Results 1 to 2 of 2

Thread: 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:

    $\displaystyle 2xy'' + y' + xy = 0$

    I'm totally confused about finding the recurrence relation when solving this when using the method of Frobenius about $\displaystyle x_0 = 0$. 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

    $\displaystyle 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:

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

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

    First we do the case when $\displaystyle k=0$.

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

    This tells us that $\displaystyle a_1=0$.

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

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

    As a consequence $\displaystyle a_1 = a_3 = a_5 = a_7 = ... = 0$

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

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

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Dec 7th 2011, 01:22 AM
  2. general solution of a recurrence relation
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Oct 17th 2010, 09:55 PM
  3. Replies: 0
    Last Post: Feb 9th 2010, 03:15 PM
  4. Replies: 4
    Last Post: May 3rd 2009, 09:31 PM
  5. Find solution to a recurrence relation...
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: Mar 20th 2009, 11:17 AM

Search Tags


/mathhelpforum @mathhelpforum