How to Show That a Function Solves a Differential Equation

To check whether a function solves a differential equation, substitute the function and its derivatives into the equation. The result must be an identity on the interval where the solution is claimed—not merely true at one point. If the problem also gives initial or boundary conditions, the function must satisfy those conditions too.

What it means to be a solution

A differential equation relates an unknown function to one or more of its derivatives. For example,

$$y''+y=0$$

is a second-order differential equation. A proposed function $y=f(x)$ is a solution on an interval if it has the required derivatives there and makes the equation true at every point of that interval.

The interval matters. A function may satisfy the equation wherever its derivatives exist but fail to be defined at some points. For instance, $1/x$ cannot be a solution on an interval containing $x=0$, even if substitution works elsewhere.

A differential equation may have many solutions. An initial condition, such as $y(0)=1$, selects functions that also pass through a specified value. A function that satisfies the differential equation but not the initial condition is not a solution to the full initial-value problem.

A step-by-step check

1. Read the equation and identify the required derivatives. For an equation involving $y''$, the candidate must be twice differentiable on the interval under consideration.
2. Differentiate the candidate. Compute only the derivatives needed, taking care with signs and constants.
3. Substitute into the equation. Replace each occurrence of the unknown function and its derivatives with the corresponding expressions.
4. Simplify. If both sides agree for every point in the interval, the differential equation is satisfied.
5. Test any extra conditions. Evaluate the function or its derivatives at the stated endpoints or initial point.

When the equation is written in a form such as $F(x,y,y',y'')=0$, substitution should reduce the left-hand side to zero. If it is written as an equality, substitute into both sides and compare them.

Worked example: a polynomial solution

Check whether $y=x^2$ solves the differential equation

$$y''=2,$$

with initial conditions $y(0)=0$ and $y'(0)=0$.

First compute the derivatives:

$$y'=2x, \qquad y''=2.$$

Substitute the second derivative into the differential equation. Its left-hand side is $2$, which equals its right-hand side, so the equation holds for every real $x$.

Now check the initial conditions:

$$y(0)=0^2=0, \qquad y'(0)=2(0)=0.$$

Both conditions hold. Therefore, $y=x^2$ is a solution of this initial-value problem. The interval is all real numbers because the polynomial and its derivatives are defined everywhere. The equation alone has other solutions, too: integrating $y''=2$ twice gives $y=x^2+C_1x+C_2$. The two initial conditions determine $C_1=C_2=0$, producing the particular solution checked above.

How to check a first-order equation

Suppose the proposed function is $y=e^{2x}$ and the equation is $y'=2y$. Differentiating gives $y'=2e^{2x}$. Substituting the candidate into the right-hand side also gives $2y=2e^{2x}$. The two sides match for all real $x$, so the function solves the equation on $\mathbb{R}$.

This same process works for nonlinear equations. For example, if the equation contains a product such as $yy'$, substitute the candidate into both factors, then simplify the product. There is no requirement that a solution be a polynomial or that the equation be linear.

Common mistakes

• Checking only one value of the independent variable. An equation that holds at $x=0$ may fail at other points. A solution must satisfy it throughout the claimed interval.
• Forgetting a derivative or using the wrong one. In a second-order equation, substituting $y'$ where $y''$ belongs will not verify the equation.
• Ignoring the domain. Check that the candidate and every derivative in the equation exist on the full interval being claimed.
• Skipping initial or boundary conditions. Satisfying the differential equation is not enough when the problem specifies additional conditions.
• Confusing verification with solving. Verification checks one proposed function. Finding the general solution requires determining all functions that satisfy the equation, which is a separate task.

If substitution does not produce an identity, the proposed function is not a solution on the claimed interval. Recheck your differentiation and algebra before deciding whether the candidate or the claimed interval is at fault.