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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?

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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:

  1. 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 $y_0, y_1$.
  2. 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.