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Thread: Differential equations - verification of homogenous solution

  1. #1
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    Unhappy Differential equations - verification of homogenous solution

    I would like to ask how can I verify that the differential equation:

    . . . . .[tex]y_n^{''}\, +\, \big(2n\, \coth(x)\big)\, y_n^{'}\, +\, \left(n^2\, -\, 1\right)\, y_n\, =\, 0[/tex]

    has the homogeneous solution:

    . . . . .[tex]y_n\, =\, \left(\dfrac{1}{\sinh(x)}\, \dfrac{d}{dx}\right)^n\, \left(Ae^x\, +\, Be^{-x}\right)[/tex]

    for degree [tex]n\, \in\, \mathbb{N}.[/tex]

    How to start this problem? Should I calulate first and second derivative from d/dn * y_n and later try to insert into equation? I donīt really understand how it can be verified. I will be grateful for all help.
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    Last edited by stapel; 02-10-2018 at 04:46 PM. Reason: Typing out the text in the graphics.

  2. #2
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    Quote Originally Posted by vid View Post
    I would like to ask how can I verify that the differential equation:

    . . . . .[tex]y_n^{''}\, +\, \big(2n\, \coth(x)\big)\, y_n^{'}\, +\, \left(n^2\, -\, 1\right)\, y_n\, =\, 0[/tex]

    has the homogeneous solution:

    . . . . .[tex]y_n\, =\, \left(\dfrac{1}{\sinh(x)}\, \dfrac{d}{dx}\right)^n\, \left(Ae^x\, +\, Be^{-x}\right)[/tex]

    for degree [tex]n\, \in\, \mathbb{N}.[/tex]

    How to start this problem? Should I calulate first and second derivative from d/dn * y_n and later try to insert into equation? I donīt really understand how it can be verified. I will be grateful for all help.
    "Verification" is a matter of calculating the derivatives and substituting.
    Last edited by stapel; 02-10-2018 at 04:47 PM. Reason: Copying typed-out graphical content into reply.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
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    Quote Originally Posted by tkhunny View Post
    "Verification" is a matter of calculating the derivatives and substituting.
    To calculate derivatives I assumed that n=1, A=1, B =1, But I am not sure if I can make assumption for A,B. (I calculated derviatives without this assumption and the results of the derivatives were very complex).

    After assumption n=1, A=1, B =1 my differential equation has form

    . . . . .[tex]y_n^{''}\, +\, \big(2\, \coth(x)\big)\, y_n^{'}\, =\, 0[/tex]

    and

    . . . . .[tex]y_n\, =\, \left(\dfrac{1}{\sinh(x)}\, \dfrac{d}{dx}\right)\, \left(e^x\, +\, e^{-x}\right)[/tex]

    Then y_n has a form of [tex]y_n\, =\, -2\, \coth\left(x^2\right)[/tex]. Then I calculated [tex]y_n^{'}[/tex] and [tex]y_n^{''}[/tex] and I inserted it into differential equation. The result which I obtained was [tex]-4\, \mbox{csch}(x)^4[/tex] which is not equal to 0 (DE doesnt have homogenous solution). Is it correct logic?
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    Last edited by stapel; 02-10-2018 at 04:52 PM. Reason: Typing out the text in the graphics.

  4. #4
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    I don't understand your need for values for these parameters. Why not just use them as they are? You have not done the exercise if you first simplify it and then do the demonstration. You must do the demonstration for ALL values of the parameters.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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