Suppose that that \(f(t)\) and \(g(t)\) are linearly dependent on an interval \(I = (a, b)\text{.}\) Then one function is a multiple of the other, say \(f(t) = c g(t)\text{.}\) Thus, \(f'(t) = cg'(t)\text{.}\)
\begin{equation*}
W(f, g)(t)
=
\det
\begin{pmatrix}
f(t) & g(t) \\
f'(t) & g'(t)
\end{pmatrix}
= f(t) g'(t) - f'(t) g(t)
= c g(t) g'(t) - cg'(t) g(t)
=
0.
\end{equation*}
Conversely, suppose that
\begin{equation*}
W(f, g)(t)
=
\det
\begin{pmatrix}
f(t) & g(t) \\
f'(t) & g'(t)
\end{pmatrix}
= 0,
\end{equation*}
for all \(t\) in \((a, b)\text{.}\) If \(g = 0\text{,}\) then \(0 f = g\) and the two functions are linearly dependent. Assume that \(g(t_0) \neq 0\) for some \(t_0\) in \((a, b)\text{.}\) Since \(g\) is differentiable, it must also be continuous and there is some interval \((c, d)\) contained in \((a, b)\) such that \(t_0 \in (c, d)\) and \(g\) does not vanish on this interval. Therefore,
\begin{equation*}
\frac{d}{dt} \left( \frac{f}{g} \right) = \frac{f' g - f g'}{g^2} = - \frac{W(f,g)}{g^2} = 0,
\end{equation*}
and \(f/g\) is constant on the interval \((c, d)\text{.}\) Thus, \(f(t_0) = c g(t_0)\) and \(f'(t_0) = c g'(t_0)\text{.}\) Since \(f\) and \(cg\) are both solutions to the differential equation \(y'' + p y' + q y = 0\) and have the same initial condition, \(f(t) = cg(t)\) for all \(t \in (a, b)\) by the existence and uniqueness theorem. Consequently, \(f\) and \(g\) are linearly dependent.
