Results 1 to 3 of 3

Math Help - Exact LDE factorizing

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    54

    Exact LDE factorizing

    Hi all,
    I'm currently working on a question that I can't fully work out. Can someone look over my working and see where I may have possibly gone wrong and maybe point me in the right direction?
    The question is,
    Given a LDE F=b_{0}y''+b_{1}y'+b_{2}y=0 and the relationship b_{0}''-b_{1}'+b_{2}=0, show that the factorization of the LDE is F=D[b_{0}D+(b_{1}-b_{0}')]y.

    From the relationship, I can use b_{2}=b_{1}'-b_{0}'' and I can factorize that into b_{2}=D[b_{1}-b_{0}'].
    Factorizing: F =  [b_{0}D^2+ b_{1}D+b_{0}]y

    Therefore, F= D[b_{0}D+ b_{1}+(b_{1}-b_{0})]y.
    I have that b_{1} that I'm not sure how to get rid of.
    Thank you for your time.
    Last edited by Silverflow; March 28th 2010 at 02:13 AM. Reason: Mispelling
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Expand F:

    F=D[b_0 D+(b_1-b_0^{'})]y

    F=D[b_0y'+(b_1-b_0^{'})y]

    F=b_0 y''+y'b_0^{'}+(b_1-b_0^{'})y'+y(b_1^{'}-b_0^{''})

    now let b_2=b_1^{'}-b_0^{''}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    54
    Thank you, I never thought about working backwards.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: September 21st 2011, 01:12 AM
  2. Why are exact ODEs called exact?
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: December 27th 2009, 10:48 AM
  3. Factorizing
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 29th 2009, 02:51 AM
  4. factorizing
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 3rd 2008, 07:33 PM
  5. factorizing
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 19th 2008, 11:25 AM

Search Tags


/mathhelpforum @mathhelpforum