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Differential equation help!

  1. May 27, 2013 #1
    Hi! I would like to ask what is the general solution of the following differential equation
    [itex] \frac{\partial X_x}{\partial t} = - \frac{\partial X_t}{\partial x}[/itex]

    Thank you very much.

    P.S. If you have some good resiource about this tyoe of equation to recommend please do so.
     
  2. jcsd
  3. May 27, 2013 #2

    CompuChip

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    What is Xx - does that stand for [itex]\frac{\partial X}{\partial x}[/itex] ?
    Because then you are basically asking about [itex]X_{xt} = -X_{tx}[/itex].

    In that case you should be looking at "weird" functions - given Clairaut's theorem at least the partial derivatives should not be continuous.
     
  4. May 27, 2013 #3
    Hi.

    If you want I can write it as [itex] \frac{\partial X}{\partial t} = -\frac{\partial T}{\partial x}[/itex] where [itex]T = T(x,t)[/itex] and [itex]X = X(x,t)[/itex] in general.

    I know they must be equal to a constant. Please correct me if I am wrong.

    Thank you.
     
  5. May 27, 2013 #4

    Office_Shredder

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    So the functions on the left and right hand side are not equal in general? And you're asking what the general form for X and T is as separate functions?
     
  6. May 28, 2013 #5
    [itex] \frac{\partial X}{\partial t} = -\frac{\partial X}{\partial x}[/itex]
    [itex] X=f(t-x) [/itex] any derivable function [itex]f [/itex]
     
  7. May 28, 2013 #6

    HallsofIvy

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    Let X(x, t)= f(x, t) be any differentiable function of x and t and define T(x, t)= -f(t, x).
    For example, take X(x, t)= x+ t^2, T(x, t)= -t- x^2. Then [itex]\partial X/\partial x= 1= -\partial T/\partial t[/itex].

    I don't know what you mean by "they must be equal to a constant".
     
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