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