Solution: Let $y_1$ and $y_2$ be linearly dependent solution of the differential equation \begin{equation}\label{eq:05July2024-1} a_2(x) y'' + a_1(x) y' + a_0 (x) y = 0. \end{equation} We need to prove that the Wronskian is identically zero. Since $y_1$ and $y_2$ are linearly dependent, there exists $c_1, c_2 in \mathbb{R} $, both are not zero together, such that \[ c_1 y_1 (x) + c y_2 (x) = 0 \implies c_ 1y_1 '(x) + c y_2'(x) = 0 \] for each $x \in (a,b)$. Thus, we got a system of homogeneous equations in the variables $c_1 $ and $c_2$ with a nontrivial solution. So, \begin{align*} \det \begin{bmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \\ \end{bmatrix} = 0 \implies W(x) = 0 , \end{align*} where $W(x)$ is the Wronskian of $y_1$ and $y_2$.

For the other direction, let the Wronskian of $y_1$ and $y_2$ be zero, that is, \[ W(x) = \begin{vmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \\ \end{vmatrix} = 0, \text{ for } x \in (a,b). \] In particular, for $x = x_0$, \[ W(x_0) = \begin{vmatrix} y_1(x_0) & y_2(x_0) \\ y_1'(x_0) & y_2'(x_0) \\ \end{vmatrix} = 0. \] This implies the system \begin{align*} c_1 y_1(x_0) + c_2 y_2(x_0) = 0 \\ c_1 y_1'(x_0) + c_2 y_2'(x_0) = 0 \end{align*} has a nontrivial solution for $c_1, c_2$. Consider the function \begin{equation}\label{eq:05July2024-2} y(x) = c_1 y_1(x) + c_2 y_2(x). \end{equation}

Since $y_1$ and $y_2$ are solutions of \eqref{eq:05July2024-1}, $y(x)$ must also be solution of \eqref{eq:05July2024-1}. This implies, \[ y(x_0) = 0 = y'(x_0). \] We have the differential equation \[ a_2(x) y'' + a_1(x) y' + a_0 (x) y = 0, \text{ with } y(x_0) = y'(x_0) = 0. \] This equation has a unique solution, namely the trivial solution Thus, $y\equiv 0$. Hence, we have \[ c_1 y_1(x) + c_2 y_2(x) = 0, \text{ for } x \in (a,b), \] with $c_1, c_2$ not zero together. Thus, $y_1$ and $y_2$ are linearly dependent.