Arriving at a particular solution of the ODEy″−2y′−2y=sin⁡x

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Skyler Newman · Accepted Answer

Step 1In principle your approach is correct, but the parametrization is wrong, especially where you use p(−1). The operator polynomial p(D) needs to be split into even and odd parts asp(D)=pe(D2)+Dpo(D2)to isolate the squares of D so that then you can transform under (D2+1)sin⁡(x)=0 as1p(D)=1pe(−1)+Dpo(−1)=pe(−1)−Dpo(−1)pe(−1)2−D2po(−1)2=pe(−1)−Dpo(−1)pe(−1)2+po(−1)2Forp(x)=x2−2x−2 one has pe(u)=u−2 and po(u)=−2, so that the last fraction becomes−3+2D(−3)2+(−2)2=2D−313.

Ayla Ward · Accepted Answer

Step 1With the help of the Laplace transform (an operator procedure) we can obtain easily a particular solution as follows. After transforming we have according to your calculationsY(s)=1(s+a)(s+b)1s2+1=c1s+a+c2s+b+c3s+c4s2+1=Yh(s)+Yp(s)here Yh(s) is the homogeneous component Laplace transform because contains the operator roots, and Yp(s) is a particular solution where{Yh(s)=c1s+a+c2s+bYp(s)=c3s+c4s2+1then inverting Yp(s) we haveyp(x)=c3cos⁡x+c4sin⁡xnow the values for c3 and c4 are obtained by substitution into the complete ODE.

Arriving at a particular solution of the ODE

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