Suppose that $y_1(t), \ldots, y_n(t)$ are solutions of $\frac{d^n y}{dt} + p_{n-1}(t) \frac{d^{n-1} y}{dt} + \cdots + p_1(t) \frac{dy}{dt} + p_0(t) y = 0$, and suppose that their Wronskian is zero for $t = t_0$, i.e.
\begin{equation*}
\left|
\begin{array}{cccc}
y_1(t_0) & y_2(t_0) & \cdots & y_n(t_0) \\
y_1'(t_0) & y_2'(t_0) & \cdots & y_n'(t_0) \\
\vdots & \vdots & \ddots & \vdots \\
y_1^{(n-1)}(t_0) & y_2^{(n-1)}(t_0) & \cdots & y_n^{(n-1)}(t_0)
\end{array}
\right| = 0.
\end{equation*}
Then the corresponding matrix is not invertible, and the system of equations
\begin{array}{c}
c_1 y_1(t_0) &+& c_2 y_2(t_0) &+& \cdots &+& c_n y_n(t_0) &=& 0 \\
c_1 y_1'(t_0) &+& c_2 y_2'(t_0) &+& \cdots &+& c_n y_n'(t_0) &=& 0 \\
\vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\
c_1 y_1^{(n-1)}(t_0) &+& c_2 y_2^{(n-1)}(t_0) &+& \cdots &+& c_n y_n^{(n-1)}(t_0) &=& 0 \\
\end{array}
has a nontrivial solution for $c_1, c_2, \ldots, c_n$ not all zero.
Let $y(t) = c_1 y_1(t) + \cdots + c_n y_n(t)$. Because $y(t)$ is a linear combination of solutions of the differential equation, $y(t)$ is also a solution of the differential equation. Additionally, because the weights satisfy the above system of equations, we have $y(t_0) = y'(t_0) = \cdots = y^{(n-1)}(t_0) = 0$.
These initial conditions and the original differential equation define an initial-value problem, of which $y(t)$ is a solution. If $p_0(t), p_1(t), \ldots, p_{n-1}(t)$ are continuous, then any initial-value problem associated with the differential equation has a unique solution. Obviously $y^*(t) = 0$ is a solution of the initial-value problem; since we know that $y(t)$ is also a solution of the same initial-value problem, it follows that $y(t) = 0$ for all $t$, not just $t = t_0$.
We now have $c_1 y_1(t) + \cdots + c_n y_n(t) = 0$ for all $t$, where $c_1, \ldots, c_n$ are not all zero. Thus the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent.
Conversely, if the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent, then the system of equations
\begin{array}{c}
c_1 y_1(t) &+& c_2 y_2(t) &+& \cdots &+& c_n y_n(t) &=& 0 \\
c_1 y_1'(t) &+& c_2 y_2'(t) &+& \cdots &+& c_n y_n'(t) &=& 0 \\
\vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\
c_1 y_1^{(n-1)}(t) &+& c_2 y_2^{(n-1)}(t) &+& \cdots &+& c_n y_n^{(n-1)}(t) &=& 0 \\
\end{array}
has a nontrivial solution for every $t$, the corresponding matrix is not invertible for any $t$, and $W[y_1, \ldots, y_n](t) = 0$.
In order for $f(x)$ and $g(x)$ to be linearly dependent functions, there have to be constants $a$ and $b$ such that
$$
af(x)+bg(x)=0
$$
for all $x$. You have found constants that work for one specific value of $x$, but they won't work for other values.
Notice that I haven't said anything about the Wronskian yet. There are two relevant facts about the Wronskian:
Fact 1: If the Wronskian of $f_1,f_2,\dots,f_n$ is nonzero at any point, then $f_1,\dots,f_n$ are linearly independent.
Fact 2: If $f_1,f_2,\dots,f_n$ are solutions to a nonsingular $n$th-order differential equation, then their Wronskian is nonzero at every point if they're linearly independent, and zero at every point if they're linearly dependent.
What this means is, if you're given any $n$ functions, there are four possible options:
- The Wronskian never vanishes, and those functions could all be linearly independent solutions to the same nonsingular $n$th-order ODE.
- The Wronskian vanishes in some places and not in others. If you got the functions as solutions to an $n$th-order ODE, either that ODE has singularities at the points where the Wronskian vanishes, or you screwed up somehow.
- The Wronskian vanishes everywhere, but the functions are linearly independent anyway. In practice, this doesn't happen very often.
- The functions are linearly dependent. Their Wronskian vanishes everywhere.
The functions $\sin 2x$ and $\cos 2x$ are examples of option 1; whatever functions you computed to have Wronskian $4x^2$ would be examples of option 2 (that is, any ODE which has them as solutions must be singular at $0$).
Best Answer
$$y_1y_2'-y_1'y_2=0$$ $$\frac{y_1y_2'-y_1'y_2}{y_1^2}=0$$ $$\frac{y_2}{y_1}=C$$
Unless I'm mistaken, this appears to imply if the Wronskian is $0$, the solutions are linearly dependent.