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AP Calculus AB/BC Unit 7 Visual ReviewA topic-by-topic visual walkthrough of Unit 7: Differential Equations — slope fields, Euler's method, separation of variables, and exponential and logistic models.
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TOPIC 7.1
Modeling with Differential Equations
An equation involving a derivative
A differential equation (DE) relates a
function to its RATE of change.
"Grows proportional to its size":
dy/dt = k·y
Translate words about rates into a DE.
Common phrasings → DEs
"proportional to y" → dy/dt = ky
"proportional to the difference from M"
→ dy/dt = k(M − y) (cooling)
"limited growth toward capacity L"
→ dy/dt = ky(1 − y/L) (logistic)
The solution is a FUNCTION, not a number
Solving a DE means finding the function y(t) whose derivative satisfies the equation.
A differential equation relates y to its rate — e.g. dy/dt = ky for proportional growth.
The Review Hub · AP Calculus AB/BC Unit 7
TOPIC 7.2
Verifying Solutions
Substitute the candidate into the equation
To check if y = g(t) solves a DE, compute its derivative and plug BOTH y and y′ in.
If the equation becomes a true statement, it's a solution.
// verify y = 3e^(2t) solves dy/dt = 2y
dy/dt = 6e^(2t)
2y = 2·(3e^(2t)) = 6e^(2t) ✓ they match
Both sides equal 6e^(2t), so y = 3e^(2t) is a solution.
You do NOT need to solve the DE to verify a given solution.
Verify a solution by substituting y and y′ and checking both sides match.
The Review Hub · AP Calculus AB/BC Unit 7
TOPIC 7.3
Sketching Slope Fields
tiny tangent segments at each point
Plug (x, y) into dy/dx
At each grid point, the DE gives a SLOPE.
Draw a short segment with that slope.
at (2,1) with dy/dx=x+y: slope 3
The picture of all solutions
A slope field shows the direction every
solution curve travels.
Segments depending only on y are constant across each row.
A slope field plots dy/dx at grid points as tiny tangent segments.
The Review Hub · AP Calculus AB/BC Unit 7
TOPIC 7.4
Reasoning Using Slope Fields
Follow the flow
Sketch a solution by starting at a point
and following the segment directions.
The curve is always TANGENT to the field.
Different starting points → different
particular solutions.
Read features from the field
Flat (horizontal) segments where dy/dx = 0
→ possible equilibrium solutions.
Steep segments → fast change.
An EQUILIBRIUM y = c is where dy/dx = 0
for all x (a constant solution).
Match a DE to its field
Test a few points: compute dy/dx and compare with the segment slopes. If dy/dx depends only on x, columns look identical.
A solution curve stays tangent to the slope field ; flat segments mark equilibria.
The Review Hub · AP Calculus AB/BC Unit 7
TOPIC 7.5
BC ONLY
Euler's Method
yₙ₊₁ = yₙ + (dy/dx)·Δx (step along the tangent)
Worked example
dy/dx = x+y, start (0,1), Δx=0.5:
slope at (0,1) = 1
y₁ = 1 + 1·0.5 = 1.5 (at x=0.5)
slope at (0.5,1.5)=2 → y₂=2.5
Repeat step by step.
A numerical approximation
Euler's method uses tangent-line steps to
estimate a solution when you can't solve
the DE exactly.
SMALLER Δx → more accurate.
For concave-up curves, Euler UNDERestimates.
Euler (BC): yₙ₊₁ = yₙ + (dy/dx)Δx — step along tangent lines; smaller Δx is better.The Review Hub · AP Calculus AB/BC Unit 7
TOPIC 7.6
Separation of Variables
get all y on one side, all x on the other, then integrate
// solve dy/dx = xy
(1/y) dy = x dx // separate
∫(1/y) dy = ∫ x dx // integrate both sides
ln|y| = x²/2 + C
y = A·e^(x²/2) // solve for y (A = ±eᶜ)
Add + C on ONE side; combine constants at the end.
Separate variables , integrate both sides, then solve for y (add +C).The Review Hub · AP Calculus AB/BC Unit 7
TOPIC 7.7
Particular Solutions (Initial Conditions)
Use the initial condition for C
The general solution has a constant C.
Plug in the given point to solve for C.
y = A·e^(x²/2), y(0) = 3
3 = A·e⁰ = A → A = 3
y = 3·e^(x²/2)
General vs. particular
GENERAL solution: family with a free C —
infinitely many curves.
PARTICULAR solution: the one curve
through the given point.
Solve for C BEFORE or after isolating y — either works.
State the domain of the solution
The particular solution is valid on the interval containing the initial point where it stays continuous.
Use the initial condition to solve for C and get the particular solution.The Review Hub · AP Calculus AB/BC Unit 7
TOPIC 7.8
Exponential Models
dy/dt = ky ⟹ y = y₀·eᵏᵗ
Proportional growth / decay
When the rate is proportional to the
amount, the solution is exponential.
k > 0 → GROWTH; k < 0 → DECAY.
y₀ = initial amount y(0).
Derived by separation of variables (7.6).
Applications
• Radioactive decay (half-life)
• Continuously compounded interest
• Unbounded population (early stage)
half-life: k = −ln2 / h
Grows without bound (no ceiling).
dy/dt = ky → y = y₀eᵏᵗ : exponential growth (k>0) or decay (k<0).The Review Hub · AP Calculus AB/BC Unit 7
TOPIC 7.9
BC ONLY
Logistic Models
y = L
fastest growth at y = L/2
dy/dt = ky(1 − y/L)
L = carrying capacity (the ceiling)
growth slows as y approaches L
Key facts
lim(t→∞) y = L (levels off at capacity).
FASTEST growth (inflection) at y = L/2.
Equilibria at y = 0 and y = L.
Logistic (BC): dy/dt = ky(1−y/L) — levels off at L, fastest growth at L/2.The Review Hub · AP Calculus AB/BC Unit 7
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How to use the visual review
Spend 30 seconds per slide before clicking next. Look at the diagram, then ask yourself: "Could I sketch this slope field or solve this equation from memory?"
Use the fullscreen button () on desktop for the best experience. Use arrow keys to navigate. Tap "Show all slides" to jump around.
This is great for review the night before the exam — fast, visual, and covers every idea you need to recognize in Unit 7.