Expressing time-dependent ODE

by CentreShifter
Tags: expressing, timedependent
 P: 24 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 $\frac{dy(t)}{dt}=x-\frac{y(t)}{z}$ as a function of time where $x$ and $z$ are known constants. I've been given a solution of $y(t)=xz(1-e^{-t/z})$ but I would very much like to remember how to get there. I do not have initial conditions, although $y(0)=0$ is a fair assumption for this problem. Thank you very much in advance. Note: this is not homework for a DE course.
 P: 350 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.
 P: 24 Excellent. Thank you.
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Expressing time-dependent ODE

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 $y= ce^{at}$, with a and c constants, $dy/dx= ace^{at}$ so the equation becomes $ace^{at}+ Bce^{at}= 0$. Since $e^{at}$ is never 0 we can divide by $ce^{at}$ to get the "characteristic equation", $a+ B= 0$ so that a= -B. That is, the general solution to the associated homogeneous equation is $y= ce^{-Bt}$ 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 $ce^{Bt}$ while $y(t)= C/B$ satisfies the entire equation. Because this differential equation is linear, it is easy to show that adding them gives $y(t)= ce^{Bt}+ C/B$ is the general solution to the entire equation.

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

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

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