Exercises 6.4The auxiliary equation,r2(r2−2) = 0, has rootsr=±√2 and a double rootr= 0. Hence,yh(t)=c1+c2t+c3e√2t+c4e−√2tis a general solution to the homogeneous equation coresponding to (6.22). A particular solutionto (6.22) has the formyp(tAe3t. Substitution yieldsD2(D2−2)±Ae3x²=(D4−2D2Ae3x²=81Ae3x−(2)9Ae3x=63Ae3x=8e3x⇒yp(tAe3x=8e3x63,and soy(typ(t)+yh(t8e3x63+c1+c2t+c3e√2t+c4e−√2tis a general solution to (6.22). We Fndx(t) from the second equation in the original system.x(te3t+y(t)−y0(t)=e3t+³8e3x63+c1+c2t+c3e√2t+c4e−√2t´−³72e3x63+2c3e√2tc4e−√2t´=−e3x63+c1+c2t−c3e√2t−c4e−√2t.EXERCISES 6.4:Method of Variation of Parameters, page 3411.To apply the method of variation of parameters, Frst we have to Fnd a fundamental solutionset for the corresponding homogeneous equation, which isy0−3y0+4y
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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 University of California, Berkeley.