But, v(x) can be described by a linear combination of u 1 (x) and u 2 (x) as in the case of
(10.6). Hence, the general solution of (10.2) should be expressed as
u x
ð Þ ¼ c 1 u 1 x
ð Þ þ c 2 u 2 x
ð Þ þ u p x
ð Þ,
ð10:22Þ
where c 1 and c 2 are arbitrary (complex) constants.
10.2 First-Order Linear Differential Equations (FOLDEs)
In a discussion that follows, first-order linear differential equations (FOLDEs)
supply us with useful information. A general form of FOLDEs is expressed as
a x
ð Þ
du
dx
þ b x
ð Þu ¼ d x
ð Þ a x
ð Þ 6 ¼ 0
½
:
ð10:23Þ
An associated boundary condition is given by a boundary functional B(u) such that
B u
ð Þ ¼ αu a
ð Þ þ βu b
ð Þ ¼ σ,
ð10:24Þ
where α, β, and σ are real constants; u(x) is defined in an interval [a, b]. If in (10.23) d
(x) 0, (10.23) can readily be integrated to yield a solution. Let us multiply both
sides of (10.23) by w(x). Then we have
w x
ð Þa x
ð Þ
du
dx
þ w x
ð Þb x
ð Þu ¼ w x
ð Þd x
ð Þ:
ð10:25Þ
We define p(x) as
p x
ð Þ w x
ð Þa x
ð Þ,
ð10:26Þ
where w(x) is called a weight function. As mentioned in Sect. 2.3, the weight
function is a real and non-negative function within the domain considered. Here
we suppose that
dp x
ð Þ
dx
¼ w x
ð Þb x
ð Þ:
ð10:27Þ
Then, (10.25) can be rewritten as
d
dx
p x
ð Þu
½
¼w x
ð Þd x
ð Þ:
ð10:28Þ
384
10 Introductory Green’s Functions
(10.6). Hence, the general solution of (10.2) should be expressed as
u x
ð Þ ¼ c 1 u 1 x
ð Þ þ c 2 u 2 x
ð Þ þ u p x
ð Þ,
ð10:22Þ
where c 1 and c 2 are arbitrary (complex) constants.
10.2 First-Order Linear Differential Equations (FOLDEs)
In a discussion that follows, first-order linear differential equations (FOLDEs)
supply us with useful information. A general form of FOLDEs is expressed as
a x
ð Þ
du
dx
þ b x
ð Þu ¼ d x
ð Þ a x
ð Þ 6 ¼ 0
½
:
ð10:23Þ
An associated boundary condition is given by a boundary functional B(u) such that
B u
ð Þ ¼ αu a
ð Þ þ βu b
ð Þ ¼ σ,
ð10:24Þ
where α, β, and σ are real constants; u(x) is defined in an interval [a, b]. If in (10.23) d
(x) 0, (10.23) can readily be integrated to yield a solution. Let us multiply both
sides of (10.23) by w(x). Then we have
w x
ð Þa x
ð Þ
du
dx
þ w x
ð Þb x
ð Þu ¼ w x
ð Þd x
ð Þ:
ð10:25Þ
We define p(x) as
p x
ð Þ w x
ð Þa x
ð Þ,
ð10:26Þ
where w(x) is called a weight function. As mentioned in Sect. 2.3, the weight
function is a real and non-negative function within the domain considered. Here
we suppose that
dp x
ð Þ
dx
¼ w x
ð Þb x
ð Þ:
ð10:27Þ
Then, (10.25) can be rewritten as
d
dx
p x
ð Þu
½
¼w x
ð Þd x
ð Þ:
ð10:28Þ
384
10 Introductory Green’s Functions
