how do you solve $y''+2y'-3y=0$? I want to solve this equation:
$y''+2y'-3y=0$
I did this:
$y' = z$ 
$y'' = z\dfrac{dz}{dy}$
$z\dfrac{dz}{dy}+2z-3y=0$
$zdz+2zdy-3ydy=0$
$zdz=(3y-2z)dy$
$z=3y-2z$
$z=y$
$y=y'=y''$
???
now, I'm pretty sure I did something wrong. could you please correct.

You can also write it in matrix form:
$u=(y',y)$, $u'=\big(\matrix{-2 & 3 \\ \hphantom- 1 & 0}\big) u$. Find the eigenvalues and eigenvectors, turn it into a diagonal system whose solution is simple. Go back to the original coordinates.
