Differential Equations

What you'll learn

Differential equations form a critical bridge between derivatives and real-world applications in AP Calculus AB. This topic focuses on understanding relationships between functions and their rates of change, solving separable differential equations analytically, and interpreting slope fields graphically. You'll learn to model exponential growth and decay, solve initial value problems, and connect graphical, numerical, and analytical representations.

Key terms and definitions

Differential equation — an equation involving a function and one or more of its derivatives, expressing a relationship between rate of change and the function itself

Separable differential equation — a differential equation that can be written in the form dy/dx = g(x)h(y), allowing variables to be separated for integration

Initial value problem (IVP) — a differential equation together with a specified value of the function at a particular point, used to find the particular solution

General solution — the family of all solutions to a differential equation, containing arbitrary constants

Particular solution — a specific solution to a differential equation obtained by applying initial conditions to determine constants

Slope field (direction field) — a graphical representation showing small line segments at grid points with slopes determined by the differential equation

Exponential growth/decay model — a differential equation of the form dy/dt = ky, where k > 0 represents growth and k < 0 represents decay

Separation of variables — the method of rewriting a differential equation so all terms involving one variable are on one side and all terms involving the other variable are on the opposite side

Core concepts

Understanding differential equations

A differential equation describes how a quantity changes. In AP Calculus AB, you work primarily with first-order differential equations involving dy/dx or dy/dt.

Standard forms you'll encounter:

• dy/dx = f(x) — simplest form, solved by direct integration
• dy/dx = ky — exponential growth/decay model
• dy/dx = g(x)h(y) — separable form requiring variable separation

The order of a differential equation refers to the highest derivative present. AP Calculus AB focuses exclusively on first-order equations.

Verification of solutions:

To verify a function is a solution to a differential equation:

1. Find the derivative of the proposed solution
2. Substitute both the function and its derivative into the differential equation
3. Confirm both sides are equal

Separation of variables

This technique solves differential equations where variables can be algebraically separated.

Step-by-step process:

1. Write the equation in the form dy/dx = g(x)h(y)
2. Rewrite as 1/h(y) dy = g(x) dx
3. Integrate both sides: ∫[1/h(y)] dy = ∫g(x) dx
4. Solve for y if possible
5. Apply initial conditions to find the particular solution

Key points:

• Both sides must be integrated with respect to their respective variables
• Don't forget the constant of integration (typically written on one side only)
• Check your domain — division by zero may create restrictions
• Implicit solutions (not solved for y) are acceptable in many contexts

Common separable forms:

• dy/dx = ky leads to y = Ce^(kx)
• dy/dx = y/x leads to y = Cx
• dy/dx = (y + 2)(x - 1) separates to dy/(y + 2) = (x - 1) dx

Initial value problems

An initial value problem provides both a differential equation and a point through which the solution curve must pass, written as y(x₀) = y₀ or y|ₓ₌ₓ₀ = y₀.

Solution strategy:

1. Find the general solution containing constant C
2. Substitute the initial condition coordinates
3. Solve for C
4. Write the particular solution with the specific C value

Example structure:

Given dy/dx = 2x with y(1) = 5:

• General solution: y = x² + C
• Apply initial condition: 5 = 1² + C, so C = 4
• Particular solution: y = x² + 4

The initial condition uniquely determines which member of the solution family satisfies the problem.

Exponential growth and decay

The differential equation dy/dt = ky models many natural phenomena:

When k > 0 (exponential growth):

• Population growth
• Compound interest
• Bacterial cultures
• Radioactive buildup (in breeder reactors)

When k < 0 (exponential decay):

• Radioactive decay
• Newton's Law of Cooling
• Drug metabolism
• Atmospheric pressure with altitude

Standard solution: y(t) = y₀e^(kt)

Where:

• y₀ is the initial amount at t = 0
• k is the growth/decay constant
• t is time

Finding k from given information:

If you know y(t₁) = y₁:

1. Substitute into y₁ = y₀e^(kt₁)
2. Solve: k = (1/t₁) ln(y₁/y₀)

Slope fields

A slope field provides a geometric visualization of a differential equation without solving it analytically.

Reading slope fields:

• Each small line segment has slope equal to dy/dx at that point
• Solution curves follow the flow of the segments
• Multiple solution curves exist, one through each point
• Isoclines are curves where dy/dx has the same value

Creating slope fields:

1. Create a grid of points (x, y)
2. Calculate dy/dx at each point using the differential equation
3. Draw a short segment with that slope at each point
4. Segments show the direction but not magnitude of change

Analytical skills:

• Identify where solutions are increasing (positive slopes) or decreasing (negative slopes)
• Locate equilibrium solutions where dy/dx = 0 (horizontal segments)
• Determine concavity by observing how slopes change
• Sketch solution curves by following the field pattern

Common exam questions:

• Sketch a solution curve through a given point
• Identify which differential equation matches a given slope field
• Determine behavior of solutions (increasing, decreasing, approaching limits)

Euler's method (numerical approximation)

Euler's method generates numerical approximations to solutions when analytical methods are impractical.

Algorithm:

Starting from (x₀, y₀) with step size Δx:

1. Calculate slope: m = f(x₀, y₀) using the differential equation
2. Move to next point: x₁ = x₀ + Δx
3. Approximate next y-value: y₁ = y₀ + m·Δx
4. Repeat from (x₁, y₁)

General formula: yₙ₊₁ = yₙ + f(xₙ, yₙ)·Δx

Key understanding:

• Smaller step sizes increase accuracy but require more calculations
• The method approximates the solution curve with tangent line segments
• Error accumulates with each step
• Euler's method underestimates when solutions are concave up, overestimates when concave down

Worked examples

Example 1: Separable differential equation with initial condition

Question: Solve the differential equation dy/dx = xy² with initial condition y(0) = 1/2.

Solution:

Step 1: Separate variables

• dy/dx = xy²
• (1/y²) dy = x dx
• y⁻² dy = x dx

Step 2: Integrate both sides

• ∫y⁻² dy = ∫x dx
• -y⁻¹ = (1/2)x² + C
• -1/y = (1/2)x² + C

Step 3: Apply initial condition y(0) = 1/2

• -1/(1/2) = (1/2)(0)² + C
• -2 = C

Step 4: Write particular solution

• -1/y = (1/2)x² - 2
• -1/y = (x² - 4)/2
• y = -2/(x² - 4)

Answer: y = -2/(x² - 4) or y = 2/(4 - x²)

Example 2: Exponential decay model

Question: A radioactive substance decays at a rate proportional to the amount present. Initially there are 100 grams, and after 5 years, 80 grams remain.

(a) Write a differential equation modeling this situation. (b) Find the amount present after 10 years.

Solution:

(a) Let Q(t) = amount in grams at time t years

Since decay rate is proportional to amount: dQ/dt = -kQ (negative because decreasing)

With Q(0) = 100

(b) Step 1: Solve the differential equation

• General solution: Q(t) = Q₀e⁻ᵏᵗ = 100e⁻ᵏᵗ

Step 2: Find k using Q(5) = 80

• 80 = 100e⁻⁵ᵏ
• 0.8 = e⁻⁵ᵏ
• ln(0.8) = -5k
• k = -ln(0.8)/5 ≈ 0.04463

Step 3: Find Q(10)

• Q(10) = 100e⁻⁰·⁰⁴⁴⁶³⁽¹⁰⁾
• Q(10) = 100e⁻⁰·⁴⁴⁶³
• Q(10) = 100(0.64)
• Q(10) = 64 grams

Answer: (a) dQ/dt = -kQ, Q(0) = 100 (b) 64 grams

Example 3: Slope field interpretation

Question: The slope field for dy/dx = x - y is shown. If the solution curve passes through (0, 2), determine whether y is increasing or decreasing at that point, and describe the long-term behavior.

Solution:

Step 1: Evaluate dy/dx at (0, 2)

• dy/dx = x - y = 0 - 2 = -2

Since dy/dx < 0, the function is decreasing at (0, 2).

Step 2: Analyze long-term behavior

Find equilibrium solutions where dy/dx = 0:

• x - y = 0
• y = x

This line y = x represents equilibrium.

For points above y = x: dy/dx < 0 (decreasing) For points below y = x: dy/dx > 0 (increasing)

Since the solution starts at (0, 2), which is above the line y = x, it will decrease toward the equilibrium line.

Answer: The function is decreasing at (0, 2). Long-term behavior: the solution curve approaches the line y = x asymptotically.

Common mistakes and how to avoid them

• Forgetting the constant of integration: Always include "+ C" when finding the general solution, even if you'll determine C later with an initial condition. The constant appears after integrating, not before.

• Incorrect separation of variables: Don't separate multiplicative terms incorrectly. For dy/dx = xy, the correct form is (1/y) dy = x dx, NOT dy = y·x dx. Each side must contain only one variable.

• Sign errors in exponential decay: Growth has dy/dt = +ky, decay has dy/dt = -ky. Check whether the quantity increases or decreases over time. A negative k in the exponent indicates decay: y = y₀e⁻ᵏᵗ with k > 0.

• Misinterpreting slope fields: The direction of line segments shows whether solutions increase or decrease, but the length of segments is arbitrary — they don't indicate speed of change. Focus on direction and pattern, not segment length.

• Dividing by zero during separation: When separating dy/dx = y(y - 2), you divide by y(y - 2). This is only valid when y ≠ 0 and y ≠ 2. Always check for equilibrium solutions (constant solutions) separately where h(y) = 0.

• Losing track of units in applied problems: In exponential models, ensure k has units of (time)⁻¹. If t is in years, k is in year⁻¹. Dimensional analysis prevents errors in multi-step problems.

Exam technique for "Differential Equations"

• Show separation of variables explicitly: Write out the separated form before integrating. Examiners award marks for correct method even if subsequent algebra contains errors. Write "(1/y²) dy = x dx" as a separate line before "∫(1/y²) dy = ∫x dx".

• Verify solutions when instructed: If asked to verify y = 3e²ˣ satisfies dy/dx = 2y, calculate dy/dx = 6e²ˣ, then show 2y = 2(3e²ˣ) = 6e²ˣ, concluding with a statement: "Since both expressions equal 6e²ˣ, the solution is verified." This clear structure earns full marks.

• Apply initial conditions carefully: Substitute coordinates precisely. If y(3) = 5, replace x with 3 AND y with 5 simultaneously. Show this substitution step explicitly: "5 = 3² + C" earns method marks even if C is miscalculated.

• Interpret slope field questions precisely: Command words matter. "Sketch" requires drawing a curve; "determine" requires calculation; "describe" needs words. For "sketch a solution curve through (1, 2)," draw a smooth curve following the field direction through that exact point—use a ruler if needed for accuracy.