Since the RHS of the DE is a part of the homogeneous solution, for a particular solution you should actually try $\displaystyle \begin{align*} y_p = A\,x\cos{(10t)} + B\,x\sin{(10t)} \end{align*}$.
Hi, I have the equation y''+100y = 0.5cos(10t) and I have to find the general solution and the IVP.
For the general solution I made the y's into s's and figured out s^2 + 100 = 0 so S 1 & 2 = 0 +/- 10i
From here I found the y-homogeneous part = k1(cos(10t))+k2(sin(10t))
so y-particular = Acos(10t), and I found the second derivitive to equal -100Acos(10t)
The problem is I don't know what to do next. If I plug the A's back in I get -100A+100A = 0.5 and I know this cannot be
Since the RHS of the DE is a part of the homogeneous solution, for a particular solution you should actually try $\displaystyle \begin{align*} y_p = A\,x\cos{(10t)} + B\,x\sin{(10t)} \end{align*}$.