Answer to Question #222722 in Differential Equations for enrik

Question

y''-2y'-3y=2e^-x-10sinx

Answer 1

Let us solve the differential equation $y''-2y'-3y=2e^{-x}-10\sin x.$ The characteristic equation $k^2-2k-3=0$ is equivalent to $(k+1)(k-3)=0,$ and hence has the roots $k_1=-1,k_2=3.$

The general solution of the differential equation is $y=C_1e^{-x}+C_2e^{3x}+y_p,$ where the particular solution is $y_p=axe^{-x}+b\cos x+c\sin x.$ Then $y'_p=ae^{-x}-axe^{-x}-b\sin x+c\cos x, y''_p=-ae^{-x}-ae^{-x}+axe^{-x}-b\cos x-c\sin x.$

It follows that

$-2ae^{-x}+axe^{-x}-b\cos x-c\sin x-2(ae^{-x}-axe^{-x}-b\sin x+c\cos x)-3(axe^{-x}+b\cos x+c\sin x)=2e^{-x}-10\sin x.$

Then

$-4ae^{-x}+(-4b-2c)\cos x+(2b-4c)\sin x=2e^{-x}-10\sin x.$

We conclude that $-4a=2,-4b-2c=0, 2b-4c=-10.$ It follows that $a=-\frac{1}{2},c=-2b, 10b=-10.$ Therefore, $a=-\frac{1}{2}, b=-1,c=2.$

We conclude that the general solutiion of the differential equation is

$y=C_1e^{-x}+C_2e^{3x}-\frac{1}{2}xe^{-x}-\cos x+2\sin x.$