# Second Order Differential Equations

1. Oct 30, 2007

### Illusionist

1. The problem statement, all variables and given/known data
Consider the second order differential equation y'' - 4y' + 4y = f(x)
Find a particular solution if f(x) = 25cos(x)

2. Relevant equations
I believe for this type of question I should let y = Asin(x) + B cos(x)
Hence y' = Acos(x) - Bsin(x) and
y'' = -Asin(x) - Bsin(x)

3. The attempt at a solution
I think everything above is right, and for the rest of the question I should just be able to substitute back into the original formula and find values for A and B.
The problem is when I do substitute back in all I get is a mess and a headache!

Here is what I got by substitution:
-Asin(x) - Bcos(x) - 4[Acos(x) - Bsin(x)] + 4[Asin(x) + Bcos(x)] = 25cos(x)
Hence
-Asin(x) + 4[Asin(x) + Bsin(x)] - Bcos(x) - 4[Acos(x) - Bcos(x)] = 25cos(x)

That's about all I think I can do and I have no idea how to obtain values for A and B. I know the answer is y(p) = 3cos(x) - 4sin(x).
How to get this answer from the above I have no idea, I'm obviously missing something.
Thanks in advance for any advice or help.

2. Oct 30, 2007

### Integral

Staff Emeritus
You are just about there. Gather your sinx terms then factor the sin, same for the cosx terms, now, since sin x does not appear in the particular solution set its coefficient to zero, solve for A in terms of B. Set the coefficient for the cos x term = 25 and find B.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook