# Separable Differential Equations: Exponential Decay

When I was in high school, my chemistry teacher presented me with a radioactive decay problem, and a formula that read:

$$Q=Q_{0} * e^{-r t}$$

Where \(Q\) represented the current amount of radioactive material, \(Q_0\) represented the starting amount of material, and \(r\) was some unknown â€śdecay rateâ€ť. I was greatly confused by this. â€śWhy do we raise the time function to the power of e?â€ť I asked. â€śWhy not raise the time function to the power of \(\frac{1} {2}\) ? We are working with half-lives after allâ€ť. It turns out that the answer to this question is surprisingly nuanced, and requires a deep dive into differential equations. So letâ€™s take a look at a radioactive decay problem through the lens of differential equations.

Now I know this problem has four parts to it, but donâ€™t panicâ€”you wonâ€™t be needing your brown pants today. In fact, donâ€™t even read Parts A-D yet. Letâ€™s just look at the differential equation, and solve it before doing anything else:

$$\frac{d Q}{d t}=-r Q$$

You get bonus points if you recognized this as a **Separable Differential Equation**, which means that you can move the two variables (Q and t) to opposite sides of the equation. The best way to succeed in a differential equations class is by always knowing what type of equation youâ€™re working with. If youâ€™re not sure what type of equation youâ€™re looking at, you can always try to separate it. In this case, we multiply both sides by \(dt\), and divide both sides by \(Q\).

$$\frac{d Q}{Q}=-r * d t$$

The point of separating the equation is so we can take the integral of both sides, thereby getting rid of the differential terms \(dQ\) and \(dt\).

$$\int \frac{d Q}{Q}=-r * \int d t$$

Note that we can leave the \(-r\) term outside the integral for \(t\), since \(r\) is a constant, and does not depend on \(t\). We solve both integrals to get

$$\ln (Q)=-r t+C_{1}$$

Never forget that \(+C\). Hopefully you kept those good habits from MAT21B. Also donâ€™t forget why weâ€™re doing this. â€śSolvingâ€ť a differential equation means finding detailed temporal information about the time-sensitive variable, \(Q(t)\). When I write \(Q(t)\), Iâ€™m saying that â€ś\(Q\) is a function of \(t\)â€ť. Letâ€™s finish up this equation:

$$Q(t)=e^{\left(-r t+C_{1}\right)}$$

Note that \(e^{(-r t+C)}\) can be written as: [\(e^{-r t} * e^{C_{1}}\)]. Also note that \(e^{C_{1}}\) is just another constant, which Iâ€™ll call \(C_2\). That was a bit fastâ€”let me do it again in slow motion:

$$Q(t)=e^{\left(-r t+C_{1}\right)}=e^{-r t} * e^{c_{1}}=C_{2} e^{-r t}$$

Great, now weâ€™re almost done! The last thing that you have to deal with in any differential equation is getting rid of any lingering \(C\) The way that we do this is through **Initial Conditions**. Initial conditions allow us to plug in predetermined values for \(Q\) and \(t\) that reveal the value of \(C_2\). If you take a close look at the problem, you might notice that the wording for Part A gives us an initial condition right away. Letâ€™s use it to find \(C_2\):

From this phrasing, and our equation: \(Q(t)=C_{2} e^{-r t}\), we get:

$$Q(0)=C_{2} e^{-r(0)}=C_{2}=100 \mathrm{mg}$$

And there you have it. Now that we know \(C_2\), we can write our solved differential equation.

$$\boldsymbol{Q}(\boldsymbol{t})=\mathbf{1} \mathbf{0} \mathbf{0} \boldsymbol{m} \boldsymbol{g} * \boldsymbol{e}^{-\boldsymbol{r} t}$$

I know I havenâ€™t solved any of Parts A-D yet, but now that you have the finished equation, they will basically solve themselves! Here we go:

### Part A

Here we are asked to find \(r\) if \(Q(1 week) = 84 \mathrm(mg)\). Letâ€™s plug it in:

$$\begin{array}{c}{Q(1)=84 m g=100 m g * e^{-r}} \\ {e^{-r}=0.84} \\ {r=-\ln (0.84)=0.1744}\end{array}$$

To make sure that units are consistent, the units of \(r\) must be \(\frac {1} {weeks}\). Otherwise, \(e^{-rt}\) would have non-cancelling units, and that would be awkward.

### Part B

We practically already solved this one; just plug in \(r\) from Part A:

$$Q(t)=100 m g * e^{-0.1744 t}$$

### Part C

This time, weâ€™re looking for \(t\) when \(Q(t) = 50mg \). Again, just plug it in:

\begin{array}{c}{Q(t)=50 m g=100 m g * e^{-0.1744 t}} \\ {e^{-0.1744 t}=0.5} \\ {t=-\frac{\ln (0.5)}{0.1744}=3.974}\end{array}

Quick, what units does \(t\) have? Hopefully you said â€śweeksâ€ť. We have to remain consistent with our given values.

### Part D

The only tricky thing going on here is labelling the two points. We know 3 points easily: \(Q(0)=100 m g, Q(1)=84 m g\), and \(Q(3.974)=50 m g\). If you want to be fancy you include: \(Q\left(\frac{1}{0.1774}\right)=\frac{100}{e} m g\), or \(Q(2 * 3.974)=25 m g\). The following graph is a reasonable solution:

**What Did We Learn? **

Aside from how to solve the above problem, we learned two important things today: first of all, that solving a differential equation should always be your first step in tackling a problem, regardless of how many parts the question has. Additionally, we learned that the equation [ \(Q=Q_{0} * e^{-r t}\) ] is the ultimate solution to exponential decay problemsâ€”problems in which the rate of change of a substance, \(\frac{d Q}{d t}\), is proportional to how much of that substance you have. When I was given this equation in high school, I didnâ€™t understand why we used the natural number \(e\). But knowing that the equation represents the *general case *of this differential equation completely explains why mathematicians prefer it over all other potential solutions for such a problem.