Another differential equation for growth and decay

Consider the differential equation

\[ \dfrac{dx}{dt} = kx + m, \]

where \(k, m\) are real constants and \(k \neq 0\). We can again think of this equation as describing the growth or decay of a quantity \(x\) with respect to time \(t\). Now the growth or decay of \(x\) is not only at a rate proportional to itself; it also has a constant component \(m\).

We solve this differential equation by a method similar to that used for our first differential equation. In the Appendix, we give an alternative, perhaps more elegant, method.

For simplicity, we consider the case that \(kx + m\) is positive. Taking the reciprocal of both sides of the differential equation yields

\[ \dfrac{dt}{dx} = \dfrac{1}{kx+m}, \]

so that

\[ t = \int \dfrac{1}{kx + m} \, dx. \]

Recalling that \(k\) and \(m\) are just constants, we can perform this integration and obtain

\[ t = \dfrac{1}{k} \log_e (kx+m) + c, \]

where \(c\) is a constant of integration. Rearranging this equation gives

\begin{align*} x &= \dfrac{e^{k(t-c)} - m}{k} \\ &= \dfrac{1}{k} e^{-kc} \, e^{kt} - \dfrac{m}{k}. \end{align*}

Now the factor \(\dfrac{1}{k} e^{-kc}\) of the first term is a constant, which we call \(C\).

In general (making no simplifying assumptions), any solution is of the form

\[ x(t) = C e^{kt} - \dfrac{m}{k}, \]

where \(C\) is a real constant. You can easily check that any such \(x\) is a solution.


Given that \(x=3\) when \(t=0\), find the solution to the differential equation

\[ \dfrac{dx}{dt} = 2x - 5. \]


The general solution of the differential equation is

\[ x(t) = C e^{2t} + \dfrac{5}{2}, \]

where \(C\) is any constant. Substituting \(t=0\), \(x=3\) gives \(C = \dfrac{1}{2}\), so the unique solution is

\[ x(t) = \dfrac{1}{2} e^{2t} + \dfrac{5}{2}. \]

If \(k < 0\) (that is, if \(x\) decays), then \(e^{kt}\) approaches 0 as \(t\) becomes very large, so

\begin{align*} \lim_{t \to \infty} x &= \lim_{t \to \infty} \Bigl(C e^{kt} - \dfrac{m}{k}\Bigr) \\ &= - \dfrac{m}{k}. \end{align*}

Thus \(x\) approaches \(-\dfrac{m}{k}\), which is called the equilibrium value. On the other hand, if \(k>0\) (that is, if \(x\) grows), then \(x\) approaches the equilibrium value \(-\dfrac{m}{k}\) as \(t \to -\infty\).


Suppose \(x\) is given by the differential equation

\[ \dfrac{dx}{dt} = -3x + 7, \]

and \(x=11\) when \(t=3\). Find the equilibrium value of \(x\) as \(t \to \infty\).


The general solution for \(x\) has the form

\[ x(t) = C e^{-3t} + \dfrac{7}{3}. \]

The first term approaches 0 for large \(t\), so the equilibrium value of \(x\) is \(\dfrac{7}{3}\).

Note that we do not need to compute \(C\) or use the initial condition in order to calculate the equilibrium value.

The equilibrium value can also be computed by setting \(\dfrac{dx}{dt} = 0\). So, for the previous example, the differential equation reduces to \(0 = -3x+7\). At equilibrium, \(x\) is in a 'steady state' and the rate of change of \(x\) is 0.

Population growth with migration

The constant term \(m\) in the differential equation can be considered as the rate of migration of a population, as in the following example — remembering, as always, that sustained exponential population growth is usually unrealistic.


Suppose again that there are 1000 birds on an island, breeding with a constant continuous growth rate of 10% per year. But now birds migrate to the island at a constant rate of 100 new arrivals per year. To three significant figures, how many birds are on the island after seven years?


Let \(x\) be the number of birds on the island after \(t\) years. With \(k=0.1\) and \(m = 100\), we have

\[ \dfrac{dx}{dt} = 0.1 \, x + 100. \]

The general solution to this differential equation is

\[ x(t) = C e^{0.1 \, t} - \dfrac{100}{0.1} = C e^{0.1 \, t} - 1000. \]

Substituting \(t = 0\), \(x = 1000\) gives \(C = 2000\), so we have

\[ x(t) = 2000 \, e^{0.1 \, t} - 1000. \]

We obtain \(x(7) = 2000 \, e^{0.7} - 1000 \approx 3030\) birds, to three significant figures.

Similar differential equations can be used to model quantities which decay but are replenished at a constant rate. In the History and applications section we will discuss the example of the HIV virus.

Exercise 5

We've now seen two examples of birds on an island, with initial population 1000 and constant continuous growth rate 10%. Consider the following three scenarios:

  1. 700 birds migrate to the island all at once, at time \(t=0\), and no further migration occurs.
  2. Birds migrate to the island at a constant rate of 100 per year, for seven years (as in the previous example).
  3. 700 birds migrate to the island all at once, at time \(t=7\), and no further migration occurs.

For each scenario, calculate the number of birds on the island after seven years, to three significant figures. (You may use the two relevant previous examples.) Discuss the differences between the answers.

Exercise 6

Nicotine patches are used by people who wish to discontinue smoking. The patches are applied to the skin and deliver nicotine to the blood stream. The concentration \(A\) of nicotine in blood plasma (in \(\mu\)g/L) can be modelled by differential equations:

\begin{alignat*}{2} \dfrac{dA}{dt} &= R_0 + k A & \qquad &\text{(when the patch is on)} \\ \dfrac{dA}{dt} &= kA & &\text{(when the patch is off)}, \end{alignat*}

where \(t\) is time (in hours), \(R_0\) is the infusion rate of nicotine, and \(k\) is the (negative) continuous decay rate of nicotine in the bloodstream. A particular brand of nicotine patch has infusion rate \(R_0 = 1\). It is to be applied for 16 hours, then removed for 8 hours (while asleep) each day.

In this exercise, assume \(k = -0.12\) and give answers to three significant figures.

  1. Assume at time \(t=0\) there is no nicotine in the bloodstream; then the patch is applied for 16 hours. What is the concentration of nicotine at \(t=16\)?
  2. The patch is then removed for 8 hours. What is the nicotine concentration at \(t=24\)?

This exercise was suggested by Geoffrey Kong. See the References section of this module for details of a relevant textbook and the research paper.

Growth and decay with input and output

The population of a country is not really ever just growing or decaying. People are always being born, dying, immigrating and emigrating. In general, a quantity may change according to various influences which cause it to grow or decay.


A cell culture in a biology lab currently holds 1 million cells. The cells have a constant continuous birth rate of \(1.5\%\) and death rate of \(0.5\%\) per hour. Cells are extracted from the culture for an experiment at the rate of 5000 per hour.

How many cells will be in the culture 10 hours from now?

Let \(p\) be the population, and let \(t\) be the number of hours from now. Births, deaths and extraction respectively contribute \(0.015 \, p\), \(-0.005 \, p\) and \(-5000\) to \(\dfrac{dp}{dt}\). So we have

\[ \dfrac{dp}{dt} = 0.015 \, p - 0.005 \, p - 5000 = 0.01 \, p - 5000. \]

The general solution to this differential equation is

\begin{align*} p(t) &= C e^{0.01 \, t} + \dfrac{5000}{0.01} \\ &= C e^{0.01 \, t} + 500\,000. \end{align*}

Since \(p=1\,000\,000\) when \(t=0\), we have \(C = 500\,000\), so

\[ p(t) = 500\,000 \, e^{0.01 \, t} + 500\,000. \]

Substituting \(t=10\), the population in 10 hours will be \(500\,000 \, e^{0.1} + 500\,000 \approx 1\,052\,585\).

Exercise 7

  1. In the previous example, when will the population double (that is, reach 2 million)?
  2. If no cells were extracted from the culture, but the birth and death rate remained constant, what would be the population doubling time?

Give answers in hours to two decimal places.

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