y1′=−3y1+y2+u(t−1)et,y2′=−4y1+2y2+u(t−1)et,y1(0)=0,y2(0)=21Enclose arguments of functions, numerators, and denominators in parentheses.

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Margot Mill · Accepted Answer

Given: y1′=−3y1+y2+u(t−1)et,y2′=−4y1+2y2+u(t−1)et,y1(0)=0,y2(0)=21Apply Laplace transform:sY1(s)−y1(0)=−3Y1(s)+Y2(s)+e−(s−1)s−1(s+3)Y1(s)−Y2(s)=e−(s−1)s−1sY2(s)−y2(0)=−4Y1(s)+2Y2(s)+e−(s−1)s−1sY2(s)−21=−4Y1(s)+2Y2(s)+e−(s−1)s−1(s+4)Y1(s)−2Y2(s)=e−(s−1)s−1+21Solve the equations:(s+3)Y1(s)−Y2(s)=e−(s−1)s−1⋅(−2)(s+4)Y1(s)−2Y2(s)=e−(s−1)s−1+21[−2(s+3)+(s+4)]Y1(s)=−e−(s−1)s−1+21Y1(s)=e−(s−1)(s−1)(s+2)−21(s+2)That is, Y1(s)=e−(s−1)s−1−e3(e−(s+2)s+2)−21(s+2)Apply inverse Laplace transform:y1(t)=[u(t−1)]∗et−[u(t+2)]∗et+3−21∗e−2tSolution:(s+3)(e−(s−1)s−1−e3(e−(s+2)s+2)−21s+2)−Y2(s)=e−(s−1)s−1Y2(s)=(s+3)e−(s−1)(s−1)−e3(s+3)(e−(s+2)s+2)−(s+3)21(s+2)−e−(s−1)s−1

y_1'=-3y_1+y_2+u(t-1)e^t, y_2'=-4y_1+2y_2+u(t−1)e^t, y_1(0)=0, y_2(0)=21

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