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Pointers for the solution of 2nd order DE with variable coefficient

  1. Jun 28, 2011 #1
    Hi,

    I am looking at the following 2nd order DE with variable coefficient:

    y''(x)-(1/x+7)y(x)=0

    I would be grateful for any help in regard to methods which may be applied to such an equation.

    Many thanks in advance!

    C.
     
  2. jcsd
  3. Jun 28, 2011 #2

    HallsofIvy

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    Typically such an equation can be solved by looking for a series solution.

    The first thing I would do is multiply both sides by x:
    [tex]x\frac{d^2y}{dx^2}- (7x+ 1)y= 0[/tex]

    Because of that x in the leading term you will need to use "Frobenius' method"- we look for a solution of the form
    [tex]y(x)= \sum_{n=0}^\infty a_nx^{n+c}[/tex]
    where c is not necessarily a positive integer.

    Then
    [tex]y'(x)= \sum_{n=0}^\infty (n+c)a_nx^{n+c- 1}[/tex]
    [tex]y''(x)= \sum_{n= 0}^\infty (n+c)(n+c-1)x^{n+c- 2}[/tex]

    Put that into the equation to get sums of powers of x equal to 0.

    Choose c by looking at the coefficient of [itex]x^0= 1[/itex] and asserting that [itex]a_x[/itex] (n=0) is not 0.

    Once you know c, combine like powers and set the coeficients equal to 0. You will get a recursive formula for the coefficients, [itex]a_n[/itex].
     
  4. Jul 3, 2011 #3
    Thanks for your help HallsofIvy!! Much appreciated.

    C.
     
  5. Jul 3, 2011 #4
    Hello !

    This ODE can be analytically solved. The closed form for the general solutions involves the Kummer function and the Tricomi function ( i.e. the confluent hypergeometric functions)
     
  6. Jul 10, 2011 #5
    Hi JJacquelin,
    Thanks for your pointer; I got a series solution together ok, though it seems to blow up strangely. Anyway, I would certainly be interested in seeing the analytical form of the solution to said de. I’ll have a look at those functions that you suggested.
    Cheers,
    C
     
  7. Jul 11, 2011 #6
    Hello crawf_777

    The analytical solution is :
     

    Attached Files:

  8. Jul 11, 2011 #7
    Hi JJacquelin,
    Thanks for that! Very much appreciated!!
    C
     
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