Decay Plus Input at a Constant Rate Revisited

In Chapter 2 of Intermediate Physics for Medicine and Biology, Russ Hobbie and I discuss the problem of decay plus input at a constant rate.

Suppose that in addition to the removal of y from the system at a rate –by, y enters the system at a constant rate a, independent of y and t. The net rate of change of y is given by

Then we go on to discuss how you can learn things about a differential equation without actually solving it.

It is often easier to write down a differential equation describing a problem than it is to solve it… However, a good deal can be learned about the solution by examining the equation itself. Suppose that y(0) = 0. Then the equation at t = 0 is dy/dt = a, and y initially grows at a constant rate a. As y builds up, the rate of growth decreases from this value because of the –by term. Finally when a – by = 0, dy/dt is zero and y stops growing. This is enough information to make the sketch in Fig. 2.13.

The equation is solved in Appendix F. The solution is

… The solution does have the properties sketched in Fig. 2.13, as you can see from Fig. 2.14.

Figure 2.13 looks similar to this figure

Sketch of the initial slope a and final value a/b of y when y(0) = 0. In this figure, a=b=1.

 And Fig. 2.14 looks like this

A plot of y(t) using Eq. 2.26, with a=b=1.

However, Eq. 2.26 is not the only solution that is consistent with the sketch in Fig. 2.13. Today I want to present another function that is consistent with Fig. 2.13, but does not obey the differential equation in Eq. 2.25.

Let’s examine how this function behaves. When bt is much less than one, the function becomes y = at, so it’s initial growth rate is a. When bt is much greater than one, the function approaches a/b. The sketch in Fig. 2.13 is consistent with this behavior.

Below I show both Eqs. 2.26 and 2.26’ in the same plot.

A plot of y(t) using Eq. 2.26 (blue) and Eq. 2.26' (yellow), with a=b=1.

The function in Eq. 2.26 (blue) approaches its asymptotic value at large t more quickly than the function in Eq. 2.26’ (yellow).

The moral of the story is that you can learn a lot about the behavior of a solution by just inspecting the differential equation, but you can’t learn everything (or, at least, I can’t). To learn everything, you need to solve the differential equation. 

By the way, if Eq. 2.26’ doesn’t solve the differential equation in Eq. 2.25, then what differential equation does it solve? The answer is

 How did I figure that out? Trial and error.