Thus, we can immediately integrate (10.28) to obtain a solution
u ¼
1
p x
ð Þ
Z x
w x
0
ð Þd x
0
ð Þdx
0
þ C
!
,
ð10:29Þ
where C is an arbitrary integration constant.
To seek w(x), from (10.26) and (10.27) we have
p
0
¼ wa
ð Þ
0 ¼ wb ¼ wa
b
a
:
ð10:30Þ
This can easily be integrated for wa to be expressed as
wa ¼ C
0 exp
Z
b
a
dx
or w ¼
C
0
a
exp
Z
b
a
dx
,
ð10:31Þ
where C
0 is an arbitrary integration constant. The quantity
C
0
a must be non-negative so
that w can be non-negative.
Example 10.1 Let us think of a following FOLDE within an interval [a, b]; i.e.,
a x b.
du
dx
þ xu ¼ x:
ð10:32Þ
A boundary condition is set such that
u a
ð Þ ¼ σ:
ð10:33Þ
Notice that (10.33) is obtained by setting α ¼ 1 and β ¼ 0 in (10.24). Following the
above argument, we obtain a solution described as
u ¼
1
p x
ð Þ
Z x
a
w x
0
ð Þd x
0
ð Þdx
0
þ u a
ð Þp a
ð Þ
!
:
ð10:34Þ
Also, we have
p x
ð Þ ¼ w x
ð Þ ¼ exp
Z x
a
x
0 dx
0
¼ exp
1
2
x
2
À a
2
À
Á
h
i
:
ð10:35Þ
The integration of RHS can be performed as follows:
10.2 First-Order Linear Differential Equations (FOLDEs)
385