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Second order differential equation using substitution

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]\sin\theta\frac{d^2y}{d\theta^2}-\cos\theta\frac{dy}{d\theta}+2y\sin^3\theta=0[/tex]


    2. Relevant equations
    Use the substitution [itex]x=\cos\theta[/itex]


    3. The attempt at a solution
    I started off by listing:

    [tex]
    x=\cos\theta\\

    \frac{dx}{d\theta}=-\sin\theta\\

    \frac{d^2x}{d\theta^2}=-\cos\theta\\
    [/tex]

    But don't know whether this helps, or where to go next. Could someone please give me a hint at how to approach this, I'd prefer not to have a full solution, but really am desperate for a starting point!

    With very many thanks,

    Froskoy.
     
  2. jcsd
  3. Mar 29, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since you are differentiating with respect to [itex]\theta[/itex], you need to use the chain rule:
    [tex]\frac{dy}{d\theta}= \frac{dy}{dx}\frac{dx}{d\theta}= -sin(\theta)\frac{dy}{dx}[/tex]
    Then
    [tex]\frac{d^2y}{d\theta^2}[/tex]
    [tex]= \frac{d}{d\theta}(\frac{dy}{d\theta})[/tex]
    [tex]= \frac{d}{d\theta}(cos(\theta)\frac{dy}{dx}[/tex]
    [tex]= -sin(\theta)\frac{dy}{dx}+ cos(\theta)\frac{d}{d\theta}\frac{dy}{dx}[/tex]
    [tex]= -sin(\theta)\frac{dy}{dx}+ cos^2(\theta)\frac{d^2y}{dx^2}[/tex]
     
    Last edited: Mar 29, 2012
  4. Mar 31, 2012 #3
    Thanks very much! Have got it now!
     
  5. Apr 7, 2012 #4
    Hi! I was wondering why
    d/dθ (dy/dθ ) does not =d/dθ (-sin(θ) dy/dx)?

    Also, once you have the relevant expressions how do you solve the differential equation? Do you have to form the auxilliary equation? I tried to do that but it doesn't work out nicely.

    Thanks!
     
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