I have the following PDEs that I would like to solve for $f(x,\omega)$ and $g(x,\omega)$:
\begin{align*} \dfrac{\partial^{2}f}{\partial x^{2}}-\omega^{4}f & =-\omega\dfrac{\partial{\cal B}}{\partial x}-a\omega^{3},\\ \dfrac{\partial^{2}g}{\partial x^{2}}-\omega^{4}g & =-\omega{\cal B}(x) \end{align*}
where ${\cal B}(x)$ is an unknown function of $x$.
Is the correct way to use variation of parameters to write down the solution of these equations \begin{align*} f(x,\omega) & =\dfrac{1}{2\omega^{2}}e^{\omega^{2}x}\int_{p_{1}(\omega)}^{x}e^{-\omega^{2}\xi}\left(-\omega\dfrac{\partial{\cal B}}{\partial\xi}+a\omega^{3}\right){\rm d}\xi-\dfrac{1}{2\omega^{2}}e^{-\omega^{2}x}\int_{p_{2}(\omega)}^{x}e^{\omega^{2}\xi}\left(-\omega\dfrac{\partial{\cal B}}{\partial\xi}+a\omega^{3}\right){\rm d}\xi\qquad(\dagger)\\ g(x,\omega) & =\dfrac{\omega}{2}e^{\omega^{2}x}\int_{p_{3}(\omega)}^{x}e^{-\omega^{2}\xi}{\cal B}(\xi){\rm d}\xi+\dfrac{\omega}{2}e^{-\omega^{2}x}\int_{p_{4}(\omega)}^{x}e^{\omega^{2}\xi}{\cal B}(\xi){\rm d}\xi.\qquad(\ddagger) \end{align*}
where $p_{1}(\omega)$, $p_{2}(\omega)$, $p_{3}(\omega)$ and $p_{4}(\omega)$ are arbitrary functions of $\omega$? I am aware that when variation of parameters is used to solve ODEs, that the lower limits of the integrals are arbitrary constants, and are determined by the boundary conditions.
If we also know that $f$ is an even function of $x$ and $g$ and $\cal{B}$ are odd functions of $x$, then (correct me if I've got this wrong) one can show that $p_{1}(\omega)=-p_{2}(\omega)$ and $p_{3}(\omega)=-p_{4}(\omega)$. However, suppose we also know that $f$ and $g$ satisfy $$ \dfrac{\partial f}{\partial x}-\omega^{2}g(x,\omega)+\omega{\cal B}(x)=0\qquad(\star) $$
Then (I think) one can show that \begin{align*} 0 & =-\omega a\left(1-\cosh\omega^{2}x\right)\\ & \phantom{=}-\dfrac{1}{2\omega}\left(e^{\omega^{2}x}\int_{p_{1}(\omega)}^{x}e^{-\omega^{2}\xi}{\cal B}(\xi){\rm d}\xi-e^{-\omega^{2}x}\int_{-p_{1}(\omega)}^{x}e^{\omega^{2}\xi}{\cal B}(\xi){\rm d}\xi\right)\\ & \phantom{=}-\dfrac{\omega^{3}}{2}\left(e^{\omega^{2}x}\int_{p_{3}(\omega)}^{x}e^{-\omega^{2}\xi}{\cal B}(\xi){\rm d}\xi-e^{-\omega^{2}x}\int_{-p_{3}(\omega)}^{x}e^{\omega^{2}\xi}{\cal B}(\xi){\rm d}\xi\right). \end{align*}
It is far from obvious what the relationship between $p_{1}(\omega)$ and $p_{3}(\omega)$ is in this case - despite the (almost) symmetry between the integrals, even setting $p_{1}(\omega)=0$ doesn't really help. Furthermore, it doesn't seem $p_{1}(\omega)$ and $p_{3}(\omega)$ will be functions of $\omega$ only, based on what I have tried.
Is it possible to use equation $(\star)$ to derive a relationship between these functions?
As a more fundamental question, are $(\dagger)$ and $(\ddagger)$ the correct ways to bring in arbtirary functions in this problem? If not, what else should I try?