Question

How to find the general solution to y"'+2y"-4y'-8y=0?

Answer 1

$y'''+2y''-4y'-8y=0$

has characteristic equation

$r^3+2r^2-4r-8=r^2(r+2)-4(r+2)=(r^2-4)(r+2)=(r-2)(r+2)^2=0$

which has roots at $r=\pm2$. The negative root has multiplicity 2. So the general solution is

$y=C_1e^{2x}+C_2e^{-2x}+C_3xe^{-2x}$