Lesson 33: Slope Fields & Separable Differential Equations

(a) Opening Question

Suppose a quantity y changes so that, at every point, its rate of change equals twice the product of x and y itself:

dy/dx = 2xy

Is y = 3eˣ² a solution to this equation? How would you check — without solving anything from scratch?

Take a minute before reading on.

You do not need to "solve" the equation to test a candidate. A solution is a function, and a function is a solution exactly when plugging it in makes both sides agree. Differentiate y = 3eˣ² using the chain rule: dy/dx = 3eˣ² · (2x) = 6x eˣ². Now compute the right side using the same y: 2xy = 2x · 3eˣ² = 6x eˣ². The two sides match for every x, so yes, y = 3eˣ² is a solution. This is the heartbeat of the whole unit: a differential equation is a condition on a function's derivative, and "solving" it means finding the functions that satisfy that condition. This lesson teaches you to read those functions off a picture (slope fields) and to compute them by hand (separable equations).

(b) Core Concepts

What a differential equation is — and what a solution is

A differential equation is an equation that relates a function to its derivative(s). In AB, they almost always have the form

dy/dx = (some expression in x and/or y)

The unknown is not a number — it is a function y = f(x). A solution of the differential equation is any function that, when substituted in, makes the equation true for every x in some interval. Because antidifferentiation introduces a +C, a differential equation usually has a whole family of solutions, called the general solution. Pinning down one specific member requires an initial condition — a known point y(x₀) = y₀ the solution must pass through. The single function selected by the initial condition is the particular solution.

Verify by substitution. The most reliable thing you can do with a proposed solution is check it: differentiate the candidate, and confirm dy/dx equals the right-hand side after substituting the candidate for y. The AP exam frequently asks "show that y = ... is a solution," and the expected work is exactly this substitution.

Mini-example. Show y = e^(−x²/2) solves dy/dx = −xy.

Differentiate: dy/dx = e^(−x²/2) · (−x) = −x·e^(−x²/2). Substitute into the right side: −xy = −x·e^(−x²/2). The sides agree, so y = e^(−x²/2) is a solution. ✓

Slope fields

You can often see the solutions before computing them. A slope field (or direction field) is a grid of short line segments. At each point (x, y), the segment is drawn with slope equal to dy/dx evaluated there. Since dy/dx is the slope of a solution curve passing through that point, the segments are tiny tangent lines: a solution curve is any curve that flows tangent to the segments everywhere.

How to read one. Pick a point, evaluate dy/dx there, and that number is the segment's slope. Three patterns are worth memorizing:

• If dy/dx depends only on x (e.g., dy/dx = x), then all segments in the same vertical column are parallel — the slope does not change as you move up or down.
• If dy/dx depends only on y (e.g., dy/dx = y), then all segments in the same horizontal row are parallel.
• Where dy/dx = 0, segments are horizontal; where dy/dx is large, they are steep.

Matching a slope field to a differential equation is a signature AB skill. Use these tests, in order:

1. Find where the slope is zero. Set dy/dx = 0 and identify that set of points — they will show as horizontal segments. For dy/dx = x − y, the zero-slope set is the line y = x. For dy/dx = xy, it is the two axes (x = 0 or y = 0).
2. Check dependence. Does the field's slope change as you move horizontally only, vertically only, or both? dy/dx = x → parallel along verticals; dy/dx = y → parallel along horizontals.
3. Check signs in regions. Pick a test point in each region and confirm the sign (up-tilt vs. down-tilt) matches.

These three checks usually eliminate every wrong option without sketching the whole field.

Beyond AB (not tested here). BC students also learn Euler's method, a numerical scheme that steps along a slope field to approximate a solution. Euler's method is not on the AP Calculus AB exam — you will not be tested on it in this course. Everything graded in this lesson stays on slope fields and separable equations.

Separable differential equations

A differential equation is separable if you can algebraically write it so that all the y's (and dy) are on one side and all the x's (and dx) are on the other:

dy/dx = g(x)·h(y)   ⟶   (1/h(y)) dy = g(x) dx

Then integrate both sides. You get an antiderivative on each side; combine the two constants of integration into a single + C on one side (conventionally the x side). This produces the general solution, often implicitly. Apply the initial condition to find C, then solve for y explicitly if possible.

Fully worked separable equation, start to finish. Solve dy/dx = 2xy with the initial condition y(0) = 3.

Step 1 — Separate. Divide both sides by y and multiply by dx:

(1/y) dy = 2x dx

Step 2 — Integrate both sides.

∫ (1/y) dy = ∫ 2x dx
ln|y| = x² + C

(Only one + C is needed.)

Step 3 — Solve for y. Exponentiate both sides: |y| = e^(x² + C) = e^C · eˣ². Writing the positive constant e^C (with the sign of y absorbed) as a single constant A, the general solution is

y = A eˣ²

Step 4 — Apply the initial condition y(0) = 3: 3 = A e⁰ = A·1, so A = 3. The particular solution is

y = 3 eˣ²

Step 5 — Verify by substitution. Differentiate: dy/dx = 3 eˣ² · (2x) = 6x eˣ². The right side is 2xy = 2x·(3eˣ²) = 6x eˣ². They match, and y(0) = 3e⁰ = 3. ✓ Solution confirmed.

Applying the initial condition early vs. late

A clean shortcut: you may apply the initial condition right after integrating, before solving for y, to pin down C. For ln|y| = x² + C with y(0) = 3: ln 3 = 0 + C, so C = ln 3, giving ln|y| = x² + ln 3, i.e. y = e^(x²)·e^(ln 3) = 3eˣ². Same answer. On the AP exam, finding C before fully solving for y is often less error-prone.

Domain of a particular solution

A particular solution lives on the largest interval containing x₀ on which the function is defined, differentiable, and satisfies the equation — and where it does not cross any point that breaks the algebra (such as y = 0 when you divided by y). For example, y = 3eˣ² is defined for all real x, so its domain is (−∞, ∞). But a solution like y = 1/(1−x) with y(0) = 1 is only valid on (−∞, 1), the interval containing x₀ = 0 before the function blows up at x = 1. State the domain as the interval through the initial point, stopping at any vertical asymptote or undefined point.

(c) Worked Examples

Example 1 — Verify a proposed solution (NO CALC)

Problem. Show that y = x² + (2/x) is a solution of the differential equation x·(dy/dx) + y = 3x² for x > 0.

Strategy. Differentiate the candidate, substitute both y and dy/dx into the left side, and check it simplifies to the right side.

Solution. With y = x² + 2x⁻¹, differentiate: dy/dx = 2x − 2x⁻². Substitute into the left side:

x·(dy/dx) + y = x(2x − 2x⁻²) + (x² + 2x⁻¹)
              = 2x² − 2x⁻¹ + x² + 2x⁻¹
              = 3x²

The left side equals 3x², the right side. ✓

Justification. Because substituting y = x² + 2/x and its derivative into x·(dy/dx) + y yields 3x² for all x > 0, the function satisfies the differential equation and is therefore a solution on (0, ∞).

Example 2 — Match a slope field to a differential equation (NO CALC)

Problem. A slope field has horizontal segments exactly along the line y = x, segments tilting upward below that line and downward above it. Which differential equation generated it?

(A) dy/dx = x + y (B) dy/dx = x − y (C) dy/dx = y − x (D) dy/dx = xy

Strategy. Use the zero-slope test first, then check the sign in one region.

Solution. Horizontal segments occur where dy/dx = 0. The field is horizontal along y = x, so we need dy/dx = 0 precisely when y = x. Test each: for (B), x − y = 0 ⟺ y = x. ✓ For (C), y − x = 0 ⟺ y = x also — so check signs. Below the line means x > y. For (B): x − y > 0, slope positive (up-tilt). ✓ matches. For (C): y − x < 0, slope negative — wrong. (A) is zero on y = −x, and (D) is zero on the axes — neither matches.

Answer: (B) dy/dx = x − y.

Justification. The differential equation must have dy/dx = 0 exactly on y = x (forcing (B) or (C)), and must give positive slope where x > y (below the line). Only dy/dx = x − y satisfies both.

Example 3 — Solve a separable equation with an initial condition (NO CALC)

Problem. Solve dy/dx = x/y with y(0) = 2. Give the explicit particular solution and its domain.

Strategy. Separate (y with dy, x with dx), integrate, apply the initial condition, solve for y.

Solution. Separate: y dy = x dx. Integrate both sides:

∫ y dy = ∫ x dx
y²/2 = x²/2 + C

Apply y(0) = 2: (2)²/2 = 0 + C, so C = 2. Then y²/2 = x²/2 + 2, i.e. y² = x² + 4. Solve for y. Since y(0) = 2 > 0, take the positive root:

y = √(x² + 4)

Domain. x² + 4 > 0 for all real x, so the domain is (−∞, ∞).

Verify. dy/dx = (1/2)(x² + 4)^(−1/2)·(2x) = x/√(x² + 4) = x/y. ✓ And y(0) = √4 = 2. ✓

Example 4 — Exponential-type separable, dy/dx = ky preview (NO CALC)

Problem. Solve dy/dx = 6x²y with y(0) = 1. Then note how this relates to the form dy/dx = ky.

Strategy. Separate variables, integrate, exponentiate, apply the initial condition.

Solution. Separate: (1/y) dy = 6x² dx. Integrate:

ln|y| = 2x³ + C

Exponentiate: y = A e^(2x³) (with A = ±e^C). Apply y(0) = 1: 1 = A e⁰ = A, so A = 1. Thus

y = e^(2x³)

Verify. dy/dx = e^(2x³)·(6x²) = 6x² e^(2x³) = 6x²y. ✓ And y(0) = e⁰ = 1. ✓

Connection to dy/dx = ky. When the rate of change of y is proportional to y with a constant k — dy/dx = ky — the same separation gives ln|y| = kx + C and y = y₀ eᵏˣ, exponential growth/decay. Here the "proportionality" multiplier 6x² is not constant, so the exponent picks up its antiderivative 2x³ instead of kx. Lesson 34 develops the constant-k case (dy/dt = ky) in full as the exponential model.

(d) Common Mistakes

Forgetting + C before applying the initial condition. Students integrate ∫(1/y)dy = ∫2x dx to get ln|y| = x² and skip + C. With no constant, there is nothing for the initial condition to adjust, and the particular solution comes out wrong. Fix: write + C the instant you integrate, then substitute the initial condition to solve for it. On the AP rubric, the constant of integration is a required, separately-scored step.

Not separating completely. A common error is trying to integrate dy/dx = 2xy as if it were ∫ 2xy dx with y treated as a constant. You cannot integrate the right side while it still contains the unknown y. Fix: the equation is solvable only because it factors as g(x)·h(y); you must get every y onto the left with dy and every x onto the right with dx before integrating. If you cannot fully separate, the method does not apply.

Sign/algebra errors solving for y. After y² = x² + 4, students write y = ±√(x² + 4) and stop, or pick the wrong root. Fix: the initial condition chooses the branch. Since y(0) = 2 > 0, keep the positive root. Also, √(x² + 4) ≠ x + 2 — you cannot distribute a square root over a sum. Keep it as √(x² + 4).

Mismatching a slope field. Students eyeball a field and guess without testing. Fix: always start with the zero-slope set (set dy/dx = 0), then test the sign of dy/dx at one point in a region. If two equations share the same zero set (like x − y and y − x), the sign test breaks the tie. Also remember: slope depending only on x ⇒ columns of parallel segments; only on y ⇒ rows of parallel segments.

Dropping the absolute value / mishandling e^C. From ln|y| = x² + C, students write y = eˣ² + C (adding the constant) instead of y = e^C·eˣ² (multiplying). Fix: exponentiating a sum gives a product: e^(x²+C) = e^C·eˣ². Rename ±e^C as a single constant A, then apply the initial condition.

(e) Practice Problems

12 (short answer). [NO CALC] Solve dy/dx = 6x²y with y(0) = 1. Give the explicit solution and verify it by substitution.

13 (short answer, justification). [NO CALC] A student claims y = eˣ² + C is the general solution of dy/dx = 2xy. Identify the algebra error and give the correct general solution. Justify why your function works.

14 (short answer, slope-field matching, justification). [NO CALC] A slope field has horizontal segments only along the x-axis (y = 0) and the slopes get steeper as |x| grows. Determine whether the differential equation is dy/dx = xy or dy/dx = x + y, and justify using the zero-slope set.

15 (short answer, justification/interpretation). [NO CALC] Solve dy/dx = −y/x with y(1) = 2 for x > 0. State the domain of the particular solution and explain why x = 0 cannot be in it.

(f) AP Exam Focus

This is a non-calculator free-response question (AP Section II, Part B style). Budget about 15 minutes. Show all separation and integration steps, label your constant, and use precise justification language.

FRQ. Consider the differential equation

`

dy/dx = 3x² / (2y)

`

Let y = f(x) be the particular solution with f(0) = 1.

(a) On the axes provided, a slope field for the differential equation is drawn at the twelve indicated points. At the points (0, 1), (1, 1), and (2, 1), find dy/dx and describe the relative steepness of the segments there. (3 points)

(b) Write an equation for the line tangent to the solution curve y = f(x) at the point (0, 1), and use it to approximate f(0.2). (2 points)

(c) Find the particular solution y = f(x) to the differential equation with the initial condition f(0) = 1, and state its domain. (4 points)

(d) Use the particular solution from part (c) to find the exact value of f(2). (1 point)

Total: 10 points

Model Solution

(a) Evaluate dy/dx = 3x²/(2y) at each point:

At (0, 1):  dy/dx = 3(0)²/(2·1) = 0
At (1, 1):  dy/dx = 3(1)²/(2·1) = 3/2
At (2, 1):  dy/dx = 3(2)²/(2·1) = 12/2 = 6

At (0, 1) the segment is horizontal (slope 0). At (1, 1) it has a moderate positive slope of 3/2. At (2, 1) it is much steeper (slope 6). The segments grow steeper as x increases (for a fixed y > 0), since dy/dx is proportional to x².

(b) The tangent line at (0, 1) uses slope dy/dx|₍₀,₁₎ = 0 from part (a):

y − 1 = 0·(x − 0)   ⟹   y = 1

Linear (tangent-line) approximation: f(0.2) ≈ 1 + 0·(0.2) = 1.

(c) The equation is separable. Separate variables (y with dy, x with dx):

2y dy = 3x² dx

Integrate both sides:

∫ 2y dy = ∫ 3x² dx
y² = x³ + C

Apply the initial condition f(0) = 1: (1)² = (0)³ + C, so C = 1. Then y² = x³ + 1. Since f(0) = 1 > 0, take the positive square root:

y = √(x³ + 1)

Domain. We need x³ + 1 ≥ 0, i.e. x ≥ −1; and for the equation dy/dx = 3x²/(2y) to be valid we need y ≠ 0, i.e. x³ + 1 > 0, so x > −1. The interval containing x₀ = 0 is therefore (−1, ∞).

(d) Using f(x) = √(x³ + 1):

f(2) = √(2³ + 1) = √(8 + 1) = √9 = 3

Scoring Commentary — where students lose points

• (a) 3 points: 1 for each correct slope value (0, 3/2, 6). The steepness description is folded into earning the (2,1) point — you must convey that larger x gives steeper segments. Trap: arithmetic slips on 3(2)²/2; it is 6, not 12 (the 2 in the denominator is part of 2y).
• (b) 2 points: 1 for the tangent line y = 1 (slope must come from part (a)), 1 for f(0.2) ≈ 1. Trap: using a nonzero slope, or confusing tangent-line approximation with the actual solution.
• (c) 4 points: 1 for separating (2y dy = 3x² dx), 1 for antidifferentiating both sides with a constant (y² = x³ + C), 1 for using f(0) = 1 to get C = 1 and solving y = +√(x³ + 1) (positive root justified by the initial value), 1 for the domain (−1, ∞). Traps: omitting + C (loses the constant point and breaks the initial-condition step); writing ±√(x³+1) without selecting the positive branch; giving the domain as all reals (it must stop at x = −1).
• (d) 1 point: the exact value f(2) = 3. Trap: leaving it as √9 is fine, but a decimal-only answer without the exact value can lose credit on a no-calculator part; here √9 = 3 is exact.