Page 39 - Advanced engineering mathematics

1.2 Linear Equations 19

1/8 lb/gal

3 gal/min

3 gal/min

FIGURE 1.5 Storage tank in Example 1.10.

is pumped into the tank at a rate of 3 gallons per minute, and the mixture is continuously stirred.
Brine also is allowed to empty out of the tank at the same rate of 3 gallons per minute (see
Figure 1.5). How much salt is in the tank at any time?
Let Q(t) be the amount of salt in the tank at time t. The rate of change of Q(t) with respect
to time must equal the rate at which salt is pumped in minus the rate at which it is pumped out:
dQ
= (rate in) − (rate out)
dt
1 pounds gallons Q(t) pounds gallons

= 3 − 3
8 gallon minute 200 gallon minute
3 3
= − Q(t).
8 200
This is the linear equation
3 3

Q (t) + Q = .
200 8

An integrating factor is e (3/200)dt = e 3t/200 . Multiply the differential equation by this to get
3 3t/200
) = e
(Qe 3t/200 .
8
Integrate to obtain
3 200
Qe 3t/200 = e 3t/200 + c.
8 3
Then

Q(t) = 25 + ce −3t/200 .
Now use the initial condition
Q(0) = 100 = 25 + c,

so c = 75 and
Q(t) = 25 + 75e −3t/200 .

Notice that Q(t) → 25 as t →∞.Thisisthe steady-state value of Q(t). The term 75e −3t/200 is
called the transient part of the solution, and it decays to zero as t increases. Q(t) is the sum of a
steady-state part and a transient part. This type of decomposition of a solution is found in many

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October 14, 2010 14:9 THM/NEIL Page-19 27410_01_ch01_p01-42