Deriving the Exponential Solution of dP/dt = kP

From Constant Relative Rate to an Exponential Formula

A differential equation can describe how a quantity changes without giving its value at every time. For a quantity \(P(t)\) whose rate of change is proportional to its current value, the equation is

where \(k\) is a constant. The equation says that the rate \(dP/dt\) is \(k\) times the amount \(P\). The goal here is to derive the formula for solutions, rather than guess it.

The key move is to separate the \(P\)-terms from the \(t\)-terms and integrate. This requires care: dividing by \(P\) is valid only where \(P\ne 0\). In particular, the logarithmic step below applies directly on an interval where \(P(t)>0\), as is appropriate for a positive population or amount.

Separate and Integrate

Suppose \(P(t)>0\) throughout an interval \(I\). Divide both sides of the differential equation by \(P(t)\), then multiply by \(dt\). In differential notation, this gives

Integrating both sides gives a logarithm on the left. The reason is that the derivative of \(\ln P\) with respect to \(P\) is \(1/P\). Since \(P\) depends on \(t\), the chain rule gives \(d(\ln P(t))/dt=P'(t)/P(t)\). Thus the separated equation integrates to

where \(A\) is an arbitrary real constant of integration. Exponentiate both sides. Using \(e^{u+v}=e^u e^v\),

Set \(C=e^A\). Because the exponential of any real number is positive, \(C>0\). Therefore every solution obtained on a positive interval has the form

The integration constant has not disappeared. Its form has changed: integration produces an additive constant \(A\) in the logarithmic equation, and exponentiation turns it into a positive multiplicative constant \(C=e^A\).

Why the Derivation Works

The separation-and-integration steps can be justified directly for any differentiable positive solution. Since \(P(t)>0\), the function \(\ln P(t)\) is defined. Applying the chain rule and then using the differential equation gives

A function whose derivative is the constant \(k\) differs from \(kt\) by a constant. Hence \(\ln P(t)=kt+A\) on the interval, and exponentiating yields \(P(t)=Ce^{kt}\) with \(C>0\). This shows why the formula follows from the original equation, rather than merely matching a few computed values.

The argument applies on an interval where \(P\) stays positive. It does not permit division by \(P\) at a point where \(P=0\). That case must be considered separately: the constant function \(P(t)=0\) satisfies the original equation, since both sides are zero. It is not lost from the differential equation; it is only excluded from the division step.

Worked Derivations

What the Arbitrary Constant Means

The phrase “arbitrary constant” can be misleading if its role is not made explicit. After integration, \(A\) may be any real number. After exponentiation, the resulting constant \(C=e^A\) may be any positive real number, but not zero or negative. Every \(C>0\) is possible because \(A=\ln C\) is real for every positive \(C\).

For a positive solution, the value of \(C\) controls which member of the family is obtained. The differential equation alone specifies the relationship between the amount and its rate of change; it does not specify one particular amount at one particular time. A further condition would be needed to select one constant from the family. The family itself is the result of the separation and integration performed here.

The zero solution is not represented by \(Ce^{kt}\) when \(C>0\), but it can be included in the formula by allowing \(C=0\). This does not mean it was derived by dividing by \(P\). It is included separately because direct substitution into the original equation confirms that \(P(t)=0\) satisfies it.

A Common Mistake: Dividing by Zero

A tempting but invalid argument is to start with \(P'=kP\), divide by \(P\), and claim that this operation is valid for every solution. For the zero solution, that produces an undefined expression. For example, if \(P=0\), then \(P'/P=0/0\), which is not defined. The separation step therefore cannot establish a formula for that solution.

The correction is to state the condition before dividing: work on an interval where \(P\ne0\), and for the positive-amount model, where \(P>0\). Derive the exponential formula on that interval. Then check the excluded case directly in the original equation: if \(P=0\) for every \(t\), then \(P'=0=kP\). Treating this case separately avoids silently discarding a solution or using an undefined operation.

Another error is to write \(e^{kt+A}=e^{kt}+e^A\). Exponentiation does not turn a sum in the exponent into a sum. The correct identity is

That product is why the integration constant becomes the multiplier \(C\).

Why This Derivation Matters

The derivation explains the link between proportional change and exponential form. When the rate is proportional to the current amount, dividing the rate by the amount produces the constant \(k\). Integrating that relative rate leads to a logarithm, and undoing the logarithm leads to an exponential.

It also illustrates a general habit in differential equations: before separating variables, check that division is legitimate; after integrating, account for the constant; and distinguish solutions covered by the division from cases that must be checked in the original equation. These checks make the reasoning complete without changing its central simplicity.