Applications of Differential Equations

In short

Differential equations arise naturally when the rate of change of a quantity depends on the quantity itself. Exponential growth and decay, Newton's law of cooling, logistic population models, and geometric problems about curves are all solved by setting up the right ODE and integrating.

A cup of tea sits on your desk. It is too hot to drink — say 90°C. The room is at 25°C. Five minutes later, the tea has cooled to 70°C. When will it reach a drinkable 50°C?

You could guess, but guessing is unreliable. The tea cools fast when it is very hot (the temperature difference with the room is large) and slowly when it is almost at room temperature (the difference is small). The cooling rate depends on the current temperature itself. That feedback loop — rate depends on state — is exactly the situation where a differential equation appears.

This article covers four families of applications: growth and decay, Newton's law of cooling, population models, and geometrical applications. Each one starts the same way: translate the physical or geometric statement into an equation involving a derivative, then solve.

Growth and decay

The simplest and most important differential equation in applications is

\frac{dN}{dt} = kN

This says: the rate at which N changes is proportional to N itself. When k > 0, the quantity grows (more of it means faster growth — compound interest, bacterial division, nuclear chain reactions). When k < 0, it decays (more of it means faster shrinkage — radioactive decay, drug metabolism).

Separate variables: \dfrac{dN}{N} = k\,dt. Integrate both sides: \ln|N| = kt + C. Exponentiate: N = N_0 e^{kt}, where N_0 = N(0) is the initial amount.

This is exponential growth (when k > 0) or exponential decay (when k < 0).

Radioactive decay

A radioactive substance decays at a rate proportional to the amount present. If you start with N_0 atoms and after time T (the half-life) exactly half remain, then N_0/2 = N_0 e^{kT}, giving k = -\ln 2/T.

Take carbon-14, with a half-life of about 5730 years. Starting with 1000 atoms:

N(t) = 1000 \cdot e^{-(\ln 2/5730)\,t} = 1000 \cdot 2^{-t/5730}

After 5730 years: N = 500. After 11{,}460 years: N = 250. After 17{,}190 years: N = 125. Each half-life halves the amount, forever.

Carbon-14 decay: starting from $1000$ atoms, the quantity halves every $5730$ years. The curve never reaches zero — it gets arbitrarily close, but never arrives.

Compound interest (continuous)

If a bank account with principal P earns interest at annual rate r, compounded continuously, then dP/dt = rP, giving P(t) = P_0 e^{rt}. With P_0 = 10{,}000 rupees and r = 0.08 (8% per annum):

P(t) = 10{,}000 \cdot e^{0.08t}

After 10 years: P(10) = 10{,}000 \cdot e^{0.8} \approx 22{,}255 rupees. The exponential model is the limiting case of the compound interest formula (1 + r/n)^{nt} as n \to \infty.

Newton's law of cooling

Back to the cup of tea. Newton's law of cooling says:

\frac{dT}{dt} = -k(T - T_s)

where T is the temperature of the object, T_s is the surrounding (ambient) temperature, and k > 0 is a constant that depends on the object's material and surface area. The negative sign ensures that when T > T_s (object is hotter than surroundings), the temperature decreases.

Let \theta = T - T_s (the temperature excess above the surroundings). Then d\theta/dt = dT/dt = -k\theta, which is the exponential decay equation. Solution:

\theta(t) = \theta_0 e^{-kt}, \quad \text{i.e.,} \quad T(t) = T_s + (T_0 - T_s)e^{-kt}

where T_0 is the initial temperature.

Example 1: The cooling cup of tea

The tea starts at 90°C. The room is at 25°C. After 5 minutes, the tea is at 70°C. When does it reach 50°C?

Step 1. Set up the model. Here T_0 = 90, T_s = 25, so T(t) = 25 + 65e^{-kt}.

Why: T_0 - T_s = 90 - 25 = 65 is the initial temperature excess.

Step 2. Find k using the data point T(5) = 70.

70 = 25 + 65e^{-5k}

45 = 65e^{-5k}

e^{-5k} = \frac{45}{65} = \frac{9}{13}

k = -\frac{1}{5}\ln\frac{9}{13} = \frac{1}{5}\ln\frac{13}{9} \approx 0.0737

Why: one data point pins down one unknown constant. This is the standard pattern: the ODE gives the form of the solution, initial/boundary data gives the constants.

Step 3. Find t when T(t) = 50.

50 = 25 + 65e^{-kt}

25 = 65e^{-kt}

e^{-kt} = \frac{25}{65} = \frac{5}{13}

t = -\frac{1}{k}\ln\frac{5}{13} = \frac{\ln(13/5)}{k} = \frac{\ln(13/5)}{\frac{1}{5}\ln(13/9)} = 5 \cdot \frac{\ln(13/5)}{\ln(13/9)}

Numerically: \ln(13/5) \approx 0.9555, \ln(13/9) \approx 0.3677, so t \approx 5 \times 2.598 \approx 13.0 minutes.

Why: the tea takes about 13 minutes total to reach 50°C — it cooled 20° in the first 5 minutes but needs another 8 minutes for the next 20°, because the rate slows as the tea approaches room temperature.

Step 4. Verify the qualitative behaviour. At t = 0: T = 90. As t \to \infty: T \to 25. The temperature approaches but never quite reaches the room temperature. The curve is a decaying exponential shifted up by T_s = 25.

Result: The tea reaches 50°C after approximately 13 minutes.

Newton's law of cooling applied to the tea. The curve starts at $90°$C and decays toward the room temperature of $25°$C (dashed line). The tea reaches $50°$C (dotted line) at about $13$ minutes. The cooling is rapid at first and progressively slower.

The asymmetry is striking: the first 20° of cooling takes 5 minutes, but the next 20° takes 8 more. This is not a flaw in the model — it reflects physical reality. A hot cup of tea loses heat faster because the temperature gradient is steeper.

Population models

Malthusian (unlimited) growth

If a population of bacteria doubles every hour, and you start with 100 bacteria, the model dN/dt = kN gives N(t) = 100 \cdot 2^t (with t in hours). After 10 hours: N = 102{,}400. After 24 hours: N \approx 1.68 billion. This is clearly unsustainable — no environment has unlimited resources.

The logistic model

A more realistic model is the logistic equation:

\frac{dN}{dt} = rN\left(1 - \frac{N}{K}\right)

Here r is the intrinsic growth rate and K is the carrying capacity — the maximum population the environment can sustain. When N is small compared to K, the factor (1 - N/K) \approx 1 and you get nearly exponential growth. When N approaches K, the factor approaches 0 and growth stalls. When N > K, the factor is negative and the population shrinks back toward K.

This ODE is separable. Using partial fractions:

\frac{dN}{N(1 - N/K)} = r\,dt

\left(\frac{1}{N} + \frac{1/K}{1 - N/K}\right)dN = r\,dt

Integrating: \ln N - \ln(1 - N/K) = rt + C, which rearranges to

N(t) = \frac{K}{1 + Ae^{-rt}}, \qquad A = \frac{K - N_0}{N_0}

This is the logistic curve — an S-shaped curve that starts near-exponential and levels off at K.

Logistic growth (solid red) vs exponential growth (dashed) for $N_0 = 10$, $K = 100$, $r = 1$. Early on, both curves are nearly identical — the population has not yet felt the resource limit. But the logistic curve bends and levels off at the carrying capacity $K = 100$ (dotted line), while the exponential shoots off to infinity.

The inflection point of the logistic curve — where growth is fastest — occurs at N = K/2. Before that point, growth is accelerating; after it, growth is decelerating. This S-shape appears everywhere: disease spread, technology adoption, bacterial cultures in a petri dish.

Geometrical applications

Differential equations also arise from purely geometric questions about curves.

Finding a curve from its slope property

Suppose you want a curve y = f(x) such that the tangent line at any point (x, y) passes through the origin. The slope of the tangent is y' = dy/dx. The slope of the line from (0,0) to (x,y) is y/x. Setting them equal:

\frac{dy}{dx} = \frac{y}{x}

Separate: dy/y = dx/x. Integrate: \ln|y| = \ln|x| + C, so y = Kx for some constant K. The answer is the family of all straight lines through the origin — which makes geometric sense. Any line through the origin has the property that its tangent at every point is the line itself, passing through the origin.

Curves whose subnormal is constant

The subnormal at a point on a curve is the horizontal distance from the foot of the normal to the point directly below the point on the x-axis. For the curve y = f(x) at the point (x, y), the subnormal has length \left|y \cdot \frac{dy}{dx}\right|.

If the subnormal is constant, say equal to a, then

y\frac{dy}{dx} = a

Separate: y\,dy = a\,dx. Integrate: y^2/2 = ax + C, so y^2 = 2ax + C'. These are parabolas with axis along the x-axis.

Example 2: A curve whose subtangent is twice the abscissa

The subtangent at a point (x, y) on a curve is \left|\dfrac{y}{dy/dx}\right|. Find the curve if the subtangent equals 2x at every point.

Step 1. Set up the equation. The subtangent equals 2x:

\frac{y}{dy/dx} = 2x

Rearranging: \dfrac{dy}{dx} = \dfrac{y}{2x}.

Why: the subtangent formula y/(dy/dx) is a standard result from the geometry of tangent lines.

Step 2. Separate variables:

\frac{dy}{y} = \frac{dx}{2x}

Why: all y-terms go to one side, all x-terms to the other.

Step 3. Integrate both sides:

\ln|y| = \frac{1}{2}\ln|x| + C = \ln|x|^{1/2} + C

Why: \int dx/(2x) = (1/2)\ln|x|, and the log rule brings the 1/2 up as an exponent.

Step 4. Exponentiate:

|y| = e^C \cdot |x|^{1/2}

Writing K = \pm e^C:

y = K\sqrt{x}

Why: taking the exponential of both sides removes the logarithm. The constant K absorbs the sign.

Step 5. Verify. Take y = K\sqrt{x}: dy/dx = K/(2\sqrt{x}). Subtangent = y/(dy/dx) = K\sqrt{x} \cdot 2\sqrt{x}/K = 2x. Correct.

Result: The family of curves is y = K\sqrt{x}, i.e., y^2 = K^2 x — parabolas with vertex at the origin and axis along the x-axis.

The family of curves $y = K\sqrt{x}$ for different values of $K$. Each curve is a parabola with vertex at the origin. At every point on every curve, the subtangent equals $2x$.

The geometric property — subtangent equals 2x — characterises the parabola y^2 = K^2 x completely. The differential equation translated a geometric condition into an algebraic answer.

The general recipe

Every application in this article follows the same four-step pattern:

Translate the physical or geometric statement into a differential equation. Identify what quantity is changing, what its rate depends on, and write dy/dx = (\text{something involving } x, y).
Classify the equation. Is it separable? Linear? Exact? First-order or second-order? The classification tells you which technique to use.
Solve using the appropriate technique (separation, integrating factor, characteristic equation, etc.).
Apply initial or boundary conditions to determine the arbitrary constants.

The hardest step is always the first — translating words into an equation. The rest is technique.

Common confusions

"Growth rate k and doubling time T are the same thing." They are inversely related: T = \ln 2 / k. A larger growth rate means a shorter doubling time. For decay, the analogous quantity is the half-life: T_{1/2} = \ln 2 / |k|.
"Newton's law of cooling says the temperature drops linearly." It says the rate of temperature drop is proportional to the temperature difference — that makes the drop exponential, not linear. A linear drop would mean the same number of degrees lost per minute, regardless of the current temperature. The exponential model captures the slowing-down effect that you observe in real life.
"The logistic model is just exponential growth with a cap." The logistic model is qualitatively different. In exponential growth, the rate dN/dt = rN depends only on N. In logistic growth, dN/dt = rN(1 - N/K) — the rate also depends on how close N is to K. This makes the growth self-limiting. The cap is not imposed from outside; it emerges from the equation itself.
"Every geometric ODE is first-order." Most are, because they involve only y' (the slope). But some involve curvature, which uses y'', giving second-order equations. The family of curves with constant curvature (circles) requires a second-order ODE.

Going deeper

If you can set up and solve the standard growth/decay, cooling, and geometry problems, you have the core skill. The rest is for readers curious about extensions and edge cases.

When Newton's law of cooling fails

Newton's law works well when the temperature difference between the object and the surroundings is moderate (say, up to about 30–40°C above ambient). For very hot objects — like molten metal at 1500°C — radiative heat loss becomes dominant, and the rate of cooling is proportional to T^4 - T_s^4 (Stefan's law), not T - T_s. The resulting ODE is nonlinear and much harder to solve in closed form.

Mixing problems

A tank contains 100 litres of brine with 10 kg of salt dissolved in it. Fresh water pours in at 5 litres per minute, and the well-mixed solution drains out at 5 litres per minute. How much salt remains after t minutes?

Let S(t) be the amount of salt at time t. Salt enters at rate 0 (it's fresh water). Salt leaves at rate (S/100) \times 5 = S/20 kg per minute (concentration times outflow rate). So

\frac{dS}{dt} = -\frac{S}{20}

This is exponential decay: S(t) = 10e^{-t/20}. After 20 minutes: S = 10/e \approx 3.68 kg. After 40 minutes: S \approx 1.35 kg. The salt washes out exponentially.

The key modelling step is recognising that the concentration of the outflow equals the current concentration in the (well-mixed) tank — S(t)/V(t) — and the volume stays constant when inflow rate equals outflow rate.

The catenary

A flexible cable hanging under its own weight takes a shape called a catenary: y = a\cosh(x/a) for some constant a. The differential equation is

\frac{d^2 y}{dx^2} = \frac{1}{a}\sqrt{1 + \left(\frac{dy}{dx}\right)^2}

This is a second-order nonlinear ODE. It can be reduced to first order by the substitution p = dy/dx, giving dp/dx = (1/a)\sqrt{1+p^2}, which is separable. The result, y = a\cosh(x/a), describes the shape of telephone wires, suspension bridge cables, and chains between two posts.

Where this leads next

Differential equations model the real world. The articles that follow explore specific settings where these tools are applied.

Orthogonal and Isogonal Trajectories — finding curves that cross a given family at right angles or at a fixed angle.
Second Order Homogeneous — the characteristic equation method for ay'' + by' + cy = 0.
Second Order Non-Homogeneous — particular integrals via undetermined coefficients and variation of parameters.
Rate of Change — the derivative as a rate, the conceptual foundation for every model in this article.
Exponential Functions — the function e^{kx} that underlies growth and decay.