Imaginary Numbers in a general homogenous solution for a differential equation

[TG3]

Homework Statement
Find the general solution for:
y''+2y'+5y=3sin2t


The attempt at a solution

y''+2y'+5y=3sin2t

First step is to find the general solution to the homogenous equation, so skipping 2 steps (letting y=e^rt and dividing)
R^2+2r+5
(-2+/- sqroot(4-4*5))/2
=-1 +/- 2i

How do I put this complex number into my equation? Were it not for the 2i I'd say c1E^-T +c2Te^-T. Were it not for the -1 I'd say Acos(T) +Bsin(T).

But how do I combine them?

 

[Mark44]

Homework Statement
Find the general solution for:
y''+2y'+5y=3sin2t


The attempt at a solution

y''+2y'+5y=3sin2t

First step is to find the general solution to the homogenous equation, so skipping 2 steps (letting y=e^rt and dividing)
R^2+2r+5
(-2+/- sqroot(4-4*5))/2
=-1 +/- 2i

How do I put this complex number into my equation? Were it not for the 2i I'd say c1E^-T +c2Te^-T. Were it not for the -1 I'd say Acos(T) +Bsin(T).

But how do I combine them?

Two linearly independent solutions are y1 = e^{(-1 + 2i)t} and y2 = e^{(-1 - 2i)t}. These can be written as e^-te^i2t and e^-te^-i2t. Using the fact that e^ix = cosx + isinx, and skipping a few steps myself, a different pair of linearly independent solutions is e^-tcos(2t) and e^-tsin(2t).