Writing, Verifying and Interpreting a Differential Equation
- Translate a verbal description of a rate into a differential equation
- Verify that a proposed function is a solution by substituting it into the equation
- Identify the exponential model as the solution of dy/dt = ky and interpret k in context
From words to a rate equation
A differential equation is a sentence about a rate. "Proportional to" becomes a constant times the quantity: dy/dt = ky. "Proportional to the difference between y and 70" becomes dy/dt = k(y − 70). "Proportional to the product of y and 500 − y" becomes the logistic dy/dt = ky(500 − y). The translation is nearly word-for-word, and the exam tests it directly — often awarding a point for the equation alone, before any solving happens.
Verifying a solution is substitution, not solving
To check that y = 3e^(2t) − 5 solves dy/dt = 2y + 10, compute both sides separately and compare. Left: dy/dt = 6e^(2t). Right: 2(3e^(2t) − 5) + 10 = 6e^(2t) − 10 + 10 = 6e^(2t). They agree, so it is a solution. This is a short, fully-earnable question, and no integration is involved — students who start separating variables have misread the task.
The exponential model and the meaning of k
dy/dt = ky with y(0) = y₀ has the unique solution y = y₀e^(kt), and this is worth knowing on sight rather than re-deriving. A positive k is growth, a negative k is decay, and |k| sets the pace: doubling time is ln2/k, half-life is ln2/|k|. In context k carries units of "per unit time" — a population growing at k = 0.03 per year is growing at 3% of its current size per year, which is a statement about the rate relative to the amount, not an absolute increase.
A tank holds 200 liters. Water leaks out at a rate proportional to the amount remaining. After 3 hours 150 liters remain. Write a differential equation, solve it, and find the amount after 10 hours.
- 1.Let A(t) be the amount in liters at time t hours. "Rate proportional to the amount" gives dA/dt = kA, with k negative since the amount is decreasing.
- 2.The solution is A(t) = 200e^(kt), using A(0) = 200.
- 3.Use A(3) = 150: 150 = 200e^(3k), so e^(3k) = 0.75 and k = ln(0.75)/3 ≈ −0.09589 per hour.
- 4.Then A(t) = 200e^(−0.09589t).
- 5.At t = 10: A(10) = 200e^(−0.9589) ≈ 200(0.3834) ≈ 76.7 liters.
"Decreasing at a rate proportional to the amount" and "decreasing at a constant rate" are different equations — dA/dt = kA versus dA/dt = −c. The second gives a straight line, not an exponential. Read for the word "proportional."
A cup of coffee cools at a rate proportional to the difference between its temperature T and the room temperature 68°F. Which equation models this?
Is y = 4e^(3t) a solution of dy/dt = 3y?
On free response, write the differential equation and the initial condition as a labeled pair before solving. Both are separately scored, and a correct equation salvages partial credit even when the integration goes wrong.
Answer the 2 checkpoints as you read.
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