The first thing you need to do is solve the homogeneous DE y'' + 4y = 0. Once you have done that you should see a problem when you try to solve for the non-homogeneous y'' + 4y = cos^2(t).
I need help with the following problem:
y'' + 4y = cos^2(t)
Here is what I have done so far
y'' + 4y = (1/2) + (1/2)cos(2t) (power reduction formula)
Y(t) = A + B*sin(2t) + C*cos(2t)
Y'(t) = 2B*cos(2t) - 2C*sin(2t)
Y''(t) = -4B*sin(2t) - 4C*cos(2t)
Pluging in Y(t) and Y''(T) into the original equation, I get:
[-4B*sin(2t) - 4C*cos(2t)] + 4[A + B*sin(2t) + C*cos(2t)] = (1/2) + (1/2)cos(2t)
4A = (1/2) + (1/2)cos(2t)
As you can see, the B and C terms cancel each other out, leaving only an A term. What should I do from here? Is there anything I did wrong?
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Please include D*t*sin(2t) and E*t*cos(2t) in your definition of Y(t) and you'll get it. If too much cancels out, this is always a good next step. Complicate the original definition.