate the right hand side using integration by parts (twice).