Name________________________
Student I.D.___________________
Math 2250−4
Quiz 2 SOLUTIONS
January 20, 2012
1) Consider the differential equation for x t :
1
x 30 .
2
a) Is this differential equation separable, linear, neither separable nor linear, or both? Explain.
x=
(2 points)
dx
Both:
= .5 x
dt
30 separates as
dx
= .5 dt .
x 30
But we can also write the DE in linear form as
1
x t
x = 15 .
2
b) Solve the following initial value problem:
1
x 30
2
x 0 = 70 .
x=
(6 points)
Separable:
dx
= .5 dt
x 30
ln x 30 = .5 t C1
C
30 = e 1 e .5 t
30 = C e .5 t .
x
x
x 0 = 70
C = 40
x t = 30
40 e
.5 t
.
Linear:
x t
Integrating factor is e
P t dt
1
x = 15 .
2
= e.5 t:
e.5 t x t
1
x = e.5 t15
2
d .5 t
e x t = 15 e.5 t
dt
.5 t
e x t = 15 e.5 tdt = 30e.5 t
Divide by e.5 t:
x t = 30
Ce
.5 t
C.
x 0 = 70
C = 40
x t = 30
40 e
.5 t
.
2) A cooling object is found outside, where the ambient temperature is a constant 30o (Farenheit). At the
time the object is found, its temperature is found to be 70o, and the object is cooling off at a rate of 20o
per hour. Assuming that the object is cooling according to Newton’s Law of Cooling, i.e. at a rate
proportional to the difference its temperature and the ambient temperature, show that the temperature of
the object satisfies the initial value problem in (1b).
(2 points)
Model assumption is that T t = k T A . From problem at t = 0, T 0 = 70, T 0 = 20 . So
from the DE,
20 = k 70 30
k = .5 .
Thus the IVP is
T = .5 T 30
T 0 = 70
which is the DE in problem (1), if we write x t for T t .