Differential Equations

What this unit covers

The topics below follow the published Calculus BC course framework for Unit 7. This unit is worth 5–10% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Lessons in this unit

• Slope Fields13 min · 3 objectivesInterpret a slope field as a picture of a differential equation · Sketch a solution curve through a given point on a slope field · Match a differential equation to its slope field
• Writing, Verifying and Interpreting a Differential Equation14 min · 3 objectivesTranslate 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
• Separation of Variables15 min · 3 objectivesSolve separable differential equations by separating and integrating · Apply an initial condition to find the particular solution · Recognize exponential growth and decay as separable models
• Euler’s Method14 min · 3 objectivesApproximate a solution numerically using Euler’s method · Carry out repeated steps with a fixed step size · Explain why Euler’s method produces an approximation, not an exact value
• The Logistic Model14 min · 3 objectivesInterpret the logistic differential equation and its carrying capacity · Identify the population level of fastest growth · Determine the long-run limit of a logistic solution

Formulas in Unit 7

Every term in Unit 7

All 24 terms we publish for Differential Equations, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Logistic differential equation

dP/dt = kP(1 − P/L). Growth is nearly exponential when P is small and slows to zero as P approaches the carrying capacity L.

Euler's method

Step forward along the tangent: y_{n+1} = y_n + h·f(x_n, y_n). It underestimates a concave-up solution because tangent lines lie below the curve.

Choosing the step size

Smaller h gives a better approximation and more arithmetic. Halving h roughly halves the error for this first-order method.

Fastest growth in a logistic model

Occurs at P = L/2, which is the inflection point of the solution curve. A standard BC free-response question.

Logistic solution behavior

As t→∞, P approaches L from whichever side it started, so L is a horizontal asymptote of the solution.

Identifying the carrying capacity

Set dP/dt = 0 and solve; the non-zero equilibrium is L. It can be read directly from the logistic form.

Separation of variables with initial conditions

Integrate both sides, add one constant, apply the initial condition immediately, then solve for the dependent variable.

Euler recomputes the slope

yₙ₊₁ = yₙ + h·f(xₙ, yₙ), with the slope recalculated at each new point. Reusing the first slope is the standard error.

Show the Euler table

Points are awarded for setup and intermediate values. A bare final number risks losing most of them.

Euler and concavity

Concave up puts tangent lines below the curve, so Euler underestimates. Smaller h shrinks the error without changing its direction.

Finding concavity without the solution

Differentiate the differential equation implicitly to get d²y/dx², then check its sign at the point.

Logistic carrying capacity

The nonzero y making dy/dt = 0. Every positive initial value approaches it as t → ∞.

Logistic equilibria

y = 0 and y = M. Solutions cannot cross an equilibrium, which is why a solution starting below M stays below it.

Fastest logistic growth at M/2

dy/dt as a function of y is a downward parabola with roots 0 and M, so its maximum is at the midpoint — also the inflection point of the solution.

AP does not require solving the logistic equation

Recognizing carrying capacity, equilibria and fastest growth is what is tested.

Match a slope field by structure

Slopes constant along vertical lines means dy/dx depends only on x; constant along horizontal lines means only on y.

Zero slopes locate the factors

Horizontal segments along y = 2 point to a factor of (y − 2) in dy/dx.

Solutions cannot cross

A line of zero slopes is itself a solution, and distinct solutions of a well-posed equation do not intersect.

Resolve C before exponentiating

Apply the initial condition while the equation is still in log form. Far easier than carrying C through an exponential.

Exponentiating a sum gives a product

From ln|y| = 3x² + C you get |y| = e^C·e^(3x²). The constant becomes multiplicative, never additive.

Choose the branch from the initial condition

Solving |y| = something leaves two branches; a negative initial value selects the negative one.

State the domain of a particular solution

The largest interval containing the initial x on which the solution is defined and differentiable. Rubrics award this separately.

Separation needs a product form

dy/dx must factor as g(x)h(y). If x and y are entangled additively, separation does not apply.

Verify a candidate solution twice

It must satisfy the differential equation AND the initial condition. Satisfying the equation alone gives a general solution, not the particular one.

What examiners penalize here

• When a free-response question shows a slope field, you can sketch a particular solution without solving the equation — just start at the given point and follow the segments. Full credit comes from a curve tangent to the field, passing through the initial condition.
• 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.
• On free response you must show the separation and the integration of both sides, then apply the initial condition explicitly. A bare final formula, even if correct, typically earns only partial credit — the graders reward the separated equation and the antiderivatives.
• Organize Euler’s method in a small table with columns for x, y, slope f(x,y), and Δy = slope·Δx. The table structure prevents arithmetic slips across steps and is exactly what free-response graders expect to see.
• You can often answer logistic questions without solving the differential equation. Read M directly from the equation as the P that zeroes the (1 − P/M) factor, then use the two facts: fastest growth at M/2, and the long-run limit is M.

Practice Calculus BC

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

Questions about this unit

All 10 units of AP Calculus BC

1. Unit 1 · Limits & Continuity
2. Unit 2 · Differentiation: Definition & Fundamental Properties
3. Unit 3 · Differentiation: Composite, Implicit & Inverse Functions
4. Unit 4 · Contextual Applications of Differentiation
5. Unit 5 · Analytical Applications of Differentiation
6. Unit 6 · Integration & Accumulation of Change
7. Unit 7 · Differential Equations
8. Unit 8 · Applications of Integration
9. Unit 9 · Parametric, Polar & Vector-Valued Functions
10. Unit 10 · Infinite Sequences & Series