y''+4y'+3y=65cos(2t)
Auxiliary eqn is
m^2 + 4m + 3 = 0
(m + 1)(m + 3) = 0
m = -1, m = -3
Solution to the homogeneous eqn is
y1 = Ae^(-t) + Be^(-3t)
2nd solution
Let y2 = C.cos(2t) + D.sin(2t)
Then
y2’ = -2C.sin(2t) + 2D.cos(2t)
y2’’ = -4C.cos(2t) – 4D.sin(2t)
Recreating the DE with y2, and its derivatives, now gives us
y2’’ + 4y2’ + 3y2 = (-4C.cos(2t) – 4D.sin(2t)) + 4(-2C.sin(2t) + 2D.cos(2t)) + 3(C.cos(2t) + D.sin(2t))
y2’’ + 4y2’ + 3y2 =(-4C + 8D + 3C).cos(2t) + (-4D – 8C + 3D).sin(2t)
Comparing coeffts of the trig terms with the original DE, we get
-4C + 8D + 3C = 65
-4D – 8C + 3D = 0
8D - C = 65
D + 8C = 0
From which C = -1, D = 8
Therefore 2nd solution is.
y2 = -cos(2t) + 8sin(2t)
The general solution is: y(t) = y1(t) + y2(t)
Answer: y(t) = Ae^(-t) + Be^(-3t) - cos(2t) + 8sin(2t)