Expressing time-dependent ODE

CentreShifter

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.

Vargo

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.

CentreShifter

Excellent. Thank you.

HallsofIvy

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].