Chapter 5Thus, a general solution to the given system isu(t) =c1−12c2e−t+12et+53t,v(t) =c1+c2e−t+53t.9.Expressed in operator notation, this system becomes(D+ 2)[x] +D[y]=0,(D−1)[x] + (D−1)[y]=sint.In order to eliminate the functiony(t), we will apply the operator (D−1) to the first equationabove and the operator−Dto the second one. Thus, we have(D−1)(D+ 2)[x] + (D−1)D[y] = (D−1) = 0,−D(D−1)[x]−D(D−1)[y] =−D[sint] =−cost.Adding these two equations yields the differential equation involving the single functionx(t)given by(D2+D−2)−(D2−D)[x] =−cost⇒2(D−1)[x] =−cost.(5.3)This is a linear first order differential equation with constant coeﬃcients and so can be solvedby the methods of Chapter 2.(See Section 2.3.)
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