mL^(2)θ’’ + cθ’ + mgLθ + kLθ = Asin(bt)
Ynh= Asin(bt) + Bsin(bt)
Help need to find the derivatives of Ynh and sub into the DE then use the method of Undetermined coefficients.
Your original equation can be written$$
mL^2\Theta'' +c\Theta' +(mgL+kL)\Theta = A\sin(bt)$$It will simplify things if you rename the complicated expressions, for example call $u = mL^2,~v =mgL + kL$ so your equation becomes$$
u\Theta'' + c \Theta' + v\Theta = A\sin(bt)$$You can put the $u$ and $v$ back in at the end. Now, when you try for a particular solution, you don't want to use an $A$ because it will be confused with the $A$ on the right side. So try something like $P\cos(bt) + Q\sin(bt)$ for a particular solution. That way you won't get the capital $C$ or $B$ letters confused with the $c$ and $b$ that are already in the equation. Try that.