1. (TCO 4) Solve the initial value problem of the system y1\' = 3y1 + 5y2 y\'2=

Question

1. (TCO 4) Solve the initial value problem of the system y1' = 3y1 + 5y2 y'2= -5y1 - 3y2 given y1(0) = 0, y2(0) =0. Verify the result using MATLAB.

1. (TCO 4) Solve the initial value problem of the system
y1' = 3y1 + 5y2
y'2= -5y1 - 3y2
given y1(0) = 0, y2(0) = 2.Verify the result using MATLAB. (Points : 10)

a. y1 = (5/2)sin(3t),
y2= -2cos(4t) + 1.5sin(4t)

b. y1 = (5/2)sin(4t),
y2= 2cos(4t) - 1.5sin(4t)

c. y1 = (-5/2)sin(3t),
y2= -2cos(4t) + 1.5sin(4t)

d. y1 = (5/2)cos(3t),
y2= -2sin(4t) + 1.5cos(4t)

Answer is b I just need to see the work and the matlab solution.

Explanation / Answer

y1' = 3y1 + 5y2 -> y2 = (y1' - 3y1)/5 -> y2' = (y1" - 3y1')/5
y2' = -5y1 - 3y2 -> (y1" - 3y1')/5 = -5y1 - 3 * (y1' - 3y1)/5   

-> y1" - 3y1' = -25y1 - 3y1' + 9y1

-> y1" + 16y1 = 0

The characteristic equation is:

s^2 + 16 = 0 -> (s-4i)(s+4i) = 0 -> y1 = Asin(4t) + Bcos(4t)

y1(0) = 0 -> B = 0 -> y1 = Asin(4t)

y2 = (y1' - 3y1)/5 -> y2(t) = (4A cos(4t) - 3A sin(4t))/5

y2(0) = 2 -> 4A/5 = 2 -> A = 5/2 = 2.5

Therefore:

y1(t) = 2.5 sin(4t)

y2(t) = 2cos(4t) - 1.5 sin(4t)

answer (b) is correct.

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