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Frobenius solution to a diff-eq

  1. May 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Find 2 Frobenius series solutions to the following differential equation:

    2xy'' + 3y' - y = 0

    2. Relevant equations



    3. The attempt at a solution

    I got r = -1/2 and 0 as roots.

    Recurrence relation:

    http://image.cramster.com/answer-board/image/cramster-equation-200951755626337813656283812509812.gif

    I got the following solutions:

    http://image.cramster.com/answer-board/image/cramster-equation-2009517550266337813622621312505598.gif
    http://image.cramster.com/answer-board/image/cramster-equation-2009517551276337813628747875004691.gif

    According to mathematica, these aren't the right solutions.

    What am I doing wrong?

    Thanks!
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. May 18, 2009 #2

    HallsofIvy

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    It's easy to see by plugging into the equation that those are NOT solutions to equation. Further since one of the solutions for r is 0, I don't see how you can get [itex]\sqrt{2x}[/itex] in both solutions- one should be a standard power series.
     
    Last edited: May 18, 2009
  4. May 18, 2009 #3
    It is a power series, I pulled out a radical so as to make it fit the form of sinh's power series.

    Do you disagree with my recurrence relation?

    Thanks for the response.
     
  5. May 18, 2009 #4

    gabbagabbahey

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    Really?

    [tex]y_1(x)=\frac{a_0}{\sqrt{2x}}\sinh(\sqrt{2x})\implies y_1'(x)=\frac{-a_0}{2\sqrt{2}x^{3/2}}\sinh(\sqrt{2x})+\frac{a_0}{2x}\cosh(\sqrt{2x}) [/tex]

    [tex] \implies y''(x)=\frac{-3a_0}{4x^2}\cosh(\sqrt{2x})+\frac{3a_0}{4\sqrt{2}x^{5/2}}\sinh(\sqrt{2x})+\frac{a_0}{2\sqrt{2}x^{3/2}}\sinh(\sqrt{2x})[/tex]

    [tex]\implies 2xy_1''(x)+3y_1'(x)-y_1(x)=\left(\frac{-3a_0}{2x}\cosh(\sqrt{2x})+\frac{3a_0}{2\sqrt{2}x^{3/2}}\sinh(\sqrt{2x})+\frac{a_0}{\sqrt{2x}}\sinh(\sqrt{2x})\right)[/tex]

    [tex]+\left(\frac{-3a_0}{2\sqrt{2}x^{3/2}}\sinh(\sqrt{2x})+\frac{3a_0}{2x}\cosh(\sqrt{2x})\right)-\left(\frac{a_0}{\sqrt{2x}}\sinh(\sqrt{2x})\right)=0[/tex]


    And y_2(x) also satisfies the DE

    [tex]y_1(x)=\frac{a_0}{\sqrt{2x}}\sinh(\sqrt{2x})=a_0\left(1+\frac{x}{3}+\frac{x^2}{630}+\ldots\right)[/tex]

    Is a regular power series:wink: (just multiply the series for [itex]sinh(\sqrt{2x})[/itex] by [itex]\frac{a_0}{\sqrt{2x}}[/itex] )
     
  6. May 18, 2009 #5

    gabbagabbahey

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    Mathematica probably gave the total general solution as [tex]y(x)=\frac{e^{\sqrt{2} \sqrt{x}} C[1]}{\sqrt{x}}-\frac{e^{-\sqrt{2} \sqrt{x}} C[2]}{\sqrt{2} \sqrt{x}}[/tex]....correct?

    Which is actually equivalent to the linear combination of your two solutions with [itex]C[1]=\frac{a_0}{2\sqrt{2}}+\frac{b_0}{2}[/itex] and [itex]C[2]=\frac{b_0}{\sqrt{2}}-\frac{a_0}{2}[/itex], since, by definition, [itex]\sinh(u)=\frac{1}{2}(e^u-e^{-u})[/itex] and [itex]\cosh(u)=\frac{1}{2}(e^u+e^{-u})[/itex]
     
    Last edited by a moderator: Apr 24, 2017
  7. May 18, 2009 #6

    HallsofIvy

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    Oh, dear, oh, dear!:blushing:
     
  8. May 18, 2009 #7
    Thanks!

    http://img507.imageshack.us/img507/7167/71268782.jpg [Broken]

    I couldn't make heads or tails of that, and couldn't get the confirmation I needed.

    I then tried plugging the result straight into the diff-eq, and mathematica likewise wasn't returning 0 as the answer, which spooked me.
     
    Last edited by a moderator: May 4, 2017
  9. May 18, 2009 #8

    gabbagabbahey

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    In the above DSolve command, you need to use a "*" between x and y''[x] (i.e. [itex]2x*y''[x][/itex]), Mathematica interprets xy''[x] as the second derivative of a function called "xy"

    As for checking that your solution satisfies the DE, try using FullSimplify[2x*y''[x]+3y'[x]-y[x]] after defining your y[x]
     
    Last edited by a moderator: May 4, 2017
  10. May 18, 2009 #9
    Ah, now that makes perfect sense! Thanks for your help.
     
  11. May 18, 2009 #10

    gabbagabbahey

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    You're Welcome!:smile:
     
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