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### Messages - Christopher Mohanlall

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##### Term Test 1 / Re: TT1 Problem 4 (main)
« on: October 16, 2018, 08:46:34 AM »
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 yc= 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)

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