Just to make it explicit which theorem i am considering here, let me first state the Existence and Uniqueness theorem for a second order linear differential equation.
Theorem: Consider an equation of the form
$$a_0 y'' + a_1 y' + a_2 y = r(x),$$
where $a_0(x)$, $a_1(x)$, $a_2(x)$ and $r(x)$ are continuous functions on an interval $(a,b)$ and $a_0(x) \ne 0$ for each $x \in (a,b)$. Let $c_1$ and $c_2$ be arbitrary real numbers and $x_0 \in (a,b)$. Then there exists a unique solution $y(x)$ defined over $(a,b)$ for the above equation satisfying $y(x_0) = c_1$ and $y'(x_0) = c_2$.
Now the author also mentions below corollary to the above mentioned theorem.
Corollary: If $y(x)$ be a solution to
$$a_0 y'' + a_1 y' + a_2 y = 0$$
(note $r(x) = 0$ here) satisfying $y(x_0) = 0$ and $y'(x_0) = 0$ for some $x_0 \in (a,b)$, then $y(x)$ is identically zero on $(a,b)$.
I am not able to convince myself as to why this corollary is valid. Need help with an informal proof to understand why this corollary holds good.
Thanks in advance.
Given that the stated Theorem holds, the Corollary follows as follows:
Consider any solution $y(x)$ to
$$a_0 y''(x) + a_1 y'(x) + a_2 y(x) = 0 \tag 1$$
on $(a, b)$ such that
$y(x_0) = y'(x_0) = 0; \tag 2$
by our theorem, $y(x)$ the unique--the only--solution to (1) such that (2) binds.
Now consider the function
$z(x) = 0, \; x \in (a, b); \tag 3$
clearly
$z(x_0) = z'(x_0) = 0, \tag 4$
and obviously
$a_0 z''(x) + a_1 z'(x) + a_2 z(x) = a_0 \cdot 0 + a_1 \cdot 0 + a_2 \cdot 0 = 0; \tag 5$
so $z(x)$ satisfies (1) as well; furthermore, by (2) and (4), $z(x)$ and $z'(x)$ agree with $y(x)$ and $y'(x)$, respectively, at $x_0$; but $y(x)$ is the only solution satisfying (2), so we must have
$z(x) = y(x), \; \forall x \in (a, b); \tag 6$
the requisite conclusion now follows by virtue of (3).