Answer to Question #104255 in Differential Equations for Ajay

Question

Using the method of undetermined coefficients, write the trial solution of the equation d^2y/dx^2+2dy/dx+5y=x e^(–1) cos2x and hence solve it.

Answer 1

$y''+2y'+5y=e^{-x}\cos 2x$

Characteristic equation: $\lambda^2+2\lambda+5=0$, which has roots $\lambda_1=-1-2i$ and $\lambda_2=-1+2i$

Trial solution is $y=Axe^{-x}\cos 2x+Bxe^{-x}\sin 2x$

$y'=Ae^{-x}\cos 2x-Axe^{-x}\cos 2x-2Axe^{-x}\sin 2x+$

$+Be^{-x}\sin 2x-Bxe^{-x}\sin 2x+2Bxe^{-x}\cos 2x$

$y''=Axe^{-x}\cos 2x-4Axe^{-x}\cos 2x-2Ae^{-x}\cos 2x-$

$-4Ae^{-x}\sin 2x+4Axe^{-x}\sin 2x+Bxe^{-x}\sin 2x-$

$-4Bxe^{-x}\sin 2x-2Be^{-x}\sin 2x+4Be^{-x}\cos 2x-$

$-4Bxe^{-x}\cos 2x$

$e^{-x}\cos 2x=y''+2y'+5y=4Be^{-x}\cos 2x-4Ae^{-x}\sin 2x$

$B=\frac{1}{4}, A=0$

Answer:

Trial solution is $\frac{1}{4}xe^{-x}\sin 2x$

General solution is $Ce^{-x}\cos 2x+De^{-x}\sin 2x+\frac{1}{4}xe^{-x}\sin 2x$