What are the conditions for a second order nonhomogenous ODE with $q(x) = A_1\sin(Bx)+A_2\cos(Bx)$? In my notes it states that it 
$y_1$ $\neq$ $\mathbf e$sin(Bx) 
as one of the conditions and this is the one I am confused about. There is no $\mathbf e$ in the original equation so is this a typo in my lecture?

I will interpret the question as follows:
Given that $y=c_1y_1+c_2y_2+y_p$ is a solution of 
\begin{equation}
y^{\prime\prime}+ay^\prime+by=A_1\sin(Bx)+A_2\cos(Bx)
\end{equation}
and that there is no solution containing a term of the form $e^{\alpha x}\cos(Bx)$, find a general solution.
Since if $a^2<4b$ the general solution of the corresponding homogeneous equation
\begin{equation}
y_c^{\prime\prime}+ay_c^\prime+by_c=0
\end{equation}
will be
\begin{equation}
y_c=e^{-\frac{a}{2}x}\left[ c_1\sin\left(\frac{\sqrt{4b-a^2}}{2}x\right)+c_2\cos\left(\frac{\sqrt{4b-a^2}}{2}x\right)\right]
\end{equation}
the condition would suggest that either

* $a^2>4b$ or

* $a^2=4b$
Therefore the general form of the particular solution should be
\begin{equation}
y_p=\gamma_1\sin(Bx)+\gamma_2\cos(Bx)
\end{equation}
