1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]
    [tex](1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0[/tex]
    [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


    User Avatar
    Science Advisor
    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


    User Avatar
    Homework Helper

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook