Appendix: All Solutions to the Problems
897
1
0 {[(x + 2)y 2
]
+ y 2 − 3x}ϕ 1 dx = 0
1
0 {[(x + 2)y 2
]
+ y 2 − 3x}ϕ 2 dx = 0
Integrating, we get
3776
45
a 1 +
63428
315
a 2 − 1 = 0
63428
315
a 1 +
1595359
3465
a 2 −
17
7
= 0
or
26432a 1 + 63428a 2 = 315
697708a 1 + 1595359a 2 = 8415
Solve for a 1 and a 2
a 1 =
31208535
2085693936
=
10402845
695231312
= 0.0149631
a 2 =
−2647260
2085693936
= −
220605
173807828
= −0.0012692
Thus, the second approximate solution is
y 2 =
10402845
695231312
(x
4
− 4x + 3) −
220605
173807828
(x
5
+ x
4
− 9x + 7)
8.15 Find the first, second characteristic value λ and the characteristic function of
the boundary value problem by the Galerkin method
y
+ λ(1 + x
2
)y = 0
y(−1) = y(1) = 0
Solution Let the second approximate solution be
y 2 = (1 − x
2
)(a 1 + a 2 x
2
)
Calculating the second derivative and the residual
y 2
= −2a 1 + (2 − 12x
2
)a 2
R = y 2
+ λ(1 + x
2
)y 2 = [λ(1 − x
4
) − 2]a 1 + [2 − 12x
2
+ λ(x
2
− x
6
)]a 2
The Galerkin equations are
1
−1 {[λ(1 − x
4
) − 2]a 1 + [λ(x
2
− x
6
) + 2 − 12x
2
]a 2 }(1 − x
2
)dx = 0
1
−1 {[λ(1 − x
4
) − 2]a 1 + [λ(x
2
− x
6
) + 2 − 12x
2
]a 2 }(x
2
− x
4
)dx = 0
Integrating the Galerkin equations and arranging, we get