AP Calculus BC - Unit 7 Study Guide
AP Calculus BC • Unit 7

Differential Equations

Exam Weight: 6–12% • Topics 7.1 – 7.9
7.1

Modeling with Differential Equations

A Differential Equation (DE) is an equation containing a derivative, like y′ = 2x or dy/dx = x + y. The big idea: it describes how a quantity changes.

Modeling Language (AP Loves These Words)
  • Rate of change = derivative (how fast something changes)
  • Proportional to means multiply by a constant (ex: dy/dt = ky)
  • Initial condition gives a starting point (ex: y(0)=5)
Modeling idea Linear: constant slope Exponential: slope increases as y increases linear exponential t y
If the rate is proportional to the amount (dy/dt = ky), the curve bends upward (growth) or downward (decay).
7.2

Verifying Solutions for Differential Equations

This is “free points” if you’re clean and organized: differentiate, substitute, simplify, conclude.

How to Verify a Solution (Protocol)
  1. Compute y′ (and y″ if needed).
  2. Substitute y and y′ into the DE.
  3. Simplify. If LHS = RHS, it is verified.
Example

Verify that y = e3x is a solution to y′ − 3y = 0.

  1. Differentiate: y′ = 3e3x.
  2. Substitute: 3e3x − 3(e3x).
  3. Simplify: 0 = 0 ✅
FRQ language: End with a conclusion sentence: “Therefore, y = ... satisfies the differential equation.”
7.3

Sketching Slope Fields

A Slope Field represents a DE by drawing tiny line segments with slope dy/dx at many points (x,y).

Sample solution curve Example DE: dy/dx = x
Quick check: if dy/dx = x, slopes are identical in vertical columns and flat on the y-axis (x=0).

Drawing Them

Plug (x,y) into the DE → get slope → draw a short segment.

Reading Them

Solution curves “flow” along the segments like a leaf in a stream.

7.4

Reasoning Using Slope Fields

Use the field to predict long-term behavior, compare solutions, and spot equilibrium solutions.

Equilibrium Solutions

If the slope is zero along y=k, then y=k is an equilibrium solution.

Compare Two Solutions

If solutions don’t cross in the field, the one that starts higher stays higher.

y = 1 (equilibrium) y = 0 (equilibrium) Example: dy/dx = y(1 − y) x y
For dy/dx = y(1 − y): solutions between 0 and 1 increase toward 1. Above 1, they decrease toward 1.
Multiple Choice Hack:
  • Zero slope check: where segments are flat, dy/dx = 0.
  • x-only DE: identical slopes in columns. y-only DE: identical slopes in rows.
7.5

Euler's Method (欧拉方法) BC Only

Euler's Method approximates a solution curve by stepping forward using the slope: start → find slope → take a step → repeat.

True solution Euler approximation
Euler step: ynew = yold + (Δx) · f(xold, yold)
The Formula (Iterative Step)
yn+1 = yn + (Δx) · f(xn, yn)

"New y = Old y + (step size × slope)"

The Table Strategy (Must Use!)

Problem: Given dy/dx = x + y and y(0)=1. Approximate y(2) with 2 steps (Δx=1).

Current (x, y) Slope y + (Δx)·slope New Point
(0, 1) 0 + 1 = 1 1 + (1)(1) = 2 (1, 2)
(1, 2) 1 + 2 = 3 2 + (1)(3) = 5 (2, 5)

Answer: y(2) ≈ 5.

7.6

Separation of Variables

Main algebraic method: move all y to one side and all x to the other, then integrate.

The 5-Step Protocol (Memorize!)
  1. Separate: move y-stuff with dy, x-stuff with dx.
  2. Integrate: integrate both sides.
  3. + C: add constant on the x-side.
  4. Use IC: plug in initial condition to find C.
  5. Solve for y: isolate y.
Worked Example

Solve: dy/dx = xy2 with y(0)=1.

  1. Separate: y−2 dy = x dx
  2. Integrate: ∫ y−2 dy = ∫ x dx → −1/y = x2/2 + C
  3. Find C: plug in (0,1): −1 = 0 + C → C = −1
  4. Solve: −1/y = x2/2 − 1 → y = −1 / (x2/2 − 1)
7.7

Particular Solutions (Initial Conditions)

General solutions include +C. Initial conditions choose the one curve you actually want.

AP Must-Say Sentence

“Using the initial condition y(x0) = y0, substitute into the general solution to solve for C. Therefore, the particular solution is …”

Mini Example

If y = Cekt and y(0)=5, then 5 = Ce0 → C=5, so y = 5ekt.

7.8

Exponential Growth & Decay

When the rate of change is proportional to the amount present, you get the exponential model.

dy/dt = ky

Solution:

y = y0ekt
k > 0 (growth) k < 0 (decay) t y
Growth curves rise faster over time; decay curves drop quickly then level off toward 0.
Growth (k > 0)

Population, bacteria, interest.

Decay (k < 0)

Half-life, cooling, depreciation.

7.9

Logistic Growth Models BC Only

Logistic growth models limited resources: fast growth early, then it slows as the population approaches the carrying capacity M.

M Carrying capacity M/2 Fastest growth t P(t)
Logistic curves are S-shaped: concave up then concave down, with fastest growth at P = M/2.
The Differential Equation
dP/dt = kP(M − P)

Where M is the carrying capacity.

Limit

P(t) → M as t → ∞

Fastest Growth

P = M/2

Inflection Point

Occurs at P = M/2.

Exam Hack: You usually don’t need to solve it. Interpret:
If dP/dt = 0.2P(100 − P), then M = 100 and max growth occurs at P = 50.
← Unit 6 Unit 8: Applications →