Chapter 4Therefore, the auxiliary equation has a double root−2 and a root 2. The functionse−2t,te−2t,ande2tform a linearly independent solution set. Therefore, a general solution in thisproblem isz(t)=c1e−2t+c2te−2t+c3e2t.45.By inspection, we see thatr= 2 is a root of the auxiliary equation,r3+3r2−4r−12 = 0.Dividing the polynomialr3r2−4r−12 byr−2 yieldsr3r2−4r−12 = (r−2)(r2+5r+6)=(r−2)(r+2)(r+3).Hence, two other roots of the auxiliary equation arer=−2andr=−3. The functionse−3t,e−2te2tare three linearly independent solutions to the given equation, and a generalsolution is given byy(tc1e−3t+c2e−2t+c3e2t.47.First we ±nd a general solution to the equationy0−y0= 0. Its characteristic equation,r3−r= 0, has roots
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