Page 36 - Advanced engineering mathematics

16 CHAPTER 1 First-Order Differential Equations

increments from 1790 through 1980. Taking 1790 as model is about 197,300,000, which the United States
year zero to determine p 0 , show that the logistic model actually exceeded in 1970.
for the United States population is Sometimes an exponential model Q (t) = kQ(t)

is used for population growth. Use the census data
123,141.5668 0.03134t
P(t) = e . (again with 1790 as year zero) to solve for Q(t).
0.03072 + 000062e 0.03134t
Compute Q(t) for the years of the census data and
Calculate P(t) in ten year increments from 1790 to the percentage error in this exponential prediction of
fill in the P(t) column in the table. Remember that population. Plot the census data and the exponential
(with 1790 as the base year) 1800 is year t = 10 in the model predicted data on the same set of axes. It should
model, 1810 is t = 20, and so on. Also, calculate the be clear that Q(t) diverges rapidly from the actual
percentage error in the model and fill in this column. census figures. Exponential models are useful for very
Plot the census figures and the numbers predicted by simple populations (such as bacteria in a dish) but
the logistic model on the same set of axes. You should are not sophisticated enough for human or (in gen-
observe that the model is fairly accurate for a long eral) animal populations, despite occasional claims by
period of time, then diverges from the actual census experts that the population of the world is increasing
numbers. Show that the limit of the population in this exponentially.

1.2 Linear Equations

A first-order differential equation is linear if it has the form

y + p(x)y = q(x)
for some functions p and q.

There is a general approach to solving a linear equation. Let

p(x)dx
g(x) = e
and notice that

g (x) = p(x)e p(x)dx = p(x)g(x). (1.3)
Now multiply y + p(x)y = q(x) by g(x) to obtain

g(x)y + p(x)g(x)y = q(x)g(x).
In view of equation (1.3), this is

g(x)y + g (x)y = q(x)g(x).

Now we see the point to multiplying the differential equation by g(x). The left side of the new
equation is the derivative of g(x)y. The differential equation has become
d
(g(x)y) = q(x)g(x),
dx
which we can integrate to obtain

g(x)y = q(x)g(x)dx + c.

If g(x) = 0, we can solve this equation for y:
1 c
y(x) = q(x)g(x)dx + .
g(x) g(x)

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October 14, 2010 14:9 THM/NEIL Page-16 27410_01_ch01_p01-42