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Differential Equation Resolve for f(x)

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Resolve for f(x).

    [tex]\frac{d^{2}f}{dx^{2}} = sinf [/tex]


    2. Relevant equations



    3. The attempt at a solution

    I haven't studied differential equations yet, so I have no idea what to do...
     
    Last edited: Oct 8, 2011
  2. jcsd
  3. Oct 8, 2011 #2
    The derivative of Sinx = Cosx
    and the The derivative of Cosx = - Sinx
    df^2/ dx^2 means you derived it twice so
    after deriving for sin you will derive for cos
     
  4. Oct 8, 2011 #3

    LCKurtz

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    That is a nonlinear equation which I doubt has a simple closed form solution. Sometimes such equations are linearized as an approximation. For small values of f you could use the approximation sin(f) ≈ f giving f'' - f = 0.
     
  5. Oct 8, 2011 #4

    Mark44

    Staff: Mentor

    If f(x) = sin(x), then d2f/dx2 = -sin(x), but this is very different from the problem in the original post. In that problem, the right side of the equation is sin(f), not sin(x). See LCKurtz's post.
     
  6. Oct 9, 2011 #5
    Yea I thought of that, but f might not be small so I need more terms on taylor's series. If I go to the 3rd term in Taylor's series:

    gif.latex?sinf%20=%20f%20-%20\frac{f^{3}}{3!}.gif

    Does it have a closed form solution this way?
     
  7. Oct 9, 2011 #6

    Mark44

    Staff: Mentor

    I don't think so. Adding that cubic term makes the equation nonlinear.
     
  8. Oct 9, 2011 #7

    LCKurtz

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    Maple gives a solution in terms of JacobiSN functions. These involve inverses of elliptic integrals and doubly periodic functions, for what it's worth.
     
  9. Oct 9, 2011 #8
    That equation is analogous to the non-linear pendulum which can be solved analytically. First start by integrating it:

    [tex]\int \frac{d^2f}{dx^2}=\int \sin(f)[/tex]

    [tex]1/2 \left(\frac{df}{dx}\right)^2=-\cos(f)+c[/tex]

    we're then left with a first order ODE, the solution of which can be expressed in terms of elliptic functions.
     
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