An initial value problem fixes $y$ and $y'$ at one point, and it always has exactly one solution. A boundary value problem fixes conditions at two different points, such as the two ends of a string or a rod, and then anything can happen: one solution, none, or infinitely many. The equations are the ones from real roots and complex roots; the new part is the conditions.
Key idea
The eigenvalue problem $y'' + \lambda y = 0$, $y(0) = 0$, $y(L) = 0$ has nonzero solutions only for $$\lambda_n = \left(\frac{n\pi}{L}\right)^{2},$$ $$y_n = \sin\frac{n\pi x}{L},$$ $$n = 1, 2, 3, \dots$$ These $\lambda_n$ are the eigenvalues and $y_n$ the eigenfunctions. With $y'(0) = y'(L) = 0$ instead, the eigenvalues are $\left(\frac{n\pi}{L}\right)^{2}$ for $n = 0, 1, 2, \dots$, with $y_n = \cos\frac{n\pi x}{L}$ (and $y_0 = 1$).
Why it works
Check all three signs of $\lambda$. If $\lambda = -\mu^{2} < 0$: $y = c_1\cosh\mu x + c_2\sinh\mu x$; $y(0) = 0$ gives $c_1 = 0$, and $c_2\sinh\mu L = 0$ forces $c_2 = 0$, since $\sinh$ is zero only at 0. If $\lambda = 0$: $y = c_1 + c_2x$, and both ends being 0 forces $y = 0$. If $\lambda = \mu^{2} > 0$: $y = c_1\cos\mu x + c_2\sin\mu x$; $y(0) = 0$ gives $c_1 = 0$, and $c_2\sin\mu L = 0$ allows $c_2 \ne 0$ exactly when $\mu L = n\pi$. So only a sequence of special values works, much as only special vectors are eigenvectors of a matrix.
Worked example: Fixed ends
Find the eigenvalues and eigenfunctions of $y'' + \lambda y = 0$, $y(0) = y(\pi) = 0$.
- $\lambda \le 0$ gives only $y = 0$ (as above).
- $\lambda = \mu^{2}$: $c_1 = 0$ and $\sin\mu\pi = 0$, so $\mu = n$.
Answer: $\lambda_n = n^{2}$, $y_n = \sin nx$, $n = 1, 2, \dots$
Worked example: One fixed end, one free end
Find the eigenvalues of $y'' + \lambda y = 0$, $y(0) = 0$, $y'(\pi) = 0$.
- $\lambda = 0$: $y = c_1 + c_2x$; $y(0) = 0$ gives $c_1 = 0$ and $y'(\pi) = c_2 = 0$. Not an eigenvalue. $\lambda < 0$ fails the same way, since $\cosh$ is never 0.
- $\lambda = \mu^{2}$: $y = c_2\sin\mu x$ and $y'(\pi) = c_2\mu\cos\mu\pi = 0$, so $\mu = \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \dots$
Answer: $\lambda_n = \frac{(2n - 1)^{2}}{4}$, $y_n = \sin\frac{(2n - 1)x}{2}$
Worked example: A boundary problem with data
Solve $y'' + 4y = 0$, $y(0) = 2$, $y\left(\frac{\pi}{4}\right) = 3$. Then try $y(0) = 0$, $y(\pi) = 1$.
- $y = c_1\cos 2x + c_2\sin 2x$ and $y(0) = c_1 = 2$.
- At $\frac{\pi}{4}$: $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2} = 1$, so $c_2 = 3$.
- For the second problem, $c_1 = 0$ and $y(\pi) = c_2\sin 2\pi = 0 \ne 1$.
Answer: $y = 2\cos 2x + 3\sin 2x$; the second problem has no solution
Common mistake
Jumping straight to $\lambda = \mu^{2} > 0$ and missing the cases $\lambda = 0$ and $\lambda < 0$. Each case has its own general solution ($\cosh$ and $\sinh$, a line, or $\cos$ and $\sin$), and for Neumann conditions $y'(0) = y'(L) = 0$ the case $\lambda = 0$ is a real eigenvalue, with the constant eigenfunction.