For the following system, find a correct expression for the scalar function x2(t

Question

For the following system, find a correct expression for the scalar function x2(t). The initial conditions are x1(0) = 1, x2(0) = –1 and x3(0) = –1 and x4(0) = 3. (Hint: Expanding the determinant along an appropriate row/column makes calculation of eigenvalues easy.)

x1’ = 2x1.

x2’ = –21x1 – 5x2 – 27x3 – 9x4.

x3’ = 5x3.

x4’ = –21x3 – 2x4.

x2(t) = e2t – 4e–5t.

x2(t) = –3e2t – 3e–5t.

x2(t) = –3e2t + 2e–5t.

x2(t) = 3e2t + 4e–5t.

x2(t) = e2t – 4e–5t.

x2(t) = –3e2t – 3e–5t.

x2(t) = –3e2t + 2e–5t.

x2(t) = 3e2t + 4e–5t.

Explanation / Answer

INtegrating x3 and x1 gives

x1=A exp(2t),x3=B exp(5t)

Initial conditions gives, A=1,B=-1

x1=exp(2t),x3=-exp(5t)

x4'+2x4=21 exp(5t)

(x4'+2x4)exp(2t)=21 exp(7t)

(x4 exp(2t))'=21 exp(7t)

Integrating gives

x4 exp(2t)=3 exp(7t)+C

x4=3 exp(5t)+C exp(-2t)

x4(0)=3=3+C, C=0

x4=3 exp(5t)

x2'=-21 exp(2t)-5x2+27 exp(5t)-27 exp(5t)=-21 exp(2t)-5x2

x2'+5x2=-21 exp(2t)

(x2'+5x2)exp(5t)=-21 exp(7t)

(x2 exp(5t))'=-21 exp(7t)

x2 exp(5t)=-3 exp(7t)+D

x2=-3exp(2t)+D exp(-5t)

x2(0)=-1=-3+D

D=2

x2=-3 exp(2t)+2 exp(-5t)

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