Environmental Engineering Reference
In-Depth Information
Polynomials of Fourth- and Fifth-Order
a
2
x
2
a
3
x
3
a
4
x
4
For polynomials of fourth- and fifth- order,
P
5
(
x
)=
a
0
+
a
1
x
+
+
+
+
a
5
x
5
(
P
5
(
a
2
x
2
a
3
x
3
when
a
5
=
x
)
reduces to
P
4
(
x
)=
a
0
+
a
1
x
+
+
0). By Eqs. (6.176),
(6.191) and (6.192), we have
P
5
(
1
τ
0
P
(
4
)
5
(
x
)
1
x
)
t
t
e
−
2
A
e
−
(
,
)=
τ
(
)
−
+
(
)+
(
)
,
u
0
x
t
0
P
5
x
2
τ
0
G
2
t
G
4
t
2!
4!
t
x
+
A
(
t
−
τ
)
1
2
A
t
−
τ
2τ
0
d
e
−
u
1
=
S
(
u
0
)=
τ
I
0
0
x
−
A
(
t
−
τ
)
P
5
(
ξ
)
d
1
e
−
τ
0
2
A
P
(
4
)
·
+
(
ξ
)
g
2
(
τ
)
ξ
5
t
τ
0
−
(
τ
0
+
τ
0
P
(
4
)
5
(
x
)
1
2
A
t
t
+
τ
2
e
−
e
−
=
τ
0
P
5
(
x
)
t
)
+
τ
0
G
2
(
t
−
τ
)
d
τ
2
0
t
0
τ
1
τ
0
d
P
(
4
)
5
(
x
)
1
2
A
t
−
τ
e
−
+
(
τ
)
−
τ
.
0
g
2
2
By Eq. (6.195), we obtain
B
2
τ
0
1
τ
0
t
P
(
4
)
5
e
−
S
2
u
2
=
S
(
u
1
)=
(
u
0
)=
(
x
)
−
e
−
τ
0
t
2
2
t
0
P
(
4
)
2
=
τ
(
x
)
τ
0
−
τ
0
+
t
+
.
5
τ
0
The solution of PDS (6.198) is thus
S
2
2
u
=
u
0
+
S
(
u
0
)
ε
+
(
u
0
)
ε
5
n
=
0
a
n
x
n
1
−
e
−
τ
0
=
τ
0
τ
0
2
a
2
+
12
a
4
x
2
20
a
5
x
3
1
6
a
3
x
+
+
24
a
4
+
120
a
5
x
4!
t
2
A
e
−
+
(
)+
(
)
2
G
2
t
G
4
t
2!
0
2
a
2
20
a
5
x
3
τ
0
24
a
4
+
120
a
5
x
4
A
t
e
−
12
a
4
x
2
+
τ
+
6
a
3
x
+
+
τ
−
(
τ
+
t
)
+
0
0
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