A differential equation usually has infinitely many solutions. The equation $y' = y$ is solved by $y(x) = Ce^x$ for every constant $C$, and the equation $y'' = -y$ is solved by $y(x) = C_1\cos x + C_2\sin x$ for every pair of constants $C_1, C_2$. These sets of all solutions are the \textbf{general solutions}, and typically the general solution to an $n^{\text{th}}$ order DE depends on $n$ arbitrary constants. To single out one \textbf{particular solution}, we need $n$ extra conditions. This page is about where those conditions are placed, and why it matters.
In the graph, the blue curve is the solution, and the faint curves are the other members of the family it was picked from. Drag the red point (and the orange point, for the second order problems) to change the conditions.
\begin{definition}
An \textbf{initial value problem (IVP)} is a differential equation together with conditions on the solution and its derivatives, all at a single initial point $x_0$. For a second order equation,
$$\begin{cases} y'' = -y \\ y(x_0) = k_1 \\ y'(x_0) = k_2. \end{cases}$$
\end{definition}
So the initial conditions specify the \emph{value} of the solution at $x_0$ and its \emph{slope} at $x_0$. In the first order case, there is just one constant to determine, and so just one condition: $y' = y$, $y(0) = 2$ gives $C = 2$, so the solution is $y(x) = 2e^x$.
\begin{definition}
A \textbf{boundary value problem (BVP)} is a differential equation together with conditions at two different points, such as
$$\begin{cases} y'' = -y \\ y(x_0) = k_1 \\ y(x_1) = k_2. \end{cases}$$
\end{definition}
This looks like a small change, but the behavior is quite different. Flip between the two second order problems and watch the family: in both, the curves shown are the solutions passing through the red point. Choosing a slope always picks out exactly one of them. Choosing a second point may not. As it turns out, this comes down to linear algebra. Imposing the conditions on $y(x) = C_1\cos x + C_2\sin x$ gives a system of two linear equations in the unknowns $C_1, C_2$, and everything we know about $A\mathbf{x} = \mathbf{b}$ applies.
\begin{proposition}
For $y'' = -y$, the initial value problem above has exactly one solution for all $x_0, k_1, k_2$. The boundary value problem has
\begin{itemize}
\item exactly one solution, if $x_1 - x_0$ is not an integer multiple of $\pi$;
\item otherwise, with $x_1 - x_0 = n\pi$, infinitely many solutions if $k_2 = (-1)^n k_1$, and no solutions if not.
\end{itemize}
\end{proposition}
\begin{proof}
Using $y'(x) = -C_1\sin x + C_2\cos x$, the two problems ask us to solve
$$\underbrace{\begin{bmatrix} \cos x_0 & \sin x_0 \\ -\sin x_0 & \cos x_0 \end{bmatrix}}_{\text{IVP}} \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} = \begin{bmatrix} k_1 \\ k_2 \end{bmatrix},$$
$$\underbrace{\begin{bmatrix} \cos x_0 & \sin x_0 \\ \cos x_1 & \sin x_1 \end{bmatrix}}_{\text{BVP}} \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} = \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}.$$
The IVP matrix has determinant $\cos^2 x_0 + \sin^2 x_0 = 1$, so it is always nonsingular and there is always exactly one solution. The BVP matrix has determinant
$$\cos x_0 \sin x_1 - \sin x_0 \cos x_1 = \sin(x_1 - x_0),$$
which is nonzero unless $x_1 - x_0 = n\pi$. In that case, $\cos x_1 = (-1)^n\cos x_0$ and $\sin x_1 = (-1)^n\sin x_0$, so the second row of the matrix is $(-1)^n$ times the first. The system is then consistent exactly when $k_2 = (-1)^n k_1$. If so, there is a free variable and so infinitely many solutions. If not, there are none.
\end{proof}
\begin{remark}
The picture agrees. Every solution through the red point has the form $y(x) = k_1\cos(x - x_0) + c\sin(x - x_0)$, and the sine term vanishes at $x = x_0 + n\pi$. So the whole family pinches together at the single point $\big(x_0 + n\pi,\ (-1)^n k_1\big)$. Move the orange point onto that pinch and every curve is a solution. Move it straight up or down from there and no curve can reach it. The buttons under \emph{Try a case} set up the three situations with $x_0 = 0$.
\end{remark}