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A second order ODE

  1. Oct 22, 2011 #1

    lavinia

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    d[itex]^{2}[/itex]u/ds[itex]^{2}[/itex]= cosu[(du/ds)[itex]^{2}[/itex] - k[itex]^{2}[/itex]]
     
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  3. Oct 22, 2011 #2

    HallsofIvy

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    "cosu" is cos(u)? Since the independent variable, s, does not appear explicitely in that equation you can use a technique called "quadrature".

    Let v= du/ds so that [itex]d^2u/ds^2=dv/ds[/itex] but, by the chain rule, dv/ds= (dv/du)(du/ds)= v(dv/du) so your equation becomes [itex]v dv/du= cos(u)(v^2+ k^2)[/itex]

    That's a separable first order equation. Once you have solved it for v, integrate to find u.
     
  4. Oct 22, 2011 #3
    That lends itself to interpretation:

    [tex]\frac{d^2 u}{ds^2}=\cos(u(v))\biggr|_{v=(u')^2-k^2}[/tex]

    that's doable right?
     
    Last edited: Oct 22, 2011
  5. Oct 23, 2011 #4

    lavinia

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    thanks - pretty cool
     
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