Question #294208

Solve the differential equation (D^3+D^2+D+1)y=Sinx

1
Expert's answer
2022-02-08T12:08:13-0500

Corresponding homogeneous equation

(D3+D2+D+1)y=0(D^3+D^2+D+1)y=0

Characteristic (auxiliary)equation


r3+r2+r+1=0r^3+r^2+r+1=0

r2(r+1)+(r+1)=0r^2(r+1)+(r+1)=0

(r+1)(r2+1)=0(r+1)(r^2+1)=0

r1=1,r2=i,r3=ir_1=-1, r_2=i, r_3=-i

The general solution of the homogeneous differential equation is


yh=c1ex+c2sinx+c3cosxy_h=c_1e^x+c_2\sin x+c_3\cos x

The particular solution of the non homogeneous equation is


yp=Axsinx+Bxcosxy_p=Ax\sin x+Bx\cos x

yp=Asinx+Axcosx+BcosxBxsinxy_p'=A\sin x+Ax\cos x+B\cos x-Bx\sin x

yp=2AcosxAxsinx2BsinxBxcosxy_p''=2A\cos x-Ax\sin x-2B\sin x-Bx\cos x

yp=3AsinxAxcosx3Bcosx+Bxsinxy_p'''=-3A\sin x-Ax\cos x-3B\cos x+Bx\sin x

Substitute


3AsinxAxcosx3Bcosx+Bxsinx-3A\sin x-Ax\cos x-3B\cos x+Bx\sin x

+2AcosxAxsinx2BsinxBxcosx+2A\cos x-Ax\sin x-2B\sin x-Bx\cos x

+Asinx+Axcosx+BcosxBxsinx+A\sin x+Ax\cos x+B\cos x-Bx\sin x

+Axsinx+Bxcosx=sinx+Ax\sin x+Bx\cos x=\sin x

xsinx:BAB+A=0x\sin x:B-A-B+A=0

xcosx:AB+A+B=0x\cos x:-A-B+A+B=0

sinx:3A2B+A=1\sin x:-3A-2B+A=1


cosx:3B+2A+B=0\cos x:-3B+2A+B=0


A=B=14A=B=-\dfrac{1}{4}

The general solution of the given non homogeneous differential equation is


y=c1ex+c2sinx+c3cosxx4sinxx4cosxy=c_1e^x+c_2\sin x+c_3\cos x-\dfrac{x}{4}\sin x-\dfrac{x}{4}\cos x




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