Solving a second order homogenous ODE with double roots I'm working on solving the following homogenous equation:
$$ y'' - 8y' + 16y = 0 $$
Seems like a straight forward $y=e^{rx}$ substitution and then solve for r1 and r2:
$$ y=e^{rx}=0 $$
$$ y=re^{rx}=0 $$
$$ y=r^2e^{rx}=0 $$
$$ r^2e^{rx} - 8re^{rx} + 16e^{rx}=0 $$
$$ e^{rx}(r^2-8r+16)=0 $$
Since $e^{rx}$ can't equal zero:
$$ r^2-8r+16 = 0 $$
$$ (r-4)(r-4)=0 $$
$$ r_{1,2} = 4 $$
How do I express the generaal solution from here? I've tried the following but it is incorrect:
$$ y=c_1e^{4x}+c_2e^{4x} $$

Because of the repeated root, we have to express it as:
$$ y(x) =c_1e^{4x}+c_2 x e^{4x} $$
See Paul's Online Notes for theory as to why.
