Second Order Differential Equations

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SUMMARY

The discussion focuses on solving the second order differential equation y'' - 4y' + 4y = f(x) with f(x) = 25cos(x). The user proposes a particular solution of the form y = Asin(x) + Bcos(x) and derives the first and second derivatives. However, they struggle with substituting these derivatives back into the original equation to find the coefficients A and B. The correct particular solution is y(p) = 3cos(x) - 4sin(x), which can be obtained by setting the coefficients of sin(x) and cos(x) terms to zero and solving the resulting equations.

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Homework Statement


Consider the second order differential equation y'' - 4y' + 4y = f(x)
Find a particular solution if f(x) = 25cos(x)

Homework 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)

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.
 
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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.
 

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