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2nd order differential equation

  1. Jun 25, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]y^{2}\frac{d^{2}y}{dx^2} [/tex] + ay = b(cx-d)

    Find y as a function of x,a,b,c & d (a,b,c & d are all constant(!))

    2. Relevant equations

    -

    3. The attempt at a solution

    Not a clue, this is actually how far I got with my own take on an orbital mechanics problem I made up. Not my homework (I swear!). My guess would be some kind of substitution, but I don't know how to go about it.

    P.S. Sorry about the alignment, but my mastery of LaTeX is minimal :wink:. As is probably obvious, the unformatted bit should be on a level with the y^2 at the start.
     
  2. jcsd
  3. Jun 25, 2010 #2

    hunt_mat

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    Nothing immediately screams out at me, the think I might be tempted to try is write $u=dy/dx$ and then treat it as a dynamical system. Other than that, solve it numerically?
     
  4. Jun 26, 2010 #3

    HallsofIvy

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    That is a very non-linear equation and, like all non-linear equations, will be very difficult to solve, even for specified initial conditions. In fact, there might not be a single formula that will give all solutions and, even if there is, I would not expect it to be in terms of elementary functions. You say this came from "an orbital mechanics problem". Those tend to give elliptic function solutions.
     
  5. Jun 28, 2010 #4
    Yes, I think that's probably the case, although this was to do with radius/time, rather than cartesian. I was just wondering if there was a standard method for solving 2nd order differentials involving powers of y as a coefficient?
     
  6. Jun 28, 2010 #5

    HallsofIvy

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    No, there isn't. As I said before, non-linear equations tend to be very difficult. There is no general way of solving even the simplest non-linear equations.

    Sometimes it helps, not so much to "solve" the equation, but to get information about the solution, to write the single equation as a system of equations. If you let v= dy/dx, then [itex]d^2y/dx^2= dv/dx[/itex] so your equation becomes [itex]y^2 dv/dt= -y+ bcx- bd[/itex] so you have the system of equations
    [tex]\begin{pmatrix} \frac{dy}{dx}= v \\ \frac{dv}{dx}= -\frac{1}{y}+ \frac{bcx- bd}{y^2}\end{pmatrix}[/tex]
     
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