# 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:

$$y=c_1e^{4x}+c_2e^{4x}$$

$$y(x) =c_1e^{4x}+c_2 x e^{4x}$$