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

If $p,q$ are continuous then the initial value problem for $y''+p(x)y'+q(x)y=0$ has a unique solution for each initial data set in $[a,b]$. Consider then, the zero solution $y=f(x)=0$ for all $x \in [a,b]$ is a solution since it is clear that $f''(x)+p(x)f'(x)+q(x)f(x)=0$ for all $x \in [a,b]$. Moreover, if $x_o \in (a,b)$ such that $y(x_o)=0$ and $y'(x_o)=0$ as the solution must be a horizontal tangent since the $x$-axis is horizontal. But, $y(x_o)=0, y'(x_o)=0$ is a set of initial values hence it uniquely determines a solution on the interval $[a,b]$. It follows that this solution must be the null solution by uniqueness of solution to the initial value problem. This theorem I'm using is usually called the "fundamental existence and uniqueness theorem for ODEs". It ought to be in whatever text you're using.
