Construct the Green's function for the following boundary value problem and use it to find the solution $$y''-y'=t^2,\quad y(0)=0,\: y(1)=0$$
I know that,
$$ \begin{aligned} &\langle L u, G\rangle=\langle f, G\rangle \\ &\left\langle u, L^{\dagger} G\right\rangle=\langle f, G\rangle \\ &\langle u, \delta(x-\xi)\rangle=\langle f, G\rangle \\ &u(\xi)=\langle f, G\rangle \end{aligned} $$ The Green's Function Solution $$ u(x)=\int_{0}^{l} f(\xi) G(\xi, x) d \xi $$
Okay, the linear operator $L=\frac{d^2}{dx^2}-\frac{d}{dx}$ with boundary conditions $y(0)=0,\: y(1)=0$. I get the adjoint operator with Green function $L^{\dagger}G=\frac{d^2}{dx^2}G+\frac{d}{dx}G=\delta(x-\xi) \quad G(0)=0,\: G(1)=0$ where the bondary conditions come from Conjunct or Bilinear Concomitant (set to zero). But I am facing issue to solve the $L^{\dagger}G=\delta(x-\xi)$.