Solving 2nd order differential equation with non-constant coefficients

  1. Hi~

    I'm having trouble with solving a certain differential equation.

    x2y'' + x y'+(k2x2-1)y = 0

    I'm tasked to find a solution that satisfies the boundary conditions: y(0)=0 and y(1)=0

    I have tried solving this using the characteristic equation, but i arrived at a solution that is unable to satisfy the boundary conditions except for when y(x)=0 (which is trivial).

    Any pointers on how i should go about this?

    Actually, i am trying to find the Green's function for this differential equation, which is why I need the solution to the said equation first. Thanks so much for any help :)
     
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,372
    Staff Emeritus
    Science Advisor

    There is at least one obvious solution: y= 0 for all x. Since that is a regular singular equation at 0, you probably will want to use "Frobenius' method", using power series. That is much too complicated to explain here- try
    http://en.wikipedia.org/wiki/Frobenius_method
     
    Last edited: Sep 13, 2011
  4. See the "Bessel" differential equation. Change variables to convert yours to that. So what you need to do is select k so that your solution has a zero at 1.
     
  5. Thanks for the tip!

    Are there any possible way to solve this? Like possibly an out of this world change in variables (ex let u = lnx)? or series substitution?
     
  6. Hi every body,

    I have another kind of equation which seems rather difficult to solve

    (1 + a Sin(x)) y'' + a Cos(x) y' + b Sin(x) y = 0

    Should I first expand sin and cos in their series and then try Laplace or serious solution methods? Or there is a method that can solve it directly?
    Please help?
     
  7. Try substitution

    u=(1+aSin(x))
     
  8. Thanks its really helping.
     
  9. Please help me to solve this DE: y''=ysinx
    (I think i should multiply both sides with 2y' but I don't know how to do next)

    thanks in advance ^^
     
  10. (y')² = y²sin(x)+C
    You can find the solutions in the particular case C=0 in terms of exponential of Incomplete elliptic integral of the second kind.
     
  11. Mute

    Mute 1,391
    Homework Helper

    I'm afraid that sin(x) ruins the integration, and as such you're missing a term ##-\int dx y(x) \cos(x)##. After multiplying by 2y' you get

    $$\frac{d}{dx}(y')^2 = \left[\frac{d}{dx}(y^2)\right] \sin(x),$$
    which can't be integrated exactly - you get ##(y')^2 = y^2\sin(x) + C - \int dx~y(x) \cos(x)##.
     
  12. You are right. My mistake !
    Damn ODE !
     
  13. Attached Files:

    Last edited: Nov 30, 2012
  14. HallsofIvy

    HallsofIvy 40,372
    Staff Emeritus
    Science Advisor

    You have already been given two methods, Frobenius and a change of variables that changes this to a Bessel equation, and a complete solution: y(x)= 0. What more do you want?
     
  15. HallsofIvy

    HallsofIvy 40,372
    Staff Emeritus
    Science Advisor

    In future, however, please do not "hi-jack" someone else's thread to ask a new question. Use the "new thread" button to start your own thread.
     
  16. Thanks for your warning. I won't do it again :D
     
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