# Find W of 2 Solutions to a Differential EQ

1. Oct 27, 2007

### stargazer2190

How do I find the Wronskian of 2 solutions to

(1-x^2)y" -2xy' + α (α+ 1) without solving the Differential EQ?

I'm not given the 2 solutions to the equation so how do find the 2 solutions?

2. Oct 28, 2007

### HallsofIvy

Staff Emeritus
You don't find the two solutions! You find the Wronskian.

Suppose you have a general linear, homogeneous, second order differential equation:
p(x)y"+ q(x)y+ r(x)y= 0 (you don't have an "=" in your equation. I presume you meant "= 0".) And that y1 and y2 are two solutions. There Wronskian is W= y1y2'- y1'y2. The derivative of that is W'= y1'y2'+ y1y2"- (y1"y2+ y1'y2') The two y1'y2' terms cancel: W'= y1y2"- y1"y2. But y1"= (-q(x)y1'-r(x)y1)/p(x) and y2"= (-q(x)y2'-r(x)y2)/p(x) so W'=y1y2"- y1"y2= y1(-q(x)y2'-r(x)y2)/p(x))- (-q(x)y1'-r(x)y1)/p(x))y2. Again, the terms with a derivative, y1(-r(x)y2/p(x)) and (r(x)y1/p(x))y2 cancel leaving
W'= (-q(x)/p(x))(y1y2'-y1'y2)= (-q(x)/p(x))W. Solve THAT first order differential equation for the Wronskian.

Thanks, coomast, I have corrected (I think!) the errors.

Last edited: Oct 29, 2007
3. Oct 29, 2007

### coomast

I think there are a few typo-errors in here.

The derivative of the Wronskian has a minus sign in it that should be a "+".
The substitution after that is not y1 and y2, but y1" and y2".

Only making sure there is no confusion, I looked at it for 15 min. before I saw it...

Remark for stargazer2190, in case you don't know what equation this is, it is the differential equation of Legendre. It is a very important one if you study solutions of partial differential equations in sperical coordinate systems. More on this can be found on the www.