2.4
53
(d)
ote that
ylx
=
(1 
cos
8)/(8

sin 8).
Setting
x
=
1,
Y
=
2, the solution
of
the
equation
(1 
cos
8)/(8

sin 8)
=
2 is
8 ~
1.401.
Substitution
into
either
of
the expressions
yields
k
~
2.193.
2.4
2.
Rewrite
the
differential
equation
as
I
1
= a
y
+
t(t

4)
y
.
It
is evident
that
the
coefficient
1/t(t

4)
is continuous
everywhere
except
at
t
=
0,4.
Since the
initial
condition
is specified
at
t
=
2,
Theorem
2.4.1
assures
the existence
of a unique
solution
on the interval
a
<
t
<
4.
3.
The
function
tan
t
is discontinuous
at odd
multiples
of ~.
Since
~
<
It
<
3; ,
the initial
value problem
has a unique
solution
on the interval
(~ , 3;).
5. p(t)
=
2tl(4

t
2
)and
g(t)
=
3t
2
/(4

t
2
).
These
functions
are discontinuous
at
x
=
±2.
The initial
value problem
has a unique
solution
on the interval
(2,2).
6. The
function
In
t
is defined
and
continuous
on the
interval
(0,00).
At
t
= 1,
In
t
=
0,
so the
normal
form
of the
differential
equation
has
a singularity
there.
Also, cot
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 Fall '08
 BINEGAR
 Derivative, Stability theory, Theorem 2.4.1

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