I have a differential equation of the form below. How to solve this. It is actually a fourth order beam equation with a derivative of the delta function. I have solved for just delta function in the right-hand side. But I never solved the derivative of the delta function on the right-hand side of the equation.
\begin{align} \begin{split} \frac{\mathrm{d}^4\psi(\eta)}{\mathrm{d}\eta^4}-\beta^4\psi(\eta)=\sum_{j=1}^{n} K_{r,j}\psi'(\eta)\,\delta'(\eta-\zeta_{3,j}) \label{eq:Governin equation in w} \end{split} \end{align}
Sketch: For convenient, let us take $\beta = K_{r, j} = 1$. Then you have \begin{align} \psi^{(4)}(t) - \psi(t) = \sum^n_{j=1}\psi'(t)\delta(t-\zeta_j). \end{align} Using the fact that \begin{align} \psi'(t)\delta(t-\zeta_j) = -\psi''(\zeta_j)\delta(t-\zeta_j) + \psi'(\zeta_j) \delta'(t-\zeta_j) \end{align} then we have \begin{align} \psi^{(4)}(t) - \psi(t) = \sum^n_{j=1} \left\{-\psi''(\zeta_j)\delta(t-\zeta_j) + \psi'(\zeta_j) \delta'(t-\zeta_j)\right\}. \end{align} Now take the Laplace tranform and solve for $\mathcal{L}[\psi](s)$. Finally, apply the inverse transform to recover $\psi$.