I am having difficulty finding the right approach to solving the following differential equation, $$ y''(t)+\int_t^Tg(s-t)y'(s)\,ds=f(t), $$ with the boundary conditions, $$y(0)=y_0\,,\quad y(T)=0.$$ I considered representing $y(t)$ using a Fourier series, but term-wise differentiation is not guaranteed because the periodic extension of $y(t)$ isn't differentiable at the boundaries $t=0,T$.
What approach is the smartest to solve this equation? Any help will be appreciated.
Set $v(t) = y'(t)$. Thus $y(t) = - \int_t^T v(s) \, ds$.
Now solve the first order integrodifferential equation $$ v'(t)+\int_t^Tg(s-t)v(s)\,ds=f(t), \; v(T) = \alpha $$ for general $\alpha \in \mathbb{R}$. The solution may be expressed in the form $$ v(t) = \alpha r(t) + \int_t^T r(s-t)f(s)\, ds = \alpha r(t) + \tilde f(t) $$
where $r$ is known as differential resolvent (see monographs on integral equations, e.g. Gripenberg - Londen - Staffans, CUP 1990, Theorem 3.1). Therefore $$ y(t) = - \int_t^T v(s) ds = - \alpha \int_t^T r(s) ds - \int_t^T \tilde f(s) \, ds $$ Set $t = 0$ to find the right value of $\alpha$. This gives a unique solution if $\int_0^T r(t) \, dt \ne 0$. If $\int_0^T r(t) \, dt = 0$, there either are infinitely many solutions or there is no solution, depending on whether $\int_0^T \tilde f(s) + y_0 = 0$ or not.