Solutions to selected problems in Section 2.1
1. The ODE is linear because it can be put in the form
y
0
+
p
(
t
)
y
=
g
(
t
). Another way to see
why this ODE is linear is because
y
and its derivatives appear by themselves or as multiples
of themselves and functions of
t
.
2. The ODE is nonlinear because of the
y
2
term.
3. The ODE is nonlinear because of the
y
0
/y
term.
4.
The equation is autonomous.
The equilibria are the roots of
y
2

y
= 0, i.e.
0 and
1.
5. Nonlinear because of the division by
y
.
6. Linear, homogeneous
7. Nonlinear because of
yy
0
.
8. Nonlinear.
9. Linear, inhomogeneous (multiply by
y
).
10. Linear, homogeneous.
13. (a). Note that for the given ODE, we have
p
(
t
) =
t
t
2

4
, g
(
t
) =
e
t
t

3
.
Discontinuous points of
p
(
t
):
t
= 2
,

2.
Discontinuous points of
g
(
t
):
t
= 3.
These discontinuous points, together with
∞
and
∞
, can form the following intervals:
(
∞
,

2)
,
(

2
,
2)
,
(2
,
3)
,
(3
,
∞
)
.
From the initial condition, we have
t
0
= 5.
The only interval found above that contains
t
0
= 5 is (3
,
∞
); hence this is the interval on which Theorem 2.1 guarantees the existence of
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 PRATHER
 Following, Order theory, Octave

Click to edit the document details