892
Appendix: All Solutions to the Problems
−
5
6
a 1 +
2
3
= 0
Solving for a 1 =
4
5
. Thus, the first approximate solution is y 1 =
4
5
(1 − x
2
).
Let the second approximate solution be y 2 = a 1 (1 − x
2
) + a 2 (x
2
− x
3
), then
y 2
= −2a 1 x + a 2 (2x − 3x
2
), y 2
= −2a 1 + a 2 (2 − 6x)
y 2
− y 2
+ 1 = (−2 + 2x)a 1 + (2 − 8x + 3x
2
)a 2 + 1
The Galerkin equations are
1
0 [(−2 + 2x)a 1 + (2 − 8x + 3x
2
)a 2 + 1](1 − x
2
)dx = 0
1
0 [(−2 + 2x)a 1 + (2 − 8x + 3x
2
)a 2 + 1](x
2
− x
3
)dx = 0
Integrating, we get
−
5
6
a 1 −
4
15
a 2 +
2
3
= 0
−
1
15
a 1 −
2
15
a 2 +
1
12
= 0
Solving for a 1 =
5
7
, a 2 =
15
56
. Therefore, the second approximate solution is
y 2 =
5
7
(1 − x
2
) +
15
56
(x
2
− x
3
) =
5
56
(8 − 5x
2
− 3x
3
)
8.12 Using the Galerkin method to find the first and second approximate solutions
of the boundary value problem of the following differential equation, and comparing
with the analytical solutions at x = 0.1n (n = 1, 2, . . . , 10).
y
+ y = x
2
(0 < x < 1)
y(0) = 0, y
(1) = 1
Solution The general solution of the homogeneous equation is
Y = c 1 cos x + c 2 sin x
Let the particular solution be
y
∗
= b 0 + b 1 x + b 2 x
2
Substituting the particular solution into the differential equation, we get
2b 2 + b 0 + b 1 x + b 2 x
2
= x
2
Solving for b 2 = 1, b 1 = 0, b 0 = −2. The general solution is