Toronto Math Forum

First find the complementary solution for the homogenous equation:
y''(t)+2y'(t)+5y(t) = 8e^-t +34sin(2t)

We use the characteristic equation:
r²+2r+5=0

Finding the roots we get:
-1+/-2i

Therefore our y_c= c₁e^-tcos(2t) + c₂e^-tsin(2t)

Particular Solutions:
8e^-t:

Assume: y=Ae^-t then:
y'=-Ae^-t
y'' =Ae^-t

Then:
Ae^-t-2Ae^-t+5Ae^-t=8e^-t
A-2A+5A=8
4A=8
A=2


34sin(2t):
Assume y=K₁cos(2t)+K₂sin(2t)
y'=-2K₁sin(2t)+2K₂cos(2t)
y''=-4K₁cos(2t)-4K₂sin(2t)

Then:
-4K₁cos(2t)-4K₂sin(2t)-4K₁sin(2t)+4K₂cos(2t)+5K₁cos(2t)+5K₂sin(2t)=34sin(2t)


Break into components:
Cosine:

-4K₁+4K₂+5K₁=0
4K₂+K₁=0

K₁=-4K₂

Sine:

-4K₂-4K₁+5K₂=34
K₂-4K₁=34


K₂-4(-4K₂)=34
17K₂=34
K₂=34/17
K₂=2

now:
K₁=-4(2)
K₁=-8

Therefore:
y= c₁e^-tcos(2t) + c₂e^-tsin(2t) + 2e^-t + -8cos(2t)+2sin(2t)