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Change of independent variables

  1. Jan 2, 2008 #1
    [SOLVED] Change of independent variables

    1. The problem statement, all variables and given/known data
    x = cos[tex]\theta[/tex]
    Show
    [tex](1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0[/tex]
    becomes
    [tex]\frac{d^2y}{d\theta ^2}+ cot(\theta) \frac{dy}{d\theta} + 2y = 0[/tex]

    2. Relevant equations



    3. The attempt at a solution

    dx = -sin([tex]\theta[/tex]) d[tex]\theta[/tex]
    ...
    As a total stab in the dark I've tried substituting -sin([tex]\theta[/tex]) d[tex]\theta[/tex] in for dx, but that didn't seem to work... and I don't think that is mathematically correct to begin with. How do I get the relationship between dy/dx and dy/dtheta? if y were given I think it would be a lot easier
     
  2. jcsd
  3. Jan 2, 2008 #2

    CompuChip

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    Homework Helper

    Use this (let me call it "inverse chain rule")

    [tex]\frac{dy}{dx} = \frac{dy}{d\theta} \frac{d\theta}{dx}[/tex]
    where you can explicitly calculate the latter factor from the equation you wrote under (3).

    Once you have that, you can do the same trick again:
    [tex]\frac{d^2y}{dx^2} = \frac{d}{dx} \frac{dy}{dx} = \left( \frac{d}{d\theta} \frac{dy}{dx} \right) \frac{d\theta}{dx}[/tex]
     
    Last edited: Jan 2, 2008
  4. Jan 2, 2008 #3

    Defennder

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    Did you leave out a factor of 2 for the coefficient of cot (theta) in the answer?

    Should it be [tex]\frac{d^2y}{d\theta ^2}+ 2cot(\theta) \frac{dy}{d\theta} + 2y = 0[/tex] ?
     
  5. Jan 2, 2008 #4
    Hey guys thanks for your replies, I'm trying it out right now.
    Defennder: Nope, it's just cot
     
  6. Jan 2, 2008 #5
    Got it, thanks
     
    Last edited: Jan 2, 2008
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