Modelling continuous compound interest with differential equations (Intuition)

Starting with the equation for continuously compounded interest, we can derive the differential equation. Let $$A$$ be the amount accumulated, $$P$$ be the principal amount and $$r$$ the rate.

\begin{align*} A &= Pe^{rt}\\ \frac{dA}{dt} &= Pre^{rt}\\ \frac{dA}{dt} &= Ar, \text{ since } A =Pe^{rt}\\ \end{align*}

How are you supposed to work out the rate of change of $$A$$ with respect to $$t$$ intuitively?

I don't know whether there is a lapse in my understanding of a derivative, however I only see how to start with the equation for compound interest and then derive the differential equation. I don't see how to model in terms of the differential equations first.

My question more generally is how to model in terms of differential equations, however I would love to know with specific reference to compound interest.

• Feel free to ask if something is still unclear May 22 '20 at 17:35

How do you perceive $$\frac{dA}{dt} = r$$? Rate of change is constant at all $$t$$ right?

Which in return gives the that $$A = rt + c$$ where $$c$$ is an arbitrary constant.

Now what about $$\frac{dA}{dt} = rA$$? The rate of change is depending on $$A$$ itself with a constant multiple of $$r$$? What does it tell us? If $$A$$ is larger, then $$\frac{dA}{dt}$$ will be larger too. This is purely the concept of exponential - the rate of change depends on previous $$A$$.

To solve the differential equation: \begin{align} \frac{dA}{dt}&=Ar \\ \frac{1}{A}dA &= r dt \\ \int \frac{1}{A}dA &= \int r dt\\ ln(A) &= rt + C,\qquad \text{where C is a constant} \\ A &= e^{rt+C} \\ A &= A_0 e^{rt}, \qquad \text {where e^C = A_0} \end{align} The $$A_0$$ is the initial condition.

• Intuitively, you can think that the interest is compounding infinitely many times in a period of $t$ rather than the usual $A= P(1+r)^t$ Apr 17 '20 at 8:43

Let $$A(t)=P\cdot (1+r)^t$$. Then firstly we can calculate the interest rate in the discrete case. This is

$$\frac{A(t+1)-A(t)}{A(t)}=\frac{P\cdot (1+r)^{t+1}-P\cdot (1+r)^{t}}{P\cdot (1+r)^{t}}=\frac{P\cdot (1+r)^{t}\cdot (1+r-1)}{P\cdot (1+r)^{t}}=r$$

We can additionaly divide the most left fraction by 1 without changing the value.

$$\frac{\color{blue}{\frac{A(t+1)-A(t)}1}}{A(t)}=r$$

On the LHS we have a difference quotient (blue term) with $$\Delta t=1$$. Next we multiply the equation by $$A(t)$$ and we let $$\Delta t \to 0$$

$$\lim_{\Delta t \to 0}\frac{A(t+\Delta t)-A(t)}{\Delta t}=A(t)r$$

Now we have a differential quotient on the LHS which is the same as a derivative.

$$\frac{dA}{dt}=A\cdot r$$

I hope it is comprehnsible how the differential equation has been derived.

• The initial condition is $A(0)=P$ Apr 17 '20 at 12:14