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Differential equation and substitution

  1. Aug 6, 2009 #1
    This is a step from my notes that I don't follow. I have sin[tex]\theta[/tex][tex]\frac{d}{d\theta}[/tex](sin[tex]\theta[/tex][tex]\frac{d\Theta}{d\theta}[/tex]) and that, when substituting u=cos[tex]\theta[/tex] and writing that [tex]\Theta[/tex]([tex]\theta[/tex])=P(u), [tex]\frac{d}{du}[/tex]((1-u[tex]^{2}[/tex])[tex]\frac{dP}{du}[/tex]) is obtained.

    I can see [tex]\frac{d\Theta}{d\theta}[/tex]=[tex]\frac{dP}{du}[/tex] for the far RHS but can't get the 1-u[tex]^{2}[/tex] to come out. I don't really remember how to do this though I'm sure I have been able to at some point so a starting tip would be much appreciated! The main trap I keep falling in is the temptation to try something with chain rule to get rid of all those d[tex]\theta[/tex] bits in the original equation. I know I'm not supposed to as I'm just supposed to be substituting, not solving, but I'm not sure how else to go about it.
     
  2. jcsd
  3. Aug 6, 2009 #2

    tiny-tim

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    Hi Pomico! :smile:

    (have a theta: θ :wink:)
    Using the chain rule, sinθ d/dθ = sin2θ d/du.

    (but I don't see how to get the whole thing :redface: … are you sure it isn't 1/sinθ on the left?)
     
  4. Aug 6, 2009 #3

    Mark44

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    tiny-tim,
    Maybe I'm missing something here, but the line above doesn't make sense to me.

    Pomico,
    Is there really a need to use uppercase and lowercase versions of the same letter? I.e., [itex]\Theta[/itex] and [itex]\theta[/itex].
     
  5. Aug 6, 2009 #4

    tiny-tim

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    hmm … u = cosθ, so du/dθ = -sinθ,

    so d/du = dθ/du d/dθ = -1/sinθ d/dθ,

    so sin2θ d/du = -sinθ d/dθ

    … ok, i missed out a minus sign :redface:
     
  6. Aug 6, 2009 #5
    Hi!
    It's definitely not a reciprocal in the notes, that was one of the things bothering me

    Yes there is, this is only part of the problem covered in the notes. Earlier on there was a function F(θ,[tex]\phi[/tex]) which was split into variables as
    F(θ,[tex]\phi[/tex]) = [tex]\Theta[/tex](θ)[tex]\Phi[/tex]([tex]\phi[/tex])

    The equation in my first post isn't the full equation, just the first term. I don't think the rest will change anything but in case I've added an attachment of the lecture - the derivation I'm working from covers mostly pages 3 and 4 of the document. I don't know how to just include these two pages so sorry for adding the whole thing!
     

    Attached Files:

  7. Aug 6, 2009 #6
    Try this. Consider the substitution [tex]\Theta(\theta)=P(u)[/tex]

    You want to find [tex]\frac{d\Theta}{d\theta}[/tex]

    [tex]\frac{d\Theta}{d\theta} = \frac{dP(u)}{d\theta} = \frac{dP(u)}{du} \frac{du}{d\theta}[/tex]

    You can get [tex]\frac{du}{d\theta}[/tex] from your substitution of u=cos([tex]\theta[/tex]). Also sin([tex]\theta[/tex]) = "something with u", using a well known trig identity.
     
  8. Aug 6, 2009 #7

    Mark44

    Staff: Mentor

    I didn't catch the minus sign, either. I guess what you were doing was equating operators. I think that's what threw me--having the operator without an operand for it to work on.
     
  9. Aug 7, 2009 #8

    tiny-tim

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    D'oh! I was right! :rolleyes:

    You're referring to equations (2.26) and (2.27) …

    but you can see that (2.26) is divided by sin2θ before the u-substitution, so then it does start with 1/sinθ !!
     
  10. Aug 7, 2009 #9
    Aah yes I see, thanks for spotting that :p
    I'll have another go!

    Edit: Got it, thanks again :)
     
    Last edited: Aug 7, 2009
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