Solve $$\frac{d^2U}{dx^2} - sU = A, \qquad (0<x<1)$$ subject to the boundary conditions $$\frac{dU}{dx}(0) = 0, \qquad U(1) = 0.$$
The part I'm stuck on is that after taking Laplace transforms, I have a $U(0)$ term that I cannot evaluate because the boundary condition gave me $U(1)$. What should I do in this case?
I already know that the correct answer is $$U =\frac{A}{s} \left( \frac{\cosh\left(\sqrt s x\right)}{\cosh \left(\sqrt s \right)} - 1 \right).$$
This is a track not a complete answer
For clarity lets denote $V=\mathcal{L}(U)$ the Laplace transfrom of $U$.
We have :
$$ \mathcal{L}(U'') : p \to p^2V(s)−pU(0)−U′(0)$$
So the equation rewrites with $U'(0)=0$ :
$$ p^2V(s)−pU(0) + sV = \dfrac{A}{p} $$
So
$$ (p^3+sp)V(s)=p^2U(0)+A$$
Hence
$$ V(s)=\dfrac{A+p^2U(0)}{p^3+sp}$$
$$ V(s)=\dfrac{A+p^2U(0)}{p^3+sp}$$
Integrating back (inverse Laplace) and considering $B\triangleq U(0)$ with the know formulas online here for example https://en.wikipedia.org/wiki/Laplace_transform