Exercises 5.2Therefore, let’s try the substitutionz(t)=x(t)+y(t). We want a functionz(t) that satisFesthe two equationsz0(t)−z(t−3e−2tandz0(t)+2z(t)=3et,(5.9)simultaneously. We start by solving the Frst equation given in (5.9). This is a linear di±erentialequation with constant coeﬃcients which has the associated auxiliary equationr−1=0.Hence, the solution to the corresponding homogeneous equation iszh(tCet.By the method of undetermined coeﬃcients, we see that a particular solution will have theformzp(tAe−2t⇒z0p=−2Ae−2t.Substituting these expressions into the Frst di±erential equation given in (5.9) yieldsz0p(t)−zp(t−2Ae−2t−Ae−2t=−3Ae−2t=−3e−2t⇒A=1.Thus, the Frst equation given in (5.9) has the general solution
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