Let \(f\) and \(g\) be solutions of
\begin{equation*}
y''+p(t)y'+q(t)y=0
\end{equation*}
on \(I=(a,b)\text{,}\) and define the Wronskian
\begin{equation*}
W(t)=W[f,g](t)=f(t)g'(t)-g(t)f'(t).
\end{equation*}
(Only if) If \(f\) and \(g\) are linearly dependent on \(I\text{,}\) then \(g=Cf\) for some constant \(C\text{.}\) Then \(g'=Cf'\text{,}\) so
\begin{equation*}
W(t)=f\cdot(Cf')-(Cf)\cdot f'=0
\end{equation*}
for all \(t\in I\text{.}\) Hence linear dependence implies \(W\equiv 0\text{.}\)
(If) Assume \(W(t)\equiv 0\) on \(I\text{.}\) Fix \(t_0\in I\) and define
\begin{equation*}
h(t)=g(t_0)f(t)-f(t_0)g(t).
\end{equation*}
Since the differential equation is linear and homogeneous, \(h\) is also a solution. Moreover,
\begin{equation*}
h(t_0)=g(t_0)f(t_0)-f(t_0)g(t_0)=0.
\end{equation*}
Also,
\begin{equation*}
h'(t)=g(t_0)f'(t)-f(t_0)g'(t),
\end{equation*}
so using \(W(t_0)=0\text{,}\) i.e. \(f(t_0)g'(t_0)-g(t_0)f'(t_0)=0\text{,}\) we obtain
\begin{equation*}
h'(t_0)=g(t_0)f'(t_0)-f(t_0)g'(t_0)=0.
\end{equation*}
By the existence and uniqueness theorem for the second-order initial value problem, the only solution satisfying \(h(t_0)=0\) and \(h'(t_0)=0\) is the zero solution. Hence \(h(t)\equiv 0\) on \(I\text{,}\) so
\begin{equation*}
g(t_0)f(t)-f(t_0)g(t)=0\quad\text{for all }t\in I.
\end{equation*}
If \(f(t_0)\neq 0\text{,}\) then
\begin{equation*}
g(t)=\frac{g(t_0)}{f(t_0)}\,f(t),
\end{equation*}
so \(g\) is a constant multiple of \(f\) and they are linearly dependent. If \(f(t_0)=0\text{,}\) choose a point in \(I\) where \(f\) is nonzero (since \(f\) is not identically zero), and the same argument applies.
Therefore,
\(f\) and
\(g\) are linearly dependent on
\(I\) if and only if
\(W[f,g](t)\equiv 0\text{.}\)
