For u"+4u'+8u=15cos(t)+10sin(t), find the steady-state solution and identify its amplitude and phase shift.

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).

## Comments

I don't understand how you got the amplitude and the phase shift. And how did you get sqrt(5)cos(t-?).