Ratio of solutions of differential equation 
Prove, that the ratio of two linearly independent solutions of
$$
x''+p(t)x'+q(t)x=0,
$$
where $p(t)$, $q(t)$ are continuous coefficients, can not have a local maximum.


I rewrote this as a system of equations:
$$
\begin{cases}
y_1'=y_2 \\
y_2'+p(t)y_2+q(t)y=0
\end{cases}
$$
I multiplied the second equation by $e^{\int pdt}$ and got:
$$
(y_2e^{\int pdt})'=-q(t)e^{\int pdt}y_1
$$
and after integrating both sides:
$$
y_2e^{\int pdt}=\int-q(t)e^{\int pdt}y_1dt
$$
Unfortunately, I don't know what to do next.
Also, I had an idea to assume one of the solutions to be $V=\begin{pmatrix} v_1\\ v_2 \end{pmatrix}$ and then try to get the oter one from the Liouville's formula, however I don't know whether that would work.
 A: Here is an outline of a solution.  Let $y_1, y_2$ be two linearly independent solutions and write $z= y_1/y_2$.  By symmetry, the sam argument can be applied to $y_2/y_1$.  Then
\begin{align*}
z' = \frac{y_1'y_2 - y_2'y_1}{y_2^2} \tag{1}\label{A}
\end{align*}
and, using the original differential equation
\begin{align*}
z'' &= -2y_2'\frac{y_1'y_2-y_2'y_1}{y_2^3}+\frac{y_1''y_2+y_1'y_2'-y_1'y_2'-y_2''y_1}{y_2^2} \\
&=-2y_2'\frac{y_1'y_2-y_1y_2'}{y_2^3}+\frac{(-p y_1'-qy_1)y_2-(-py_2'-qy_2)y_1}{y_2^2} \\
&=-2y_2'\frac{y_1'y_2-y_1y_2'}{y_2^3}-p\frac{y_1'y_2-y_2'y_1}{y_2^2} \tag{2}\label{B}
\end{align*}
so that
\begin{align*}
z'' + \left( 2\frac{y_2'}{y_2}-p \right) z' = 0
\end{align*}
and the function $z'$ satisfies a first order linear differential equation.  Now, if $y_1 / y_2$ has a local maximum or minimum then $(y_1/y_2)' = 0$ at some point, which is to say $y_1'y_2 - y_1y_2' = 0$ at that point.  The equation \eqref{A} for $z'$ implies $z' = 0 $ at the same point.  We now have $z'$ satisfies a first order linear differential equation with $z'=0$ at a point, which we can take to be the initial condition, so that the solution $z'$ must be identically zero.  The function $z$ is therefore constant and $y_1$ is a multiple of $y_2$, contradicting their linear independence.
