# Math Insight

### Exponential growth and decay: a differential equation

This little section is a tiny introduction to a very important subject and bunch of ideas: solving differential equations. We'll just look at the simplest possible example of this.

The general idea is that, instead of solving equations to find unknown numbers, we might solve equations to find unknown functions. There are many possibilities for what this might mean, but one is that we have an unknown function $y$ of $x$ and are given that $y$ and its derivative $y'$ (with respect to $x$) satisfy a relation $$y'=ky$$ where $k$ is some constant. Such a relation between an unknown function and its derivative (or derivatives) is what is called a differential equation. Many basic ‘physical principles’ can be written in such terms, using ‘time’ $t$ as the independent variable.

Having been taking derivatives of exponential functions, a person might remember that the function $f(t)=e^{kt}$ has exactly this property: $${d\over dt}e^{kt}=k\cdot e^{kt}$$ For that matter, any constant multiple of this function has the same property: $${d\over dt}(c\cdot e^{kt})=k\cdot c\cdot e^{kt}$$ And it turns out that these really are all the possible solutions to this differential equation.

There is a certain buzz-phrase which is supposed to alert a person to the occurrence of this little story: if a function $f$ has exponential growth or exponential decay then that is taken to mean that $f$ can be written in the form $$f(t)=c\cdot e^{kt}$$ If the constant $k$ is positive it has exponential growth and if $k$ is negative then it has exponential decay.

Since we've described all the solutions to this equation, what questions remain to ask about this kind of thing? Well, the usual scenario is that some story problem will give you information in a way that requires you to take some trouble in order to determine the constants $c,k$. And, in case you were wondering where you get to take a derivative here, the answer is that you don't really: all the ‘calculus work’ was done at the point where we granted ourselves that all solutions to that differential equation are given in the form $f(t)=ce^{kt}$.

First to look at some general ideas about determining the constants before getting embroiled in story problems: One simple observation is that $$c=f(0)$$ that is, that the constant $c$ is the value of the function at time $t=0$. This is true simply because $$f(0)=ce^{k \cdot 0}=ce^{0}=c\cdot 1=c$$ from properties of the exponential function.

More generally, suppose we know the values of the function at two different times: $$y_1=ce^{kt_1}$$ $$y_2=ce^{kt_2}$$ Even though we certainly do have ‘two equations and two unknowns’, these equations involve the unknown constants in a manner we may not be used to. But it's still not so hard to solve for $c,k$: dividing the first equation by the second and using properties of the exponential function, the $c$ on the right side cancels, and we get $${y_1\over y_2}=e^{k(t_1-t_2)}$$ Taking a logarithm (base $e$, of course) we get $$\ln y_1-\ln y_2=k(t_1-t_2)$$ Dividing by $t_1-t_2$, this is $$k={\ln y_1-\ln y_2\over t_1-t_2}$$ Substituting back in order to find $c$, we first have $$y_1=ce^{{\ln y_1-\ln y_2\over t_1-t_2}t_1}$$ Taking the logarithm, we have $$\ln y_1=\ln c+{\ln y_1-\ln y_2\over t_1-t_2}t_1$$ Rearranging, this is $$\ln c=\ln y_1-{\ln y_1-\ln y_2\over t_1-t_2}t_1= {t_1\ln y_2-t_2\ln y_1\over t_1-t_2}$$ Therefore, in summary, the two equations $$y_1=ce^{kt_1}$$ $$y_2=ce^{kt_2}$$ allow us to solve for $c,k$, giving $$k={\ln y_1-\ln y_2\over t_1-t_2}$$ $$c=e^{t_1\ln y_2-t_2\ln y_1\over t_1-t_2}$$

A person might manage to remember such formulas, or it might be wiser to remember the way of deriving them.

#### Example 1

A herd of llamas has $1000$ llamas in it, and the population is growing exponentially. At time $t=4$ it has $2000$ llamas. Write a formula for the number of llamas at arbitrary time $t$.

Solution: Here there is no direct mention of differential equations, but use of the buzz-phrase ‘growing exponentially’ must be taken as indicator that we are talking about the situation $$f(t)=ce^{kt}$$ where here $f(t)$ is the number of llamas at time $t$ and $c,k$ are constants to be determined from the information given in the problem. And the use of language should probably be taken to mean that at time $t=0$ there are $1000$ llamas, and at time $t=4$ there are $2000$. Then, either repeating the method above or plugging into the formula derived by the method, we find $$c=\hbox{ value of f at t=0 } = 1000$$ $$k={\ln f(t_1)-\ln f(t_2)\over t_1-t_2}={\ln 1000-\ln 2000\over 0-4}$$ $$={ \ln {1000\over 2000}}{-4}={\ln {1\over 2} \over -4 } = (\ln 2)/4$$ Therefore, $$f(t)=1000\;e^{{\ln 2\over 4}t}=1000\cdot 2^{t/4}$$ This is the desired formula for the number of llamas at arbitrary time $t$.

#### Example 2

A colony of bacteria is growing exponentially. At time $t=0$ it has $10$ bacteria in it, and at time $t=4$ it has $2000$. At what time will it have $100,000$ bacteria?

Solution: Even though it is not explicitly demanded, we need to find the general formula for the number $f(t)$ of bacteria at time $t$, set this expression equal to $100,000$, and solve for $t$. Again, we can take a little shortcut here since we know that $c=f(0)$ and we are given that $f(0)=10$. (This is easier than using the bulkier more general formula for finding $c$). And use the formula for $k$: $$k={\ln f(t_1)-\ln f(t_2)\over t_1-t_2}= {\ln 10 -\ln 2,000\over 0-4}={ \ln {10\over 2,000} \over -4 }= {\ln 200\over 4}$$ Therefore, we have $$f(t)=10\cdot e^{{\ln 200\over 4}\;t}=10\cdot 200^{t/4}$$ as the general formula. Now we try to solve $$100,000=10\cdot e^{{\ln 200\over 4}\;t}$$ for $t$: divide both sides by the $10$ and take logarithms, to get $$\ln 10,000={\ln 200\over 4}\;t$$ Thus, $$t=4\,{\ln 10,000\over \ln 200}\approx 6.953407835.$$

#### Exercises

1. A herd of llamas is growing exponentially. At time $t=0$ it has $1000$ llamas in it, and at time $t=4$ it has $2000$ llamas. Write a formula for the number of llamas at arbitrary time $t$.
2. A herd of elephants is growing exponentially. At time $t=2$ it has $1000$ elephants in it, and at time $t=4$ it has $2000$ elephants. Write a formula for the number of elephants at arbitrary time $t$.
3. A colony of bacteria is growing exponentially. At time $t=0$ it has $10$ bacteria in it, and at time $t=4$ it has $2000$. At what time will it have $100,000$ bacteria?
4. A colony of bacteria is growing exponentially. At time $t=2$ it has $10$ bacteria in it, and at time $t=4$ it has $2000$. At what time will it have $100,000$ bacteria?