898
Appendix: All Solutions to the Problems
(48λ − 105)a 1 + (8λ − 21)a 2 = 0
(88λ − 231)a 1 + (32λ − 363)a 2 = 0
In order to make a 1 and a 2 have the non-null solution, then the values of the
determinant composed of their coefficients should be zero, namely
48λ − 105 8λ − 21
88λ − 231 32λ − 363
= 0
Expanding the determinant and arranging, we get
13λ
2
− 267λ +
2079
4
= 0
Solve for λ, two characteristic values can be obtained
λ 1 =
267 −
√
44262
26
= 2.177486
λ 2 =
267 +
√
44262
26
= 18.360975
Integrating from the Galerkin equations, we get
a 1 =
21 − 8λ
48λ − 105
a 2
Substituting λ 1 and λ 2 into the above equation respectively, we get
a 1 =
21 − 8λ 1
48λ 1 − 105
a 2 =
26663 + 364
√
4918
−7007
a 2
and
a 1 =
21 − 8λ 2
48λ 2 − 105
a 2 =
26663 − 364
√
4918
−7007
a 2
The corresponding characteristic function is
y 21 = (1 − x
2
)
26663 + 364
√
4918
−7007
+ x
2
a 2
y 22 = (1 − x
2
)
26663 − 364
√
4918
−7007
+ x
2
a 2
Appendix: All Solutions to the Problems
(48λ − 105)a 1 + (8λ − 21)a 2 = 0
(88λ − 231)a 1 + (32λ − 363)a 2 = 0
In order to make a 1 and a 2 have the non-null solution, then the values of the
determinant composed of their coefficients should be zero, namely
48λ − 105 8λ − 21
88λ − 231 32λ − 363
= 0
Expanding the determinant and arranging, we get
13λ
2
− 267λ +
2079
4
= 0
Solve for λ, two characteristic values can be obtained
λ 1 =
267 −
√
44262
26
= 2.177486
λ 2 =
267 +
√
44262
26
= 18.360975
Integrating from the Galerkin equations, we get
a 1 =
21 − 8λ
48λ − 105
a 2
Substituting λ 1 and λ 2 into the above equation respectively, we get
a 1 =
21 − 8λ 1
48λ 1 − 105
a 2 =
26663 + 364
√
4918
−7007
a 2
and
a 1 =
21 − 8λ 2
48λ 2 − 105
a 2 =
26663 − 364
√
4918
−7007
a 2
The corresponding characteristic function is
y 21 = (1 − x
2
)
26663 + 364
√
4918
−7007
+ x
2
a 2
y 22 = (1 − x
2
)
26663 − 364
√
4918
−7007
+ x
2
a 2