1. The problem is to understand how the boundary conditions $X'(0)=0$ and $X'(\frac{\pi}{2})=0$ lead to the general solution for $X(x)$. 2. The general form of the solution to a second-order linear differential equation with constant coefficients is $$X(x) = A \cos(kx) + B \sin(kx)$$ where $k = \frac{\sqrt{\lambda}}{2}$. 3. The derivative is $$X'(x) = -A k \sin(kx) + B k \cos(kx)$$ 4. Apply the first boundary condition $X'(0) = 0$: $$X'(0) = -A k \sin(0) + B k \cos(0) = B k = 0$$ which implies $$B = 0$$ 5. Apply the second boundary condition $X'(\frac{\pi}{2}) = 0$: $$X'(\frac{\pi}{2}) = -A k \sin\left(k \frac{\pi}{2}\right) + B k \cos\left(k \frac{\pi}{2}\right) = -A k \sin\left(k \frac{\pi}{2}\right) = 0$$ 6. Since $A \neq 0$ for a nontrivial solution and $k \neq 0$, we require $$\sin\left(k \frac{\pi}{2}\right) = 0$$ which means $$k \frac{\pi}{2} = n \pi, \quad n = 0,1,2,\ldots$$ 7. Solving for $k$: $$k = \frac{2n}{1} = 2n$$ 8. Therefore, the general solution satisfying the boundary conditions is $$X(x) = A \cos(2n x)$$ with $B=0$. This explains how the boundary conditions lead to the form of the general solution.