MTH306 Differential Equations Recitation Notes
2012.5.292012.6.27
Yin Su
Section 1.6 Bernoulli equations
•
Differential equations of form
d y
d x
=
F
(
ax
+
b y
+
c
)
.
– Use substitution
v
=
ax
+
b y
+
c
.
•
Homogeneous equations
d y
d x
=
F
(
y
x
)
.
– Use substitution
v
=
y
x
.
•
Bernoulli equations
d y
d x
+
P
(
x
)
y
=
Q
(
x
)
y
n
.
–
n
=
0: Standard linear first order equation. (Section 1.5)
–
n
=
1:
d y
d x
= (
Q
(
x
)

P
(
x
))
y
. Separable equation. (Section 1.4)
– all the other
n
: do the substitution
v
=
y
1

n
.
Steps:
(1) Substitution:
v
=
y
1

n
. Then
dv
d x
= (
1

n
)
y

n
d y
d x
(2) Multiply the original equation by
(
1

n
)
y

n
. We have
dv
d x
+ (
1

n
)
P
(
x
)

{z
}
new
P
(
x
)
v
= (
1

n
)
Q
(
x
)

{z
}
new
Q
(
x
)
(3) Solve (2) for
v
(
x
)
and get the implicit solution for
y
.
Example. §2.6 # 7 Solve the equation
x y
2
y
0
=
x
3
+
y
3
.
x y
2
+
y
3
(3) Solve this equation and get the solution.
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Equation
dv
d x
+ (

3
x
)
v
=
3
x
2
. Then
P
(
x
) =

3
x
,
Q
(
x
) =
3
x
2
.
(1) Integrating factor:
ρ
(
x
) =
e
R

3
x
d x
=
e

3ln
x
=
x

3
=
1
x
3
(2) Multiply both sides by the integrating factor, we have
1
x
3
v
(
x
) =
Z
1
x
3
·
3
x
2
d x
=
Z
3
x
d x
=
3ln

x

+
C
(3) Solve for
v
:
y
3
=
v
=
x
3
(
3ln

x

+
C
)
Section 2.2 Equilibrium solutions and stability.
•
Autonomous equation:
d x
dt
=
f
(
x
)
. (
x
is a function of
t
.)
•
Equilibrium solutions: the solution of
f
(
x
) =
0. (Critical points)
•
Stability: Suppose
x
=
c
is a critical point.
 Stable: Solution curves approach
x
=
c
as
t
→
+
∞
.
 Unstable: Solution curves diverge away from
x
=
c
as
t
→
+
∞
.
•
Steps to find the equilibrium solutions and stability:
(1) Find critical points of autonomous 1st order ODE;
(2) Classify critical points by analyzing sign of
f
(
x
)
around critical points.
(Pick points around the critical points.)
(3) Draw the phase diagram.
Example. §2.2 # 1
d x
dt
=
x

4.
Solution:
f
(
x
) =
x

4. This is an autonomous equation.
(1) Critical points:
f
(
x
) =
x

4
=
0
⇒
x
=
4.
(2) Stability at
x
=
4:
If
x
>
4 and
x
is close to 4,
f
(
x
) =
x

4
>
0. Then
x
(
t
)
%
; (We can pick
x
=
5.)
If
x
<
4 and
x
is close to 4,
f
(
x
) =
x

4
<
0. Then
x
(
t
)
&
. (We can pick
x
=
3.)
Then
x
=
4 is unstable.
(3) Phase diagram:
x
=
4 is unstable:
Example. §2.2 # 6
d x
dt
=
9

x
2
.
Solution:
f
(
x
) =
9

x
2
. This is an autonomous equation.
(1) Critical points:
f
(
x
) =
9

x
2
=
0
⇒
x
=
3,

3.
(2) Stability at
x
=
3:
If
x
>
3 and
x
is close to 3,
f
(
x
) =
9

x
2
<
0. Then
x
(
t
)
&
; (We can pick
x
=
4.)
If
x
<
3 and
x
is close to 3,
f
(
x
) =
9

x
2
>
0. Then
x
(
t
)
%
. (We can pick
x
=
2.)
Then
x
=
3 is stable.
Stability at
x
=

3:
If
x
>

3 and
x
is close to

3,
f
(
x
) =
9

x
2
>
0. Then
x
(
t
)
%
; (We can pick
x
=

2.)
If
x
<

3 and
x
is close to

3,
f
(
x
) =
9

x
2
<
0. Then
x
(
t
)
&
. (We can pick
x
=

4.)
Then
x
=

3 is unstable.
(3) Phase diagram:
x
=
3 is stable;
x
=

3 is unstable.