Exercises 4.4To chooses, we find the roots of the characteristic equation, which isr2+ 9 = 0.Sincer=±3iare its simple roots, we takes= 1. Thusyp(t) =t(A3t3+A2t2+A1t+A0)cos 3t+t(B3t3+B2t2+B1t+B0)sin 3t.29.The characteristic equationr2−6r+ 9 = (r−3)2= 0 has a double rootr= 3. Therefore, aparticular solution is of the formyp(t) =t2(A6t6+A5t5+A4t4+A3t3+A2t2+A1t+A0)e3t.31.From the form of the right-hand side, we conclude that a particular solution should be of theformyp(t) =ts(A3t3+A2t2+A1t+A0)cost+(B3t3+B2t2+B1t+B0)sint e−t.Sincer=−1±iare simple roots of the characteristic equation,r2+ 2r+ 2 = 0, we shouldtakes= 1. Therefore,yp(t) =t(A3t3+A2t2+A1t+A0)cost+(B3t3+B2t2+B1t+B0)sint e−t.33.The right-hand side of the equation suggests thatyp(t) =ts(Acost+Bsint). By inspection,we see thatr=
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