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.

Start by assuming a solution of the form
$y(t)=e^{ct}$
Hence $y'(t)=ce^{ct}$ and $y''(t)=c^2e^{ct}$
Substitute into the original equation. You will find a quadratic for $c$. Solve it, and note that both solutions work, and as the ODE is linear and homogeneous, the sum of solutions also works. 
