Unit 7: Differential Equations
Unit 7 turns derivatives into models. It covers what a differential equation says about a changing quantity, how to check whether a function satisfies one, how slope fields picture every solution at once, how Euler's method approximates a solution step by step, and how to solve by separating variables. It closes with the two models the exam returns to again and again: exponential and logistic growth.
How to use this guide
Read it in order the first time because the topics build on each other. The equation tells you how a quantity changes, the slope field shows all possible solutions at once, Euler's method walks along one solution numerically, and separation of variables solves the equation exactly. The exponential and logistic models at the end are applications of everything before them.
After the first read, use the trap boxes and the confusions table to review the distinctions that exam questions test most often. Finish with the practice questions, then complete the recall check on the last page out loud and note any items you cannot explain yet.
What this unit is worth. Differential equations appear regularly in the free-response section, often as a multi-part question that moves from a slope field to Euler's method to an exact solution. The separation-of-variables technique also shows up inside later units whenever a rate equation needs solving.
7.1 Modeling Situations with Differential Equations
A differential equation is a relation that ties a function to its derivatives. The equation dy/dx = ky does not give you y directly. It tells you how y changes: at every point, the rate of change of y is proportional to y itself. When a quantity grows or shrinks in proportion to its current size, like a population or a cooling object, the situation is modeled by an equation of this form.
Reading the equation is a skill worth practicing. The left side names the rate, dy/dx or dy/dt. The right side names the rule the rate follows. In dy/dx = 2xy, the rate depends on both x and y. In dy/dt = ky(a − y), the rate depends on y and on how far y is from the value a. Before solving anything, make sure you can say in words what the right side claims about the quantity.
Trap. The differential equation is not the solution. The equation dy/dx = 2x describes how y changes. The function y = x2 + 3 is a solution. Exam questions sometimes ask which of two statements is the equation and which is the solution, so keep the roles straight.
7.2 Verifying Solutions for Differential Equations
A solution of a differential equation is a function that satisfies the equation, along with its derivatives, when substituted in. To verify a proposed solution, compute its derivative, plug both the function and the derivative into the equation, and check that the left side equals the right side.
Check y = 5ex2 against dy/dx = 2xy. The derivative is y' = 5ex2 · 2x, by the chain rule. The right side is 2x · y = 2x · 5ex2. Both sides equal 10xex2, so the function is a solution. The check has two parts: differentiate correctly, then confirm the substitution balances.
Trap. A function that satisfies the equation at one point is not automatically a solution. Verification means the equation holds for every x in the domain, because you substitute the whole function and its whole derivative, not a single value.
7.3 Sketching Slope Fields
A slope field is a grid of short line segments showing the value of dy/dx at many points. At each point (x, y), you evaluate the right side of the differential equation and draw a tiny segment with that slope. A solution curve is any curve that follows the segments, staying tangent to them as it passes through.
To sketch one by hand, pick a few y values and compute the slope at each. For dy/dx = x + y, at (0, 0) the slope is 0, at (0, 1) the slope is 1, at (1, 0) the slope is 1, and at (−1, 1) the slope is 0. Draw short segments with those slopes at those points. Where the right side is zero, the segments are horizontal, and those points mark where solution curves flatten out.
7.4 Reasoning Using Slope Fields
Reading a slope field is just as important as drawing one. If the segments get steeper as y increases, solutions grow faster at larger y. If the segments are horizontal along some line y = c, that line may be a constant solution. When a question gives you a slope field and asks which differential equation it matches, test a point: compute the right side at that point and check it against the segment you see.
Trap. A slope field shows the slopes of solutions, not the solutions themselves. The segments are tangent lines, not pieces of the curve. A solution curve passes through the grid following the segments, but it is not drawn as a connect-the-dots path through the segment endpoints.
7.5 Approximating Solutions Using Euler's Method
Euler's method approximates a solution curve numerically. Start from a known point, compute the slope there from the differential equation, and step forward along that slope to a new point. Then repeat: compute the new slope, step forward again. Each step uses the formula ynew = y + (Δx)(dy/dx), evaluated at the current point.
Work it through for dy/dx = x + y with y(0) = 1, approximating y(0.2) with step size Δx = 0.1. Start at (0, 1). The slope is 0 + 1 = 1, so y1 = 1 + 0.1(1) = 1.1, and x1 = 0.1. At (0.1, 1.1), the slope is 0.1 + 1.1 = 1.2, so y2 = 1.1 + 0.1(1.2) = 1.1 + 0.12 = 1.22, and x2 = 0.2. The approximation is y(0.2) ≈ 1.22.
Notice what each step needs: the current x, the current y, and the slope from the equation at that point. Smaller step sizes give better approximations but require more steps. On the exam, show every step in a table or a list so the reader can follow the arithmetic.
Trap. The most common error is reusing the old slope instead of recomputing it at the new point. After the first step you are at (0.1, 1.1), not (0, 1), so the slope for the second step is 1.2, not 1. Update both x and y before computing the next slope.
7.6 Finding General Solutions Using Separation of Variables
Separation of variables solves equations of the form dy/dx = g(x) · h(y). Rewrite the equation so that all the y terms are on one side with dy and all the x terms are on the other side with dx: (1/h(y)) dy = g(x) dx. Then integrate both sides.
Solve dy/dx = 2xy. Divide both sides by y to get (1/y) dy = 2x dx. Integrate: the left side gives ln|y|, and the right side gives x2 + C. So ln|y| = x2 + C. Exponentiate both sides: |y| = ex2 + C = eC · ex2. Since eC is a positive constant, absorbing the sign gives y = Aex2, where A is an arbitrary constant.
The general solution is the whole family of functions satisfying the equation, parameterized by the arbitrary constant C (or A). Every choice of the constant gives a different member of the family. The absolute value in ln|y| is handled by letting the constant absorb the sign, which is why the final form drops the bars.
Trap. Do not forget the constant of integration, and do not add it twice. One constant is enough because combining C1 and C2 from the two sides into a single C covers every possibility. Also, the constant goes in as soon as you integrate, not at the end.
7.7 Finding Particular Solutions Using Initial Conditions
An initial condition is a specified solution value, such as y(0) = 3. It picks one member out of the general solution's family. A particular solution is the single function satisfying both the equation and the initial condition, with no arbitrary constant remaining.
Finish the example from 7.6: dy/dx = 2xy with the initial condition y(0) = 5. The general solution is y = Aex2. Substitute x = 0 and y = 5: 5 = Ae0 = A, so A = 5. The particular solution is y = 5ex2. As a check, y' = 5ex2 · 2x = 2x · y, which matches the original equation, and y(0) = 5e0 = 5, which matches the condition.
There are two places to apply the condition: after solving for y in terms of the constant, or right after integrating by using the condition to evaluate a definite integral. Either way works. The key step is substituting the known x and y values and solving for the constant before writing the final answer.
Trap. A particular solution has no arbitrary constant left in it. If your final answer still contains C or A, you have written the general solution, not the particular one. The initial condition exists precisely to determine that constant.
7.8 Exponential Models with Differential Equations
The exponential model is the equation dy/dt = ky. It says the rate is proportional to the quantity itself. The solutions are y = Cekt: they grow when k is greater than 0 and decay when k is less than 0. This is the same family you met in 7.6, now read as a model of real quantities.
Solve dy/dt = 3y with y(0) = 4. Separate: (1/y) dy = 3 dt. Integrate: ln|y| = 3t + C, so y = Ae3t. The initial condition gives 4 = Ae0, so A = 4. The particular solution is y = 4e3t. Checking: y' = 12e3t and 3y = 12e3t, and y(0) = 4. The constant k = 3 is the relative growth rate: the quantity grows at 3 per unit of itself per unit time.
Exponential models fit situations with unlimited growth or decay: unrestricted populations, continuously compounded interest, radioactive decay. The sign of k tells the whole story. Positive k means the quantity increases without bound, negative k means it decreases toward zero.
Trap. In y = Cekt, the constant C is the initial value y(0), not a free parameter to ignore. And k is not the growth itself, it is the proportional rate. A question asking for the population after 5 years needs both numbers substituted, not just the form of the solution.
7.9 Logistic Models with Differential Equations
The logistic growth model is dy/dt = ky(a − y). The rate is jointly proportional to the current size y and the remaining room (a − y). Early on, when y is small, (a − y) is nearly a, and growth looks exponential. As y approaches a, the remaining room shrinks and growth slows to a stop.
The carrying capacity is the limiting value a. As the independent variable goes to infinity, solutions approach a without crossing it. The point of fastest logistic growth is the inflection point of the solution curve, at y = a/2, where the rate is greatest. You can find it without solving the equation: it is the y value that maximizes the product y(a − y).
For dy/dt = 0.1y(100 − y), the carrying capacity is 100. The fastest growth happens at y = 50, where the rate is 0.1 · 50 · 50 = 250. At y = 25 the rate is 0.1 · 25 · 75 = 187.5, and at y = 75 it is 0.1 · 75 · 25 = 187.5, confirming that the rate peaks in the middle and falls off symmetrically on both sides.
Trap. The carrying capacity is a horizontal asymptote of the solution, not a value the solution reaches. A logistic curve approaches a as time goes to infinity but never touches it. Also, the fastest growth is at half the carrying capacity, not at the carrying capacity itself, where the rate is zero.
Confusions That Cost Points
| Pair | How to keep them straight |
|---|---|
| General vs particular solution | The general solution keeps the arbitrary constant and describes the whole family. The particular solution uses the initial condition to fix the constant and names one function. |
| Slope field vs solution curve | The field is the grid of tangent segments. A solution curve is one path through the grid, following the segments. The field contains every solution at once. |
| Euler's method vs exact solution | Euler's method approximates by walking along tangent lines in steps. Separation of variables gives the exact formula. Euler is for when you cannot or need not solve exactly. |
| Exponential vs logistic model | Exponential, dy/dt = ky, grows without bound. Logistic, dy/dt = ky(a − y), levels off at the carrying capacity a. The (a − y) factor is what stops the growth. |
| Carrying capacity vs fastest growth point | The carrying capacity a is the ceiling the solution approaches. The fastest growth happens at y = a/2, the inflection point, where the rate is greatest. |
Practice Questions
Original questions written for this guide in the style of the AP exam. Answers and explanations are on the next page, so complete the questions before checking them.
1. Let dy/dx = x + y and y(0) = 1. Using Euler's method with step size Δx = 0.1, what is the approximation for y(0.2)?
- 1.10
- 1.20
- 1.22
- 1.24
2. Which of the following is the general solution of dy/dx = 2xy?
- y = ex2 + C
- y = Aex2
- y = 2x2 + C
- y = Ae2x
3. A population P satisfies dP/dt = 0.2P(500 − P). What is the carrying capacity, and at what population size is the growth rate greatest?
- Carrying capacity 500; fastest growth at P = 500
- Carrying capacity 500; fastest growth at P = 250
- Carrying capacity 250; fastest growth at P = 250
- Carrying capacity 0.2; fastest growth at P = 100
4. If dy/dt = 3y and y(0) = 4, then y(1) =
- 4e3
- 12
- 3e4
- 4 + 3e
Answer Key
1. C. Step 1: at (0, 1), the slope is 0 + 1 = 1, so y1 = 1 + 0.1(1) = 1.1 and x1 = 0.1. Step 2: at (0.1, 1.1), the slope is 0.1 + 1.1 = 1.2, so y2 = 1.1 + 0.1(1.2) = 1.22. A stops after one step. B comes from reusing the old slope for both steps, computing 1 + 0.1(1) + 0.1(1) = 1.20 instead of updating the slope to 1.2 at the second step. D may come from using the wrong slope at the second step.
2. B. Separate: (1/y) dy = 2x dx. Integrate to get ln|y| = x2 + C, then exponentiate: y = Aex2. A adds the constant outside the exponential, which does not come from integrating correctly. C integrates the right side as if it were the whole equation, ignoring the separation step. D treats 2x as a constant multiplier, giving the wrong exponent.
3. B. In the logistic form dy/dt = ky(a − y), the carrying capacity is a = 500, and the fastest growth is at y = a/2 = 250. A confuses the fastest growth point with the carrying capacity itself, where the rate is actually zero. C halves the carrying capacity by mistake. D reads the rate constant 0.2 as the capacity.
4. A. The equation dy/dt = 3y has solutions y = Ce3t. The initial condition gives 4 = Ce0, so C = 4 and y = 4e3t. At t = 1, y(1) = 4e3. B multiplies the rate constant by the initial value, which is the derivative at t = 0, not the value at t = 1. C swaps the roles of the constant and the rate. D adds terms instead of using the exponential form.
One-Page Recall Check
- Explain what a differential equation models and give an example in words.
- Verify that y = 5ex2 satisfies dy/dx = 2xy, showing both steps.
- Describe how to read a slope field and how to sketch one for dy/dx = x + y.
- State the Euler's method step formula and explain why the slope must be recomputed each step.
- Use Euler's method with Δx = 0.1 to approximate y(0.2) for dy/dx = x + y, y(0) = 1.
- Solve dy/dx = 2xy by separation of variables, showing the integration and the constant.
- Explain the difference between a general solution and a particular solution.
- Apply y(0) = 5 to y = Aex2 and write the particular solution.
- Solve dy/dt = 3y with y(0) = 4 and verify the answer by substitution.
- State the logistic equation and identify the carrying capacity in dy/dt = 0.1y(100 − y).
- Explain why the fastest logistic growth happens at y = a/2.
- Give the rate of change at y = 25, y = 50, and y = 75 for dy/dt = 0.1y(100 − y).
Where to go next. Turn every missed item above into flashcards and drill them spaced out over several days rather than in one sitting. In Rycal, open the Differential Equations deck under AP Calculus BC. The deck covers the terms in this guide, and its practice questions target the same traps named here. If you have a test date, add it in the Test Planner. You can also start your next review with a Brain Dump, then check what you missed against this guide.
Key terms for this unit
Differential equation, Solution of a differential equation, Slope field, Euler's method, Separation of variables, General solution, Particular solution, Initial condition, Exponential model, Logistic growth model, Carrying capacity, Point of fastest logistic growth.
About this guide. Written for Rycal and aligned to the College Board AP Calculus BC course framework, Unit 7. All questions and explanations are original Rycal writing. Rycal is independent and is not affiliated with or endorsed by the College Board.