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Differential equation

  1. Nov 4, 2008 #1
    solve this differential equation:

    y''-a(y'^2)-b(siny-a*cosy)=0

    (a&b are cte.)
     
  2. jcsd
  3. Nov 5, 2008 #2

    Mark44

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    cte. - ??
     
  4. Nov 5, 2008 #3

    gabbagabbahey

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    I'm assuming that "cte." means constants?

    The first step is to classify the differential equation: Is it ordinary? Is it linear? Is it homogeneous?

    Once you classify it, think of which methods you know that can be used to solve this type of DE. Which methods come to mind in this case?
     
  5. Nov 6, 2008 #4
    a&b are constant,
    i found the way of solving this !:

    v=y'
    => y''=v(dv/dy)

    after solving, i get this equation:

    y'^2=(-0.82b*cosy-0.9b*siny)/1.36

    (a=0.3)

    I'm not noob! I'm professor of university!

    please help me to solve this!:
    y'^2=(-0.82b*cosy-0.9b*siny)/1.36
     
  6. Nov 6, 2008 #5

    gabbagabbahey

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    Umm...if y is function of x, then y''=v(dv/dx) not vdv/dy...What variable(s) is y actually a function of here?
     
  7. Nov 6, 2008 #6

    HallsofIvy

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    No, gabbagabbahey, that is a perfectly valid method. If v= dy/dx, then
    [tex]\frac{d^2y}{dx^2}= \frac{dv}{dx}[/tex]
    NOT "v(dv/dx)" and, by the chain rule,
    [tex]= \frac{dy}{dx}\frac{dv/dy}= v\frac{dv}{dy}[/tex]

    That's a method called "quadrature" and is often used where the independent variable does not appear explicitely in the equation.

    Unfortunately, once you have solved for v as a function of y, the integral for y as a function of x tends to be wicked! Often leading to an "elliptic integral" which I suspect is the case here.
     
  8. Nov 6, 2008 #7

    gabbagabbahey

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    Oops, yes, I think I need some coffee :redface:
     
  9. Nov 6, 2008 #8

    HallsofIvy

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    By the way, here's an interesting application of "quadrature". Suppose you have a particle moving, on the real number line, under the influence of a force that depends only on position x and not t.

    The equation of motion, "force= mass times acceleration", is
    [tex]m \frac{d^2 x}{dt^2}= f(x)[/tex]
    where f(x) is the force function. Since t does not appear explicitely (the equation is "autonomous") we can let v= dx/dt and, as before, the equation becomes
    [tex]mv\frac{dv}{dt}= f(x)[/tex]
    a separable equation. Write it as mv dv= f(x)dx and integrate both sides:
    [tex]\frac{1}{2}mv^2= \int f(x)dx+ C[/tex]
    or
    [tex]\frac{1}{2}mv^2- \int f(x)dx= C[/tex]

    Do you see that that is "conservation of energy"? [itex](1/2)mv^2[/itex] is the kinetic energy and [itex]-\int f(x)dx[/itex] is the potential energy.
     
  10. Nov 6, 2008 #9
    ???
    y'=v=>(dy/dx)=v *
    y''=dv/dx=(dv/dx)(dy/dx)=v(dv/dy) =_*_=> y"=v(dv/dy)
     
  11. Nov 6, 2008 #10

    HallsofIvy

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    Hey, he said he was a University Professor and Universtiy Professors NEVER make mistakes!
     
  12. Nov 6, 2008 #11

    gabbagabbahey

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    Of course not, but apparently they do get "homework" :wink: :rolleyes:
     
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