8.8 Interpreting differential-equation solutions in context — Learn

Prerequisites: solving separable differential equations from Section 8.7; the exponential model and logarithms from Chapter 6.

By the end of this section you can: build a differential equation from a written description; solve it; find a required value or time from the solution; describe the long-term behaviour; and name the limitations of the model.

1. From words to an equation

A rate proportional to a quantity becomes a differential equation. The words tell you the sign and the constant.

"grows at a rate proportional to the amount present" gives $\dfrac{dP}{dt} = kP$ with $k > 0$ .
"decays at a rate proportional to the amount present" gives $\dfrac{dm}{dt} = -km$ with $k > 0$ .
"cools at a rate proportional to the difference from the room" gives $\dfrac{dT}{dt} = -k(T - T_s)$ , where $T_s$ is the surrounding temperature.

Example 1. A cup of tea cools according to $\dfrac{dT}{dt} = -0.1(T - 20)$ , with $T = 90$ when $t = 0$ and $t$ in minutes. Find $T$ in terms of $t$ , and describe the temperature in the long run.

Separate the variables:

\int \frac{1}{T - 20}dT = \int -0.1dt.

Integrate:

\ln|T - 20| = -0.1t + C.

Exponentiate, with $A = e^C$ :

T - 20 = Ae^{-0.1t}, \qquad T = 20 + Ae^{-0.1t}.

Substitute $t = 0$ , $T = 90$ : $90 = 20 + A$ , so $A = 70$ . Therefore

T = 20 + 70e^{-0.1t}.

In the long run the term $70e^{-0.1t}$ tends to zero, so the temperature approaches the room temperature of $20$ °C. This is the limiting value, or steady state.

2. Finding a value or a time

The solution gives the quantity at any time. To find when the quantity reaches a given value, substitute the value and solve for the time.

Example 2. A population satisfies $\dfrac{dP}{dt} = 0.02P$ with $P = 5000$ when $t = 0$ and $t$ in years. Find the population after $10$ years, and find the time for the population to double.

Separate and integrate:

\ln P = 0.02t + C, \qquad P = 5000e^{0.02t}.

After $10$ years:

P = 5000e^{0.2} \approx 6107 \text{ to the nearest whole number}.

To double, set $P = 10000$ :

10000 = 5000e^{0.02t}, \qquad 2 = e^{0.02t}, \qquad 0.02t = \ln 2, \qquad t = \frac{\ln 2}{0.02} \approx 34.7 \text{ years}.

The doubling time is about $34.7$ years. It does not depend on the starting population, because every value doubles in the same time.

3. Links to kinematics

In kinematics the rate of change of velocity is acceleration, and the rate of change of displacement is velocity. A separable equation gives the velocity as a function of time.

Example 3. A small ball is dropped in a thick fluid. Its velocity $v$ satisfies $\dfrac{dv}{dt} = 9.8 - 2v$ with $v = 0$ when $t = 0$ , and $t$ in seconds. Find $v$ in terms of $t$ and describe the motion.

Separate:

\int \frac{1}{9.8 - 2v}dv = \int 1dt.

Integrate the left side. The derivative of $\ln|9.8 - 2v|$ is $\dfrac{-2}{9.8 - 2v}$ , so a factor of $-\dfrac{1}{2}$ is needed:

-\frac{1}{2}\ln|9.8 - 2v| = t + C.

Rearrange and exponentiate:

9.8 - 2v = Ae^{-2t}.

Substitute $t = 0$ , $v = 0$ : $9.8 = A$ . Therefore

v = \frac{9.8 - 9.8e^{-2t}}{2} = 4.9\left(1 - e^{-2t}\right).

As $t$ grows, $e^{-2t} \to 0$ and $v \to 4.9$ m/s. This is the terminal velocity, where the drag balances gravity and the acceleration is zero.

4. Limitations of a model

A solution describes the model, not the world. The model has limits, and a good answer names them.

The rate constant is assumed constant. It may change as conditions change.
The model is assumed valid for the whole range of the variable. A population model may fail once food or space runs out.
A limiting value may only be approached. The cooling model predicts that the temperature equals room temperature only after infinite time.
Real data has variation. A smooth solution is an idealised curve through noisy measurements.

Watch out. Three mistakes are common.

Putting the sign in the wrong place. Growth has a positive constant and decay has a negative one. Decide from the words whether the quantity rises or falls before you write the equation.

Ignoring the surroundings in a cooling model. The rate depends on $T - T_s$ , not on $T$ alone. Using $\dfrac{dT}{dt} = -kT$ gives a limiting temperature of $0$ °C, not the room temperature.

Reporting the limiting value as reached. The model approaches the limiting value but never attains it in finite time. Re-check: after solving, identify the initial value, the direction of change, and the limiting value, and check that all three match the context.

Progress check

Return to the cooling drink in the chapter problem. The coffee starts at $90$ °C in a room at $20$ °C, and after $20$ minutes it is $55$ °C.

Write the differential equation for the temperature $T$ at time $t$ .
Solve it to show that $T = 20 + 70e^{-kt}$ .
Use the temperature at $20$ minutes to find $k$ .
Find the temperature after $40$ minutes.
Explain what happens to the temperature in the long run, and give one limitation of the model.

You should now be able to

translate a proportional rate into a differential equation with the correct sign;
solve the equation and apply an initial condition;
find a quantity at a given time, or the time for a given value;
describe the limiting value and the long-term behaviour;
use a differential equation in a kinematics context;
state the limitations of a model.