Find the steady-state solution?

Question

For u"+2u'+6u=15cos(3t), find the steady-state solution and identify its amplitude and phase shift.

u=A cos 3t+B sin 3t

15 cos 3t= -9A cos 3t-9B sin 3t+2(-3A sin 3t+3B cos 3t)+6(A cos 3t+B sin 3t)

= -9A cos 3t-9B sin 3t-6A sin 3t+6B cos 3t+6A cos 3t+6B sin 3t

= -3A cos 3t-3B sin 3t-6A sin 3t+6B cos 3t

=A cos 3t+B sin 3t+2A sin 3t-2B cos 3t

Now what? Help me step by step.

Answer 1

r^2+2r+6 = (r+1)^2 + 5 = 0

r = -1 +/- sqrt(5) i

General solution for homogeneous equation: u = e^(-t)(C1 cos(sqrt(5)t) + C2 sin(sqrt(5)t))

Steady state solution: u* = Acos(3t) + Bsin(3t)

Substitute y = u* in the original differenital equation, u"+2u'+6u=15cos(3t),

-9Acos(3t) - 9Bsin(3t) - 6Asin(3t) + 6Bcos(3t) + 6Acos(3t) + 6Bsin(3t) = 15cos(3t)

Compare coefficients:

-3A+6B = 15 => -A+2B = 5

-3B-6A = 0 => B = -2A

A = -1, B = 2
Steady state solution:

u* = - cos(3t) + 2sin(3t) = sqrt(5) cos(3t - pi + φ)

So, the amplitude = sqrt(5), and the phase shift = (pi - φ)/3 to the right, where φ = tan^-1(2).