Appendix: All Solutions to the Problems
907
Making the substitution, x + 1 = e
T , then Eq. (2) can be transformed into the
second-order linear differential equation with constant coefficients
d
2 y
dt 2 +
dy
dt
+ (3 − λ)y = 0
( 3 )
The characteristic equation of the equation is
r
2
+ r + (3 − λ) = 0
( 4 )
The two roots of Eq. (4) are 为r 1,2 = −
1
2
±
1
2
√
−(11 − 4λ). When 11 − 4λ > 0
namely λ <
11
4
, the general solution of Eq. (3) is
y(t) = e
−
1
2 t
c 1 cos
√
11 − 4λ
2
t + c 2 sin
√
11 − 4λ
2
t
(5)
Reducing the variable t to the variable x, we obtain the general solution of Eq.
(2)
y(x) = c 1
cos
1
2
√
11 − 4λ ln(x + 1)
√
x + 1
+ c 2
sin
1
2
√
11 − 4λ ln(x + 1)
√
x + 1
(6)
According to the boundary conditions y(0) = y(1) = 0, there are
c 1 + 0 = 0
c 1
cos(
1
2
√
11−4λ ln 2)
√
2
+ c 2
sin(
1
2
√
11−4λ ln 2)
√
2
= 0
(7)
If Eq. (7) have the nonzero solution, then the coefficient determinant of c 1 and c 2
should be zero, namely
1
0
cos(
1
2
√
11−4λ ln 2)
√
2
sin(
1
2
√
11−4λ ln 2)
√
2
=
sin
1
2
√
11 − 4λ ln 2
√
2
= 0
(8)
Thus
1
2
√
11 − 4λ ln 2 = nπ. The characteristic values are λ n =
11
4
−
n
2 π
2
ln
2 2
, where
n = 1, 2, . . .. Obviously, It follows 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(x+1)
ln 2
√
x + 1
(9)