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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.

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  • $\begingroup$ Feel free to ask if something is still unclear $\endgroup$ May 22 '20 at 17:35
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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.

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

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

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