Why do '2-point boundary conditions' provide a less powerful theorem for telling us that a fixed solution to a DE always exists? Looking through my current lecture notes, it states a theorem for existance and uniqueness, stating the following:
Consider a second order linear ODE of the form
$$
y'' + p(x)y' + q(x)y = 0
$$
Now, if the real functions $p(x), q(x)$ are continuous on some interval $I$, given by $a \leq x \leq b$, then this equation has some general solution defined on the interval $I$.
The notes then go on to say that for any $\alpha, \beta \in \mathbb{R}$ and any $x_{0} \in I$, there exists a solution to satisfy the below boundary conditions $$
y(x_{0}) = \alpha, y'(x_{0}) = \beta
$$
This much I understand/accept, but the next part states that if the boundary conditions are given in the form
$$
y(x_{1}) = y(x_{2}) = 0
$$
then we have a 'much less powerful theorem for proving that a nontrivial solution always exists'.
What exactly is meant by this, and why is this the case?
 A: Two-point boundary value problems for second-order differential equations can fail to have solutions, and when solutions exist they may be nonunique.  Consider for example 
$$ y'' + y = 0$$
for which the general solution is
$$ y = a \cos(x) + b \sin(x)$$
Thus all solutions are periodic: $y(0) = y(2\pi)$.  If you specify boundary conditions $y(0) = y_0$, $y(2\pi) = y_1$, then you either have no solution (if $y_0 \ne y_1$) or infinitely many solutions (if $y_0 = y_1$).
More generally, for the d.e. 
$$ y'' + p(x) y' + q(x) y = 0 $$
the general solution is of the form
$$ y = a Y_1 + b Y_2 $$
where $Y_1, Y_2$ are linearly independent functions.
Consider boundary conditions $y(x_0) = y_0$, $y(x_1) = y_1$.
In terms of the general solution, this requires solving
the linear system 
$$ \pmatrix{Y_1(x_0) & Y_2(x_0)\cr Y_1(x_1) & Y_2(x_1)\cr} \pmatrix{a\cr b\cr} = \pmatrix{y_0\cr y_1\cr} $$
Then there are two possible cases, depending on the rank
of the coefficient matrix:


*

*If $Y_1(x_0) Y_2(x_1) - Y_2(x_0) Y_1(x_1) \ne 0$, the rank is $2$.  The two-point boundary problem has a unique solution for all $y_0, y_1$.

*If $Y_1(x_0) Y_2(x_1) - Y_2(x_0) Y_1(x_1) = 0$, the rank is $1$. There is some  $(v_0,v_1)$ (with $v_0, v_1$ not both $0$) such that the two-point boundary problem has no solution unless $(y_0, y_1)$ is a scalar multiple of $(v_0,v_1)$; if $(y_0, y_1)$ is a scalar multiple of $(v_0,v_1)$ then the solution is non-unique. 

