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

vazelinahS 2020-11-01 Answered

y1=3y1+y2+u(t1)et,y2=4y1+2y2+u(t1)et,y1(0)=0,y2(0)=21
Enclose arguments of functions, numerators, and denominators in parentheses.

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Expert Answer

Margot Mill
Answered 2020-11-02 Author has 106 answers
Given: y1=3y1+y2+u(t1)et,y2=4y1+2y2+u(t1)et,y1(0)=0,y2(0)=21
Apply Laplace transform:
sY1(s)y1(0)=3Y1(s)+Y2(s)+e(s1)s1
(s+3)Y1(s)Y2(s)=e(s1)s1
sY2(s)y2(0)=4Y1(s)+2Y2(s)+e(s1)s1
sY2(s)21=4Y1(s)+2Y2(s)+e(s1)s1
(s+4)Y1(s)2Y2(s)=e(s1)s1+21
Solve the equations:
(s+3)Y1(s)Y2(s)=e(s1)s1(2)
(s+4)Y1(s)2Y2(s)=e(s1)s1+21
[2(s+3)+(s+4)]Y1(s)=e(s1)s1+21
Y1(s)=e(s1)(s1)(s+2)21(s+2)
That is, Y1(s)=e(s1)s1e3(e(s+2)s+2)21(s+2)
Apply inverse Laplace transform:
y1(t)=[u(t1)]et[u(t+2)]et+321e2t
Solution:
(s+3)(e(s1)s1e3(e(s+2)s+2)21s+2)Y2(s)=e(s1)s1
Y2(s)=(s+3)e(s1)(s1)e3(s+3)(e(s+2)s+2)(s+3)21(s+2)e(s1)s1

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