Differential Equations

What this unit covers

The topics below follow the published Calculus AB 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 the geometry of a differential equation · Sketch short segments from dy/dx at sample points · Match a differential equation to its slope field and trace solution curves
• Separation of Variables15 min · 3 objectivesSeparate a differential equation so each variable is on its own side · Integrate both sides and solve for the general solution · Apply an initial condition to find the particular solution
• Exponential Growth & Decay14 min · 3 objectivesRecognize dy/dt = ky as the exponential growth/decay model · Write its solution y = y₀·e^(kt) and interpret the constants · Distinguish growth (k > 0) from decay (k < 0)
• Modeling with Differential Equations14 min · 3 objectivesTranslate a verbal description of a rate into a differential equation · Solve applied growth and decay problems, including half-life and doubling time · Interpret solutions and their long-term behavior in context
• Matching Slope Fields, and Using Initial Conditions Honestly15 min · 3 objectivesMatch a slope field to its differential equation by testing structural features · Sketch a particular solution through a given point on a slope field · Apply an initial condition to resolve the constant, the sign, and the domain

Formulas in Unit 7

Every term in Unit 7

All 28 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.

Separation of variables

Rewrite dy/dx = g(x)h(y) as dy/h(y) = g(x) dx, integrate both sides, then use the initial condition to find C before solving for y.

Slope field

Short segments showing dy/dx at each point. Solution curves follow the segments without crossing one another.

Why the constant matters

Solve for C before simplifying further. Substituting the initial condition after exponentiating is a frequent source of error.

Exponential growth and decay

dy/dt = ky has solution y = y₀e^(kt). Growth for positive k, decay for negative.

Interpreting k

The constant relative rate of change: k = 0.05 means the quantity grows by about 5% per unit time.

Half-life and doubling time

From y = y₀e^(kt), the half-life is ln(2)/|k| and the doubling time is ln(2)/k.

Newton's law of cooling

dT/dt = k(T − T_env), separable, with solution approaching ambient temperature exponentially.

Verifying a proposed solution

Differentiate the candidate and substitute into the differential equation; both sides must agree identically, and the initial condition must hold.

Sketching a slope field solution

Start at the given point and follow the segments in both directions. Solution curves never cross one another.

Reading equilibrium solutions

Set dy/dx = 0 and solve for y. Those constant solutions appear as horizontal rows of flat segments in the slope field.

Domain of a particular solution

The interval containing the initial condition on which the solution is defined and differentiable. Free-response questions ask for it explicitly.

Sign of the constant in exponential models

Positive k grows, negative k decays. Solve for k from a second data point rather than guessing.

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 occur where dy/dx = 0, so horizontal marks along y = 2 point to a factor of (y − 2).

Sketching a particular solution

Start at the given point, stay tangent to nearby segments, and extend across the full field. The curve must pass through the point.

Solutions cannot cross an equilibrium

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

Equilibrium solution

A constant solution where dy/dx = 0 for all x, found by setting the right side to zero — for dy/dx = y(3 − y), the lines y = 0 and y = 3.

Resolve C before simplifying

Apply the initial condition while the equation is still in log form. It is far easier than carrying C through an exponentiation.

Exponentiating a sum gives a product

From ln|y| = 3x² + C you get |y| = e^C·e^(3x²), so the constant becomes a multiplicative factor, never an additive one.

Choose the branch from the initial condition

Solving |y| = something or y² = 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.

dy/dt = ky solution

y = y₀e^(kt). Positive k gives growth, negative decay. Worth memorizing rather than re-deriving.

k as a continuous rate

k is a CONTINUOUS relative rate, so k = 0.05 does not mean exactly 5% growth per unit time — the actual factor is e^0.05 ≈ 1.0513, about 5.13%.

Half-life from k

Set e^(kt) = 1/2, so t = ln(1/2)/k = −ln2/k, positive because k is negative for decay.

Two checks on a candidate solution

A function satisfying the differential equation but not the initial condition is a general solution, not the particular one asked for. Both checks are required.

Newton’s law of cooling

dT/dt = k(T − Tₑ). The temperature difference decays exponentially, so T approaches ambient asymptotically rather than reaching it.

Logistic behavior qualitatively

For dy/dt = ky(M − y), growth is fastest at y = M/2 and slows as y approaches the carrying capacity M. AB is not required to solve it.

Separation requires a product form

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

What examiners penalize here

• When asked to sketch a solution curve on a given slope field, start at the initial point and keep your curve tangent to the nearby segments, never crossing them at odd angles. Follow the flow smoothly in both directions from the starting point.
• Free-response separable-equation problems award points for the separated equation, the antiderivatives of both sides, and finding C from the initial condition. Show each stage explicitly — a correct final formula with no work in between leaves points on the table.
• For half-life or doubling-time questions, set y = y₀/2 (or 2y₀) in y = y₀·e^(kt); the y₀ cancels, leaving an equation you solve with a natural log. The half-life depends only on k, never on the starting amount.
• 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.

Practice Calculus AB

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 8 units of AP Calculus AB

1. Unit 1 · Limits & Continuity
2. Unit 2 · Differentiation: Definition & Rules
3. Unit 3 · Composite & Implicit Differentiation
4. Unit 4 · Contextual Applications
5. Unit 5 · Analytical Applications
6. Unit 6 · Integration & Accumulation
7. Unit 7 · Differential Equations
8. Unit 8 · Applications of Integration