1550078481-Ordinary_Differential_Equations__Roberts_

70 Ordinary Differential Equations

Hence, if T(t) is the temperature of the body at time t and A is the temper-

ature of the surrounding medium, then according to Newton's law of cooling,

T(t) satisfies the differential equation

( 40)

dT

- = k(T - A) = kT - kA

dt

where k is the constant of proportionality. Equation ( 40) is linear in the

dependent variable T - that is, T' = a(t)T + b(t) where a(t) = k and b(t) =

-kA. Since k and A are constants, a(t) and b(t) are constant functions which

are defined on ( -oo, oo). Integrating, we find

T1(t) = ef'a(s)ds = ef'kds = ekt+C.

Setting C = 0, substituting into v(t), and integrating, yields

f

v(t) = t --b(s) ds = f t ---kA ds = -kA f t e-ks. ds = Ae-kt. + D.

T1(s) eks

Setting D = 0 and substituting T 1 (t) and v(t) into equation (36) where y has

been replaced by T and x has been replaced by t , we find the solution of the

linear differential equation ( 40) is

(41)

where K is an arbitrary constant.

EXAMPLE 6 An Application of Newton's Law of Cooling

A cup of coffee whose temperature is 190° F is poured in a room whose

temperature is 65° F. Two minutes later the temperature of the coffee is

175° F. How long after the coffee is poured does it reach a temperature of

150° F?

SOLUTION

For this problem, A = 65° F and the constants K and k of equation (41)

must be chosen to satisfy the two conditions T(O) = 190 ° F and T(2) = 175° F.

Evaluating equation ( 41) at t = 0 and imposing the first condition, we find K

must satisfy

T(O) = K +A or 190° F = K + 65° F.

Solving for K, we find

K = 190° F - 65 ° F = 125 ° F.

Substituting the value 125° F for Kin equation ( 41), evaluating the resulting

equation at t = 2, and imposing the second condition, we obtain the equation

T(2) = 125 ° F e^2 k +A or 17 5° F = 125 ° Fe^2 k + 65° F.