Answer to Question #302578 in Differential Equations for haru

Question

Find the general solution of the following differential equations using method of undetermined coefficients: (iii) y''− 2y'+y = xsinx.

Answer 1

Corresponding homogeneous differential equation

$y''−2y'+y =0$

Characteristic (auxiliary) equation

$r^2-2r+1=0$

$r_1=r_2=1$

The general solution of the homogeneous differential equation is

$y_h=c_1e^x+c_2xe^x$

Find the particular solution of the non homogeneous differential equation

$y_p=Ax\sin x+Bx\cos x+C\sin x+D\cos x$

$y_p'=A\sin x+Ax\cos x+B\cos x-Bx\sin x$

$+C\cos x-D\sin x$

$y_p''=2A\cos x-Ax\sin x-2B\sin x$

$-Bx\cos x-C\sin x-D\cos x$

Substitute

$2A\cos x-Ax\sin x-2B\sin x$

$-Bx\cos x-C\sin x-D\cos x$

$-2A\sin x-2Ax\cos x-2B\cos x+2Bx\sin x$

$-2C\cos x+2D\sin x+Ax\sin x+Bx\cos x$

$+C\sin x+D\cos x=x\sin x$

$2B=1$

$-2A=0$

$-2B-C-2A+2D+C=0$

$2A-D-2B-2C+D=0$

$A=0, B=1/2, C=-1/2, D=1/2$

$y_p=\dfrac{x}{2}\cos x-\dfrac{1}{2}\sin x+\dfrac{1}{2}\cos x$

The general solution of the non homogeneous differential equation is

$y=c_1e^x+c_2xe^x+\dfrac{x}{2}\cos x-\dfrac{1}{2}\sin x+\dfrac{1}{2}\cos x$