The discussion focuses on finding the Wronskian of two solutions to the differential equation (1-x^2)y" - 2xy' + α(α+1) = 0 without explicitly solving it. The Wronskian is defined as W = y1y2' - y1'y2, and its derivative can be expressed as W' = (-q(x)/p(x))W, where p(x) and q(x) are coefficients from the general linear homogeneous second-order differential equation. The equation discussed is identified as the differential equation of Legendre, which is significant in the study of partial differential equations in spherical coordinate systems.
PREREQUISITESMathematicians, physicists, and engineering students who are studying differential equations, particularly those interested in the applications of Legendre polynomials and the Wronskian in solving complex problems in spherical coordinates.
HallsofIvy said: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.