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Chapter 5
By simplifying these equations, we observe that this cooling problem satisFes the system
4
x
0
(
t
)+3
x
(
t
)
−
y
(
t
)
=
200
,
−
x
0
(
t
)+4
y
0
(
t
)+2
y
(
t
)=5
2
.
In operator notation, this system becomes
(4
D
+3)[
x
]
−
[
y
]
=
200
,
−
[
x
]+(4
D
+2)[
y
]=5
2
.
Since we are interested in the temperature in the attic,
x
(
t
), we will eliminate the function
y
(
t
) from the system above by applying (4
D
+2) to the Frst equation and adding the resulting
equations to obtain
{
(4
D
+ 2)(4
D
+3)
−
1
}
[
x
]=(4
D
+ 2)[200] + 52 = 452
⇒
(
16
D
2
+20
D
+5
)
[
x
] = 452
.
(5.23)
This last equation is a linear equation with constant coeﬃcients whose corresponding homo
geneous equation has the associated auxiliary equation 16
r
2
r
+ 5 = 0. By the quadratic
formula, the roots to this auxiliary equation are
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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 University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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