Originally Posted by

**danny arrigo** Here's an answer for 1). Let us assume that your functions, $\displaystyle \phi_1 , \phi_2 $ satisfy some ODE

$\displaystyle a y'' + by' + cy = 0,\;\;\; a \ne 0$

So $\displaystyle a \phi_1'' + b\phi_1' + c\phi_1 = 0,$ and $\displaystyle a \phi_2'' + b\phi_2' + c\phi_2 = 0.$ Then multiplying the first by $\displaystyle \phi_2$ and the second by $\displaystyle \phi_1$ and subtracting gives

$\displaystyle a\left(\phi_2 \phi_1'' - \phi_1 \phi_2'' \right) + b \left( \phi_2 \phi_1' - \phi_1 \phi_2' \right) = 0$ or $\displaystyle a\left(\phi_2 \phi_1' - \phi_1 \phi_2' \right)' + b \left( \phi_2 \phi_1' - \phi_1 \phi_2' \right) = 0$.

If $\displaystyle W_{\phi} = W(\phi_1,\phi_2)$ is the usual Wronskian, then the above equation becomes

$\displaystyle a W_{\phi} ' + b W_{\phi} = 0$

Similarly, for $\displaystyle \psi_1\; \text{and}\, \psi_2$

$\displaystyle a W_{\psi} ' + b W_{\psi} = 0$

Thus,

$\displaystyle \frac{W_{\phi} ' }{W_{\phi} } = - \frac{b}{a} = \frac{W_{\psi} ' }{W_{\psi} }$ or $\displaystyle \frac{W_{\phi} ' }{W_{\phi} } = \frac{W_{\psi} ' }{W_{\psi} }$

from which we obtain $\displaystyle W_{\phi} = c W_{\psi}.$