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posted by  johnmc on 7/23/2008 9:39:19 PM  |  status: Live  

ODE's Need Help!

Course Textbook Chapter Problem
Differential Equations N/A N/A N/A
Question Details:
Please work this out step by step so I can see how you work it out and I will rate lifesaver.
 
Solve y'' + y' = sin 2t             when y(0) = 0 and y'(0) = 0 use the method of undetermined coefficients.
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posted by Altair on 7/23/2008 9:41:52 PM  |  status: Live
Asker's Rating: Lifesaver   
johnmc's comment:
"THANK YOU!!!"
Response Details:
first consider: y"+y'=0
 
m2 + m = 0
m(m+1)=0
 
m=0, -1
 
the general solution for the homogeneous part of the DE is:
 
y=c1 + c2 e-x
 
now we need to find the particular solution.
 
Let Y = Asin2t + Bcos2t
 
Y' = 2Acos2t - 2Bsin2t
Y"=-4Asin2t - 4Bcos2t
 
 
Y"+Y' = -4Asin2t - 4Bcos2t+2Acos2t - 2Bsin2t = (-4A-2B)sin2t + (2A-4B)cos2t = sin2t
 
=> -4A-2B = 1
     2A-4B = 0   => A=2B
 
solve:
B=-1/10
A= -1/5
 
Y=(-1/5)sin2t - (1/10)cos2t
 
the general solution is:
 
y=c1 + c2 e-x -(1/5)sin2t - (1/10)cos2t
 

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