Math 2280-1
Quiz 2 Solutions
January 23, 2015
1a) Find the solution x t to the initial value problem
1
x# t =K x K 30
2
x 0 = 70
(7 points)
Solution: this DE is separable and linear, and we could use either algorithm. Using the separable
algorithm:
dx
1
= K x K 30
dt
2
dx
1
=K dt
x K 30
2
1
ln x K 30 =K t C C1
2
(If x t ! 30 would need to use ln x K 30 instead.) In any case, next step is to exponentiate:
1
K tCC
2
1
x K 30 = e
1
K t
2
=Ce
1
K t
2
x t = 30 C C e
Use IC to find C : x 0 = 70 = 30 C C 0 C = 40.
.
1
K t
2
x t = 30 C 40 e
.
1b) A cooling object is found outside, where the constant ambient temperature is 30o (Fahrenheit). At the
time the object is found it has temperature 70o and is cooling off at a rate of 20o per hour. Assuming the
object is cooling according to Newton's law of cooling - i.e. its temperature is changing at a rate
proportional to its difference with the ambient temperature - explain why the temperature x t of the object
satisfies the initial value problem in part 1a.
(3 points)
"its temperature (x t is changing at a rate proportional to its difference with the ambient temperature"
and sign considerations for temperature change imply the DE
x# t = k x K 30 .
We are told that at t = 0, x 0 = 70 and that x# 0 =K20. If we plug this into the differential equation we
get
x# 0 =K20 = k 70 K 30 = 40 k
1
which implies k =K . Thus the IVP is
2
1
x# t =K x K 30
2
x 0 = 70.