Polynomial annihilator method $y''+4y=\sin^2(2x)$ The question asks to solve the equation by this method. I know how to annihilate $\sin(2x)$ by $(D^2+4)$ however i don't know for the case $\sin^2(2x)$. Thanks!

**Hint.** If you know how to annihilate $\sin(2x)$, you surely know how to annihilate $\sin^2(2x)$, you just write $$ \sin^2(2x)=\frac{1-\cos (4x)}{2} $$ then you annihilate $ \cos (4x)$ as you did for $\sin(2x)$, obtaining $$ (D^2+16)(\cos (4x))\equiv0 $$