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Solving 2nd order DE HELP!

  1. Oct 24, 2006 #1
    can any1 help me solve this? d2x/dt2 = -(k/m)*x -g
    at x= 0, t=0, v=sqrt(u^2-2gh)

    what is x?? can show the steps?
     
  2. jcsd
  3. Oct 24, 2006 #2

    radou

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    This DE can be solved by direct integration. After calculating [tex]\frac{dx}{dt} = v(t)[/tex] and x(t), use the initial conditions to get the constants of integration.

    P.S. x(t) is a function of position in time.
     
  4. Oct 24, 2006 #3

    HallsofIvy

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    One of the reasons we ask that people show what they have already done is so we will have a better idea of what kind of help you need and what kinds of suggestions you would understand. I can think of several ways of solving this but I'm not sure what "tools" you have to use.

    The method radou is suggesting is called "quadrature". Since t does not explicitely appear in the equation, you can eliminate it. Let v= dx/dt. The d2x/dt2= dv/dt= (dv/dx)(dx/dt)= vdv/dx. That is, the original equation can be written as a first order equation.
    Now we have vdv/dx= -(k/m)x- g. In fact, that is separable and, so, easily integrable: vdv= (-(k/m)x- g)dx so (1/2)v2= -(k/2m)x2- 2x+ C (the v2 is the reason for the name "quadrature"). Once you have found v as a function of t, integrate dx/dt= v to find x.

    But that is not the method I would use. This is a linear d.e. with constant coefficients. I has "characteristic equation" r2= -(k/m) which has roots [itex]r= \pm i\sqrt{k/m}[/itex]. The general solution to the associated homogeneous equation is
    [tex]x(t)= C_1cos(\sqrt{k/m}t)+ C_2sin(\sqrt{k/m}t)[/tex]
    Now you need to find a "specific" solution to the entire equation. Since the "non-homogeneous part" is the constant, -g, I would suggest assuming y is a constant: y(t)= A, and seeing if you can pick A to make the equation true. Does any of that make sense? That was what I meant by "not sure what "tools" you have to use".
    By the way, is "sqrt(u^2-2gh)" simply a constant?
     
  5. Oct 24, 2006 #4
    ya sqrt(u^2- 2gh) is a constant how should i go about solving it to satisfy this condition?
     
  6. Oct 25, 2006 #5

    HallsofIvy

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    Do what I suggested before. Assume y= A, a constant, and plug that into the equation. What does A have to be in order for the equation to be true?
     
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