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Differentiate hyperbolic function

  1. Dec 7, 2011 #1

    sharks

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    The problem statement, all variables and given/known data
    If lny = sinh^(-1)(x), prove that

    (1+ x^2)y'' + xy' - y = 0

    The attempt at a solution

    I have tried various (unsuccessful) ways of doing this, but the basic procedure that i've done is:

    D.w.r.t.x for lny = sinh^(-1)(x)

    This gives: (1/y)y' = 1/(1 + x^2)

    To obtain y'', d.w.r.t.x for equation above.

    This is the part where i'm not sure if i should use product rule for differentiating (1/y)y'

    Is is correct?
    -y^(-2)y' + y^(-1)y''

    Or is this the right way of doing it?
    -y^(-2)y''
     
  2. jcsd
  3. Dec 7, 2011 #2

    LCKurtz

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    You must use the product (or quotient) rule.
     
  4. Dec 7, 2011 #3
    Well, you should be aware that:
    [itex]sinh^{-1}x=log(x+\sqrt{x^2+1})[/itex]

    so you have
    [itex]\log y=\log(x+\sqrt{x^2+1})[/itex]

    from now on it should be simple.
     
  5. Dec 7, 2011 #4

    vela

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    You differentiate sinh-1 x incorrectly. You should have
    [tex]\frac{y'}{y} = \frac{1}{\sqrt{1+x^2}}[/tex]Then I'd get rid of the quotients to make things simpler:
    [tex]y'\sqrt{1+x^2} = y[/tex]before differentiating. (I just never really liked the quotient rule.)
     
    Last edited: Dec 7, 2011
  6. Dec 8, 2011 #5

    sharks

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    I have taken into account all of your advices. OK, now here's the evolution of this problem from my side.

    So, making y' the subject of formula, i get y' = y/[(1 + x^2)^(1/2)]

    I further differentiate w.r.t.x the equation by vela above (i apologize for not using the better-looking LaTex, but i still have a lot to learn on how to use it):

    So, y''(1 + x^2)^(1/2) + y'/[2(1 + x^2)^(1/2)] = y'

    Now, i make y'' the subject of formula.

    y'' = {y' - y'/[2(1 + x^2)^(1/2)]}/[(1 + x^2)^(1/2)]

    Then, i replace y' and y'' above, into the original equation: (1+ x^2)y'' + xy' - y, hoping that it will all reduce to 0. After substituting, i will end up with a (rather long) equation involving x, y and y' (which i have to substitute again). Is this the right direction to solve this problem?
     
  7. Dec 8, 2011 #6

    vela

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    You need to use the chain rule when calculating the second term.
    You're making more work for yourself than you need to.

    Hint: After you differentiated, the y'' term almost looks like what you want.
     
  8. Dec 8, 2011 #7

    sharks

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    OK, i got it. Thank you all very much.
     
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