Page 73 - Schaum's Outline of Differential Equations

56 APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAR 7

7.5. The population of a certain country is known to increase at a rate proportional to the number of people
presently living in the country. If after two years the population has doubled, and after three years the
population is 20,000, estimate the number of people initially living in the country.
Let N denote the number of people living in the country at any time t, and let N Q denote the number of people
initially living in the country. Then, from (7.1),

which has the solution

k(
At t = 0, N = N 0; hence, it follows from (1) that N 0 = ce °\ or that c = N 0. Thus,

Substituting these values into (2), we have
At t= 2, N= 2N 0. Substituting these values into (2), we have
from which

Substituting this value into (2) gives

At t = 3, N= 20,000. Substituting these values into (3), we obtain
Substituting these values into (3), we obtain

7.6. A certain radioactive material is known to decay at a rate proportional to the amount present. If initially
there is 50 milligrams of the material present and after two hours it is observed that the material has lost
10 percent of its original mass, find (a) an expression for the mass of the material remaining at any time t,
(b) the mass of the material after four hours, and (c) the time at which the material has decayed to one
half of its initial mass.

(a) Let N denote the amount of material present at time t. Then, from (7.1),

This differential equation is separable and linear; its solution is

k(
At t = 0, we are given that N= 50. Therefore, from (1), 50 = ce °\ or c = 50. Thus,

At t = 2, 10 percent of the original mass of 50 mg, or 5 mg, has decayed. Hence, at t = 2, N= 50 - 5 = 45.
Substituting these values into (2) and solving for k, we have

Substituting this value into (2), we obtain the amount of mass present at any time t as

where t is measured in hours.
(b) We require N at t = 4. Substituting t = 4 into (3) and then solving for N, we find that