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Answer Guide 3
Math 427K: Unique Number 55160
Wednesday, September 14, 2011
1. After barbecuing a beef brisket, it is left to cool in a light breeze. Let
x
(
t
)
be its temperature
after
t
x
with time
is proportional with factor
k
to the di/erence between
x
and the temperature
A
(
t
) =
30 + 5 cos (2
(
t=
24))
of the ambient air measured in degrees Celsius. Assuming that
k
= 1
and
x
(0) = 120
C
, ±nd a formula for
x
(
t
)
.
(The solution requires some tedious calculations,
but this is typical of realworld problems.)
Notice the hysteresis in the graphs of
x
(
t
)
(black) and
A
(
t
)
(red) below.
0
5
10
15
20
25
30
35
40
45
0
20
40
60
80
100
120
t
x
Solution:
The initial value problem we need to solve is
x
0
(
t
) =
A
(
t
)
x
(
t
)
x
(0) = 120
:
An integrating factor is
±
(
t
) =
e
t
. So we need to antidi/erentiate
Z
e
t
h
30 + 5 cos
12
t
±i
d
t:
It is easier not to substitute those constants until later in the problem. We ±rst recall that
for any constant
a
, we can integrate by parts twice to see that
Z
e
t
cos(
at
) d
t
=
e
t
1 +
a
2
[
a
sin(
at
) + cos(
at
)]
:
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View Full DocumentPut
a
=
12
. It follows that the general solution is thus
x
gen
(
t
) =
Ce
t
+ 30 +
5
1 +
a
2
[
a
sin(
at
) + cos(
at
)]
:
Using the initial conditions, one can with some e/ort determine
C
and write the particular
solution in the (nicer?) form
x
(
t
) = 30
1 +
2
sin(
12
t
) + 24 cos(
12
t
) + (408 + 3
2
)
e
t
144 +
2
±
:
(
F
)
why not. If it is exact, ±nd an implicit solution.
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 Spring '11
 DELALLAVE
 Differential Equations, Equations, Rate Of Change

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