The Initial Value Problem y' = f(x, y); y(c) = d 71
Solving the right-hand equation for e^2 k, we find2 k^175 -^65 llO^22e = 125 = 125 = 25 ·
Hence,k = (22) 1/2
e 25It is not necessary to determine the value of the constant of proportionality k
explicitly, since the expression ekt = (ek)t appears in equation (41). Substi-
tuting the expression above for ek in equation ( 41), we find the temperature
of the coffee as a function of the time after it is poured is(22) t/2
T(t) = 125° F
25- 65° F.
The coffee reaches the temperature of 150° F when t satisfies the equation(22)t/2
150° F = 125° F
25+ 65° F.
Solving for t , we find21 (150-65)
n 125 2ln (~) 125-~-~ = 6.034 minutes.
t = ln (~~) ln (~~)EXERCISES 2.3.3
In exercises 1-7 solve the given initial value problem by finding the
solution of the differential equation and then determining the value
of the constant of integration which satisfies the initial condition.
Then specify the interval on which each solution is exists.
- y' = 4y + 1; y(O) = 1
- y' = xy + 2; y(O) = 1
- y' = y/x; y(-1) = 2