Biomedical Engineering Reference
In-Depth Information
Þ
e
K
21
t
u
ð
t
Þ:
Since Eq. (7.54) involves only
q
2
, it is easily solved as
q
2
¼
q
2
ð
0
By substituting the
solution for
q
2
into Eq. (7.53), we now have one equation, giving
q
1
¼
q
2
ð
Þ
K
21
e
K
21
t
K
1
M
þ
K
1
U
0
ð
Þ
q
1
and after rearranging
Þ
K
21
e
K
21
t
q
1
þ
K
1
M
þ
K
1
U
ð
Þ
q
1
¼
q
2
ð
0
ð
7
:
55
Þ
Þ
K
21
e
K
21
t
. The natural solu-
This is a first-order differential equation with a forcing function
q
2
ð
0
q
1
n
¼
B
1
e
K
1
M
þ
K
1
U
ð
Þ
t
and the forced response is
q
1
f
¼
B
2
e
K
21
t
. To determine
q
1
f
¼
B
2
e
K
21
t
tion is
B
2
,
is substituted into Eq. (7.55), which gives
K
21
B
2
e
K
21
t
þ
K
1
M
þ
K
1
U
Þ
B
2
e
K
21
t
¼
q
2
ð
Þ
K
21
e
K
21
t
ð
0
Solving for
B
2
gives
Þ
K
21
K
1
M
þ
K
1
U
K
21
q
2
ð
0
B
2
¼
The complete response is
q
1
¼
q
1
n
þ
q
1
f
¼
B
1
e
K
1
M
þ
K
1
U
ð
Þ
t
þ
B
2
e
K
21
t
Þ
K
21
K
1
M
þ
K
1
U
K
21
e
K
21
t
q
2
ð
0
ð
7
:
56
Þ
¼
B
1
e
K
1
M
þ
K
1
U
ð
Þ
t
þ
B
1
is found using the initial condition
q
1
(0)
¼
0
Þ
K
21
K
1
M
þ
K
1
U
K
21
e
K
21
t
q
2
ð
0
Þ
K
21
K
1
M
þ
K
1
U
K
21
q
2
ð
0
¼
B
1
e
K
1
M
þ
K
1
U
ð
Þ
t
q
1
ð
0
Þ¼
0
þ
0
¼
B
1
þ
t
¼
giving
Þ
K
21
K
1
M
þ
K
1
U
K
21
q
2
ð
0
B
1
¼
and
Þ
K
21
K
1
M
þ
K
1
U
K
21
q
2
ð
0
e
K
21
t
e
K
1
M
þ
K
1
U
ð
Þ
t
q
1
¼
u
ð
t
Þ
ð
7
:
57
Þ
or in terms of concentration,
1
V
q
2
ð
0
Þ
K
21
e
K
21
t
e
K
1
M
þ
K
1
U
ð
Þ
t
c
1
¼
u
ð
t
Þ
ð
7
:
58
Þ
ð
K
1
M
þ
K
U
K
Þ
1
1
21
To determine the time when the maximum solute is in compartment 1 in Example Problem
7.8, Eq. (7.57) is differentiated with respect to
t
, set equal to zero, and solved as follows:
0
@
1
A
q
1
¼
d
dt
q
2
ð
0
Þ
K
21
e
K
21
t
e
K
1
M
þ
K
1
U
ð
Þ
t
K
M
þ
K
U
K
1
1
21
ð
7
:
59
Þ
q
2
ð
0
Þ
K
21
K
1
M
þ
K
1
U
K
21
K
21
e
K
21
t
þ
K
1
M
þ
K
1
U
Þ
e
K
1
M
þ
K
1
U
ð
Þ
t
¼
ð
