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I'm not sure how to find the first derivative of the Wronskian. I have the equation of the Wronskian for two functions where I only use the functions and their first derivatives.

I have the following:

$$\underline{\overline{X}}(t) = [x^{(1)}(t), x^{(2)}(t)]$$

is the solution to

$$\frac{d\underline{\overline{X}}}{t} = A(t)\underline{\overline{X}}(t)$$

where

\begin{equation*} A(t) = \begin{pmatrix} a_{11}(t) & a_{12}(t) \\ a_{21}(t) & a_{22}(t) \end {pmatrix} \end{equation*}

I calculated the Wronskian by taking the determinant of the derivative matrix of $\underline{\overline{X}}(t)$ and got: $$x^{1}(x^{(2)})' - x^{2}(x^{(1)})'$$

We were told in class that the first derivative of the Wronskian will always be

$$trace(A)W$$

but I'm not sure how to get this result from differentiating what I got for the Wronskian in this case.

Any help is appreciated!

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Given the differential equation

$\dot X = AX, \tag 0$

where

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}, \tag 1$

and

$X = \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}, \tag 2$

we have

$\dot X = \begin{bmatrix} \dot x_{11} & \dot x_{12} \\ \dot x_{21} & \dot x_{22} \end{bmatrix} = AX = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} = \begin{bmatrix} a_{11}x_{11} + a_{12}x_{21} & a_{11}x_{12} + a_{12}x_{22} \\ a_{21}x_{11} + a_{22}x_{21} & a_{21}x_{12} + a_{22}x_{22} \end{bmatrix}; \tag 3$

we define the Wronskian

$W = \det \left ( \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} \right ) = x_{11}x_{22} - x_{12}x_{21}; \tag 4$

we compute, using (3):

$\dot W = \dot x_{11}x_{22} + x_{11} \dot x_{22} - \dot x_{12}x_{21} - x_{12} \dot x_{21}$ $= (a_{11}x_{11} + a_{12}x_{21})x_{22} + x_{11}(a_{21}x_{12} + a_{22}x_{22}) - (a_{11}x_{12} + a_{12}x_{22})x_{21} - x_{12}(a_{21}x_{11} + a_{22}x_{21})$ $= a_{11}x_{11}x_{22} + a_{22}x_{11}x_{22} - a_{11}x_{12}x_{21} - a_{22}x_{12}x_{21}$ $= a_{11}(x_{11}x_{22} - x_{12}x_{21}) + a_{22}(x_{11}x_{22} - x_{12}x_{21}) = (a_{11} + a_{22})(x_{11}x_{22} - x_{12}x_{21}) = \text{Tr}(A) W. \tag 5$

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    $\begingroup$ Amazing. I didn't even think of using the given differential equation. Is it always the case that Xdot has 4 components? In the case where we have A as a 2x2 matrix? Is it because the right hand side is a 2x2 matrix? $\endgroup$ – Ced Apr 1 at 3:47
  • $\begingroup$ @Ced: yes, when $A$ is $2 \times 2$, $X$ will always be $2 \times 2$ as well, since each column of $X$ is a vector solution of the differential equation. We want $X$ to be $2 \times 2$ so each column can be a solution, usually linearly independent from the other column! $\endgroup$ – Robert Lewis Apr 1 at 3:50
  • $\begingroup$ @Ced: if you like my answer and/or find it helpful, you might consider green-checking (i.e. accepting) it. Cheers! And thanks! $\endgroup$ – Robert Lewis Apr 1 at 3:56
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    $\begingroup$ If you don't mind checking your email, that'd be great! $\endgroup$ – Ced Apr 1 at 4:03

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