Tangent solution of differential equation of second degree Consider the following ODE on the interval $[a,b]$
$ y''+p(x)y'+q(x)y=0 $
such that $p$ and $q$ are continuous functions in $[a,b]$.
Show that if a solution of the above ODE is tangent to the $x$ axis, then is null in $[a,b]$
What I did say was that if $y_{1}$ is a solution of the ODE and it is tangent to the $x$ axis at a point, then 
$y_{1} (x_{0}) = 0$, $y'_{1} (x_{0}) = 0$, $a <x_{0} <b$. 
Then replacing it in the differential equation can be deduced that $y_{1}'' (x_{0})$ = 0, 
then use a theorem of existence and uniqueness. 
Can anyone give me a hand 
thank you very much

The solution the OP gave is basically correct, and the reasoning not fundamentally flawed, though introduction of the fact that $y_1''(x_0) = 0$ is perhaps a trifle redundant.  This problem indeed illustrates an important principle in the theory of linear ordinary differential equations, that being as follows:  let $\vec r(t)$ satisfy the linear, time dependent system
$\dot{\vec r} = A(t) \vec r, \tag{1}$
where $A(t)$ is a continuous $n \times n$ matrix function of $t$; then the only solution which takes the value $\vec r(t_0) = 0$ at some $t_0 \in \Bbb R$ the identically zero solution $r(t)  = 0$ for all $t \in D \subset \Bbb R$, where $D$ is the domain (interval) on which the system is defined; this assertion indeed follows from the uniqueness of solutions which always applies to a linear system with continuous c0efficients, since then $A(t)\vec r$ is easily seen to Lipschitz continuous in $\vec r$ and continuous in $t$, which are the necessary hypotheses to insure existence and uniqueness of solutions.  So, if $\vec r(t_0) = 0$, $\vec r(t_0)$ agrees with the zero solution at $t_0$, hence everywhere in $D$ by uniqueness.
This concept applies to the OP's problem in the following manner:  with
$y''+p(x)y'+q(x)y=0 \tag{2}$
we can set
$z = y' \tag{3}$
so that (2) becomes
$z' = -p(x)z - q(x) y, \tag{4}$
and then setting
$\vec r = \begin{pmatrix} y \\ z \end{pmatrix} \tag{5}$
we then write (2) as the two-dimensional system
$\vec r'(x) = A(x)r(x) \tag{6}$
where
$A(x) = \begin{bmatrix} 0 & 1 \\ -q(x) & -p(x) \end{bmatrix}. \tag{7}$
The condition that the graph of $y_1(x)$ be tangent to the $x$-axis at $x_0$ translates to $y_1(x_0) = 0$ (since the graph must meet the axis) and $z_1(x_0) = y_1'(x_0) = 0$ (since the graph is tangent to the axis at the point of meeting).  Thus we have
$\vec r_1(x_0) = \begin{pmatrix} y_1(x_0) \\ z_1(x_0) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \tag{8}$
and hence $\vec r_1(x) = 0$ for all $x \in [a, b]$ in light of the general situation described in the previous paragraph.
Hope this helps.  Cheerio,
and as always, 
Fiat Lux!!!
