Appendix: All Solutions to the Problems
909
y(x) = c 1
cos
1
2
√
λ ln(2x + 3)
√
2x + 3
+ c 2
sin
1
2
√
λ ln(2x + 3)
√
2x + 3
(6)
According to the boundary conditions y(0) = y(3) = 0, there are
c 1 cos
1
2
√
λ ln 3
+ c 2 sin
1
2
√
λ ln 3
= 0
c 1 cos(
√ λ ln 3) + c 2 sin(
√
λ ln 3) = 0
(7)
If Eq. (7) have the nonzero solution, then the coefficient determinant of c 1 and c 2
should be zero, namely
cos
1
2
√
λ ln 3
sin
1
2
√
λ ln 3
cos(
√ λ ln 3) sin(
√
λ ln 3)
= sin
1
2
√
λ ln 3
= 0
( 8 )
Therefore
1
2
√
λ ln 3 = nπ. The characteristic values are λ n =
4n
2 π
2
ln
2 3
, where n =
1, 2, . . .. It can be seen from the first equation of Eq. (7) that c 1 = 0. Substituting
these relations into Eq. (6), the characteristic functions of the given problem can be
obtained
y n =
c 2 sin
nπ ln(2x+3)
ln 3
√
2x + 3
(9)
where, the constant c 2 can be determined by the normal characteristic function namely
the isoperimetric condition of the problem, there is
1
0
⎧
⎨
⎩
c 2 sin
nπ ln(2x+3)
ln 3
√
2x + 3
⎫
⎬
⎭
2
dx =
1
2
c
2
2
1
0
1 − cos
nπ ln(2x+3)
ln 3
2x + 3
dx =
1
4
c
2
2 ln 3 = 1
(10)
Getting c 2 = ±
2
√
ln 3
. Therefore the characteristic functions are
y n = ±
2 sin
nπ ln(2x+3)
ln 3
√ (2x + 3) ln 3
(n = 1, 2, . . .)
(11)