Question #302582

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


1
Expert's answer
2022-03-02T11:09:52-0500

Corresponding homogeneous differential equation


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

Characteristic (auxiliary) equation


r2+2r+2=0r^2+2r+2=0

r1=1i,r2=1+ir_1=-1-i,r_2=-1+i

The general solution of the homogeneous differential equation is


yh=c1excosx+c2exsinxy_h=c_1e^{-x}\cos x+c_2e^{-x}\sin x

Find the particular solution of the non homogeneous differential equation


yp=Aexcos2x+Bexsin2xy_p=Ae^x\cos 2x+Be^x\sin 2x

yp=Aexcos2x2Aexsin2xy_p'=Ae^x\cos 2x-2Ae^x\sin 2x

+Bexsin2x+2Bexcos2x+Be^x\sin 2x+2Be^x\cos 2x

yp=Aexcos2x2Aexsin2xy_p''=Ae^x\cos 2x-2Ae^x\sin 2x


2Aexsin2x4Aexcos2x-2Ae^x\sin 2x-4Ae^x\cos 2x

+Bexsin2x+2Bexcos2x+Be^x\sin 2x+2Be^x\cos 2x

+2Bexcos2x4Bexsin2x+2Be^x\cos 2x-4Be^x\sin 2x

Substitute


3Aexcos2x4Aexsin2x3Bexsin2x-3Ae^x\cos 2x-4Ae^x\sin 2x-3Be^x\sin 2x

+4Bexcos2x+2Aexcos2x4Aexsin2x+4Be^x\cos 2x+2Ae^x\cos 2x-4Ae^x\sin 2x

+2Bexsin2x+4Bexcos2x+2Aexcos2x+2Be^x\sin 2x+4Be^x\cos 2x+2Ae^x\cos 2x

+2Bexsin2x=excos2x+2Be^x\sin 2x=e^x\cos 2x

A+8B=1A+8B=1

8A+B=0-8A+B=0

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


The general solution of the non homogeneous differential equation is


y=c1excosx+c2exsinxy=c_1e^{-x}\cos x+c_2e^{-x}\sin x+165excos2x+865exsin2x+\dfrac{1}{65}e^x\cos 2x+\dfrac{8}{65}e^x\sin 2x


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