MHB Use Wronskian to show only 2 indepndent solutions of 2nd order ODE

[ognik]

Given standard ODE $ y'' + P(x)y' + Q(x)y=0 $, use wronskian to show it cannot have 3 independent sltns. Assume a 3rd solution and show W vanishes for all x.

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?

 

[Ackbach]

Assuming your Wronskian is correct, maybe you could substitute $y_2''=-P(x) y_2'-Q(x) y_2$, and the same for the other two solutions?

 

[ognik]

... and they all cancel neatly, I wish I could think of these obvious-in-retrospect approaches.

 

[Ackbach]

Well, in this case, what led me to that suggestion was that, so far, you had not used the fact that the functions were solutions. That had to be important.

 

[ognik]

I also thought it important, but instead thought that suggested using the Wronskian, I think with a lot more practice it gets better, its just frustrating when missing something obvious costs me many hours ... without this forum I suspect I would have given up by now, so I really appreciate all the help!