Exercises 4.537.Clearly,r= 0 is not a root of the auxiliary equation,r3−2r2−r+ 2 = 0. (One can findthe roots, say, using the factorizationr3−2r2−r+ 2 = (r−2)(r−1)(r+ 1), but they arenot needed for the form of a particular solution: the only important thing is that they aredifferent from zero.) Therefore, a particular solution has the formyp(t) =A2t2+A1t+A0.Substitution into the original equation yieldsyp−2yp−yp+ 2yp=(0)−2(2A2)−(2A2t+A1) + 2(A2t2+A1t+A0)=2A2t2+ (A1−2A2)t+ (A0−A1−4A2) = 2t2+ 4t−9.Equating the coeﬃcients, we obtain2A2= 2,2A1−2A2= 4,2A0−A1−4A2=−9⇒A2= 1,A1= 3,A0=−1.Therefore,yp(t) =t2+ 3t−1.39.The auxiliary equation in this problem isr3+r2−2 = 0. By inspection, we see that
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