Biomedical Engineering Reference
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into
(7.83)
results in:
d
2
T
d
t
2
¼
c
2
X
d
2
X
d
x
2
þ
c
2
Y
d
2
Y
1
T
d
y
2
¼ u
2
:
(7.85)
Separating and solving the temporal term:
d
2
T
d
t
2
þ
u
2
T
¼
0
(7.86)
results in
T
ð
t
Þ¼
E
sin
ut
þ
f
cos
ut
:
(7.87)
The spatial terms become:
d
2
X
d
x
þ
u
2
c
2
¼
d
2
Y
d
y
2
¼ k
2
1
X
1
Y
:
(7.88)
Solving
u
2
c
2
k
2
X
d
2
X
d
x
2
þ
¼
0
(7.89)
results in
q
ð
q
ð
2
2
k
2
k
2
X
ð
x
Þ¼
A
sin
u
=
c
Þ
þ
B
cos
u
=
c
Þ
:
(7.90)
Solving
d
2
Y
d
y
2
þ
k
2
Y
¼
0
(7.91)
results in
Y
ð
y
Þ¼
C
sin
ky
þ
D
cos
ky
:
(7.92)
Applying the boundary conditions of
z
¼
0at
x
¼
0 and
x
¼
L
leads to
B
¼
0 and
q
ð
2
k
2
L
A
sin
u
=
c
Þ
¼
0
:
(7.93)
Thus
A
sin
mpx
L
X
ð
x
Þ¼
(7.94)
q
ð
2
k
2
L
where
u
=
c
Þ
¼
mp
and
m
¼
1, 2,
. Applying the boundary conditions of
z
¼
0at
y
¼
0
.
and
y
¼
W
leads to
D
¼
0 and
C
sin
kW
¼
:
0
(7.95)
Thus
C
sin
mpy
W
Y
ð
y
Þ¼
:
(7.96)
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