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Find the steady-state solution?

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1 Answer

Let u = e^r,

r^2+4r+8 = (r+2)^2 + 4 = 0

r = -2 + i or  -2 - i

For steady-solution, the homogeneous part approaches 0 as t -> oo.

Let the steady-solution be u* = Acos(t) + Bsin(t).

u*"+4u*'+8u*=15cos(t)+10sin(t) becomes,

cos(t)(-A+4B+8A) + sin(t)(-B-4A+8B) = 15cos(t)+10sin(t)

Compare coefficients:

4B+7A = 15 ......(1)

-4A+7B = 10 ......(2)

Solve for the system,

A = 1

B = 2

u* = cos(t) + 2sin(t) = sqrt(5)cos(t - θ), where θ = arctan(2)

So, the amplitude is sqrt(5), and the phase shift to the right by arctan(2).