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Chapter 3
Integrating, we obtain
Bu(
t
)=2
Z
(
e
−
t
−
e
−
2
t
)
dt
=
−
2
e
−
t
+
e
−
2
t
+
C,
and the initial condition Bu(0) = 0 gives
C
= 1. Therefore,
Bu(
t
)=
−
2
e
−
t
+
e
−
2
t
+1
.
EXERCISES 3.3:
Heating and Cooling of Buildings, page 107
1.
Let
T
(
t
) denote the temperature of coFee at time
t
(in minutes). According to the Newton’s
Law (1) on page 102 of the text,
dT
dt
=
K
[21
−
T
(
t
)]
,
wherewehavetaken
H
(
t
)
≡
U
(
t
)
≡
0,
M
(
t
)
≡
21
◦
C, with the initial condition
T
(0) = 95
◦
C.
Solving this initial value problem yields
dT
21
−
T
=
Kdt
⇒−
ln

T
−
21

=
Kt
+
C
1
⇒
T
(
t
)=21+
Ce
−
;
95 =
T
(0) = 21 +
−
K
(0)
⇒
C
=74
⇒
T
(
t
)=21+74
e
−
.
To ±nd
K
, we use the fact that after 5 min the temperature of coFee was 80
◦
C. Thus
80 =
T
(5) = 21 + 74
e
−
K
(5)
⇒
K
=
ln(74
/
59)
5
,
and so
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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