430
7.5. MODELING WITH DIFFERENTIAL EQUATIONS
value problem
dP
dt
=
1
100
(1.5 · 10
5 − P), P(0) = 0.
Separating variables, we find that
1
1.5 · 10 5 − P
dP
dt
=
1
100
.
Integrating with respect to t, we have
1
1.5 · 10 5 − P
dP
dt
dt =
1
100
dt,
and thus changing variables on the left and antidifferentiating on both sides, we find that
dP
1.5 · 10 5 − P
=
1
100
dt
− ln |1.5 · 10
5 − P| =
1
100
t + C
Finally, multiplying both sides by −1 and using the definition of the logarithm, we find that
1.5 · 10
5 − P = Ce
−t/100 .
(7.1)
This is a good time to determine the constant C. Since P = 0 when t = 0, we have
1.5 · 10
5 − 0 = Ce
0 = C.
In other words, C = 1.5 · 10 5 .
Using this value of C in Equation (7.1) and solving for P, we arrive at the solution
P(t) = 1.5 · 10
5 (1 − e
−t/100 ).
Superimposing the graph of P on the slope field we saw in Figure 7.3, we see, as shown
in Figure 7.4 We see that, as expected, the amount of plastic pollution stabilizes around
1.5 · 10 5 cubic meters.
There are many important lessons to learn from Example 7.3. Foremost is how we
can develop a differential equation by thinking about the “total rate = rate in - rate out”
model. In addition, we note how we can bring together all of our available understanding
(plotting
dP
dt vs. P, creating a slope field, solving the differential equation) to see how the
differential equation describes the behavior of a changing quantity.
Of course, we can also explore what happens when certain aspects of the problem
change. For instance, let’s suppose we are at a time when the plastic pollution entering
7.5. MODELING WITH DIFFERENTIAL EQUATIONS
value problem
dP
dt
=
1
100
(1.5 · 10
5 − P), P(0) = 0.
Separating variables, we find that
1
1.5 · 10 5 − P
dP
dt
=
1
100
.
Integrating with respect to t, we have
1
1.5 · 10 5 − P
dP
dt
dt =
1
100
dt,
and thus changing variables on the left and antidifferentiating on both sides, we find that
dP
1.5 · 10 5 − P
=
1
100
dt
− ln |1.5 · 10
5 − P| =
1
100
t + C
Finally, multiplying both sides by −1 and using the definition of the logarithm, we find that
1.5 · 10
5 − P = Ce
−t/100 .
(7.1)
This is a good time to determine the constant C. Since P = 0 when t = 0, we have
1.5 · 10
5 − 0 = Ce
0 = C.
In other words, C = 1.5 · 10 5 .
Using this value of C in Equation (7.1) and solving for P, we arrive at the solution
P(t) = 1.5 · 10
5 (1 − e
−t/100 ).
Superimposing the graph of P on the slope field we saw in Figure 7.3, we see, as shown
in Figure 7.4 We see that, as expected, the amount of plastic pollution stabilizes around
1.5 · 10 5 cubic meters.
There are many important lessons to learn from Example 7.3. Foremost is how we
can develop a differential equation by thinking about the “total rate = rate in - rate out”
model. In addition, we note how we can bring together all of our available understanding
(plotting
dP
dt vs. P, creating a slope field, solving the differential equation) to see how the
differential equation describes the behavior of a changing quantity.
Of course, we can also explore what happens when certain aspects of the problem
change. For instance, let’s suppose we are at a time when the plastic pollution entering