I'm trying to find the solution, which is supposed to be
$y(t)=c_2\sin(2t)+c_1\cos(2t)+\frac{t\sin(2t)}{8}+\frac{\cos^2(t)}{4}$,
but I'm doing something wrong along the way and I can't figure out what I did. I was hoping someone would be able to find my mistake for me. I'll show what I've gotten:
First I got the complementary solution $y_c(t)=c_1\cos(2x)+c_2\cos(2x)$
I then used the particular solution $y_p=A+B\cos(2t)+C\sin(2t)$ – I'm guessing this is where my error occurs, but I don't see what's wrong with it.
I derived it two times and got $y_p''=-4B\cos2t-4C\sin(2t)$
I plugged it into the original equation (assuming $\cos^2t$ is $\frac{1}{2}+\frac{\cos(2x)}{2}$) and then tried to solve the problem, at which point I found myself running in circles:
$\sin2t(-4C-4C)+\cos2t(-4B+4B)+4A=\frac{1}{2}+\frac{\cos(2x)}{2}$.
Can anyone help me find my mistake(s)? Thanks a lot.
Best Answer
You figured out the issue with:
$$y_c(t) = c_1 \cos 2t + c_2 \sin 2t$$
For the particular solution, some care is needed given the homogeneous (complementary) solution. We expand the $\cos^2 2t$ term as:
$$\cos^2 2t = \dfrac{1}{2}\left(1 +\cos 2t\right)$$
Now, we notice that this has a common term with the homogeneous term, so we multiply the particular solution by $t$ to account for that. Thus, we choose a particular solution as:
$$y_p(t) = a + b~ t \cos 2t + c~ t \sin 2t$$
Now, substitute and solve for $a$, $b$ and $c$.
You should get:
$$a = \dfrac{1}{8}, b = 0, c = \dfrac{1}{8}$$
Your final solution will be:
$$y(t) = y_c(t) + y_p(t)$$