Proving Wronskian of $(\star)$ System using Hint

[evinda]

Hello! (Wave)

Prove that the determinant of the following system $(\star)$ is the Wronskian.$$(\star) \begin{pmatrix}
y_1(s) & -y_2(s)\\
-y_1'(s) & y_2'(s)
\end{pmatrix} \begin{pmatrix}
c_1(s)\\
c_2(s)
\end{pmatrix}=\begin{pmatrix}
0\\
\frac{1}{p(s)}
\end{pmatrix}$$Hint: Write the equation $(py')'+qy=0$ in the form $y'=z, z'=-\frac{(p'z+qy)}{p}$.Could you explain to me how we can use the hint to show that the determinant is indeed the Wronskian?How does it help to write the equation $(py')'+qy=0$ in the form $y'=z, z'=-\frac{(p'z+qy)}{p}$?

 

[gtoguy97]

The hint helps to solve the system $(\star)$ by writing it in the form of a first-order linear system. Specifically, we can express the two equations of the system as:$$y_1'(s) = -\frac{p'(s) y_2(s) + q(s)y_1(s)}{p(s)} $$$$y_2'(s) = \frac{p'(s) y_1(s) + q(s)y_2(s)}{p(s)}$$These equations can be written in matrix form as follows:$$\begin{pmatrix}y_1'(s) \\ y_2'(s)\end{pmatrix} = \begin{pmatrix}-\frac{p'(s)}{p(s)} & \frac{q(s)}{p(s)} \\\frac{q(s)}{p(s)} & \frac{p'(s)}{p(s)}\end{pmatrix} \begin{pmatrix}y_1(s) \\ y_2(s)\end{pmatrix} $$Now, we can multiply both sides of this equation by the inverse of the coefficient matrix and rearrange terms to obtain the desired system $(\star)$.The determinant of the coefficient matrix in the first equation is the Wronskian. This is because the Wronskian of two solutions $y_1(s)$ and $y_2(s)$ of the second-order ODE $(py')' + qy = 0$ is defined as$$W(y_1, y_2) = y_1(s) y_2'(s) - y_2(s)y_1'(s).$$Since the determinant of the coefficient matrix is equal to the Wronskian, it follows that the determinant of the system $(\star)$ is also the Wronskian.