ODE Applications · Visual Guide

Malthus' Law: Exponential Growth

The Model

Each individual in the population produces offspring at a constant rate $r$.

Malthus (1798) observed that populations grow proportionally to their own size: the bigger the population, the faster it grows. Passing to a continuous limit gives a first-order ODE.

Discrete intuition: each time-step, every individual produces one new individual. $\Delta N = N(t) - N(t-1)$, so $\dfrac{\Delta N}{N} = 1$

Continuous limit ($\Delta t \to 0$): $$\frac{dN}{dt} = rN$$

$$\frac{dN}{dt} = rN, \qquad N(0)=N_0$$

$$N(t) = N_0\,e^{rt}$$

r > 0 growth r < 0 decay

Malthus' model predicts infinite growth, which is unrealistic for real populations with limited resources.

Doubling Time

If the growth rate is $r=3\%$/year, when does the population double?

Set $N(t^*)=2N_0$ and use the solution:

$$2N_0 = N_0 e^{0.03\,t^*}$$

Take the natural log:

$$\ln 2 = 0.03\,t^* \;\Longrightarrow\; t^* = \frac{\ln 2}{0.03} \approx 23.1 \text{ years}$$

Rule of 70: doubling time $\approx 70 / r\%$. At $r=3\%$: $70/3 \approx 23.3$ years.

U.S. Population (pre-Civil War). $\Delta N/N$ is nearly constant at $\approx 0.35$, consistent with exponential growth.

Year	N (millions)	ΔN	ΔN/N
1790	3.929	–	–
1800	5.308	1.379	0.351
1810	7.240	1.932	0.364
1820	9.638	2.398	0.331
1830	12.866	3.228	0.335
1840	17.069	4.203	0.327
1850	23.192	6.123	0.359
1860	31.443	8.251	0.356

Newton's Cooling Law

The Law

The rate of heat loss is proportional to the temperature difference between the body and its surroundings.

Let $T(t)$ be the temperature of the body and $T_s$ the surroundings temperature (constant). The bigger the gap, the faster cooling occurs.

$$\frac{dT}{dt} = c\,(T_s - T)$$

$$T(t) = T_s - (T_s - T_0)\,e^{-ct}$$

Stationary solution: $T^* = T_s$. When the body reaches the surroundings temperature, $dT/dt = 0$ and it stays there.

cooling ($T_0 > T_s$) warming ($T_0 < T_s$)

Freezer Problem

Food at $25°C$ enters a freezer at $-10°C$. After $8$ min it is at $0°C$. When does it reach $-5°C$?

Write the solution with $T_s = -10$, $T_0 = 25$:

$$T(t) = -10 + 35\,e^{-ct}$$

Find $c$ using $T(8)=0$:

$$0 = -10 + 35\,e^{-8c} \;\Rightarrow\; e^{-8c} = \tfrac{2}{7} \;\Rightarrow\; c = -\tfrac{1}{8}\ln\tfrac{2}{7} \approx 0.1566\;\text{min}^{-1}$$

Solve for $T(t^*)=-5$:

$$-5 = -10 + 35\,e^{-ct^*} \;\Rightarrow\; e^{-ct^*} = \tfrac{1}{7} \;\Rightarrow\; t^* = 8\cdot\frac{\ln(1/7)}{\ln(2/7)} \approx 12.4\;\text{min}$$

Stationary solution: $T^* = -10°C$. The food asymptotically approaches the freezer temperature and never goes below it.

Compound Bank Interest

Discrete Compounding

Interest is added $n$ times per year at annual rate $k$.

Each compounding event multiplies the capital by $(1 + k/n)$. Over $t$ years there are $nt$ such events.

$$A(t) = P\!\left(1 + \frac{k}{n}\right)^{nt}$$

Compounding	n	Formula (P=1, k=6%, t=1 yr)	Amount
Annual	1	$(1.06)^1$	1.0600
Semi-annual	2	$(1.03)^2$	1.0609
Monthly	12	$(1.005)^{12}$	1.0617
Daily	365	$(1+0.06/365)^{365}$	1.0618
Continuous	∞	$e^{0.06}$	1.0618…

Continuously Compounded Interest

Taking $n \to \infty$ and recovering the ODE.

Take the limit $n\to\infty$ using $\lim_{x\to\infty}(1+1/x)^x = e$:

$$A(t) = P\lim_{n\to\infty}\!\left(1+\frac{k}{n}\right)^{nt} = P\,e^{kt}$$

Differentiate. Capital grows proportionally to itself:

$$\frac{dA}{dt} = Pke^{kt} = k\cdot A$$

This is Malthus' Law with $r=k$. Population growth and continuous interest share the exact same ODE structure.

$$\frac{dA}{dt} = k\,A, \quad A(0)=A_0 \;\Rightarrow\; A(t)=A_0 e^{kt}$$

continuous $A_0 e^{kt}$ annual steps

Unified view: Malthus (population), Newton's cooling, radioactive decay, and continuous interest all satisfy $\dot{y} = r\,y$. The sign of $r$ determines growth or decay.

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Summary

Model	ODE	Solution	Stationary point
MalthusPopulation	$\dot N = rN$	$N(t)=N_0 e^{rt}$	$N^*=0$ (unstable)
CoolingTemperature	$\dot T = c(T_s-T)$	$T(t)=T_s-(T_s-T_0)e^{-ct}$	$T^*=T_s$ (stable)
InterestCapital	$\dot A = kA$	$A(t)=A_0 e^{kt}$	$A^*=0$ (unstable)