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Exercises 5.4
that solutions to the system ﬂow into (0
,
0) on one of them and ﬂow away from (0
,
0) on
the other. This is the case of a saddle (unstable) point.
(f)
The only remaining picture is the asymptotically stable spiral.
(One can also get a
diagram
.

%
for solution’s ﬂows with just one matching picture in Figure 5.7.)
31. (a)
Setting
y
0
=
v
and so
y
0
=
v
0
, we transform given equation to a ±rst order system
dy
dx
=
v,
dv
dx
=
f
(
y
)
.
(b)
By the chain rule,
dv
dy
=
dv
dx
·
dx
dy
=
dv
dx
±
dy
dx
=
f
(
y
)
v
⇒
dv
dy
=
f
(
y
)
v
.
This equation is separable. Separation variables and integration yield
vdv
=
f
(
y
)
dy
⇒
Z
vdv
=
Z
f
(
y
)
dy
⇒
1
2
v
2
=
F
(
y
)+
K,
where
F
(
y
) is an antiderivative of
f
(
y
). Substituting back
v
=
y
0
gives the required.
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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