1550078481-Ordinary_Differential_Equations__Roberts_

144 Ordinary Differential Equations

If we assume that the body is falling freely in air instead of in a vacuum,

then we must make some assumption regarding the retarding force due to
air resistance. Let us assume that the retarding force is proportional to the
velocity of the body. Then the initial value problem for the velocity becomes

dv/dt = g - cv; v(O) = v 0 , where c > 0 is the constant of proportionality.

Separating variables res ults in

Integrating, we find

_±! = dt.

g-CV

_-ln---'-1 g-_cv--'-1 = t + A ,

c
where A is an arbitrary constant. Multiplying by - c and exponentiating, we
get
lg -cvl = e-c(t+A) = e-cAe-ct = Be-ct

where B is an arbitrary positive constant. By letting B be an arbitrary
constant- positive, negative, or zero- we can remove the absolute value ap-

pearing in this equation. When t = 0 , v = vo, so B = g - cvo and therefore

g - cv = (g - cvo)e-ct.

Solving for the velocity v, we find

v(t) = :2.(1-e-ct) +voe-ct.

c

Since c > 0 , as t approaches oo the velo city v(t) approaches the value g/c, the

terminal velocity- the maximum velocity that the falling body can attain.

EXAMPLE 1 Velocity of a Body Falling Freely in Air

Assume the equation for the velocity of a body falling freely in air is

dv

dt = g - cv; v(O) = 20 ft/sec,

where g = 32 ft/sec^2 and c = .25. Compute numerically and graph the
solution to this initial value problem on the interval [O, 10]. What is the value

of v(lO)? How does this compare with the terminal velocity of v= = g/c =

128 ft/sec?

SOLUTION
Associating v(t) with y(x), we ran SOLVEIVP by setting f(x, y) = 32-.25y,
specifying the interval of integration to be [O, 10], and specifying the initial
condition to be y(O) = 20. A graph of the velocity as a function of time over
the first 10 seconds is shown in Figure 3 .13. From the table of values for the

numerical solution, we found that v(lO) = 117.4931 ft/sec, which is within

8.213 of the terminal velocity.