Results 1 to 2 of 2

Math Help - Finding Recurrence relation for Series Solution of Differential Equation

  1. #1
    Member Jason Bourne's Avatar
    Joined
    Nov 2007
    Posts
    132

    Finding Recurrence relation for Series Solution of Differential Equation

    I'm trying to solve the following differential equation with non constant coefficients:

    2xy'' + y' + xy = 0

    I'm totally confused about finding the recurrence relation when solving this when using the method of Frobenius about x_0 = 0. I know it's a regular singular point and all that but I'm just rubbish at finding the recurrence relation, apparently the right answer is

    a_n = - \frac{a_{n-2}}{(n+r)(2(n+r) -1)} where r is one of the roots of the indicial equation

    Can anyone help me get unconfused about finding the recurrence relation as I'm rubbish at it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Jason Bourne View Post
    I'm trying to solve the following differential equation with non constant coefficients:

    2xy'' + y' + xy = 0 for x>0.
    The point 0 is a regular singular point.
    Therefore, solutions shall have the form y= x^k \sum_{n=0}^{\infty} a_n x^n.

    The equation associated with finding k is: 2k(k-1)+k=0\implies k = 0 \text{ or }\tfrac{1}{2}.

    First we do the case when k=0.

    We looking for a solution y=\sum_{n=0}^{\infty} a_n x^n.
    Substituting that into the equation we get,
    2x\sum_{n=2}^{\infty}n(n-1) a_n x^{n-2} + \sum_{n=1}^{\infty}na_nx^{n-1} + x\sum_{n=0}^{\infty} a_nx^n = 0
    This becomes,
    2\sum_{n=2}^{\infty} n(n-1)a_n x^{n-1} + \sum_{n=1}^{\infty} na_n x^{n-1} + \sum_{n=0}^{\infty}a_nx^{n+1} = 0
    We can rewrite this as,
    2\sum_{n=1}^{\infty} (n+1)na_{n+1}x^n + \sum_{n=0}^{\infty}(n+1)a_{n+1} x^n + \sum_{n=1}^{\infty} a_{n-1} x^n = 0
    Evaluate the middle summation at n=0 and combine,
    a_1 + \sum_{n=1}^{\infty} [2(n+1)na_{n+1} + (n+1)a_{n+1} + a_{n-1}]x^n = 0

    This tells us that a_1=0.

    The condition in the middle says,
    2(n+1)na_{n+1} + (n+1)a_{n+1}+a_{n-1} = 0 \text{ for }n\geq 1

    Thus, a_{n+1} = - \frac{a_{n-1}}{(2n+1)(n+1)}

    As a consequence a_1 = a_3 = a_5 = a_7 = ... = 0

    Taking a_0 to be arbitrary (like a_0=1) shall produce coefficients for a_2,a_4,a_6,....

    That would be one solution to the differencial equation.
    Of course the other linearly independent solution is found with k=\tfrac{1}{2}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 7th 2011, 02:22 AM
  2. general solution of a recurrence relation
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 17th 2010, 10:55 PM
  3. Replies: 0
    Last Post: February 9th 2010, 04:15 PM
  4. Replies: 4
    Last Post: May 3rd 2009, 10:31 PM
  5. Find solution to a recurrence relation...
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: March 20th 2009, 12:17 PM

Search Tags


/mathhelpforum @mathhelpforum