Lecture Notes in Differential Equations - Bruce E. Shapiro

50 LESSON 6. EXPONENTIAL RELAXATION
Hence
− ln(400 − T ) = kt − ln 330 (6.40)
ln 400 − T = −kt (6.41)
330
From the second observation, that T (3) = 150,
we conclude that
From (6.41)
ln
400 − 150
330
k = 1 3
400 − T
330
= −3k (6.42)
33
ln ≈ .09 (6.43)
25
= e −.09t (6.44)
T = 400 − 330e −.09t (6.45)
The problem asked when the potato will reach a temperature of 350. Returning
to (6.41) again,
t ≈ − 1 400 − 350
ln ≈ 11 ln 33 .09 330
5
≈ 20 minutes (6.46)
The motion of an object subjected to external forces F 1 , F 2 , ...is given by
the solution of the differential equation
Example 6.4. Falling Objects. Suppose an object of constant mass m
is dropped from a height h, and is subject to two forces: gravity, and air
resistance. The force of gravity can be expressed as
F gravity = −mg (6.47)
where g = 9.8 meters/ sec ond 2 , and the force due to air resistance is
dy
F drag = −C D (6.48)
dt
where C D is a known constant, the coefficient of drag; a typical value of
C D ≈ 2.2. According Newtonian mechanics
Making the substitution
m d2 y
dt 2
= ∑ i
F = −mg − C D
dy
dt
v = dy
dt
(6.49)
(6.50)