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Exercises 2.3
⇒
s
Z
0
d
[
µ
(
x
)
y
]=
s
Z
0
µ
(
x
)
xdx
⇒
µ
(
x
)
y
(
x
)
±
±
±
x
=
s
x
=0
=
s
Z
0
µ
(
x
)
xdx
⇒
µ
(
s
)
y
(
s
)
−
µ
(0)
y
(0) =
s
Z
0
µ
(
x
)
xdx.
From the initial condition,
y
(0) = 2. Also, note that
µ
(0) = exp
0
Z
0
p
1+sin
2
tdt
=
e
0
=1
.
This yields
µ
(0)
y
(0) = 2 and so
µ
(
s
)
y
(
s
)=
s
Z
0
µ
(
x
)
xdx
+2
.
Dividing by
µ
(
s
) and interchanging
x
and
s
give the required.
(b)
The values of
µ
(
x
),
x
=0
.
1, 0
.
2,
...
,1
.
0, approximated by using Simpson’s rule, are
given in Table 2C.
Table 2–C
: Approximations of
ν
(
x
)=
R
x
0
√
1+sin
2
tdt
and
µ
(
x
)=
e
ν
(
x
)
using Simpson’s
rule.
x
ν
(
x
)
ν
(
x
)
ν
x
)
µ
(
x
)
µ
(
x
)
µ
(
x
)
x
ν
(
x
)
ν
(
x
)
(
x
)
µ
(
x
)
µ
(
x
)
µ
(
x
)
0.0
0.0
1.0000
0.6
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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