Results 1 to 4 of 4

Math Help - Power series solutions to ODE

  1. #1
    Junior Member
    Joined
    Nov 2007
    Posts
    48

    Power series solutions to ODE



    In an attempt to find a power series solution to an ordinary differential equation, our professor illustrated the concept with an example. As a try-at-home exercise, the following homogeneous equation was given:

    y''-5y'+6y=0

    The above differs from his example critically - a_k+2 is in terms of a_k+1 as well as a_k which yields a2,a3,a4... in terms of a1 & a0.

    My calculatios are here:

    http://johnny.sas.upenn.edu/~gsobti/power.htm

    The next step of separing a0 & a1 terms is proving to be erroneous.

    Would aprpeciate some assistance.

    Best,
    wirefree
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    It's a second order linear ODE so it's going to have two linearly independent solutions, and the general solution will be a linear combination of these. Probably the easiest way to handle the recurrence relation is to get the two lin. ind. solutions as follows: for the first one, take a_0=1 and a_1=0. For the second one, take a_0=0 and a_1=1. That will have the effect of "separating the a_0 and a_1 terms", as you wanted. You will get two solutions, and you can then form a lin. comb. of these to get the general solution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2007
    Posts
    48
    Quote Originally Posted by Opalg View Post
    Probably the easiest way to handle the recurrence relation is to get the two lin. ind. solutions as follows: for the first one, take a_0=1 and a_1=0. For the second one, take a_0=0 and a_1=1. That will have the effect of "separating the a_0 and a_1 terms", as you wanted.
    Appreciate the response, Opalg.

    As suggested, I proceeded with successively equating a_0 & a_1 to zero and calculating the resulting series.

    The resulting series, however, does not resemble the exponential series for 3x & 2x as the two solutions are found to be using characteristic equation method. Here are the results:

    http://johnny.sas.upenn.edu/~gsobti/1.htm

    The polynomial used is:

    http://johnny.sas.upenn.edu/~gsobti/2.htm

    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    I see what you mean. (I hadn't even noticed that the diff. equation has an easy analytical solution. In my defence, I never claimed that the recurrence relation would lead to an easy solution.)

    If you want to retrieve the exponential solution from the recurrence relation (n+2)(n+1)a_{n+2} -5(n+1)a_{n+1} +6a_n=0, I would start by noticing that if you write b_n = n!a_n then the relation becomes b_{n+2}-5b_{n+1}+6b_n=0. You solve this difference equation by the standard auxiliary equation method to get b_n=A.2^n+B.3^n. (Of course, that method is far more laborious than solving the diff. eqn directly by using the same auxiliary equation.)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find two power series solutions.
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: April 17th 2011, 08:41 AM
  2. Power Series Solutions of linear DE
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: March 19th 2010, 06:07 AM
  3. power series solutions
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: June 12th 2009, 08:19 AM
  4. power series solutions
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: June 6th 2009, 10:41 PM
  5. Power Series Solutions of ODE's
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 22nd 2008, 11:58 AM

Search Tags


/mathhelpforum @mathhelpforum