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)`