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Solution of 2 x'' y + 3 x' y' + x y'' = 0 where ' is d/dt

  1. May 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Given [itex]\dot{x} \equiv { \mathrm{d}x \over \mathrm{d}t }[/itex] and [itex]\ddot{x}[/itex] is [itex]{ \mathrm{d^2}x \over \mathrm{d}t^2 }[/itex]

    what is solution for x(t) and y(t) which satsifies

    [itex]2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0[/itex]

    2. Relevant equations

    [itex]2 \ddot{x} y + 3 \dot{x} \dot{y} + x \ddot{y} = 0[/itex]

    3. The attempt at a solution

    clearly, guesses could be made for x(t) and y(t) but is there a most general solution which somehow includes all possible solutions?
     
    Last edited: May 12, 2016
  2. jcsd
  3. May 12, 2016 #2

    LCKurtz

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    Don't put equations in the titles in the first place. It violates forum rules.
     
  4. May 13, 2016 #3

    Mark44

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    Let ##x = e^{rt}## and let ##y = e^{st}##.
    See where that gets you.
     
  5. May 13, 2016 #4
    I know but I said that already in "3. The attempt at a solution"
    [itex]e^{rt}[/itex] is just a guess
    I am looking for a more general more formal solution method
     
  6. May 13, 2016 #5

    Mark44

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    This is more of an "educated guess" than just a guess. This is also a standard (and formal) technique that is used in solving differential equations.
     
  7. May 13, 2016 #6

    LCKurtz

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    It's kind of an odd problem with a second order DE and two unknown functions with no second equation nor boundary conditions. Not sure what to expect, but Mark's suggestion leads to two solutions of the form $$
    (x(t),y(t)) = C(e^{rt},e^{-2rt})\text{ and } D(e^{rt},e^{-rt})$$each with a free parameter ##r## and arbitrary constants. Of course, your question involves whether or not that is all there are. That question is usually answered by existence and uniqueness theorems. But this is a nonlinear setting so all bets are off. Whether there are such theorems for this kind of setting, I don't know.
     
  8. May 14, 2016 #7

    haruspex

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    It's too easy to generate whole families of solutions. E.g. set x=tm, y=tn (not even bothering with constant multiples) gives a solution whenever neither n nor m is zero and n is either -2m or 1-m.
     
  9. May 15, 2016 #8

    pasmith

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    Let [itex]x(t)[/itex] be arbitrary. You are then left with a second-order linear ODE for [itex]y(t)[/itex] with singular points wherever [itex]x(t) = 0[/itex].

    EDIT: You have [tex]
    2\ddot x y + 3 \dot x \dot y + x \ddot y = (\ddot x y + 2 \dot x \dot y + x \ddot y) + (\ddot x y + \dot x \dot y)
    = \frac{d^2}{dt^2}(xy) + \frac{d}{dt}(\dot x y) = 0[/tex] so you can integrate once immediately.
     
    Last edited: May 15, 2016
  10. May 15, 2016 #9

    haruspex

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    taking that a bit further, we can set x to be any function of t and obtain ##y=\frac c{x^2}\int x.dt##
     
  11. May 16, 2016 #10
    Hi @Doug Brown:

    I am not sure if this idea would be of general use, but I seem to remember learning about this technique as an undergraduate many decades ago.

    Define two differential operators as follows:
    Dx = (d/dt)(∂/∂y)
    Dy = (d/dt)(∂/∂x)​
    Then the equation can be written as:
    [ 3 Dx2 + 2 Dx Dy +Dy2 ] xy = 0​
    This can be factored as follows:
    [ [ Dx + Dy ] × [ 2 Dx + Dy ] ] xy = 0​
    This produces two equations:
    [ Dx + Dy ] xy = 0
    and
    [ 2 Dx + Dy ] xy = 0​
    I don't have a way of using a dot to indicate differentiation WRT t, so I will use an apostrophe instead.
    These equations can be written as:
    (1) x'y + xy' = 0
    (2) 2x'y + xy' = 0
    I have not checked this out, but I think that if a pair P1{x(t),y(t)} satisfies (1), and P[SUB2[/SUB]{x(t),y(t)} satisfies (2), then one general solution (of possibly many) of the original equation would be
    A × P1{x(t),y(t)} + B × P2{x(t),y(t)}​
    with A and B arbitrary constants.

    Regards,
    Buzz
     
    Last edited: May 16, 2016
  12. May 16, 2016 #11

    Mark44

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    (Above) Is this really what you meant? Usually Dx is a way of writing the operator ##\frac{d}{dx}## and similarly for Dy
    This is what I did using the hint I gave back in post #3.
     
  13. May 17, 2016 #12
    Hi Mark:

    I am familiar with Dx being an alternative notation for d/dx. I don't remember what the notation was to represent (d/dt)(∂/∂y) when I learned differential equations back in the 1950s, so I borrowed the Dx notation for that.

    What I found was that (1) yields y = -1/x, and (2) y = -1/x2. One solution for (1) is x = tα, y = t. This solution seems to be valid for any value of α. Similarly for (2) x = tα, y = t-2α. I have not checked out combinations.

    Regards,
    Buzz
     
  14. May 17, 2016 #13

    haruspex

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    Seems to me nobody has grasped the significance of my post #8. It shows how you can get all possible solutions by plugging in any integrable function for x(t) that you like, then determining y(t). What more is there to say?
     
  15. May 19, 2016 #14
    Another aspect of the approach I suggested is that it allows one to directly see the possible relationships between the variables x and y in a form independent of the parameter t.

    Regards,
    Buzz
     
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