# Exponential growth and decay (Part 2): Paying off credit-card debt

The following problem in differential equations has a very practical application for anyone who has either (1) taken out a loan to buy a house or a car or (2) is trying to pay off credit card debt. To my surprise, most math majors haven’t thought through the obvious applications of exponential functions as a means of engaging their future students, even though it is directly pertinent to their lives (both the students’ and the teachers’).

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a relative rate of growth of 25% per year. If you pay the minimum amount of$50 per month (or $600 per year), how long will it take for the balance to be paid? In this post, I present the actual solution of this problem. In tomorrow’s post, I’ll give some pedagogical thoughts about how this problem, and other similar problems inspired by financial considerations, could fit into a Precalculus course. Let’s treat this problem as a differential equation (though it could also be considered as a first-order difference equation… more on that later). Let $A(t)$ be the amount of money on the credit card after $t$ years. Then there are two competing forces on the amount of money that will be owed in the future: 1. The effect of compound interest, which will increase the amount owed by $0.25 A(t)$ per year. 2. The amount that’s paid off each year, which will decrease the amount owed by $\600$ per year. Combining, we obtain the differential equation $\displaystyle \frac{dA}{dt} = 0.25 A - 600$ There are a variety of techniques by which this differential equation can be solved. One technique is separation of variables, thus pretending that $dA/dt$ is actually a fraction. (In the derivation below, I will be a little sloppy with the arbitrary constant of integration for the sake of simplicity.) $\displaystyle \frac{dA}{0.25 A - 600} = dt$ $\displaystyle \int \frac{dA}{0.25 A - 600} = \displaystyle \int dt$ $\displaystyle 4 \int \frac{0.25 dA}{0.25 A - 600} = \displaystyle \int dt$ $4 \ln |0.25A - 600| = t + C$ $\ln |0.25A - 600| = 0.25 t + C$ $|0.25A - 600| = e^{0.25 t + C}$ $|0.25 A - 600| = C e^{0.25t}$ $0.25A - 600 = C e^{0.25t}$ $0.25 A = 600 + C e^{0.25t}$ $A = 2400 + C e^{0.25t}$ To solve for the missing constant $C$, we use the initial condition $A(0) = 2000$: $A(0) = 2400 + C e^0$ $2000 = 2400 + C$ $-400 = C$ We thus conclude that the amount of money owed after $t$ years is $A(t) = 2400 - 400 e^{0.25t}$ To determine when the amount of the credit card will be reduced to$0, we see $A(t) = 0$ and solve for $t$: $0 = 2400 - 400 e^{0.25 t}$ $400 e^{0.25 t} = 2400$ $e^{0.25t} = 6$ $0.25t = \ln 6$ $t = 4 \ln 6$ $t \approx 7.2 \hbox{~years}$ 1. Stephen says: