Why Verification Matters
A formula can look familiar, satisfy an initial condition, or resemble a solution obtained by algebraic manipulation. None of those facts alone establishes that it solves the differential equation. Verification is a direct test: differentiate the proposed function, substitute it into the original equation, and check that the required equality holds throughout the interval in question.
For an initial-value problem, there are two separate requirements. The function must satisfy the differential equation, and it must satisfy the specified initial condition. A formula that passes one test may still fail the other.
The phrase “for every \(x\) in an interval” is important. Matching the equation at one input value does not verify a solution. The proposed formula must make the differential equation true at every point of the interval where it is claimed to be a solution.
A Reliable Verification Process
Write the differential equation in its original form before beginning the check. If the proposal is explicit, \(y=g(x)\), calculate \(g'(x)\), substitute \(g(x)\) into every occurrence of \(y\), and compare the two sides. Then test any initial condition by direct substitution into \(g\).
Note the original differential equation, the interval on which the proposal is claimed to work, and any initial condition.
Find \(y'\) from the proposed formula, using the appropriate derivative rules.
Replace \(y\) and \(y'\) with the proposed expressions and check whether both sides are equal throughout the interval.
Evaluate the proposed function at the stated initial input and compare its value with the required initial value.
This process tests the claim rather than how the formula was obtained. It does not require solving the differential equation again.
Worked Example: Verifying an Explicit Solution and Initial Condition
Worked Example: Verifying an Explicit Solution and Initial Condition
Test whether \(y=1+4e^{2x}\) solves \(y'=2(y-1)\) with the initial condition \(y(0)=5\).
Step 1: Differentiate the proposed function.
Step 2: Substitute the proposed \(y\) into the right-hand side.
The left-hand side \(y'\) and the right-hand side \(2(y-1)\) both equal \(8e^{2x}\). Thus the differential equation holds for every real \(x\).
Step 3: Check the initial condition.
This matches the required value. Therefore the proposed function solves the differential equation and the initial-value problem on \(\mathbb{R}\).
Check: Substituting the proposed function into the original equation gave identical expressions, and direct evaluation at \(x=0\) gave \(5\).
Why Substitution Is a Proof of the Claim
The definition of a solution is an equality involving the derivative and the function at each input in an interval. For an explicit proposal \(y=g(x)\), the derivative test produces \(g'(x)\), while substitution produces \(f(x,g(x))\). If these expressions are equal for every \(x\) in the interval, then the defining condition for a solution is satisfied. If they differ at even one point in the claimed interval, it is not a solution there.
This statement follows directly from the definition. The first equality is precisely the requirement that \(g\) satisfy the differential equation. The second equality is precisely the requirement that it take the prescribed initial value. Both are necessary, and together they are sufficient.
Checking an Implicit Solution
A proposed solution need not be written as \(y=g(x)\). It may instead be given by a relation involving \(x\) and \(y\). To check such a proposal, differentiate the relation with respect to \(x\), treating \(y\) as a function of \(x\). Then use the relation and the resulting derivative to test the original equation.
For a relation \(F(x,y)=0\), implicit differentiation gives \(F_x(x,y)+F_y(x,y)y'=0\), where \(F_x\) and \(F_y\) are the partial derivatives. If \(F_y\ne 0\), this yields \(y'=-F_x/F_y\). The condition \(F_y\ne 0\) matters: dividing by \(F_y\) is valid only where it is nonzero. Also, a relation must actually define a differentiable branch \(y\) on the interval being considered for it to be a solution function.
Worked Example: Verifying an Implicit Solution
Worked Example: Verifying an Implicit Solution
Check whether the relation \(x^2+y^2=25\), on the branch through \((3,4)\), solves \(y'=-x/y\) with \(y(3)=4\).
Step 1: Identify the relevant branch and its domain near the initial point. The relation gives \(y^2=25-x^2\). The branch through \((3,4)\) has positive \(y\), so locally it is \(y=\sqrt{25-x^2}\). Near \(x=3\), \(25-x^2>0\) and \(y\ne 0\), so this branch is defined and differentiable there.
Step 2: Differentiate the implicit relation.
Solving for \(y'\), using \(y\ne0\), gives
This is exactly the original differential equation on the chosen branch near \(x=3\).
Step 3: Check the initial condition.
The point \((3,4)\) lies on the proposed relation, and the selected branch has \(y(3)=\sqrt{25-9}=4\). Therefore the relation, interpreted as this local branch, solves the initial-value problem near \(x=3\).
Check: Differentiation gives \(y'=-x/y\), and the branch takes the required value \(4\) at \(x=3\).
The Original Equation and the Claimed Interval Matter
A verification must use the original differential equation, not merely an equation produced during an earlier manipulation. An algebraic step may have involved division by an expression that can be zero, or some other operation that changes which cases are represented. Substitution into the original equation is the decisive test of the proposed function.
The interval also matters. A formula may solve an equation where it is differentiable and where all expressions in the equation are defined, but fail to qualify on a larger interval containing a point where a denominator is zero. For an implicit relation, a curve may have points where it cannot be represented as a differentiable function \(y\) of \(x\). State the interval or branch being checked, and do not use a formula outside its domain.
Worked Example: An Initial Condition Is Not Enough
Worked Example: An Initial Condition Is Not Enough
A proposed solution for the initial-value problem \(y'=2xy\), \(y(0)=1\), is \(y=x^2+1\). Determine whether the proposal works.
Step 1: Check the initial condition.
The initial condition is satisfied.
Step 2: Differentiate the proposal.
Step 3: Substitute into the right-hand side of the original equation.
The equation would require \(2x=2x^3+2x\) for every \(x\) in the interval. This is false. For instance, at \(x=1\), the left-hand side \(y'\) is \(2\), while the right-hand side \(2xy\) is \(2(1)(2)=4\).
Conclusion and check: Although the proposal satisfies \(y(0)=1\), it does not satisfy the differential equation. It is therefore not a solution of the initial-value problem.
Common Pitfall: Checking at Just One Point
At the initial point, the differential equation describes the required slope of a solution, but matching that slope at the initial point does not establish that the formula has the right slope everywhere. For example, the proposal \(y=x^2+1\) in the previous example has \(y(0)=1\), and at \(x=0\) both sides of \(y'=2xy\) equal \(0\). That single agreement does not make the differential equation true on an interval: at \(x=1\), the sides are \(2\) and \(4\).
The correction is to compare the two sides as expressions in \(x\), and establish equality throughout the claimed interval. A convenient way to disprove a proposed solution is to find one point in the interval where the sides differ. But to verify a solution, one must establish the identity across the whole interval.
Check Your Understanding
For each question, show enough work to distinguish checking the equation from checking an initial condition.
- What two requirements must a proposed function meet to solve an initial-value problem?
- For the proposed function \(y=3e^x\), what derivative must be compared with the right-hand side when checking \(y'=y\)?
- Why must an implicit relation be interpreted as a differentiable branch before it can represent a solution function?
- When differentiating \(F(x,y)=0\), what expression results from the chain rule, and what condition is needed to solve it for \(y'\) by division?
- Explain why agreement between the two sides of a differential equation at the initial point alone does not verify a solution on an interval.
