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Verifying Solutions for
Differential Equations

Learn how to substitute a proposed function into a differential equation and confirm it satisfies the equation: the essential verification skill for AP Calculus BC.

AP Calculus BC

What Is a Differential Equation?

A differential equation (DE) is any equation that involves a function and one or more of its derivatives. Instead of solving for a number, you solve for an entire function.

Example DE
\(\dfrac{dy}{dx} = 2x\)
A Solution
\(y = x^2 + C\)

Why Verification Matters

Many DEs are hard to solve from scratch. But checking whether a given function is a solution is straightforward: just differentiate and substitute. This skill appears in two key AP settings:

Multiple Choice
"Which of the following is a solution to \(\dfrac{dy}{dx} = 3y\)?"
Free Response
"Verify that \(y = Ce^{3x}\) satisfies the given differential equation."

The Verification Process at a Glance

GIVEN DE + proposed
solution \(y = f(x)\)
 DIFFERENTIATE
Find \(y', y''\), etc.
 as needed by DE SUBSTITUTE
Plug \(y, y', y''\)
 into the DE
\(\text{LHS} = \text{RHS}\) ✓

The Verification Principle

This is not a theorem you "prove" once: it is the definition of what it means to be a solution. Every verification problem reduces to this single check.

General Solution
Contains an arbitrary constant \(C\). Represents a family of curves.
Particular Solution
Specific value of \(C\) chosen by an initial condition like \(y(0) = 5\).

General vs. Particular Solutions

x y (0, 3) General: \(y = x^2 + C\) Particular: \(y = x^2 + 3\) (with \(y(0) = 3\))

Solving vs. Verifying: Know the Difference

Solving a DE Verifying a Solution
Goal Find \(y = f(x)\) from scratch Confirm a given \(f(x)\) works
Techniques Separation, integrating factors, etc. Differentiation + substitution
Difficulty Often hard or impossible analytically Always mechanical and doable
On AP Exam Free response (multi-step) MCQ or FR sub-part (quick)

How to Verify: Step by Step

1
Identify what the DE needs. Read the equation to determine which derivatives (\(y'\), \(y''\), ...) are required.
2
Differentiate the proposed solution. Compute \(y'\), \(y''\), etc. using standard rules (chain, product, quotient).
3
Substitute everything into the DE. Replace every \(y\), \(y'\), \(y''\) with the expressions you computed.
4
Simplify both sides. Use algebra to reduce. If \(\text{LHS} = \text{RHS}\) identically, the solution is verified ✓.
5
Check initial conditions (if given). Plug in the initial \(x\)-value and confirm \(y\) matches.

Example 1: First-Order Verification

1
Differentiate: \(y' = 5 \cdot 3e^{3x} = 15e^{3x}\)
2
Compute RHS: \(3y = 3 \cdot 5e^{3x} = 15e^{3x}\)
3
Compare: LHS = \(y' = \) \(15e^{3x}\), RHS = \(3y = \) \(15e^{3x}\)

Example 2: Second-Order Verification

1
First derivative: \(y' = 3 \cdot 2\cos(2x) = 6\cos(2x)\)
2
Second derivative: \(y'' = 6 \cdot (-2)\sin(2x) = -12\sin(2x)\)
3
Substitute into DE: \(y'' + 4y = -12\sin(2x) + 4 \cdot 3\sin(2x)\)
4
Simplify: \(= -12\sin(2x) + 12\sin(2x) = 0\)   ✓

Example 3: Product Rule + Initial Condition

1
Differentiate (product rule): \(y' = xe^x + \dfrac{x^2}{2}e^x = e^x\!\left(x + \dfrac{x^2}{2}\right)\)
2
Compute \(y' - y\): \(e^x\!\left(x + \dfrac{x^2}{2}\right) - \dfrac{x^2}{2}e^x = xe^x\)   ✓
3
Check IC: \(y(0) = \dfrac{0^2}{2}e^0 = 0\)   ✓

When Verification Fails

1
Differentiate: \(y' = 3e^{3x}\)
2
Compute RHS: \(2y = 2e^{3x}\)
3
Compare: \(3e^{3x} \neq 2e^{3x}\)   \(\text{LHS} \neq \text{RHS}\)

Verifying Implicit Solutions

Sometimes a solution is given in implicit form, e.g. \(x^2 + y^2 = 25\). You cannot isolate \(y\) easily. Instead, use implicit differentiation to find \(\dfrac{dy}{dx}\), then substitute.

1
Differentiate implicitly: \(2x + 2y\dfrac{dy}{dx} = 0\), so \(\dfrac{dy}{dx} = -\dfrac{x}{y}\)
2
Compare: This matches the given DE exactly. ✓ Verified.

Solutions with Arbitrary Constants

When verifying a general solution like \(y = Ce^{kx}\), the constant \(C\) should cancel out during substitution, confirming the solution works for every value of \(C\).

1
\(y' = 3C_1 e^{3x} - 3C_2 e^{-3x}\)
2
\(y'' = 9C_1 e^{3x} + 9C_2 e^{-3x}\)
3
\(y'' - 9y = 9C_1 e^{3x} + 9C_2 e^{-3x} - 9(C_1 e^{3x} + C_2 e^{-3x}) = 0\)   ✓

Common Mistakes to Avoid

✕ Forgetting the Chain Rule
When differentiating \(e^{3x}\), many students write \(e^{3x}\) instead of \(3e^{3x}\). Always multiply by the inner derivative.
✕ Substituting Only Partially
You must replace every occurrence of \(y\), \(y'\), and \(y''\) in the DE: not just some of them.
✕ Ignoring the Initial Condition
A function can satisfy the DE but fail the initial condition. Always check both if an IC is given.

Summary: Verification Checklist

Step Action Watch For
1 Identify required derivatives Order of the DE (1st, 2nd, ...)
2 Differentiate \(y = f(x)\) Chain rule, product rule errors
3 Substitute into DE Replace all \(y, y', y''\) terms
4 Simplify both sides Factor, combine like terms
5 Check initial condition Only if IC is given in problem

Question 1 of 5

Which function is a solution to the differential equation \(\dfrac{dy}{dx} = -2y\)?

Question 2 of 5

To verify that \(y = \sin(x)\) satisfies \(y'' + y = 0\), what should you compute first?

Question 3 of 5

Given \(\dfrac{dy}{dx} = \dfrac{y}{x}\) and the proposed solution \(y = 5x\), what is the result of substituting into the DE?

Question 4 of 5

A student claims \(y = e^{x^2}\) solves \(y' = 2xy\). Is this correct?

Question 5 of 5

\(y = Ce^{-t}\) is proposed as the general solution to \(\dfrac{dy}{dt} + y = 0\). Which statement is true?