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(t) =Ae3t. Substitution yieldsD2(D2−2)Ae3x=(D4−2D2)Ae3x= 81Ae3x−(2)9Ae3x= 63Ae3x= 8e3x⇒yp(t) =Ae3x=8e3x63,and soy(t) =yp(t) +yh(t) =8e3x63+c1+c2t+c3e√2t+c4e−√2tis a general solution to (6.22). We findx(t) from the second equation in the original system.x(t)=e3t+y(t)−y(t)=e3t+8e3x63+c1+c2t+c3e√2t+c4e−√2t−72e3x63+ 2c3e√2t+ 2c4e−√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, first we have to find a fundamental solutionset for the corresponding homogeneous equation, which is
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