Answer to Question #302582 in Differential Equations for haru

Question

Find the general solution of the following differential equations using method of undetermined coefficients:, (v) y''+2y'+2y =e^x cos2x.

Answer 1

Corresponding homogeneous differential equation

$$y''+2y'+2y =0$$

Characteristic (auxiliary) equation

$$r^2+2r+2=0$$

$$r_1=-1-i,r_2=-1+i$$

The general solution of the homogeneous differential equation is

$$y_h=c_1e^{-x}\cos x+c_2e^{-x}\sin x$$

Find the particular solution of the non homogeneous differential equation

$$y_p=Ae^x\cos 2x+Be^x\sin 2x$$

$$y_p'=Ae^x\cos 2x-2Ae^x\sin 2x$$

$$+Be^x\sin 2x+2Be^x\cos 2x$$

$$y_p''=Ae^x\cos 2x-2Ae^x\sin 2x$$

$$-2Ae^x\sin 2x-4Ae^x\cos 2x$$

$$+Be^x\sin 2x+2Be^x\cos 2x$$

$$+2Be^x\cos 2x-4Be^x\sin 2x$$

Substitute

$$-3Ae^x\cos 2x-4Ae^x\sin 2x-3Be^x\sin 2x$$

$$+4Be^x\cos 2x+2Ae^x\cos 2x-4Ae^x\sin 2x$$

$$+2Be^x\sin 2x+4Be^x\cos 2x+2Ae^x\cos 2x$$

$$+2Be^x\sin 2x=e^x\cos 2x$$

$$A+8B=1$$

$$-8A+B=0$$

$$A=1/65, B=8/65$$

The general solution of the non homogeneous differential equation is

$$y=c_1e^{-x}\cos x+c_2e^{-x}\sin x$$ $$+\dfrac{1}{65}e^x\cos 2x+\dfrac{8}{65}e^x\sin 2x$$