Graphics Programs Reference
In-Depth Information
Problems 6-10
Solve the givenboundary value problemwith the finite difference
methodusing
n
=
21.
6.
y
=
xy
,
y
(1)
=
1
.
5
y
(2)
=
3.
7.
y
+
2
y
+
xe
1
−
x
.
y
=
0,
y
(0)
=
0,
y
(1)
=
1. Exact solutionis
y
=
x
2
y
+
xy
+
8.
y
=
0,
y
(1)
=
0,
y
(2)
=
0
.
638961. Exact solutionis
y
=
sin(ln
x
).
y
=
y
2
sin
y
,
y
(0)
9.
=
0,
y
(
π
)
=
1.
y
+
2
y
(2
xy
+
2,
y
(1)
10.
y
)
=
0,
y
(0)
=
1
/
=−
2
/
9. Exact
solution s
y
=
x
2
)
−
1
.
(2
+
11.
w
0
I
0
I
0
x
L
/4
L
/4
L
/2
I
1
v
The simply supportedbeam consists of three segments with the moments of
inertia
I
0
and
I
1
as shown. A uniformlydistributed load of intensity
w
0
acts over
the middle segment. Modeling only the left half of the beam, wecan show that
the differentialequation
d
2
v
dx
2
M
E I
=−
for the displacement
v
is
⎧
⎨
⎩
x
L
L
4
in 0
<
x
<
d
2
v
dx
2
w
0
L
2
4
E I
0
×
=−
x
L
−
2
in
L
4
2
x
I
0
I
1
1
4
L
2
L
−
<
x
<
Introducing the dimensionless variables
x
L
E I
0
w
0
L
4
v
I
1
I
0
ξ
=
y
=
γ
=
changes the differentialequation to
⎨
⎩
1
4
ξ
1
4
−
in 0
<ξ<
d
2
y
d
=
2
in
1
4
2
2
ξ
1
4
1
4
1
2
−
ξ
−
ξ
−
<ξ<
γ
with the boundary conditions
ξ
=
1
/
2
=
dy
d
y
|
ξ
=
0
=
0
ξ
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