From a Solution Family to a Specific Model
The differential equation \( \frac{dP}{dt}=kP \) describes a quantity whose rate of change is proportional to its current value. For a fixed real number \(k\), its exponential solution family is \(P(t)=Ce^{kt}\), where \(C\) is a constant. The differential equation alone does not determine one particular solution: different choices of \(C\) give different solutions.
An initial condition supplies the missing information. It gives the value of the quantity at a specified reference time, such as \(P(t_0)=P_0\). Applying that condition determines \(C\), so that the model describes the particular solution passing through the specified value.
Determining the Constant
Begin with the solution family \(P(t)=Ce^{kt}\). To satisfy the initial condition, the solution must have value \(P_0\) when \(t=t_0\). Substituting \(t_0\) for \(t\) gives \(P_0=Ce^{kt_0}\). Since the exponential is never zero, we can divide by \(e^{kt_0}\) and obtain the unique value of \(C\).
Substituting this constant back into the solution family gives an expression that makes the reference time especially visible. The exponent depends on the difference \(t-t_0\), not on \(t\) alone.
The expression \(t-t_0\) is the elapsed time measured from the reference time. At \(t=t_0\), elapsed time is zero, so the exponential factor equals one and the model returns \(P_0\). This form also works when \(t\) is earlier than \(t_0\): then \(t-t_0\) is negative. No separate constant needs to be found for times before the reference time.
Why the Reference-Time Formula Works
The formula can be checked directly. Differentiating \(P_0e^{k(t-t_0)}\) with respect to \(t\) multiplies the exponential by the derivative of its exponent, which is \(k\). The result is therefore \(k\) times the original function. Substituting \(t=t_0\) makes the exponent zero, and \(e^0=1\). These two checks establish that the formula solves the differential equation and meets the prescribed initial condition.
For a fixed \(k\), use \(P(t)=Ce^{kt}\).
Set \(t=t_0\) and \(P(t_0)=P_0\), giving \(P_0=Ce^{kt_0}\).
Divide by the nonzero quantity \(e^{kt_0}\), so \(C=P_0e^{-kt_0}\).
Substitute the constant into the family, then verify the differential equation and the initial value.
This is also a short uniqueness argument within the exponential family. Any solution in that family that satisfies the condition must have \(C e^{kt_0}=P_0\). Because \(e^{kt_0}\ne 0\), this equation forces \(C=P_0e^{-kt_0}\). There is no second choice of \(C\) that meets the same condition.
Worked Examples
Worked Example: Applying an Initial Value at a Positive Reference Time
Solve \( \frac{dP}{dt}=\frac12P \) subject to \(P(2)=8\). Here \(k=\frac12\), \(t_0=2\), and \(P_0=8\). Use the reference-time formula:
Equivalently, the absolute-time family is \(P(t)=Ce^{t/2}\). Substitution of the initial condition gives
Thus \(P(t)=8e^{-1}e^{t/2}=8e^{(t-2)/2}\), the same model as above. To check the differential equation, differentiate:
To check the initial condition, substitute \(t=2\):
Both required checks hold.
Worked Example: Evaluating Before the Reference Time
Solve \( \frac{dP}{dt}=-\frac13P \) subject to \(P(6)=81\), and find the value at \(t=3\). The reference time is \(t_0=6\), while the requested time \(3\) is earlier than that reference time. The model is
At \(t=3\), the time difference is \(3-6=-3\). Substitution gives
Check the differential equation by differentiating the model:
Check the initial value:
The same formula applies on either side of \(t_0\); the sign of \(t-t_0\) is handled by ordinary substitution.
Worked Example: Using a Negative Reference Time
Solve \( \frac{dP}{dt}=2P \) subject to \(P(-2)=5\). Here \(t_0=-2\), so \(t-t_0=t+2\). The reference-time form is
To find the corresponding constant in the absolute-time form, write \(P(t)=Ce^{2t}\) and substitute \(t=-2\):
Therefore the absolute-time form is \(P(t)=5e^4e^{2t}\). The exponent law gives \(5e^4e^{2t}=5e^{2(t+2)}\), confirming that the two forms agree. Differentiate the reference-time form to check the equation:
Finally, substitute the reference time:
The model satisfies the equation and the given value.
Special Cases
The formula \(P(t)=P_0e^{k(t-t_0)}\) includes important boundary cases without requiring a different method. If \(P_0=0\), then
This constant zero function satisfies \(P'(t)=kP(t)\) and \(P(t_0)=0\). If \(k=0\), the formula becomes \(P(t)=P_0e^0=P_0\), a constant function. Its derivative is zero, which agrees with \(kP(t)=0\). Thus neither a zero initial value nor a zero rate parameter makes the formula invalid.
In an application where \(P\) represents a physical amount, \(P_0\) is often nonnegative. The mathematical formula itself is valid for any real \(P_0\); the interpretation of a negative amount would depend on the application.
A Common Mistake: Setting the Constant Equal to the Initial Value
A frequent error is to see the absolute-time form \(P(t)=Ce^{kt}\) and set \(C=P_0\) automatically. That works only when the reference time is \(t_0=0\), because \(P(0)=Ce^0=C\). For a different reference time, \(C\) and \(P_0\) need not be equal.
For example, suppose \( \frac{dP}{dt}=\frac12P \) and \(P(4)=10\). The incorrect choice \(C=10\) would give \(P(t)=10e^{t/2}\). But at the stated reference time this predicts
which is not \(10\). The error is visible because the proposed function fails the initial condition. Instead, solve for \(C\):
The corrected model is \(P(t)=10e^{-2}e^{t/2}=10e^{(t-4)/2}\). Its derivative is
and its value at the reference time is
Checking both parts of the initial-value problem catches this mistake: a proposed model must satisfy the differential equation and the specified value, not merely have an exponential form.
Why the Reference-Time Form Is Useful
The absolute-time form \(Ce^{kt}\) is convenient when the constant \(C\) is already known. In an initial-value problem, however, the given information is often \(P(t_0)=P_0\), with \(t_0\) not equal to zero. The form \(P_0e^{k(t-t_0)}\) places that information directly into the model: the starting value is displayed as the factor \(P_0\), and the elapsed time is displayed in the exponent.
That arrangement also makes evaluation straightforward. To find \(P\) at a requested time \(t\), subtract the reference time \(t_0\), multiply that time difference by \(k\), and evaluate the exponential. Keep the parentheses around \(t-t_0\), especially when \(t_0\) is negative or \(t\) lies before the reference time. The model does not require \(t\) to be later than \(t_0\).
Check Your Understanding
For each question, identify the reference time and initial value before applying the formula.
- For \( \frac{dP}{dt}=3P \) with \(P(1)=4\), write the model relative to the reference time and check its initial value.
- For \( \frac{dP}{dt}=-2P \) with \(P(5)=7\), find \(P(3)\) using the reference-time form.
- In the family \(P(t)=Ce^{kt}\), why does \(C=P_0\) follow directly when \(t_0=0\), but not necessarily for another reference time?
- Show that \(P(t)=P_0e^{k(t-t_0)}\) satisfies \(P'(t)=kP(t)\).
- What model results when \(P_0=0\), and why does it satisfy both parts of the initial-value problem?
