
986 HIGHER-ORDER PAR TIAL DIFFERENTIAL EQUATIONS
For a 6= b, the solution of the problem is described by formula (1), where
f
1
(x) =
1
2(a
2
− b
2
)
−b
2
ϕ
0
(x) +
b
2
a
Φ
1
(x) + Φ
2
(x) −
1
2a
Φ
3
(x)
,
f
2
(x) =
1
2(a
2
− b
2
)
−b
2
ϕ
0
(x) −
b
2
a
Φ
1
(x) + Φ
2
(x) +
1
2a
Φ
3
(x)
,
f
3
(x) =
1
2(b
2
− a
2
)
−a
2
ϕ
0
(x) +
a
2
b
Φ
1
(x) + Φ
2
(x) −
1
2b
Φ
3
(x)
,
f
4
(x) =
1
2(b
2
− a
2
)
−a
2
ϕ
0
(x) −
a
2
b
Φ
1
(x) + Φ
2
(x) +
1
2b
Φ
3
(x)
,
Φ
n
(x) =
Z
x
x
0
ϕ
n
(ξ)(x − ξ)
n−1
dξ, n = 1, 2, 3,
and x
0
is an arbitrary constant.
For a = b, the solution of the problem is described by formula (2), where
f
1
(x) =
1
2
ϕ
0
(x) −
3
4a
Z
x
x
0
ϕ
1
(ξ) dξ +
1
8a
3
Z
x
x
0
ϕ
3
(ξ)(x − ξ)
2
dξ,
f
2
(x) =
1
2
ϕ
0
(x) +
3
4a
Z
x
x
0
ϕ
1
(ξ) dξ −
1
8a
3
Z
x
x
0
ϕ
3
(ξ)(x − ξ)
2
dξ,
f
3
(x) =
1
4
aϕ
′
0
(x) −
1
4
ϕ
1
(x) −
1
4a
Z
x
x
0
ϕ
2
(ξ) dξ ...