Question

Find the general solution to y'' − y' − 2y = 2e^3x

Answer 1

$k^2-k-2=0$

$k=\frac{1\pm\sqrt{1+8}}{2}$

$k_1=-1,\ k_2=2$

$Y=c_1e^{-x}+c_2e^{2x}$

$\tilde{y}=Ae^{3x}$

$\tilde{y}'=3Ae^{3x}$

$\tilde{y}''=9Ae^{3x}$

$9Ae^{3x}-3Ae^{3x}-2Ae^{3x}=2e^{3x}$

$9A-3A-2A=2$

$A=1/2$

$\tilde{y}=e^{3x}/2$

$y=Y+\tilde{y}=c_1e^{-x}+c_2e^{2x}+e^{3x}/2$