AP Calculus AB and BC · Exam guide

AP Calculus Differential Equations and Slope Fields: Separation of Variables Step by Step

A differential equation question usually has three parts: sketch or read a slope field, find the slope at a point, and solve the equation with an initial condition. The solving part is separation of variables, and graders give a point for each step. This guide walks one example through all three.

Slope fields in plain words

A slope field draws a short segment at points in the plane, with the slope given by the differential equation. A solution curve follows the segments. Two ideas answer most questions: look at the sign of the slope to see increasing or decreasing, and at where the slope is zero to find where the curve is level.

A worked example

Consider the differential equation \(\dfrac{dy}{dx} = xy\). (a) At the point \((1, 2)\), what is the slope of the solution curve? (b) Find the particular solution \(y = f(x)\) with \(f(0) = 2\). (c) Find \(f(1)\).

(a) Substitute: \(\dfrac{dy}{dx} = (1)(2) = 2\).

(b) Separate the variables: \(\dfrac{1}{y}\,dy = x\,dx\). Integrate both sides: \(\ln|y| = \dfrac{x^2}{2} + C\). Use \(y(0) = 2\): \(\ln 2 = C\). Then \(|y| = 2e^{x^2/2}\), and because \(y(0) = 2 \gt 0\), the solution is \(y = 2e^{x^2/2}\).

(c) \(f(1) = 2e^{1/2} \approx 3.297\).

The points, one by one

StepWhat the grader looks for
SeparateAll the \(y\) terms with \(dy\) on one side and all the \(x\) terms with \(dx\) on the other. Doing this is the first point.
AntiderivativesBoth sides integrated correctly, with a constant of integration on one side. Leaving out the constant ends the problem.
Use the initial conditionSubstitute the given point into the equation that has the constant, before solving for \(y\).
Solve for yAn explicit solution \(y = \dots\), when the question asks for one, with the sign chosen by the initial condition.
Does not earn the pointsln y = x^2/2, so y = e^(x^2/2).

It leaves out the constant of integration, so the initial condition cannot be used.

Earns the pointsln|y| = x^2/2 + C. Since y(0) = 2, C = ln 2. So y = 2e^(x^2/2).

It includes the constant, uses the condition, and solves.

Other things the question may ask

  • Match a slope field to an equation by checking where the slope is zero, and the sign in each quadrant.
  • Say whether a solution curve is concave up at a point, by differentiating the equation implicitly to find \(\dfrac{d^2y}{dx^2}\).
  • Find a limit of a solution as \(x\) grows, from the equation and the field.
  • In BC, use Euler's method for a few steps, or solve a logistic equation.

Common mistakes

  • Forgetting the constant of integration.
  • Using the initial condition before separating, or after solving for the wrong variable.
  • Dropping the absolute value around \(y\) and not checking the sign.
  • Writing the solution only as a formula in two variables when an explicit \(y = f(x)\) is requested.

Frequently asked questions

Do I need to solve for y?

If the question says to write an equation for y, yes. Otherwise an implicit solution may be accepted.

Can I use a slope field to find a solution exactly?

A slope field shows the shape of solutions but not exact values. Use separation of variables for exact answers.

Is Euler's method on both AB and BC?

No. Euler's method and logistic growth are BC topics.

What if I forget the constant of integration?

You will lose points, since the initial condition cannot be used. Add it as soon as you integrate.

Where can I find the official scoring rules?

Exam formats change a little from year to year, so check the current course and exam description on the College Board site before test day.

Keep practicing

Every lesson in the AP Calculus courses ends with practice problems and an Exam focus box that shows how the topic is tested, and every unit has a 10-problem practice set with worked answers. These free lessons also build the skills the free-response section tests:

See the AP Calculus AB and BC course Free lessons and lesson list See the AP Calculus BC course

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