Let us write the equation y′′′+y=cosx+sinx in the operator form: (D3+1)y=cosx+sinx.
The annihilator operator for the right part is D2+1. Then we get (D2+1)(D3+1)y=(D2+1)(cosx+sinx)=−cosx−sinx+cosx+sinx=0.
In the annihilator method for finding a particular solution the method of undetermined coefficients is used. Since i and −i are not roots of the characteristic equation k3+1=0 of the homogeneous differential equation y′′′+y=0, we conclude that the particular solution of the differentail equation y′′′+y=cosx+sinx is of the form yp=Acosx+Bsinx. Then yp′=−Asinx+Bcosx, yp′′=−Acosx−Bsinx, yp′′′=Asinx−Bcosx. Thus we get the equation Asinx−Bcosx+Acosx+Bsinx=cosx+sinx, which is equivalet to (A−B)cosx+(A+B)sinx=cosx+sinx. Therefore, A−B=1 and A+B=1, and hence A=1 and B=0.
We conclude that the particular solution is yp=cosx.
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