1
MA360
Handout 6b (week of Feb 20 )
Sp2006 Kallfelz
This cursory review of linear ODEs (first and second order) is entirely exampledriven.
•
Firstorder
Any garden variety first order linear ODE is of the form:
( ) ( )
( ) ( )
( )
t
f
t
x
t
a
t
x
t
a
=
+
1
0
&
,
where:
( )
( )
t
x
t
x
dt
d
≡
&
, for the unknown function
x
(
t
) and
a
0
(
t
),
a
1
(
t
),
f
(
t
) are given
functions.
We can always make this ODE simpler (by setting the leading term to be
monic
, i.e. the leading coefficient = 1) by dividing through by
a
0
(
t
).
1
Hence:
( )
( ) ( )
( )
t
g
t
x
t
p
t
x
=
+
&
, where:
( )
( )
( )
( )
( )
( )
t
a
t
f
t
a
t
a
t
g
t
p
0
0
1
,
=
=
Lemma1:
Multiplying (Eqn1) by the integrating factor
( )
( )
∫
=
dt
t
p
e
t
r
makes it
separable.
Proof:
Multiply both sides by
r
(
t
) and bringing all terms to one side:
( ) ( )
( )
[
]
( )
( ) ( )
0
=
+

t
x
t
r
t
r
t
g
t
x
t
p
&
If the above is
separable, then the following condition must hold
2
:
( ) ( )
( )
[
]
( )
{
}
( )
t
r
t
r
t
g
t
x
t
p
t
x
∂
∂
∂
∂
=

For the left hand side:
( ) ( )
( )
[
]
( )
{
}
( ) ( ) ( )
(
)
( ) ( )
(
)
t
r
t
g
t
r
t
x
t
p
t
r
t
g
t
x
t
p
x
x
x
∂
∂
∂
∂
∂
∂

=

Since
p
,
g
,
r
are all homogeneous functions depending on
t
only, they behave as
constants, with respect to taking the partial derivative with respect to
x
.
Hence the
left side simplifies to:
p
(
t
)
r
(
t
)
(Since:
( ) ( ) ( )
( ) ( )
( ) ( )
t
r
t
p
x
t
r
t
p
t
r
t
x
t
p
x
x
=
=
∂
∂
∂
∂
and
( ) ( )
0
=
∂
∂
t
r
t
g
x
)
On the other hand, the right hand side is just a total derivative with respect to
t
,
since
r
is a homogeneous function with respect to
t
.:
( )
( )
t
r
t
r
dt
d
t
=
∂
∂
So the above simplifies to:
( ) ( )
( )
t
r
t
r
t
p
dt
d
=
, which implies that
( )
( )
∫
=
dt
t
p
e
t
r
.
1
Since the leading coefficient function must be nonzero.
2
This is true because if we assume there exists an exact solution curve
φ
(
x
,
t
) =
c
(in the (
x
,
t
) plane)
for the
equation:
M
(
x
,
t
) +
N
(
x
,
t
)
x
/
(
t
) = 0, then, when taking the total differential of
φ
(
x
,
t
) =
c
:
d
φ
(
x
,
t
) = 0 =
∂
φ
/
∂
t
dt
+
∂
φ
/
∂
x
dx
⇒
0 =
∂
φ
/
∂
t
+
∂
φ
/
∂
x
d
x
/
d
t
=
M
(
x
,
t
) +
N
(
x
,
t
)
x
/
(
t
)
⇒
M
(
x
,
t
) =
∂
φ
/
∂
t
,
N
(
x
,
t
) =
∂
φ
/
∂
x
.
So because of the continuity of
φ
(
x
,
t
) :
∂
2
φ
/
∂
t
∂
x
=
∂
2
φ
/
∂
x
∂
t
⇒
∂
M
/
∂
x
=
∂
N
/
∂
t
.
In the specific case above:
N
(
x
,
t
) =
r
(
t
),
M
(
x
,
t
) = [
p
(
t
)
x
(
t
) –
g
(
t
)]
r
(
t
)
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2
•
Example
(The Bernoulli Equations)
A first order Bernoulli Equation is of the form:
( )
( ) ( )
( )
( )
t
x
t
q
t
x
t
p
t
x
n
=
+
&
(for any integer
n
> 1)
This equation is
not
linear, as one of its terms (i.e. on the right hand side) involves a
power (greater than 1) of the unknown function.
It can be made linear by first diving
both sides by
x
n
(
t
) and then using the substitution:
u
(
t
) =
x
(1n)
(
t
) =
x
(
t
)
x
n
(
t
):
( )
( ) ( )
( )
( )
q
px
x
x
t
x
t
q
t
x
t
p
t
x
n
n
n
=
+
⇒
=
+


1
&
&
(dividing through by
x
n
(
t
)).
Using substitution
u
(
t
) =
x
(1n)
(
t
) =
x
(
t
)
x
n
(
t
), note:
( )
( )
( )
(
)
( ) ( )
t
x
t
x
n
t
u
t
x
t
u
n
n
&
&



=
⇒
=
1
1
Hence the equation becomes:
( )
( ) ( )
( )
( )
( )
( ) ( )
( )
t
q
t
u
t
p
t
u
q
px
x
x
t
x
t
q
t
x
t
p
t
x
n
n
n
n
=
+
⇒
=
+
⇒
=
+



&
&
&
1
1
1
(which
is
linear)
Consider the Bernoulli Eqn:
( )
( )
( )
t
x
t
t
x
t
x
t
2
2
3
=
+
&
.
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 Spring '06
 WILLIAMM.KALLFELZ
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