Exercises 4.3So,α=−1,β= 2, and a general solution is given byy(t) =c1e−tcos 2t+c2e−tsin 2t.15.First, we find the roots of the auxiliary equation.r2+ 10r+ 41 = 0⇒r=−10±102−4(1)(41)2=−5±4i.These are complex numbers withα=−5 andβ= 4. Hence, a general solution to the givendifferential equation isy(t) =c1e−5tcos 4t+c2e−5tsin 4t.17.The auxiliary equation in this problem,r2−r+ 7 = 0, has the rootsr=1±12−4(1)(7)2=1±√−272=12±3√32i.Therefore, a general solution isy(t) =c1et/2cos3√32t+c2et/2sin3√32t.19.The auxiliary equation,r3+r2+3r−5 = 0, is a cubic equation. Since any cubic equation hasa real root, first we examine the divisors of the free coeﬃcient, 5, to find integer real roots (if
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