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Exercises 1.3
9. (a)
The function
φ
(
x
), being a solution to the given initial value problem, satisFes
dφ
dx
=
x
−
φ
(
x
)
,φ
(0) = 1
.
(1.4)
Thus
d
2
φ
dx
2
=
d
dx
±
dφ
dx
²
=
d
dx
(
x
−
φ
(
x
)) = 1
−
dφ
dx
=1
−
x
+
φ
(
x
)
,
where we have used (1.4) substituting (twice)
x
−
φ
(
x
)for
dφ/dx
.
(b)
±irst we note that any solution to the given di²erential equation on an interval
I
is
continuously diferentiable on
I
. Indeed, if
y
(
x
) is a solution on
I
,then
y
0
(
x
)doesex
ist
on
I
,andso
y
(
x
) is continuous on
I
because it is di²erentiable. This immediately implies
that
y
0
(
x
) is continuous as the di²erence of two continuous functions,
x
and
y
(
x
).
±rom (1.4) we conclude that
dφ
dx
³
³
³
x
=0
=[
x
−
φ
(
x
)]
³
³
x
=0
=0
−
φ
(0) =
−
1
<
0
and so the continuity of
φ
0
(
x
) implies that, for

x

small enough,
φ
0
(
x
)
<
0. By the
Monotonicity Test, negative derivative of a function results that the function itself is
decreasing.
When
x
increases from zero, as far as
φ
(
x
)
>x
, one has
φ
0
(
x
)
<
0andso
φ
(
x
) decreases.
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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