# Thread: Periodic solution

1. ## Periodic solution

I'm trying to find a periodic solution for the following inhomogeneous equation:
u'' − u = cos^2(t).
I'm thinking that in order for the solution to be periodic it must be equal just with the particular solution(for the complementary we put the constants C1=C2=0)
Also I can write cos^2(t) as cos^2(t)=(1+cos2t)/2 and then I can choose the particular solution to be of the form u_p(x)=-1/2 +Asin(t)+Bcos(t).
After I diff and substitute into the initial equation I get the answer -1/2 +1/3*cos2t which is not correct.
How can I solve this?Also if I can leave cos^2(t) as this and find a solution of the form
Acos^(t)+Bsin^2(t) will it still be correct?

2. Your approach is right but how did you obtain $\frac{1}{3} \cos 2t$?

3. Because is...

(1)

... imposing ...

(2)

... means that the 'periodic solution' is...

(3)

Kind regards

$\chi$ $\sigma$

4. I"m sorry but I don't see how you've obtained 1/10 cost 2t?
Danny...I know I did wrong but I diff the particular equation and substituted...and I get B=0 and A=-1/3

5. If $u = A \cos 2t$ then $u'' = - 4A \cos 2t$ so

$u''-u = -5A \cos 2t = \frac{1}{2} \cos 2t$ giving $A = - \frac{1}{10}$

6. ahhhhh...........thanks danny