[tex]y'''+2y''-4y'-8y=0[/tex]
has characteristic equation
[tex]r^3+2r^2-4r-8=r^2(r+2)-4(r+2)=(r^2-4)(r+2)=(r-2)(r+2)^2=0[/tex]
which has roots at [tex]r=\pm2[/tex]. The negative root has multiplicity 2. So the general solution is
[tex]y=C_1e^{2x}+C_2e^{-2x}+C_3xe^{-2x}[/tex]
