Checking Solutions by Differentiation

Why Check a Proposed Solution?

A formula obtained while solving a differential equation is a candidate, not yet a verified answer. Differentiation and substitution provide a direct test: does the function have the derivative required by the equation at every point of its stated interval? If an initial condition is part of the problem, does the same function also take the required value at the specified point?

This check is useful even when the formula looks plausible. A sign error, a missing constant, or a branch of an implicit relation can change whether the formula actually satisfies the equation. Verification is a matter of checking the original equation and data, not just checking that the algebra used to obtain the formula appears reasonable.

What It Means to Be a Solution

Consider a first-order differential equation written in the form \(y'=f(x,y)\), where \(f\) is defined on a set \(D\) of points \((x,y)\). A proposed explicit solution is a function \(y:I\to\mathbb{R}\) on an interval \(I\).

An initial-value problem adds a condition such as \(y(x_0)=y_0\), where \(x_0\in I\). To solve that initial-value problem, the candidate must satisfy both the differential equation throughout \(I\) and the initial condition at \(x_0\). These are separate requirements. Checking only one is not enough.

For an explicit candidate, the basic procedure is direct:

Worked Example: Verifying an Explicit Solution

Implicit Relations and the Chain Rule

A solution may be presented by a relation \(F(x,y)=C\) rather than by an explicit formula for \(y\). If a differentiable function \(y(x)\) satisfies this relation, then the composite function \(x\mapsto F(x,y(x))\) is constant. When \(F\) has continuous first partial derivatives near the graph of \(y\), the chain rule gives a way to test the differential equation.

Here \(F_x\) and \(F_y\) denote the partial derivatives of \(F\) with respect to its first and second variables. The regularity condition on \(F\) ensures the chain rule applies to the composite. The relation alone does not automatically verify an arbitrary differential equation: the derivative obtained from the relation must still agree with the right-hand side of the equation.

Why Implicit Differentiation Gives the Derivative

The key step is to differentiate the constant composite \(F(x,y(x))\). Since its value is \(C\) for every \(x\in I\), its derivative is zero. The multivariable chain rule gives

Equating this derivative to zero yields \(F_x+F_y y'=0\), evaluated along the graph. If \(F_y\ne 0\), ordinary algebra permits division by \(F_y\), giving \(y'=-F_x/F_y\). To verify that the implicit relation describes a solution of \(y'=f(x,y)\), it is therefore sufficient, under these hypotheses and wherever \(F_y\ne0\), to check that \(-F_x/F_y=f(x,y)\) along the relation. If \(F_y=0\), this division is not valid; instead, check the derivative of the actual function or use the undivided chain-rule equation together with the original differential equation.

Worked Example: Checking an Implicit Solution

Worked Example: An Equation Is Not the Whole Initial-Value Problem

A Common Mistake: Stopping After Differentiation

A frequent error is to calculate the derivative and then assume the candidate is a solution without evaluating the right-hand side of the equation. For example, for \(y=x^2+1\) and the equation \(y'=y\), a calculation might stop at

That is only the derivative; it is not a check of the equation. Substitution gives \(y=x^2+1\) on the right, so the required equality would be \(2x=x^2+1\) for every \(x\). It fails already at \(x=0\): the left side is \(0\), while the right side is \(1\). Thus the candidate is not a solution of \(y'=y\). The correction is to compare both sides as functions on the claimed interval, rather than merely to compute one side.

Another important distinction is that verification establishes that a candidate works; it does not, by itself, show that it is the only solution or that a list of candidates is complete. Those conclusions require additional arguments. For this check, the task is narrower: confirm the proposed function, its stated interval, the original differential equation, and any initial data.

Summary of the Verification Test

For an explicit solution, differentiate the candidate and compare the result with the original right-hand side after substituting the candidate. For an implicit solution, use the chain rule under appropriate differentiability assumptions, and only divide by \(F_y\) where it is nonzero. In either form, verify the domain and check each initial condition separately.