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Solving 2nd order differential equation with non-constant coefficients

  1. Sep 2, 2009 #1

    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. Sep 2, 2009 #2


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    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
    Last edited by a moderator: Sep 13, 2011
  4. Sep 2, 2009 #3
    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. Sep 2, 2009 #4
    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. Sep 8, 2011 #5
    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. Sep 8, 2011 #6
    Try substitution

  8. Sep 12, 2011 #7
    Thanks its really helping.
  9. Nov 27, 2012 #8
    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. Nov 28, 2012 #9
    (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. Nov 29, 2012 #10


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    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. Nov 30, 2012 #11
    You are right. My mistake !
    Damn ODE !
  13. Nov 30, 2012 #12

    Attached Files:

    Last edited: Nov 30, 2012
  14. Nov 30, 2012 #13
  15. Nov 30, 2012 #14


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    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?
  16. Nov 30, 2012 #15


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    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.
  17. Nov 30, 2012 #16
    Thanks for your warning. I won't do it again :D
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