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How do you check whether a given function is a solution of a differential equation?

Topic 7.2 Verifying Solutions for Differential Equations: verify that a proposed function satisfies a differential equation by substitution.

A focused answer to AP Calculus AB Topic 7.2, verifying that a proposed function solves a differential equation by differentiating and substituting into both sides, with worked checks of general and particular solutions.

Generated by Claude Opus 4.88 min answer

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  1. What this topic is asking
  2. The verification method
  3. A worked verification
  4. General versus particular solutions
  5. Why verification is a separate, scoreable skill
  6. Verifying solutions of higher-order equations
  7. Using verification as a confidence check

What this topic is asking

The College Board (Topic 7.2) asks you to verify that a given function is a solution of a differential equation. You differentiate the proposed function, substitute it (and its derivative) into the equation, and check that both sides are equal as an identity.

The verification method

A worked verification

General versus particular solutions

A differential equation usually has a family of solutions involving an arbitrary constant (the general solution); fixing the constant with an initial condition gives one particular solution. Verification works for both. In the worked example, the constant CC cancelled, confirming that the whole family solves the equation. When a specific function is claimed (no free constant) and an initial condition is stated, you verify both that it satisfies the equation and that it passes through the required point. This mirrors the distinction made when solving by separation of variables in the later topics.

Why verification is a separate, scoreable skill

Verification is examined on its own because it tests whether you understand what "solution" means: a function that, together with its derivative, makes the equation an identity. It is also a useful check after solving, since substituting your answer back catches algebra errors. The marks come from showing the substitution into both sides and the simplification, not just asserting the result. The most common error is differentiating incorrectly, especially forgetting the chain rule on an exponential like exe^{-x}, which then makes the two sides fail to match even for a correct solution. Differentiate carefully, then substitute fully.

Verifying solutions of higher-order equations

Some verification problems involve a second derivative, for example checking that y=sinxy = \sin x solves d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0. The method is unchanged: differentiate as many times as the equation requires, substitute yy and all its needed derivatives, and confirm the equation holds identically. For y=sinxy = \sin x, d2ydx2=sinx\frac{d^2y}{dx^2} = -\sin x, and sinx+sinx=0-\sin x + \sin x = 0 matches the right side, so it is a solution. The discipline is simply to compute every derivative the equation mentions before substituting, and to keep the chain rule and product rule accurate through the higher derivatives, where errors are easier to make.

Using verification as a confidence check

Beyond being its own question type, verification is the most reliable way to check work after solving a differential equation. Once you have produced a candidate solution by separation of variables, differentiating it and substituting into the original equation confirms whether the algebra was correct. If the two sides do not match, you have caught an error before it costs marks elsewhere. Because differentiation is generally more dependable than the integration and algebra involved in solving, this back-substitution is a cheap and powerful safeguard. Treat it as the final step of any solve, not only as a standalone task.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (style)1 marksSection I (multiple choice, no calculator). Which function satisfies dydx=2y\frac{dy}{dx} = 2y? (A) y=2xy = 2x (B) y=e2xy = e^{2x} (C) y=x2y = x^2 (D) y=2exy = 2e^x
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The correct answer is (B), y=e2xy = e^{2x}.

If y=e2xy = e^{2x} then dydx=2e2x=2y\frac{dy}{dx} = 2e^{2x} = 2y, which matches. For (D), y=2exy = 2e^x gives dydx=2ex=y2y\frac{dy}{dx} = 2e^x = y \neq 2y.

AP 2023 (style)3 marksSection II (free response, no calculator). Show that y=3ex+2y = 3e^{-x} + 2 is a solution of the differential equation dydx=(y2)\frac{dy}{dx} = -(y - 2), and find y(0)y(0).
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A 3-point verification question.

(2 points) Differentiate: dydx=3ex\frac{dy}{dx} = -3e^{-x}. Compute the right side: (y2)=((3ex+2)2)=3ex-(y - 2) = -((3e^{-x} + 2) - 2) = -3e^{-x}. Both sides equal 3ex-3e^{-x}, so yy satisfies the equation.
(1 point) y(0)=3e0+2=3+2=5y(0) = 3e^{0} + 2 = 3 + 2 = 5.

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