Exercises 2.3
25. (a)
This is a linear problem and so an integrating factor is
µ
(
x
)=exp
±Z
2
xdx
²
=exp
(
x
2
)
.
Multiplying the equation by this integrating factor yields
e
x
2
dy
dx
+2
xe
x
2
y
=
e
x
2
⇒
D
x
³
ye
x
2
´
=
e
x
2
⇒
x
Z
2
D
t
³
ye
t
2
´
dt
=
x
Z
2
e
t
2
dt,
where we have changed the dummy variable
x
to
t
and integrated with respect to
t
from
2 (since the initial value for
x
in the initial condition is 2) to
x
. Thus, since
y
(2) = 1,
ye
x
2
−
e
4
=
x
Z
2
e
t
2
dt
⇒
y
=
e
−
x
2
e
4
+
x
Z
2
e
t
2
dt
=
e
4
−
x
2
+
e
−
x
2
x
Z
2
e
t
2
dt .
(b)
We will use Simpson’s rule (page A.3 of the Appendix B) to approximate the deFnite
integral found in part (a) with upper limit
x
= 3.
Simpson’s rule requires an even
number of intervals, but we don’t know how many are required to obtain the desired 3
place accuracy. Rather than make an error analysis, we will compute the approximate
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, et dt, intervals yields Table

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