Regular Sturm-Liouville, one-dimensional eigenfunctspace

[usn7564]

Sturm-Liouville problem:
$$\frac{d}{dx} (r(x) \frac{dX}{dx}) + q(x)X(x) + \lambda p(x) X(x) = 0 \quad x \in \left[a, b \right]$$
$$a_1 X(a) + a_2X'(a) = 0$$
$$b_1 X(b) + b_2X'(b) = 0$$
$$r, r', q, p \in \mathbb{C} \forall x \in in \left[a, b \right]$$

Theorem:
Under the additional condition that r(a)>0 (or r(b) > 0), the S.L BVP cannot have two linearly independent eigenfunctions corresponding to the same eigenvalue.

Proof:
X, Y two eigenfunctions corresponding to the same eigenvalue. Then Z = Y(a)X(x) - X(a)Y(x) is also a solution and it can be shown that [skipping how to get z(a), z'(a) as there's no real problem there for me]

$$z(a) = z'(a) = 0 \implies z(x) = 0 \forall x \in \left[a, b \right]$$
QED, catfoots and other scribbles.

Question:
I don't get the final implication at all. Why does knowing that z(a) and z'(a) = 0 at x = a show that z(x) is identically zero over the interval in question?

 

[Orodruin]

z is a solution to the SL problem with boundary conditions z(a) and z'(a)=0. The trivial solution f(x)=0 also is. Uniqueness leads to z=0.

 

[usn7564]

Explains why it wasn't elaborated further, feel silly now. Thank you, need to remind myself of BVP's it seems.