Y '' - (TaNx) y '+ 2Y = 0 Without the first question, there is a special solution y = sin X.
Let y = C (x) SiNx, replace the original equation to get C (x), if I am not wrong, I should get it, but it is very complicated
Let y = C (x) SiNx, replace the original equation to get C (x), if I am not wrong, I should get it, but it is very complicated