Compound Interest with Differential Equations

Compound Interest with Differential Equations

Let $S$ be an initial sum of money. Let $r$ represent an interest rate. We can model the growth of an initial deposit with respect to the interest rate $r$ with differential equations. If $t$ represents time, then the rate of change of the initial deposit is $\frac{dS}{dt}$ and assuming that the initial deposit is compounded continuously, then we have that:

\begin{align} \quad \frac{dS}{dt} = rS \end{align}

We can further set up an initial value problem to this differential equation. Suppose that the initial deposit is $S_0$. Then $S(0) = S_0$. The solution to the initial value problem with the differential equation and initial condition above will give us a function $S$ which gives us the amount in the individuals account at time $t$.

The differential equation above can be easily solved as a separable differential equation. Noting that $\ln \mid S \mid = \ln S$ (since $S > 0$ ) and we have that:

\begin{align} \quad \frac{dS}{dt} = rS \\ \quad \frac{1}{S} \frac{dS}{dt} = r \\ \quad \frac{1}{S} \: dS = r \: dt \\ \quad \int \frac{1}{S} \: dS = \int r \: dt \\ \quad \ln S = rt + D \\ \quad S = e^{rt + D} \\ \quad S = Ce^{rt} \end{align}

Using the initial condition that $S(0) = S_0$ and we have that $C = S_0$. Therefore the solution to this initial value problem is:

\begin{align} \quad S(t) = S_0e^{rt} \end{align}

If you are familiar with problems regarding compound interest - this formula should be somewhat familiar.

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