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Expressing time-dependent ODE

  1. Nov 29, 2012 #1
    This is surely the simplest problem imaginable in DE, but it's been a few years and I'm having trouble recalling. The goal of my task doesn't necessitate relearning DE, so I thought I would take a shot at asking directly.

    Simply, I wish to express the time-dependent rate equation [itex]\frac{dy(t)}{dt}=x-\frac{y(t)}{z}[/itex] as a function of time where [itex]x[/itex] and [itex]z[/itex] are known constants. I've been given a solution of [itex]y(t)=xz(1-e^{-t/z})[/itex] but I would very much like to remember how to get there. I do not have initial conditions, although [itex]y(0)=0[/itex] is a fair assumption for this problem.

    Thank you very much in advance.

    Note: this is not homework for a DE course.
     
  2. jcsd
  3. Nov 29, 2012 #2
    Your equation is first order linear, so the standard technique is to solve it using an integrating factor. Here is a link that explains:
    http://en.wikipedia.org/wiki/Integrating_factor

    If you look up integrating factor on Youtube, you can see people do the technique there.
     
  4. Nov 29, 2012 #3
    Excellent. Thank you.
     
  5. Nov 30, 2012 #4

    HallsofIvy

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    Another way, though it really applies mostly to higher order equations, is to use the fact that this equation is "linear". You have dy/dt+ By= C where B= 1/a and C= x are constants.

    We start by looking at the "associated homogeneous equation- just drop the "C" to get dy/dt+ By= 0. If we "try" a solution of the form [itex]y= ce^{at}[/itex], with a and c constants, [itex]dy/dx= ace^{at}[/itex] so the equation becomes [itex]ace^{at}+ Bce^{at}= 0[/itex]. Since [itex]e^{at}[/itex] is never 0 we can divide by [itex]ce^{at}[/itex] to get the "characteristic equation", [itex]a+ B= 0[/itex] so that a= -B. That is, the general solution to the associated homogeneous equation is [itex]y= ce^{-Bt}[/itex] for c any constant.

    Now, we recognize that the derivative of any constant is 0 so if y(t)= D for some constant D, the entire equation becomes dy/dt+ By= 0+ BD= C and D must be C/B. That is, the associated homogeneous equation has general solution [itex]ce^{Bt}[/itex] while [itex]y(t)= C/B[/itex] satisfies the entire equation. Because this differential equation is linear, it is easy to show that adding them gives [itex]y(t)= ce^{Bt}+ C/B[/itex] is the general solution to the entire equation.

    Putting B= 1/z and C= x again, that gives us [itex]y(t)= ce^{t/z}+ xz[/itex] as the general solution to the entire equation.

    If we have y(0)= 0, the setting both y and t equal to 0 gives [itex]y(0)= 0= ce^{0/z}+ xz= c+ xz[/itex] so that c= -xz. Then [itex]y(t)= -xze^{t/z}+ xz= xz(1- e^{t/z})[/itex].
     
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