There cannot be two linearly independent eigenfunctions in a Sturm–Liouville problem Consider the Sturm–Liouville problem
$y'' + [λp(x) − q(x)]y = 0$,
$α_1y(a) + α_2y'(a) = 0$,
$β_1y(b) + β_2y'(b) = 0$,
with $p(x)$ and $q(x)$ continuous, $p(x) > 0$ on $[a, b]$, and $α_1β_1 > 0$. Suppose that $α_1α_2 \neq 0$. Show that for each $λ_n$ there cannot be two linearly independent eigenfunctions $φ_n(x, λ_n)$ and $ψ_n(x, λ_n)$. 
I'm very unsure of how to approach this. I was thinking about using the Wronskian but I'm wondering if there is some formula for $φ_n(x, λ_n)$ and $ψ_n(x, λ_n)$ that I would need to use to approach it that way. I'm also struggling to understand how the fact that $α_1β_1 > 0$ factors into it at all. Any help would be greatly appreciated.
 A: The following problem has a unique solution
$$
             y''+(\lambda-q)y = 0, \\
             y(a)=A,\; y'(a)=B. \tag{*}
$$
This follows from standard theories of ODEs, assuming $q$ is continuous on the interval $[a,b]$ of interest. Therefore, if you have two non-trivial solutions $y_1,y_2$ of
$$
                y''+(\lambda-q)y = 0\\
             \alpha_1 y(a)+\alpha_2 y'(a) = 0,              
$$
then it is possible to find non-zero scale factors $C$ and $D$ such that
$$
              Cy_1(a) = -\alpha_2,\;\; Cy_1'(a)= \alpha_1 \\
              Dy_2(a) = -\alpha_2,\;\; Dy_2'(a)= \alpha_1
$$
It then follows that $Cy_1-Dy_2$ is identically $0$ by uniqueness of solutions for problem $(*)$. Hence $y_1,y_2$ are linearly dependent if they satisfy the system (*). And that is true, even without imposing a second condition at $b$. So there cannot be two independent solutions of the problem with endpoint conditions at both ends; there cannot be two independent solutions with only one such endpoint condition.
