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(c) Use Galerkin's method with the trial solution
x
2
2
=
)
+
u
u
1
(
y
)
x
(
1
−
x
u
2
(
y
)
(
1
−
x
)
8035
e
−
3
.
1416
y
1965
e
−
10
.
1059
y
Answer:
u
1
(
y
)
=
0
.
+
0
.
e
−
3
.
1416
y
e
−
10
.
1059
y
(
)
=
.
(
−
)
u
2
y
0
9105
7.8 Solve the differential equation
u
+
u
=
x
for 0
≤
x
≤
1
,
with
u
(
0
)
=
u
(
1
)
=
0
.
Use Galerkin's method and compare your results with the exact solution
u
=
x
−
sin
x
/
cos 1
.
u
(
Hint:
The two term approximating solution
x
)
=
(
1
−
x
)(
u
1
+
u
2
x
)
,
with
3]
u
1
x
2
x
2
x
3
(
)
=
−
/
/
−
/
=
N
u
u
x
[
x
2
2
u
u
2
satisfies the boundary conditions. Form
R
=
L
u
−
f
,
which leads to the Galerkin
dx
2
d
2
orthogonality condition, with
L
=
/
+
1 and
f
=
x,
1
1
N
u
L
N
u
dx
N
u
u
−
fdx
=
0 r
0
0
72
−
giving
17
u
1
75
=
u
1
=−
0
.
8597 and
u
2
=−
0
.
7706
119
58
u
2
−
147
k
u
u
=
p
u
0446
x
2
2569
x
3
Note that
k
u
is not symmetric. Finally,
u
(
x
)
=−
0
.
8597
x
+
0
.
+
0
.
.
Results:
Solution
u
x
Exact
Galerkin
1/4
−
0.2079
−
0.2081
1/2
−
0.3873
−
0.3866
3/4
−
0.5115
−
0.5115
1
−
0.5574
−
0.5582
7.9 Use the method of moments to find a solution of the nonlinear differential equation
[
u
]
=
(
1
+
au
)
0 with boundary conditions
u
(
0
)
=
0
,u
(
1
)
=
1
.
Choose a polynomial
as a trial solution.
Hint:
A form of a polynomial that satisfies the boundary conditions is
m
x
i
+
1
u
(
x
)
=
x
+
1
u
i
(
−
x
)
i
=
)
u
+
(
u
)
2
The residual is
R
=
(
1
+
a
u
a
.
ψ
=
=
+
(
x
2
−
).
=
As a first approximation, choose
1
,
u
x
u
1
x
For
a
1
,
1
1
1
1
x
2
]
2
dx
Rdx
=
0
=
[1
+
x
+
u
1
(
−
x
)
]2
u
1
dx
+
[1
+
u
1
(
2
x
−
1
)
0
0
0
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