How Do You Solve a Time-Dependent Rate Equation?

[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 \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.

 

[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 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}).