1. The problem is to solve the differential equation $$y'' + \lambda y = 0$$ with boundary conditions $$y'(0) = 0$$ and $$y(1) = 0$$. 2. Start by considering the characteristic equation of the differential equation: $$r^2 + \lambda = 0$$ which gives $$r = \pm i \sqrt{\lambda}$$. 3. The general solution to the differential equation is then $$y(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x)$$ where $A$ and $B$ are constants. 4. Apply the first boundary condition: $$y'(0) = 0$$. Calculate $$y'(x) = -A \sqrt{\lambda} \sin(\sqrt{\lambda} x) + B \sqrt{\lambda} \cos(\sqrt{\lambda} x)$$. At $$x=0$$, $$y'(0) = B \sqrt{\lambda} = 0$$, therefore $$B = 0$$ (assuming $$\lambda \neq 0$$). 5. The solution reduces to $$y(x) = A \cos(\sqrt{\lambda} x)$$. 6. Apply the second boundary condition: $$y(1) = 0$$. Substitute $$x=1$$: $$y(1) = A \cos(\sqrt{\lambda}) = 0$$. For nontrivial solutions ($$A \neq 0$$), we need $$\cos(\sqrt{\lambda}) = 0$$. 7. The zeros of cosine occur at $$\sqrt{\lambda} = \frac{\pi}{2} + n\pi$$ for integer $$n \geq 0$$. Therefore, $$\lambda_n = \left(\frac{\pi}{2} + n\pi\right)^2$$. 8. The eigenfunctions corresponding to each $$\lambda_n$$ are $$y_n(x) = A_n \cos\left(\left(\frac{\pi}{2} + n\pi\right) x \right)$$. **Final answer:** The eigenvalues are $$\lambda_n = \left(\frac{\pi}{2} + n\pi\right)^2$$ for $$n=0,1,2,...$$ The eigenfunctions are $$y_n(x) = \cos\left(\left(\frac{\pi}{2} + n\pi\right) x\right)$$.