Let $q:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. If $x_1(t)$ and $x_2(t)$ are the solutions to ODE: $\ddot{x}=q(t)x$ on $(a,b)$, show that Wronskian determinate $$ \begin{vmatrix} x_1(t) & x_2(t)\\ \dot{x_1}(t)& \dot{x_2}(t) \end{vmatrix} $$
is time independent on $(a,b)$.
I am not sure how to start this problem at all and I would appreciate any help.
Edit 1:
Edit 2:
Edit 3: Edits 1 and 2 have been summarised in the answer provided down.
The answer is the summary of the hints and help from the comments and my calculation:
First, we calculate the derivative of the Wronskian:
$$ \frac{d}{dt}\begin{vmatrix} x_1(t) & x_2(t)\\ \dot{x_1}(t)& \dot{x_2}(t) \end{vmatrix} = \dot{x_1}(t) \dot{x_2}(t)+x_1(t) \ddot{x_2}(t)- \ddot{x_1}(t)x_1(t)- \dot{x_1}(t) \dot{x_2}(t) $$
$$ = \begin{vmatrix} x_1(t) & x_2(t)\\ \ddot{x_1}(t)& \ddot{x_2}(t) \end{vmatrix} + \begin{vmatrix} \dot{x_1}(t) & \dot{x_2}(t)\\ \dot{x_1}(t)& \dot{x_2}(t) \end{vmatrix}$$
but since the second determinate is equal to zero we are left with
$$ = \begin{vmatrix} x_1(t) & x_2(t)\\ \ddot{x_1}(t)& \ddot{x_2}(t) \end{vmatrix}$$
Since $x_1$ and $x_2$ are the solutions of $\ddot{x}=q(t)x$ then we can substitute them in our determinante:
$$ = \begin{vmatrix} x_1(t) & x_2(t)\\ q(t)x_1 & q(t)x_2 \end{vmatrix} = x_1q(t)x_2(t)-x_2q(t)x_1 = 0 \Rightarrow \dot{W}=0$$
Since $\dot{W}=0$, it follows that it´s time-independent.