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Also O.D.E stuck on integration step

  1. Aug 27, 2005 #1
    In the problem
    y` (cosh^2)x - (sin^2)y = 0, y(0)=pi/2
    i'm also stuck on the integration step...
    my first step is
    dy/((sin^2)y)=dx/(cosh^2)x
    however, i don't how to integrate the right side~
     
    Last edited: Aug 27, 2005
  2. jcsd
  3. Aug 27, 2005 #2

    TD

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    You could try using the definition:

    [tex]\cosh \left( x \right) \equiv \frac{{e^x + e^{ - x} }}{2}[/tex]
     
  4. Aug 27, 2005 #3
    ok, i'll try that~
    thanks!
     
  5. Aug 27, 2005 #4

    lurflurf

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  6. Aug 29, 2005 #5
    wow~ i thought that only cos and sin have that kinda identity...
    thanks! :)
     
  7. Sep 3, 2005 #6
    hmm...
    if i use [tex]\cosh \left( x \right) \equiv \frac{{e^x + e^{ - x} }}{2}[/tex] to integrate cosh^2x=(e^2x + e^(-2x) +1)/2,
    then [integrated(cosh)dx]=[(1/4)e^2x+(1/4)e^(-2x) + x/2]
    but does that equal tanhx?
     
  8. Sep 3, 2005 #7

    Fermat

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    I'm afraid not. It's the integral of sech²x that is equal to tanhx.

    let y = tanhx
    then
    y = sinhx/coshx
    y' = {coshx.(sinhx)' - sinhx.(coshx)'} / cosh²x
    y' = {cosh²x - sinh²x} / cosh²x
    y' = 1 / cosh²x
    y' = sech²x
     
  9. Sep 4, 2005 #8
    ok, thanks!
     
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