When the Proportional Rate Changes
A proportional-rate model says that the rate of change of an amount is proportional to the amount itself. In the simplest version, the proportionality factor is constant. In many situations, however, the factor changes with time. A population might grow faster during some seasons than others, or a quantity might experience growth during one interval and decline during another.
When the changing factor is known as a function of time, the differential equation has the form \(y'=r(t)y\). The function \(r(t)\) is the time-varying proportional rate. Instead of multiplying elapsed time by one constant rate, we accumulate the changing rate by integrating it.
The variable \(s\) in the integral is a dummy variable: it indicates that the rate is being accumulated from the starting time \(t_0\) to the time \(t\). The integral is dimensionless when \(r\) has units of inverse time. The exponential factor is always positive, so a solution starting positive remains positive, and a solution starting negative remains negative.
Why Integrating the Rate Works
First suppose that \(y(t)\ne 0\). Dividing the equation by \(y\) gives \(y'/y=r(t)\). Integrating from \(t_0\) to \(t\) accumulates the proportional rate over the interval. The resulting logarithm of the ratio of the amounts is the accumulated rate. Exponentiating gives the multiplicative change factor.
There is a small but important issue with dividing by \(y\): that step is not allowed for a solution that is zero. A direct argument handles zero and nonzero initial values together and confirms that the formula gives every solution.
Let \(R(t)=\int_{t_0}^{t}r(s)\,ds\). By the Fundamental Theorem of Calculus, \(R'(t)=r(t)\).
For any solution \(y\), apply the product rule to \(y(t)e^{-R(t)}\).
The derivative is \(y'e^{-R}-yR'e^{-R}=e^{-R}(y'-r(t)y)=0\).
A function with derivative zero is constant, so \(y(t)e^{-R(t)}=y(t_0)e^{-R(t_0)}=y_0\), since \(R(t_0)=0\).
Therefore \(y(t)=y_0e^{R(t)}\), which is the stated formula. This argument does not divide by \(y\), so it also proves that \(y(t)=0\) for all \(t\) when \(y_0=0\). It also shows why the initial condition must be applied at its actual starting time: the accumulated rate is measured from \(t_0\), not automatically from zero.
Interpreting the Accumulated Rate
The integral \(\int_{t_0}^{t}r(s)\,ds\) is the signed area under the graph of \(r\) between \(t_0\) and \(t\). Positive portions add to the accumulated rate; negative portions subtract from it. The final amount depends on this accumulated value, not only on the rate at the ending time.
For any two times \(t_1\) and \(t_2\) in the interval, the solution satisfies \(y(t_2)=y(t_1)\exp\left(\int_{t_1}^{t_2}r(s)\,ds\right)\). Thus a positive accumulated rate multiplies the amount by a factor greater than one, while a negative accumulated rate multiplies it by a factor between zero and one. This describes growth or decay relative to the starting amount when that amount is positive.
Instantaneous direction requires care. Since \(y'=r(t)y\), the sign of \(y'\) depends on both factors. If \(y(t)>0\), positive \(r(t)\) means the amount is increasing at that instant, while negative \(r(t)\) means it is decreasing. If \(y(t)<0\), these directions are reversed. If \(y(t)=0\), then \(y'(t)=0\).
Worked Examples
Worked Example: A Rate That Increases Over Time
A quantity begins at \(100\) milligrams when \(t=0\) hours and satisfies \(y'=r(t)y\), where \(r(t)=0.20\text{ h}^{-1}+(0.10\text{ h}^{-2})t\). Find the quantity at \(t=2\) hours.
Use the initial-value formula: $$ y(2\text{ h})=100\text{ mg}\cdot\exp\left(\int_{0\text{ h}}^{2\text{ h}}\left(0.20\text{ h}^{-1}+(0.10\text{ h}^{-2})s\right)\,ds\right). $$ An antiderivative of the rate is \(0.20s\text{ h}^{-1}+0.05s^2\text{ h}^{-2}\). Evaluating at the bounds gives $$ \left[0.20s\text{ h}^{-1}+0.05s^2\text{ h}^{-2}\right]_{0\text{ h}}^{2\text{ h}} =(0.40+0.20)-0=0.60. $$ Therefore \(y(2\text{ h})=100e^{0.60}\text{ mg}\), approximately \(182.21\text{ mg}\) to two decimal places.
Check the solution function \(y(t)=100\text{ mg}\exp(0.20t\text{ h}^{-1}+0.05t^2\text{ h}^{-2})\). Differentiating gives \(y'(t)=y(t)(0.20\text{ h}^{-1}+0.10t\text{ h}^{-2})=r(t)y(t)\). At \(t=0\), the exponent is zero, so \(y(0)=100\text{ mg}\), as required.
Worked Example: Growth Followed by Decline
A positive amount begins at \(50\) milligrams at \(t=0\) hours and has rate \(r(t)=1\text{ h}^{-1}-(1\text{ h}^{-2})t\). Find the amount at \(t=1\) hour and \(t=2\) hours, and describe its direction of change.
The accumulated rate from zero to \(t\) is $$ \int_0^t\left(1\text{ h}^{-1}-(1\text{ h}^{-2})s\right)\,ds =t\text{ h}^{-1}-\frac{1}{2}t^2\text{ h}^{-2}. $$ Thus the solution is \(y(t)=50\text{ mg}\exp(t\text{ h}^{-1}-\frac12t^2\text{ h}^{-2})\). At one hour, the exponent is \(1-\frac12=\frac12\), so \(y(1\text{ h})=50e^{1/2}\text{ mg}\), approximately \(82.44\text{ mg}\) to two decimal places. At two hours, it is \(2-\frac12(4)=0\), so \(y(2\text{ h})=50e^0\text{ mg}=50\text{ mg}\).
For \(0\le t<1\) hour, \(r(t)>0\), and because \(y(t)>0\), \(y'(t)>0\): the amount increases. At \(t=1\) hour, \(r(t)=0\), so \(y'(t)=0\). For \(t>1\) hour, \(r(t)<0\), and the positive amount decreases. The decrease brings it back to its starting value by two hours because the net accumulated rate from zero to two is zero.
Check by differentiating the proposed solution: \(y'(t)=y(t)(1\text{ h}^{-1}-(1\text{ h}^{-2})t)=r(t)y(t)\). At \(t=0\), \(y(0)=50e^0\text{ mg}=50\text{ mg}\). Both the equation and initial condition hold.
Worked Example: Applying an Initial Condition at a Later Time
A quantity is \(8\) grams at \(t=1\) day. Its rate satisfies \(r(t)=(0.4\text{ day}^{-2})t\). Find the quantity at \(t=2\) days.
Because the initial time is one day, integrate from \(1\) day rather than from zero: $$ y(2\text{ days})=8\text{ g}\cdot\exp\left(\int_{1\text{ day}}^{2\text{ days}}(0.4\text{ day}^{-2})s\,ds\right). $$ The integral is $$ \left[0.2s^2\text{ day}^{-2}\right]_{1\text{ day}}^{2\text{ days}} =0.2(4)-0.2(1)=0.8-0.2=0.6. $$ Hence \(y(2\text{ days})=8e^{0.6}\text{ g}\), approximately \(14.58\text{ g}\) to two decimal places.
The full solution is \(y(t)=8\text{ g}\exp(0.2(t/\text{day})^2-0.2)\). Its derivative is \(y'(t)=y(t)(0.4t\text{ day}^{-2})=r(t)y(t)\). At \(t=1\) day, the exponent is \(0.2-0.2=0\), giving \(y(1\text{ day})=8\text{ g}\). This verifies both the differential equation and the specified initial value.
A Common Mistake: Using the Ending Rate as if It Were Constant
In the first example, the rate at two hours is \(r(2\text{ h})=0.20+0.10(2)=0.40\text{ h}^{-1}\). A tempting but incorrect calculation is to treat this ending rate as if it had held for the full two hours: $$ y(2\text{ h})\mathrel{\not=}100\text{ mg}\cdot e^{(0.40\text{ h}^{-1})(2\text{ h})}=100e^{0.80}\text{ mg}. $$ This gives approximately \(222.55\text{ mg}\), which does not match the actual accumulated rate. The correct calculation integrates the rate as it changes: $$ \int_0^{2\text{ h}}r(s)\,ds=0.60, \qquad y(2\text{ h})=100e^{0.60}\text{ mg}\approx182.21\text{ mg}. $$ The ending value \(r(2\text{ h})\) describes the instantaneous proportional rate at that moment; it does not represent the rate over the entire interval.
Why This Model Is Useful
A time-varying proportional rate separates two questions. The rate function describes how conditions change the instantaneous proportional growth or decline, while the integral summarizes their combined effect over an interval. This makes it possible to reason about intervals with changing rates, including intervals where the rate changes sign.
For AP Calculus problems, keep the starting time, the bounds of integration, and the sign of the rate function in view. The integral gives the net proportional effect; the differential equation \(y'=r(t)y\) gives the instantaneous direction after accounting for the sign of \(y\). These are related but distinct interpretations.
Check Your Understanding
Answer each question using the time-varying proportional-rate model.
- Write the initial-value solution for \(y'=r(t)y\) with \(y(3)=12\), assuming \(r\) is continuous.
- If \(\int_0^4r(s)\,ds=-0.3\) and \(y(0)=20\), write \(y(4)\) exactly in terms of an exponential.
- For positive \(y\), what does \(r(t)=0\) imply about the instantaneous derivative at that time?
- Explain why replacing a changing rate by its value at the ending time generally gives an incorrect solution.
- If \(y(t)<0\) and \(r(t)<0\), is \(y'(t)\) positive or negative? Justify your answer from the differential equation.
