Question

Given that p(x) = −2 is a particular solution to y'' − 3y' − 4y = 8, write the general solution and verify that the general solution satisfies the equation. 

Answer 1

The characteristic equation $k^2-3k-4=0$ of the homogeneous differential equation $y'' − 3y' − 4y=0$ is equivalent to $(k+1)(k-4)=0,$ and hence has the roots $k_1=-1,\ k_2=4.$ Since $p(x) = −2$ is a particular solution, we conclude that the general solution of

$y'' − 3y' − 4y=8$ is $y=C_1e^{-x}+C_2e^{4x}-2.$

Let us verify that the general solution satisfies the equation. It follows that $y'=-C_1e^{-x}+4C_2e^{4x},\ y''=C_1e^{-x}+16C_2e^{4x}.$ Therefore,

$C_1e^{-x}+16C_2e^{4x}− 3(-C_1e^{-x}+4C_2e^{4x})− 4(C_1e^{-x}+C_2e^{4x}-2)\\ =C_1e^{-x}+16C_2e^{4x}+3C_1e^{-x}-12C_2e^{4x}− 4C_1e^{-x}-4C_2e^{4x}+8\\ =8.$

Consequently, $y=C_1e^{-x}+C_2e^{4x}-2$ indeed is the general solution of the differential equation

$y'' − 3y' − 4y=8.$