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← Rates of Change and Differential Equations›Applications of Differential Equations

Applications of Differential Equations

Key Terms

Exponential growth/decay
dN/dt = kN. Solution: N = N0ekt. If k > 0, growth; if k < 0, decay. Half-life T1/2 = −ln2/k.
Newton’s law of cooling
dT/dt = −k(T − Ta), where Ta is ambient temperature. Solution: T − Ta = (T0 − Ta)e−kt.
Logistic growth
dP/dt = rP(1 − P/K), where K is the carrying capacity. Solution: P(t) = K / (1 + Ae−rt) where A = (K − P0)/P0.
Mixing problems
Rate of change of amount = rate in − rate out. Set up: dQ/dt = cin × rin − (Q/V) × rout.
RL circuits
L dI/dt + RI = V(t). First-order linear; solved by integrating factor μ = eRt/L.
Equilibrium solutions
Set dy/dx = 0 (or dN/dt = 0) and solve. Stable if nearby solutions approach it; unstable if they diverge.
Key Models:
  • Exponential: dN/dt = kN  ⇒  N = N0ekt
  • Newton’s cooling: dT/dt = −k(T − Ta)  ⇒  T = Ta + (T0−Ta)e−kt
  • Logistic: dP/dt = rP(1 − P/K)  ⇒  P = K / (1 + Ae−rt)
  • Mixing: dQ/dt = (rate in) − (concentration out) × (flow out)
Hot Tip Always define your variables clearly: state what the dependent and independent variables represent, and include units. For Newton’s law of cooling, note that T − Ta (not T itself) decays exponentially. For logistic growth, as t → ∞, P → K regardless of the initial population P0 (provided P0 > 0).

Worked Example 1 — Newton’s Law of Cooling

A cup of coffee at 90°C is left in a room at 20°C. After 5 minutes it has cooled to 70°C. Find the temperature after 10 minutes.

dT/dt = −k(T − 20). Let u = T − 20. Then du/dt = −ku, so u = u0e−kt.

At t = 0: u0 = 90 − 20 = 70. At t = 5: u = 70 − 20 = 50.

50 = 70e−5k ⇒ k = −(1/5)ln(50/70) = (1/5)ln(7/5).

At t = 10: u = 70e−10k = 70(50/70)² = 70 × 2500/4900 = 35.7.

T(10) = 20 + 35.7 ≈ 55.7°C

Worked Example 2 — Radioactive Decay

A radioactive substance has a half-life of 30 years. If 100 g is present initially, how much remains after 90 years?

dN/dt = kN ⇒ N = 100ekt. Half-life: N(30) = 50, so 50 = 100e30k ⇒ k = −ln2/30.

N(90) = 100e(−ln2/30)(90) = 100e−3ln2 = 100 × (1/2)3 = 100/8 = 12.5 g

Worked Example 3 — Logistic Growth

A population grows logistically with r = 0.1 per year and K = 1000. Initially P0 = 100. Find P(20).

A = (1000 − 100)/100 = 9.   P(t) = 1000 / (1 + 9e−0.1t).

P(20) = 1000 / (1 + 9e−2) = 1000 / (1 + 9 × 0.1353) ≈ 1000/2.218 ≈ 451

Why Differential Equations Arise in Applications

Many physical, biological, and engineering processes have a natural description in terms of rates of change: the rate of cooling depends on the temperature difference, the rate of population growth depends on the current population, the rate of discharge of a capacitor depends on the current voltage. Each of these leads directly to a first-order ODE. The skill lies in translating the verbal description into the correct differential equation, then applying the solution technique appropriate for its type (separable or linear).

Newton’s Law of Cooling

Newton’s law states that the rate of change of temperature is proportional to the difference between the object’s temperature T and the ambient temperature Ta:

dT/dt = −k(T − Ta),   k > 0.

This is separable. Letting u = T − Ta: du/dt = −ku ⇒ u = Ce−kt. Substituting back: T = Ta + Ce−kt. Applying T(0) = T0 gives C = T0 − Ta, so T(t) = Ta + (T0 − Ta)e−kt.

Logistic Growth

Exponential growth dP/dt = rP is unrealistic in bounded environments. Logistic growth introduces a “braking factor”: dP/dt = rP(1 − P/K), where K is the carrying capacity. When P < K, growth is positive; when P > K, the right side is negative and population declines. The solution is obtained by partial fractions:

P(t) = K / (1 + Ae−rt), where A = (K − P0) / P0.

The S-shaped (sigmoidal) graph has an inflection point at P = K/2, where the growth rate dP/dt is maximised.

Mixing Problems

A tank contains a solution. Fluid flows in at rate rin (litres per minute) with concentration cin (grams per litre), and flows out at rate rout. If Q(t) is the amount (grams) of solute at time t, and V(t) is the volume:

dQ/dt = cin × rin − (Q/V) × rout.

When rin = rout = r, V is constant and the equation becomes linear: dQ/dt + (r/V)Q = cinr, solved by integrating factor μ = e(r/V)t.

RL Circuits

Kirchhoff’s voltage law for a series RL circuit gives L dI/dt + RI = V(t), where L is inductance (henries), R is resistance (ohms), I is current (amperes), and V(t) is applied voltage. This is a first-order linear ODE. For constant V, the steady-state current is Iss = V/R, and the transient decays with time constant τ = L/R.

Exam Tip: In applied problems, always state the differential equation, identify its type (separable or linear), and write the general solution before applying initial conditions. Include units in your final answer. When finding the decay constant k, leave it in exact form (e.g., k = ln2/30) if the question asks for further calculations.

Mastery Practice

  1. A bacterial culture grows at a rate proportional to its size. Initially there are 500 bacteria; after 2 hours there are 2000. Find the number after 5 hours. Fluency

  2. A radioactive substance has a half-life of 10 years. Write the differential equation and find the fraction remaining after 25 years. Fluency

  3. A body is found at 25°C in a room kept at 20°C. One hour later the body temperature is 22°C. Using Newton’s law of cooling, find the value of k. Fluency

  4. A population grows logistically with carrying capacity K = 500 and growth rate r = 0.2 per year. If P(0) = 50, state the logistic differential equation and write the solution formula. Fluency

  5. A cup of tea at 95°C is placed in a room at 22°C. After 10 minutes the temperature is 65°C. Find the temperature after 30 minutes. Understanding

  6. A 200-litre tank contains pure water. Brine with a concentration of 0.05 kg/L enters at 4 L/min, and the well-mixed solution leaves at 4 L/min. Find Q(t), the amount of salt at time t, and the limiting amount as t → ∞. Understanding

  7. A logistic population has K = 800, r = 0.3, and P(0) = 200. Find the time at which P = 600 (use exact logarithm). Understanding

  8. A series RL circuit has inductance L = 1 H, resistance R = 2 Ω, and applied voltage V(t) = 4e−t V. Find the current I(t) given I(0) = 1. Understanding

  9. A forensic investigator finds a body in a room maintained at 18°C. At 9:00 am the body temperature is 32°C; at 10:00 am it is 28°C. Normal body temperature is 37°C. Estimate the time of death (assume Newton’s law of cooling). Problem Solving

  10. Carbon-14 has a half-life of 5730 years. A charcoal sample contains 62% of its original carbon-14. Estimate the age of the sample. Problem Solving

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