$$ \begin{cases} v''(t)-v(t)=f(t) & 0<t<h\\ v(0)=0=v(h) \end{cases}, $$ where $f\in L^{2}$.
Is there a explicit solution of integral form?
For the case of $$ v''(t)-v(t)=0, $$ the solution is $v(t)=C_{0}e^{t}+C_{1}e^{-t}$ for constants $C_{0},$ $C_{1}$. But I can't go further with the general nonhomogenous term $f(t)$.
Consider the solutions of the associated homogeneous problem $$y_1(t)=\sinh(t),y_2(t)=\sinh(t-h),$$ satisfying $y_1(0)=y_2(h)=0$. Variation of parameters gives some particular solutions: $$v(t)=\sinh(t)\int_{t_1}^t \frac{-\sinh(\tau-h)f(\tau)}{\sinh(h)} \mathrm{d} \tau+ \sinh(t-h)\int_{t_2}^t \frac{\sinh(\tau) f(\tau)}{\sinh(h)} \mathrm{d} \tau .$$ In order to satisfy the boundary condition $v(0)=0$ we take $t_2=0$, and in order to satisfy $v(h)=0$ we take $t_1=h$.