Fall 2003 Math 308/501–502
1 Introduction
1.2 Solutions & Initial Value Problems
Wed, 03/Sep
c 2003, Art Belmonte
Summary
•
A general form for an
n
thorder ODE is
F
x
,
y
,
y
0
,
y
00
, . . .
y
(
n
)
=
0
.
Here
x
is the independent variable and
y
is the dependent
variable. This equation is assumed to hold on some open real
interval
I
, possibly of infinite extent.
•
Preferably, we can isolate
y
(
n
)
, the
n
th derivative of
y
, and
rewrite the ODE in the form
y
(
n
)
=
f
x
,
y
,
y
0
,
y
00
, . . . ,
y
(
n

1
)
.
•
An
explicit solution
is a function
y
=
φ(
x
)
which when
substituted into of one of the aforementioned ODEs yields an
identity for all
x
in
I
. This is shown via differentiation and
direct substitution.
•
An
implicit solution
is a relation
G
(
x
,
y
)
=
0 which yields
one or more explicit solutions on
I
, perhaps only
theoretically (say via the Implicit Function Theorem from
mulitivariable calculus or advanced calculus). This is usually
shown via implicit differentiation.
•
An
n
thorder
initial value problem
(IVP) consists of an
n
thorder differential equation together with
n
initial
conditions
(ICs)
y
(
k
)
(
x
0
)
=
y
k
,
k
=
0
,
1
, . . . ,
n

1
.
Here
x
0
and the
y
k
are constants and
y
(
k
)
denotes the
k
th
derivative of
y
, with
y
(
0
)
signifying
y
itself.
•
Therefore, a firstorder IVP is usually specified as
y
0
=
dy
dx
=
f
(
x
,
y
),
y
(
x
0
)
=
y
0
.
•
Similarly, a secondorder IVP is given as
y
00
=
d
2
y
dx
2
=
f
(
x
,
y
,
y
0
),
y
(
x
0
)
=
y
0
,
y
0
(
x
0
)
=
y
1
.
•
Existence and Uniqueness Theorem (EUT)
: Given the IVP
y
0
=
f
(
x
,
y
),
y
(
x
0
)
=
y
0
, let
f
and
f
y
=
∂
f
/∂
y
be
continuous on
R
= {
(
x
,
y
)
:
a
<
x
<
b
,
c
<
y
<
d
}
, an open
rectangle containing
(
x
0
,
y
0
)
. Then the IVP has a
unique
solution
y
=
φ(
x
)
on
J
=
(
x
0

δ <
x
<
x
0
+
δ)
, an open
interval centered at
x
0
.
Hand Examples
14/6
Determine whether the function
θ
=
2
e
3
t

e
2
t
is a solution of the
differential equation
d
2
θ
dt
2

θ
d
θ
dt
+
3
θ
= 
2
e
2
t
.
Solution
Now
d
θ/
dt
=
6
e
3
t

2
e
2
t
and
d
2
θ/
dt
2
=
18
e
3
t

4
e
2
t
.
Substituting into the DE yields
(
18
e
3
t

4
e
2
t
)

(
2
e
3
t

e
2
t
)(
6
e
3
t

2
e
2
t
)
+
3
(
2
e
3
t

e
2
t
)
=

2
e
2
t

12
e
6
t
+
10
e
5
t

2
e
4
t
+
24
e
3
t

9
e
2
t
=
0
an equation which holds on
no
open interval
I
⊂
R
. Therefore,
θ
is
not
a solution of the differential equation. NOTE: The fact that
the last equation actually holds at one isolated point (
t
≈
0
.
36) is
insufficient. It must hold over an entire open interval.
14/10
Determine whether the relation
y

ln
y
=
x
2
+
1 is an implicit
solution of the differential equation
dy
dx
=
2
xy
y

1
.
Solution
Via the Implicit Function Theorem, we assume that the
relationship does indeed define
y
as a function of
x
. Implicitly
differentiate the relation with respect to
x
, then isolate
dy
/
dx
.
y

ln
y
=
x
2
+
1
y
0

1
y
y
0
=
2
x
y
0
=
2
x
(
1

1
y
)
dy
dx
=
2
xy
y

1
Accordingly, the relationship is an implicit solution of the DE.
14/14
Show that
φ(
x
)
=
c
1
sin
x
+
c
2
cos
x
is an explicit solution to
y
00
+
y
=
0 for any constants
c
1
,
c
2
.
Solution
Now
y
0
=
φ
0
(
x
)
=
c
1
cos
x

c
2
sin
x
and
y
00
=
φ
00
(
x
)
= 
c
1
sin
x

c
2
cos
x
. Substituting into the DE
yields
(

c
1
sin
x

c
2
cos
x
)
+
(
c
1
sin
x
+
c
2
cos
x
)
=
0
,
an identity for all real
x
. Therefore
φ(
x
)
=
c
1
sin
x
+
c
2
cos
x
is
an explicit solution of the DE for any constants
c
1
,
c
2
. We say it
constitutes a
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 Spring '08
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 Math, Derivative, IVP

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