Question #230037
Annihilator method obtain a particular solution y"'+y=cosx+sinx on R
1
Expert's answer
2021-08-30T17:21:32-0400

Let us write the equation y+y=cosx+sinxy'''+y=\cos x+\sin x in the operator form: (D3+1)y=cosx+sinx.(D^3+1)y=\cos x+\sin x.

The annihilator operator for the right part is D2+1.D^2+1. Then we get (D2+1)(D3+1)y=(D2+1)(cosx+sinx)=cosxsinx+cosx+sinx=0.(D^2+1)(D^3+1)y=(D^2+1)(\cos x+\sin x)=-\cos x-\sin x+\cos x+\sin x=0.

In the annihilator method for finding a particular solution the method of undetermined coefficients is used. Since ii and i-i are not roots of the characteristic equation k3+1=0k^3+1=0 of the homogeneous differential equation y+y=0,y'''+y=0, we conclude that the particular solution of the differentail equation y+y=cosx+sinxy'''+y=\cos x+\sin x is of the form yp=Acosx+Bsinx.y_p=A\cos x+B\sin x. Then yp=Asinx+Bcosx, yp=AcosxBsinx, yp=AsinxBcosx.y_p'=-A\sin x+B\cos x,\ y_p''=-A\cos x-B\sin x,\ y_p'''=A\sin x-B\cos x. Thus we get the equation AsinxBcosx+Acosx+Bsinx=cosx+sinx,A\sin x-B\cos x+A\cos x+B\sin x=\cos x+\sin x, which is equivalet to (AB)cosx+(A+B)sinx=cosx+sinx.(A-B)\cos x+(A+B)\sin x=\cos x+\sin x. Therefore, AB=1A-B=1 and A+B=1,A+B=1, and hence A=1A=1 and B=0.B=0.

We conclude that the particular solution is yp=cosx.y_p=\cos x.

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