Principle of Superposition and Wronskian 
Assume that $p$ and $q$ are continuous and that the functions $y_1$ and $y_2$ are solutions of the differential equation $$y''+p(t) y'+q(t)y=0$$ an open interval $I$. Prove that if $y_1$ and $y_2$ are zero at the same point in $I$, then they cannot be a fundamental set of solutions on that interval.

If $y_1$ and $y_2$ are zero at the same point does this cause the determinant of Wronskian matrix is equal zero?
If so is this the reason why there cannot be a fundamental set of solutions on that interval.
Thanks in advance for any help.
 A: The Wronskian matrix $W(y_1, y_2)$ of two solutions $y_1$ and $y_2$ of
$y'' + p(t)y'(t) + q(t)y(t) = 0 \tag 1$
may be defined as
$W(y_1, y_2) = \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix}, \tag 2$
with determinant
$\Delta_W = \vert W(y_1, y_2) \vert = \det \left (\begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} \right ) = y_1y_2' - y_2 y_1'; \tag 3$
we calculate
$\Delta_W' = y_1'y_2' + y_1y_2'' - y_2'y_1' - y_2y_1'' = y_1y_2'' - y_2y_1''; \tag 4$
we may now use (1) in the form
$y_i'' = -py_i' - qy_i, \; i = 1, 2, \tag 5$
to transform (4) to
$\Delta_W' = y_1(-py_2' -qy_2) - y_2(-py_1' - qy_1) = -py_1y_2' - qy_1y_2 + py_1'y_2 + qy_1y_2 = -p(y_1y_2' - y_1'y_2) = -p \Delta_W, \tag 6$
a simple first order, linear ordinary differential equation for $\Delta_W$; the solutions of this equation are 
$\Delta_W(t) = \Delta_W(t_0) e^{-\int_{t_0}^t p(s)\; ds}; \; t_0, t \in I, \tag 7$
which the reader may easily check. It follows from this equation and the uniquenness of solutions that 
if
$\Delta_W(t_0) = 0, \tag 8$
then
$\Delta_W(t) = 0, \; \forall t \in I; \tag 9$
thus if
$y_1(t_0) = y_2(t_0) = 0, \tag{10}$
it follows that
$\Delta_W(t_0) = 0, \tag{11}$
and (9) binds as well.  Thus $y_1$ and $y_2$ cannot be a fundamental solution system for (1) on $I$ since fundamental systems are characterized by the non-vanishing of $\Delta_W$ everywhere, that is, by the linear independence of the columns of (2); but of course when the columns are linearly dependent then $W(y_1, y_2) = 0$.
Finally, as pointed out by user539887 in his comment to the question itself, it's not that a fundamental system doesn't exist but that $y_1$, $y_2$ does not form one.  A fundamental solution system always exists, as may be seen by taking
$W(t_0) = I = \begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix}, \tag{12}$
for then
$\Delta_W(t) \ne 0, \; \forall t \in I, \tag{13}$
as follows from (7) since
$\Delta_W(t_0) = 1 \tag{14}$
in this case.
