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Modeling with Differential Equations

You’ll be able to

From words to a differential equation

Modeling starts by turning a sentence about a rate into an equation for dy/dt. "Grows at a rate proportional to its size" becomes dy/dt = ky. "Grows at a constant rate" becomes dy/dt = k. "The rate is proportional to the difference between y and 70" becomes dy/dt = k(y − 70) — the form of Newton’s Law of Cooling. Identifying the right structure is the crucial first step.

Half-life and doubling time

For an exponential model, half-life is the time for a decaying quantity to fall to half its amount, and doubling time is the time for a growing quantity to double. Set y = y₀/2 or y = 2y₀ in y = y₀·e^(kt); the initial amount cancels, and solving with a natural log gives the time in terms of k alone. These times are constants — the same at every stage of the process.

Interpreting long-term behavior

A model is only useful if you can read its long-run behavior. For pure exponential decay (k < 0), y → 0 as t → ∞. For growth (k > 0), y → ∞. For a cooling model dy/dt = k(y − T_room) with k < 0, the temperature approaches the room temperature T_room, not zero. Always state limits in the units and context of the problem.

Solving for a characteristic time
Doubling: 2y₀ = y₀·e^(kt) ⟹ t = (ln 2)/k; Half-life: t = (ln 2)/|k|
The initial amount y₀ always cancels, so these times depend only on the rate constant k.
Worked example

A radioactive isotope decays with dy/dt = ky. A 100 g sample decays to 80 g in 2 years. Find k, then the amount after 10 years.

  1. 1.The model is y = 100·e^(kt). Use the data point y(2) = 80: 80 = 100·e^(2k).
  2. 2.Divide and take the log: 0.8 = e^(2k), so 2k = ln(0.8), giving k = (ln 0.8)/2 ≈ −0.1116.
  3. 3.Write the full model: y(t) = 100·e^(−0.1116t).
  4. 4.Evaluate at t = 10: y(10) = 100·e^(−1.116) ≈ 100·(0.328) ≈ 32.8.
Answer: k ≈ −0.1116 per year (negative, confirming decay), and about 32.8 g remain after 10 years.
Watch out

Distinguish "grows at a constant rate" (dy/dt = k, a linear solution) from "grows at a rate proportional to its size" (dy/dt = ky, an exponential solution). The word "proportional" is what makes the model exponential — read the phrasing carefully.

Checkpoint

A tank’s temperature changes at a rate proportional to the difference between its temperature and the 20° room. Which differential equation models this?

Checkpoint

A population follows P = 500·e^(0.2t). What is its doubling time? (Use ln 2 ≈ 0.693.)

On the exam

When a problem gives a data point to find k, substitute it into y = y₀·e^(kt), isolate the exponential, and take a natural log — do not try to guess k. Carry enough decimal places through so the final rounded answer stays accurate.

Answer the 2 checkpoints as you read.

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