Verify That Is a Solution to the Differential Equation: Step-by-Step Guide

To verify that a given function is a solution to a differential equation, substitute the function and its derivatives into the equation. If the left-hand side equals the right-hand side for all values in the domain, the function is a solution. This method does not require solving the equation from scratch—only correct differentiation, substitution, and simplification.

Step 1: Identify the Function and Equation

Begin by clearly identifying two components: the proposed solution (a function of the independent variable, usually $ y(x) $) and the differential equation it may satisfy. The equation will involve the function and one or more of its derivatives.

For example, if given $ y = e^{3x} $ and asked to verify it solves $ y’ = 3y $, you must compute $ y’ $ and check whether it equals $ 3y $. Write both the function and the equation down explicitly. Note the order of the differential equation—this tells you how many derivatives to compute. A second-order equation like $ y’’ + 4y = 0 $ requires computing up to the second derivative.

Ensure the function is defined and differentiable on the interval where the equation is considered. A function such as $ y = \ln(x) $ is only valid for $ x > 0 $, so it cannot be a solution on any interval containing zero or negative numbers. Similarly, $ y = |x| $ is not differentiable at $ x = 0 $, so it fails as a solution for equations requiring $ y’ $ at that point unless the domain excludes it.

Step 2: Compute Required Derivatives

Differentiate the function as many times as needed based on the order of the differential equation. Use standard rules: power rule, chain rule, product rule, and known derivatives of exponential, trigonometric, or logarithmic functions.

Work carefully and avoid skipping steps. For instance, if $ y = \cos(2x) $, then:

• $ y’ = -2\sin(2x) $
• $ y’’ = -4\cos(2x) $

Label each derivative clearly. Errors in differentiation are the most common source of incorrect verification. Even a small sign mistake can lead to a false conclusion.

Step 3: Substitute and Simplify

Replace every occurrence of $ y $, $ y’ $, $ y’’ $, etc., in the differential equation with the expressions you computed. Then simplify the resulting expression algebraically.

If the simplified result is an identity—such as $ 0 = 0 $ or $ 6x = 6x $—the function satisfies the equation. If not, it does not.

For example, suppose you are verifying that $ y = \sin(x) $ solves $ y’’ + y = 0 $:

1. Compute $ y’ = \cos(x) $
2. Compute $ y’’ = -\sin(x) $
3. Substitute into the left-hand side:
   $ y’’ + y = -\sin(x) + \sin(x) = 0 $
4. The right-hand side is $ 0 $, so both sides match.

Since the equation holds for all $ x $, $ y = \sin(x) $ is a solution.

This process works the same way regardless of whether the equation is linear or nonlinear, first-order or higher-order. Superposition applies in linear cases, but verification depends only on substitution, not on broader solution structure.

Avoid common errors:

• Do not forget to apply the chain rule when needed.
• Do not substitute only part of the expression.
• Do not confuse initial conditions with the verification process—those select specific solutions from a family but do not determine whether a function satisfies the equation.

Double-check signs and coefficients, especially with trigonometric or exponential functions. A miscalculated derivative invalidates the entire check.

Example: Verifying an Exponential Function

Verify that $ y = 2e^{-4x} $ is a solution to the differential equation $ y’ + 4y = 0 $.

Step 1: Identify the function and equation.
We have $ y = 2e^{-4x} $ and $ y’ + 4y = 0 $. This is a first-order equation, so we need only $ y’ $.

Step 2: Compute the derivative.
Using the chain rule:
$ y’ = 2 \cdot (-4)e^{-4x} = -8e^{-4x} $

Step 3: Substitute into the equation.
Left-hand side:
$ y’ + 4y = (-8e^{-4x}) + 4(2e^{-4x}) = -8e^{-4x} + 8e^{-4x} = 0 $

Right-hand side: $ 0 $

The two sides are equal. Therefore, $ y = 2e^{-4x} $ satisfies the differential equation for all real $ x $.

Note that the constant factor $ 2 $ remains through differentiation and cancels out during simplification. This shows that even specific members of a solution family can be verified directly.

The same method applies to higher-order equations. For instance, to verify $ y = e^{x} + e^{-x} $ satisfies $ y’’ - y = 0 $:

• $ y’ = e^{x} - e^{-x} $
• $ y’’ = e^{x} + e^{-x} $
• $ y’’ - y = (e^{x} + e^{-x}) - (e^{x} + e^{-x}) = 0 $

Again, the result is identically zero. The function is a solution.

Domain considerations remain important. In this case, $ e^{x} + e^{-x} $ is defined and infinitely differentiable for all real $ x $, so no restrictions apply.

Resources like Paul’s Online Math Notes and Khan Academy provide additional practice problems and step-by-step explanations for various types of differential equations.

Key Takeaways

• To verify a solution, substitute the function and its derivatives into the differential equation and check if both sides are equal for all values in the domain.
• Accurately compute all necessary derivatives using standard calculus rules—errors here will invalidate the verification.
• Confirm the function is defined and differentiable over the interval of interest; otherwise, it cannot be a valid solution.
• After substitution, simplify completely. The result must be an identity (like $ 0 = 0 $) for the function to be a solution.