Lesson 3: Differential Equations, Slope Fields and Writing FRQsAP Deep Dive

The last deep dive targets the differential-equations family at its hardest: logistic models (where students quote the rule "fastest at half capacity" without being able to derive or use it), Euler's method with error direction (where the point is in the justification, not the arithmetic), and separable equations whose solutions carry branches and domains that change the answer. These are the parts of recent FRQs with the lowest national scores — which makes them the highest-value practice in this course.

The logistic model, fully owned

The logistic equation dy/dt = ky(1 − y/L) models growth with carrying capacity L. Its solution is y = L/(1 + A·e^(−kt)) with A = (L − y₀)/y₀.

Worked example 1. A fish population satisfies dP/dt = 0.3P(1 − P/1200) with P(0) = 200.

• Carrying capacity: L = 1200 (read directly from the equation's structure — the factor (1 − P/1200) vanishes at P = 1200).
• A = (1200 − 200)/200 = 5, so P(t) = 1200/(1 + 5e^(−0.3t)).
• Fastest growth occurs where dP/dt is maximal. Treat the rate as a function of P: 0.3P − 0.00025P², a downward parabola in P with vertex at P = L/2 = 600. Max rate = 0.3(600)(1 − 0.5) = 90 fish per year. The "half capacity" rule is just the vertex of that parabola — derive it this way once and you will never misapply it to the *time* axis.
• When does P reach 900? 900 = 1200/(1 + 5e^(−0.3t)) → 1 + 5e^(−0.3t) = 4/3 → e^(−0.3t) = 1/15 → t = ln 15/0.3 ≈ 9.03 years. Note the population passes 600 (fastest growth) at t = ln 5/0.3 ≈ 5.36 years — the inflection point of the S-curve, where concavity changes.

Qualitative slope-field behaviour completes the picture: solutions below L rise toward it and can never cross it (uniqueness — solution curves of an autonomous equation never intersect), solutions above L decay toward it, and P = L and P = 0 are equilibrium (constant) solutions.

Euler's method, with the error argued

Worked example 2. Approximate y(1) for dy/dx = y − x, y(0) = 2, using two steps of size h = 0.5 — and then determine whether the estimate is too high or too low.

Step 1: slope at (0, 2) is y − x = 2, so y(0.5) ≈ 2 + 0.5(2) = 3. Step 2: slope at (0.5, 3) is 3 − 0.5 = 2.5, so y(1) ≈ 3 + 0.5(2.5) = 4.25.

The exact solution (verify by substitution: y = x + 1 + eˣ gives dy/dx = 1 + eˣ = y − x ✓) yields y(1) = 2 + e ≈ 4.72. The Euler estimate 4.25 is an underestimate. Why: y″ = eˣ > 0, so the solution is concave up; a concave-up curve lies *above* each of its tangent lines, and Euler follows tangent lines — so every step lands below the curve. The scored sentence pattern is always: "The solution is concave up/down on the interval because y″ …; therefore the tangent-line steps lie below/above the curve and the estimate is an under/overestimate." Memorise the pairing (concave up → underestimate; concave down → overestimate), but derive it from the picture in the exam, because questions reverse which quantity is estimated.

Separation with branches and domains

Worked example 3. Solve dy/dx = 4x/y with y(0) = −3.

Separate: y dy = 4x dx. Integrate: y²/2 = 2x² + C, so y² = 4x² + C′. The initial condition: 9 = 0 + C′ → y² = 4x² + 9. Now the step students skip: the implicit relation describes *two* functions, y = √(4x² + 9) and y = −√(4x² + 9). The initial value y(0) = −3 selects the negative branch: y = −√(4x² + 9), defined for all x. Solutions of y² = … that silently report the positive root earn the algebra points and lose the solution point. The same care is needed whenever separation produces |y| (take exponentials carefully: y = ±e^C·e^(…), then fold the sign into a single nonzero constant A, and check whether y = 0 is itself a solution that the division by y threw away — for dy/dx = 2xy it is not an issue, but for dy/dx = y² it is: y ≡ 0 is a genuine solution lost in the separation step).

Reading slope fields at speed

Three checks identify an equation from its field: (1) horizontal segments occur where dy/dx = 0 — find the curve where the proposed equation vanishes (for dy/dx = y + 2x, the line y = −2x); (2) dependence — if slopes depend only on x, the field's columns are identical (dy/dx = f(x)); only on y, rows are identical; (3) sign regions — pick one point per region and compare signs. For sketching a solution through a point: follow the segments like a streamline, never crossing another solution curve, and state long-term behaviour (approaches an equilibrium, grows without bound) in the conclusion.

Common mistakes at the top band

• Quoting "fastest growth at K/2" as a time. It is a population *level*; finding when it occurs requires solving the logistic solution for t.
• Euler arithmetic drift: using the *updated* x with the *old* y (or vice versa) when computing the next slope. The slope for a step comes from the point at the step's start, both coordinates.
• Dropping the +C, or applying the initial condition before integrating — the constant is found after the general solution exists.
• Ignoring the equilibrium solution lost when dividing by y (or by (L − y) style factors).
• Reporting y² = … as the final answer when the question asks for y explicitly, or taking the wrong branch.

AP Exam Experience

1. (No calculator.) The slope of the solution to dy/dx = x/y at the point (2, −1) is

• (A) 2
• (B) −2
• (C) −1/2
• (D) undefined

Answer: (B). Substitute the point: 2/(−1) = −2. Choice (D) tempts students who see a denominator, but y = −1 ≠ 0 at this point — the slope exists and is negative.

2. A population grows according to dP/dt = 0.5P(1 − P/800). The population grows fastest when P equals

• (A) 800
• (B) 400
• (C) 200
• (D) it grows fastest at t = 0

Answer: (B). The growth rate, as a function of P, is a downward parabola with zeros at P = 0 and P = 800; its vertex — the maximum rate — is at P = 400 = L/2. At P = 800 the rate is zero.

3. Using Euler's method with step size 0.5 for dy/dx = x + y, y(0) = 1, the first-step estimate of y(0.5) is

• (A) 1
• (B) 1.5
• (C) 2
• (D) 2.5

Answer: (B). Slope at (0, 1) is 0 + 1 = 1; y(0.5) ≈ 1 + 0.5 × 1 = 1.5. Choice (C) results from using the full step as if the slope were evaluated after moving — the classic first-step slip.

Free-response question. A cup of coffee at 200 °F is placed in a room at 70 °F. Its temperature T(t) (in °F, t in minutes) obeys Newton's law of cooling: dT/dt = −0.2(T − 70).

(a) Show that T = 70 is a constant (equilibrium) solution, and describe what it means physically. (b) Solve the differential equation with the initial condition T(0) = 200 to find T(t) explicitly. (c) Find the time at which the coffee reaches 100 °F. (d) Use Euler's method with one step of size 1 minute to estimate T(1), and determine whether this estimate is greater or less than the true value, with justification.

Scoring guideline.

• (a) 1 point. If T ≡ 70, then dT/dt = 0 and −0.2(70 − 70) = 0, so the equation is satisfied; physically, coffee already at room temperature stays there (no net heat flow). Both the verification and the interpretation are required for the point.
• (b) 3 points. 1 point for separation: dT/(T − 70) = −0.2 dt. 1 point for the integrated form ln|T − 70| = −0.2t + C, equivalently T = 70 + A·e^(−0.2t). 1 point for applying the initial condition: A = 130, giving T(t) = 70 + 130e^(−0.2t). Omitting the absolute value is acceptable if the sign of T − 70 is addressed; losing the "+70" shift forfeits the final point.
• (c) 2 points. 1 point for the equation 100 = 70 + 130e^(−0.2t), i.e. e^(−0.2t) = 30/130 = 3/13. 1 point for t = ln(13/3)/0.2 ≈ 7.33 minutes. Units (minutes) expected.
• (d) 2 points. 1 point for the Euler step: slope at t = 0 is −0.2(200 − 70) = −26, so T(1) ≈ 200 − 26 = 174 °F. 1 point for the comparison with justification: the true value is T(1) = 70 + 130e^(−0.2) ≈ 176.4 °F, so the Euler estimate is less than the true value, because T″ = −0.2·T′ > 0 (T′ < 0) makes the solution concave up, and a tangent line lies below a concave-up curve. An assertion of "underestimate" without the concavity argument earns only the first point.

A mixing problem: separation in disguise

A 100 L tank contains pure water. Brine carrying 0.5 kg of salt per litre flows in at 3 L/min, the mixture is kept uniform, and it drains at 3 L/min. Let A(t) be the mass of salt (kg) at time t. Salt enters at 3 × 0.5 = 1.5 kg/min and leaves at 3 × (A/100) kg/min, so dA/dt = 1.5 − 0.03A. This is separable: dA/(1.5 − 0.03A) = dt, giving −ln|1.5 − 0.03A|/0.03 = t + C, and with A(0) = 0, A(t) = 50(1 − e^(−0.03t)). Read the structure rather than the arithmetic: the equilibrium A = 50 kg is where inflow balances outflow (1.5 = 0.03A), the solution approaches it exponentially and never overshoots, and the "time constant" 1/0.03 ≈ 33 minutes sets the pace. Newton cooling, mixing, and drug elimination are all this same equation — d(quantity)/dt = rate in − k × quantity — and FRQs rotate the costume, not the mathematics.