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Orthogonality/infinite series solutions differential equation

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data

    This is the problem statement in the picture for exersize 14, it's rather long (pertaining to orthogonality - which I only understand what the definition of orthogonality is, which is the "(15)" on the side of the image below.

    [PLAIN]http://postimage.org/image/oxhw2uf8p/ [Broken][/PLAIN]



    2. Relevant equations

    y'' + (k^2)y = 0


    3. The attempt at a solution

    My attempted solution attached below has a hint for exercise 14 which I will write out at the top of the page

    photo_22.jpg
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 1, 2012 #2

    gabbagabbahey

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    When you multiply both sides of the differential equation [itex]y_n''(x)+k_n^2y_n(x)=0[/itex] by [itex]y_m(x)[/itex], you don't get [itex]y_m''(x)+k_n^2y_m(x)=0[/itex]. That's not how multiplication works.
     
  4. Oct 1, 2012 #3
    Oh wow, sorry careless error!

    I got this far and don't know what I'm supposed to do after this:

    trigg_ex_14.jpg
     
  5. Oct 2, 2012 #4

    gabbagabbahey

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    Does [itex]k_n=k_m[/itex] for all possible values of [itex]m[/itex] and [itex]n[/itex]? If not, how do you justify cancelling out the 2 terms that you cancelled?

    Rather, you should have [itex](k_m^2-k_n^2)y_m y_n = \frac{d}{dx}\left( y_my_n' - y_n y_m' \right)[/itex]. As for what to do next, just follow the hint (it is very explicit in its instructions) and integrate both sides of the equation over a full period of [itex]y_n[/itex] (you'll probably want to start by figuring out what the period of [itex]y_n[/itex] is :wink:)...what do you get?
     
  6. Oct 2, 2012 #5
    Thanks so much, i got the final answer !!!!!!!!!!!!!!!!!!!!!!!!
     
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