Verifying Growth and Decay Formulas by Differentiation

What Verification Requires

A proposed formula may look like a solution to a differential equation, but its appearance is not enough. To verify it, check directly that its derivative satisfies the equation and that it has the required initial value.

For the exponential model, the differential equation is

where \(k\) is a fixed real constant. The formula to be checked is

where \(P_0\) is a fixed real number. The initial condition considered here is \(P(0)=P_0\). The value \(P_0\) is the amount at the initial time \(t=0\); \(k\) is the rate constant.

Verification therefore involves two separate checks. The differential equation concerns the relationship between the function and its derivative at every time in the interval. The initial condition concerns the value of the function at one specified time. Passing one check does not automatically mean the other has passed.

The Two Checks

To check the differential equation, differentiate the candidate formula with respect to \(t\). Since \(P_0\) and \(k\) are constants, the chain rule gives

The right-hand side of the differential equation, evaluated at the candidate, is

The derivative and the right-hand side agree. For the initial condition, substitute \(t=0\) into the formula and use \(e^0=1\):

Both checks work for every real \(k\) and \(P_0\). The formula is differentiable for all real \(t\), so it satisfies the equation on any interval of real times, including an interval containing \(0\).

Why the Check Works

The verification follows from the chain rule, not just from the formula’s familiar shape. The inner function in the exponential is \(kt\), whose derivative is \(k\). Thus differentiating the exponential produces the factor \(k\), while leaving the original exponential factor in place. Multiplication by the constant \(P_0\) then gives exactly \(kP_0e^{kt}\), which is \(kP(t)\).

This argument verifies the proposed expression directly. It does not, by itself, establish that no other formula can satisfy the same initial-value problem. A uniqueness result is needed for that further conclusion; the earlier material on local existence and uniqueness provides the relevant context.

Worked Examples

What the Examples Show

The positive-rate and negative-rate examples use the same verification procedure. The sign changes the derivative factor produced by the chain rule, but the equation check still compares that derivative with \(kP(t)\). The zero-rate example also follows the same procedure: both sides of the differential equation equal zero.

For each candidate, the initial-value check depends on the exponential being evaluated at \(t=0\). Since \(e^{k\cdot 0}=1\), the initial value is the coefficient \(P_0\). This check confirms the stated initial condition; it does not determine an unknown coefficient from different initial data.

A Common Mistake: Forgetting the Inner Derivative

A frequent error is to differentiate \(e^{kt}\) as though its derivative were just \(e^{kt}\). For example, an incorrect calculation for \(P(t)=120e^{t/5}\) would be

That expression omits the derivative of the inner function \(t/5\). The chain rule requires multiplying by \(1/5\):

The incorrect derivative would not match the model’s right-hand side, which is \(\frac15P(t)=24e^{t/5}\). Keeping the inner derivative in the calculation is essential, especially when \(k\) is negative or is not equal to \(1\).

Verification Is More Than a Spot Check

Checking the equation at a single time is useful for finding arithmetic errors, but a single matching value does not verify the differential equation on an interval. The equation requires equality at every time in the interval. For the exponential candidate, the general derivative calculation proves that equality for every \(t\):

Likewise, checking only the initial value is not enough. A formula can take the correct value at \(t=0\) and still have a derivative that fails to satisfy the equation. A complete verification records both checks explicitly.

These checks establish that the candidate is a solution of the stated equation with the stated initial value. A conclusion that it is the unique solution requires an applicable uniqueness result, rather than only substitution and differentiation.