A boundary value problem (often abbreviated to "BVP") for a second-order linear ODE also entails prescribing two pieces of data,
one at each of two different points $x_0$ and $x_1$ (such as endpoints of an interval: its "boundary").
The points $\pm \infty$ arise if the interval is unbounded in some direction, and we may impose limiting behavior there.
Among the many types of condition we may impose on boundary values are Dirichlet boundary conditions and Neumann boundary conditions:
A Dirichlet boundary condition on a function $X:[0,L] \to \R$ is the requirement
$$X(0) = a,\,\,\, X(L) = b$$
for given scalars $a, b \in \R$. This is specifying values at the boundary of the interval $[0, L]$.
A Neumann boundary condition on a function $X:[0,L] \to \R$ is the requirement
$$X'(0) = a,\,\,\, X'(L) = b$$
for given scalars $a, b \in \R$.
Boundary value problems I: Dirichlet conditions.
For $L > 0$, consider the following Dirichlet BVP
$$X'' = \lambda X;\,\,\,X(0) = a, X(L)=b.$$
This example is extensively worked out in the textbook. We know it has exactly one solution except when
$\lambda = - n^2\pi^2/L^2$ for some $n = 1, 2, 3, \dots$. This unique solution is:
$X(x) = \dfrac{b-a}{L}\,x + a$ when $\lambda = 0$,
$X(x) = a\left(\cos(\sqrt{|\lambda|}x) - \dfrac{\cos(\sqrt{|\lambda|}L)}{\sin(\sqrt{|\lambda|}L)} \sin(\sqrt{|\lambda|}x)\right)
+ \dfrac{b}{\sin(\sqrt{|\lambda|}L)} \sin(\sqrt{|\lambda|}x)$ when $\lambda < 0$ with $\lambda \ne - n^2\pi^2/L^2$.
If $\lambda = - n^2\pi^2/L^2$ for some $n = 1, 2, 3, \dots$ then
there is no BVP solution unless $b = (-1)^n a$. In these latter cases, the solutions are
$X(x) = A \sin((n\pi/L)x) + a\cos((n\pi/L)x)$ for all $A \in \R$.
Below is a visualization of this Dirichlet BVP with $\lambda \ne - n^2\pi^2/L^2$.
We visualize the solutions to the following three Dirichlet BVP's:
$X_{a,0}$ satisfying $X_{a,0}(0)=a$ and $X_{a,0}(L)=0$.
$X_{0,b}$ with $X_{0,b}(0)=0$ and $X_{0,b}(L)=b$.
$X_{a,b}$ with $X_{a,b}(0)=a$ and $X_{a,b}(L)=b$.
Observe how the solutions change with $\lambda$, and also observe the additivity of solutions.
Below is a visualization of the above Dirichlet BVP with $\lambda = - n^2\pi^2/L^2$.
We shall assume here that $L = \pi$ (so $n\pi/L$ simply becomes $n$). We require $b = (-1)^n a$ since this is necessary
for the existence of BVP solutions for when $\lambda = -n^2$ for a positive integer $n$.
Here we visualize the solutions $X(x) = A \sin(n x) + a\cos(n x)$.