Find the particular solutions of $\displaystyle y_{p}(t)$ for the DEs:

$\displaystyle y''+B^{2}y=cos(wt)$ and

$\displaystyle y''+B^{2}y=sin(wt)$

by taking both the real and imaginary parts of the solution:

$\displaystyle z_{p}=\dfrac{sin((w-B)\dfrac{t}{2})}{(w-B)\dfrac{1}{2}} \dfrac{e^{i(w+B)\dfrac{t}{2}}}{i(w+B)}$

Verify the solution for for:$\displaystyle y''+B^{2}y=cos(wt)$

Just to clarify, all other variables except $\displaystyle i$ are real numbers.

Well, here is what i tried...

I started of by trying to take the Real part of $\displaystyle z_{p}$, as the directions state. I think i can do it fine until i get to the denominator on the right side. Once i take the real part of $\displaystyle i(w+b)$, it makes the demoninator zero, which leaves me stuck.

Any helps would be be GREAT!