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Solving a differential equation

  1. Mar 5, 2013 #1
    I have the following differential equation

    [itex]\frac{\partial}{\partial t}\left(\frac{a}{X}\right)+\frac{X}{b}\frac{ \partial Y}{\partial t}+\frac{c}{X}=0[/itex]

    where [itex]a[/itex], [itex]b[/itex] and [itex]c[/itex] are constants and [itex]X[/itex] is a function of
    [itex]t[/itex]. I want to solve it for [itex]Y[/itex] analytically (if possible) or numerically.
  2. jcsd
  3. Mar 5, 2013 #2


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    First, that's not really a partial differential equation because the only differentiation is with respect to the single variable, t. If X(t) is a known function of t, then [itex]\partial/\partial ta/X[/itex] is also a known function of t- call it X'(t). Then your equation can be written
    [tex]\frac{X}{b}\frac{dY}{dt}= -X'- C/X[/tex]

    [tex]\frac{dY}{dt}= (-X'- C/X)\frac{b}{X}= -\frac{X'X- C}{X^2}[/tex]
    and you solve for Y by integrating.b
    Last edited by a moderator: Mar 5, 2013
  4. Mar 5, 2013 #3
    If X(t) is a known function of t I would solve it easily. Unfortunately this is not the case.
  5. Mar 5, 2013 #4


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    Then all you can do is write
    [tex]Y(t)= -\int\frac{XX'- C}{X^2}dt[/tex]
  6. Mar 5, 2013 #5
    Two unknown functions and one equation only is not enough. You need two equations.
  7. Mar 5, 2013 #6
    Many thanks to you all!

    To close the gap, I obtained a second relation


    where [itex]d[/itex] is a constant and [itex]O[/itex] is a known function of [itex]t[/itex] with
    a closed analytical form.
  8. Mar 6, 2013 #7


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    I would be inclined to call ##\frac{d}{dt}(a/X(t))## something other than ##X'(t)##, which could be confused with ##dX/dt##. :tongue2:
  9. Mar 10, 2013 #8


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    Take the derivative w.r.t. t, then you can substitute for [itex]\frac{dY}{dt}[/itex] in HallsofIvy's equation in Post #2. Now you have a differential equation in just one unknown function, X(t).

    p.s. "d" is not the best name for a quantity in anything having to do with calculus :smile:
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