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 $W[y_1, \ldots, y_n](t) = 0$.
From the definition
$W(u, v) = \det \begin{bmatrix} u & v \\ u' & v' \end{bmatrix} = uv' - vu' \tag 1$
we find
$W'(u, v) = u'v' + uv'' - v'u' - vu'' = uv'' - vu''; \tag 2$
using the fact that $u$ and $v$ are both solutions of
$y'' + P(x)y' + Q(x)y = 0, \tag 3$
we may transform (2) to
$W'(u, v) = u(-P(x)v' - Q(x)v) - v(-P(x)u' - Q(x)u) = -P(x)uv' - Q(x)uv + P(x)vu' + Q(x)uv = -P(x)(uv' - vu') = -P(x)W(u, v); \tag 4$
the solution to this ordinary linear equation for $W(u, v)$ is readily seen to be
$W(u, v)(x) = \exp \left (-\displaystyle \int_{x_0}^x P(s) \; ds \right )W(u, v)(x_0); \tag 5$
under fairly mild conditions on $P(x)$ we may affirm that
$\forall x, \; \exp \left (-\displaystyle \int_{x_0}^x P(s) \; ds \right ) \ne 0, \tag 6$
and thus
$W(u, v)(x_0) \ne 0 \Longrightarrow \forall x, W(u, v)(x) \ne 0, \tag 7$
and likewise,
$W(u, v)(x_0) = 0 \Longrightarrow \forall x, W(u, v)(x) = 0; \tag 8$
these well-known results (6)-(8) establish a bidirectional linkage 'twixt our OP sat091's hypotheses (a) and (b); certainly (b) implies (a), and from what we have done here we have been able to conclude that the weaker (a) yields (b); thus our OP sat091 is correct on both counts.
Best Answer
There are two points which are wrong:
First, the Wronskian of $x^2$ and $1$ is not identically $0$:
$$W(y_1, y_2)(x) = \begin{vmatrix} x^2 & 1 \\ 2x & 0 \end{vmatrix} = -2x.$$
Second: It is true for all differentiable functions $y_1$ and $y_2$ that: If $W(y_1, y_2) (x_0) \neq 0$ for at least one $x_0$, then $y_1$ and $y_2$ are linearly independent. But the backwards direction is not true in general, i.e., if $y_1$ and $y_2$ are linearly independent, then their Wronskian is not necessarily non-zero. Consider as a counterexample $y_1 = x^2/2$ and $y_2 = x\lvert x \rvert/2$. The backwards direction is true if you additionally assume that $y_1$ and $y_2$ are solutions of a linear second order differential equation.