Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to solve these type of differential equations?

  1. Oct 7, 2012 #1
    find a solution of (a^2-x^2)y''-2xy'+n(n+1)y=0, a not 0 by reduction to the legendre equation???
     
  2. jcsd
  3. Oct 7, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since Legendre's differential equation starts "[itex](1- x^2)d^2y/dx^2[/itex]" an obvious first step would be to define a new variable, t= x/a. Then dy/dx= (dy/dt)(dt/dx)= 1/a dy/dt, [itex]d^2y/dx^2= 1/a^2 d^2y/dt^2[/itex], and [itex]a^2- x^2= a^2(1- (x/a)^2)= a^2(1- t^2)[/itex] so that [itex](a^2- x^2)d^2y/dx^2= (1- t^2)d^2y/dt^2[/itex].

    2xy'= 2(at)((1/a)dy/dt)= 2t dy/dt.
     
  4. Oct 7, 2012 #3
    if dx/dt=1/a then d^2 t/dx^2=0 how will it will be 1/a^2??
     
  5. Oct 10, 2012 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I have no clue what you are talking about. There is no "[itex]d^2t/dx^2[/itex]" in what I wrote.

    I did say that [itex]d^2y/dx^2= (1/a^2) d^2y/dt^2[/itex]. That is true because
    [tex]d^2y/dx^2= d/dx(dy/dx)= d/dx((1/a)dy/dt)= (1/a) d/dx( dy/dt)= (1/a)((1/a)d/dt)(dy/dt)= (1/a^2) d^2/dt^2[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to solve these type of differential equations?
Loading...