ODE : Existence and Uniqueness theorem 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.
 A: 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).
A: Suppose otherwise that the solution is nonzero, i.e. there is a $z(x)\neq 0$ that satisfies the homogenous equation. Then this would contradict uniqueness in the original theorem, because if you have a nonzero solution $y$ to the original theorem, then $y+z$ is also such a solution, since $y(x_0)+z(x_0)=y(x_0)=c_1$ and $y'(x_0)+z'(x_0)=y'(x_0)=c_2$. 
