I am trying to solve
$$ A \cdot \frac {\partial^2 y} {\partial t^2} + B \cdot \frac {\partial^4 y} {\partial x^4} + C \cdot\frac {\partial^2 y} {\partial x^2} = q(x,t)$$
by separation of variables,
$$ y(x,t) = \sum_{n=0}^\infty X_n(x) \cdot T_n(t)$$
with boundary conditions
$$ y(0,t) = \frac{\partial y}{\partial x}(0,t) = \frac{\partial^2 y}{\partial x^2}(L,t) = \frac{\partial^3 y}{\partial x^3}(L,t) = 0$$
I haven't looked at differential equations for a long time. I really don't know where to start.
Suppose the equation was homogeneous, then separation of variables would give
\begin{align} AT'' + \lambda T &= 0 \\ BX^{(4)} + CX'' - \lambda X &= 0 \end{align}
Now, the equation is not homogeneous, but we can still use the above results to decompose the solution into its eigenmodes.
Let $X_n(x)$ be solutions of the eigenvalue problem
$$ B{X_n}^{(4)} + C{X_n}'' - \lambda_n X_n = 0 $$ $$ X_n(0) = {X_n}'(0) = {X_n}''(L) = {X_n}'''(L) = 0 $$
The conditions in which these solutions exist depends on the constant $B,\ C$, which I won't go into here. If $X_n$ are mutually orthogonal and form a complete basis in $[0,L]$, then it's possible to write
\begin{align} u(x,t) &= \sum_n T_n(t)X_n(x) \\ q(x,t) &= \sum_n q_n(t)X_n(x) \end{align}
where $T_n(t)$ and $q_n(t)$ are time-dependent components of each respective function. Since $q$ is known, $q_n(t)$ can be found by applying the standard trick, i.e. multiplying through by $X_m(x)$ and integrating, which gives the result
$$ q_m(t) = \frac{\int_0^L q(x,t) X_m(t)\ dt }{\int_0^L {X_m}^2(x) \ dx} $$
Plugging the series form of $u(x,t)$ into the equation, we obtain (term-wise)
\begin{align} A{T_n}''X_n + T_n \big[ B{X_n}^{(4)} + C{X_n}''\big] &= q_n X_n \\ A{T_n}''X_n + T_n\big[\lambda_n X_n\big] &= q_n X_n \end{align}
Therefore
$$ A{T_n}''(t) + \lambda_n T_n(t) = q_n(t) $$
It remains to solve this family of ODEs. If there's no specific form of $q(t)$, variation of parameters will do the job.
If you have any initial conditions, i.e. $u(x,0) = f(x),\ \frac{\partial u}{\partial t}(x,0) = g(x)$, they can also be employed here, once you decompose them to their corresponding series form.