The general solution of homogeneous equation is $y_h=A\sin(w_0t+\phi)$ with $w_0=2\sqrt{5}$ and $A,\phi$ some parameters.
Now let's search a particular solution in the case $w\
eq w_0$.
We can hope to find it under the form $y_p=B\sin(wt)$.
$y_p''+20y_p=-Bw^2\sin(wt)+20B\sin(wt)=100\sin(wt)\iff-Bw^2+20B=100$
$y_p=\frac{100}{20-w^2}\sin(wt)$
Now you have your final solution $y(t)=y_h+y_p=A\sin(w_0t+\phi)+\frac{100}{20-w^2}\sin(wt)$
* * *
You can now apply initial conditions : $y(0)=0=A\sin(\phi)$
$A=0$ is possible, but this would be very reductive. So I guess $\sin(\phi)=0\iff \phi=k\pi$ is more appropriate.
$A\sin(w_0t+k\pi)=\pm A\sin(w_0t)$ but since $A$ is a dummy parameter, $A$ or $\pm A$ means the same.
> Finally $y(t)=A\sin(w_0t)+\frac{100}{20-w^2}\sin(wt)$
_We do not loose in generality setting $\phi$ to $0$, because it is still possible to set $A=0$, but an additional initial condition would be required to be absolutely certain of the value of $A$ anyway._