- #1

ognik

- 643

- 2

so 1st row of W = {$ {y}_{1}, {y}_{2},{y}_{3} $}, 2nd row is 1st derivatives, 3rd row is 2nd derivatives.

I can find the determinate, W (should it be written as |W| ?) = $ y_1 y'_2y''_3 - y_1y''_2y'_3 - y_2y'_1y''_3 + y_2y''_1y'_3 + y_3y'_1y''_2 - y_3y''_1y'_2 $

I know that $ y_1y'_2 - y_2y'_1 \ne 0 $(IE $y_3$ is the 'other' solution), so I grouped the above into $ y''_3(y_1y'_2 - y_2y'_1) - y''_2(y_1y'_3 - y_3y'_1) - y''_1(y_2y'_3 - y_3y'_1) $. The terms in brackets are Wronskians and if the 3 solutions are linearly independent, then they cannot=0. It would be convenient then, to have the 2nd derivatives = 0, but I can't argue that as the y(x)'s could have $x^2$ terms...

I have tried various other combinations, but most of them just transform to expansions by a different row/column. Can anyone give me a hint please?