Chapter 65.We can factor the auxiliary equation,r3+3r2+28r+ 26 = 0, as follows:r3r2r+26 = (r3+r2)+(2r2+2r)+(26r+ 26)=r2(r+1)+2r(r+ 1) + 26(r+1)=(r+1)(r2r+ 26) = 0.Thus eitherr+1=0⇒r=−1orr2r+26=0⇒r=−1±5i. Therefore, ageneral solution is given byy(x)=c1e−x+c2e−xcos 5x+c3e−xsin 5x.7.Factoring the characteristic polynomial yields2r3−r2−10r−7=(2r3r2)+(−3r2−3r−7r−7)=2r2(r+1)−3r(r−7(rr+ 1)(2r2−3r−7).Thus the roots of the characteristic equation, 2r3−r2−10r−7 = 0, arer⇒r=−1,2r2−3r−7=0⇒r=3±p32−4(2)(−7)4=3±√654,and a general solution isy(xc1e−x+c2e(3+√65)x/4+c3e(3−√65)x/4.9.In the characteristic equation,r3−9r2+27r−27 = 0, we recognize a complete cube, namely,(r−3)3= 0. Thus, it has just one root,r= 3, of multiplicity three. Therefore, a generalsolution to the given di±erential equation is given by
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.