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Find W of 2 Solutions to a Differential EQ

  1. Oct 27, 2007 #1
    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. jcsd
  3. Oct 28, 2007 #2


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    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 by a moderator: Oct 29, 2007
  4. Oct 29, 2007 #3
    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.
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