Exercises 4.2(d)By the defnition oF cosht,y3(t)=cosht=et+e−t2=12et+12e−t=12y1(t)+12y2(t),and given Functions are linearly dependent on (−∞,∞).41.The auxiliary equation For this problem isr3+r2−6r+ 4 = 0. ±actoring yieldsr3+r2−6r+4 =(r3−r2)+(2r2−2r)+(−4r+4)=r2(r−1) + 2r(r−1)−4(r−1) = (r−1)(r2+2r−4).Thus the roots oF the auxiliary equation arer=1 andr=−2±p(−2)2−4(1)(−4)2=−1±√5.ThereFore, the Functionset,e(−1−√5)t,ande(−1+√5)tare solutions to the given equation, andthey are linearly independent on (−∞,∞) (see Problem 40). Hence, a general solution toy0+y0−6y0+4y=0isgivenbyy(t)=c1et+c2e(−1−√5)t+c3e(−1+√5)t.42.The auxiliary equation associated with this di²erential equation isr3−6
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