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MA 341
Homework 3
(Nonlinear Ordinary Differential Equations)
Hoon Hong
1st-order, 1 variable (Sketch)
Problem:
y' = K y C1
2
y K2
Solution:
stable
saddle
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Problem:
y' = K y K1
y K2
y K4
2
Solution:
saddle
stable
unstable
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Problem:
y' = y2 C1
y K1
y K2
2
Solution:
saddle
unstable
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Problem:
y' = K y K2
2
y K1
Solution:
saddle
stable
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Problem:
y' = y C1
y K1
y K3
Solution:
unstable
saddle
stable
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Problem:
y' = K y2 C2
y K1
y K2
Solution:
saddle
stable
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1st-order, 2 variables (Sketch)
Problem:
y1' = y1 y2 K1
y2' = y1 Ky2
1. Equilibriums:
y1 K1
K1 y1
y12 K1 = 0
y1 = K1 , y1 = 1
----y1 = K1
Ky2 K1 = 0
Ky2 K1 = 0
y2 = K1
----y1 = 1
y2 K1 = 0
1 Ky2 = 0
y2 = 1
----K1, K1 , 1, 1
2. Around [-1, -1]:
y1 = z1 K1
y2 = z2 K1
z1' = z1 K1
z2 K1 K1
z2' = z1 Kz2
z1' = z1 z2 Kz1 Kz2
z2' = z1 Kz2
z1' = Kz1 Kz2
z2' = z1 Kz2
z' =
K1 K1
1 K1
λ = K1 CI
z
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λ = K1 KI
stable
3. Around [1, 1]:
y1 = z1 C1
y2 = z2 C1
z1' = z1 C1
z2 C1 K1
z2' = z1 Kz2
z1' = z1 z2 Cz1 Cz2
z2' = z1 Kz2
z1' = z1 Cz2
z2' = z1 Kz2
z' =
λ=
1
1
1 K1
z
2 , v = 1,
2 K1
λ = K 2 , v = 1, K1 K 2
saddle
4. Sketch:
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Problem:
y1' = y12 Ky22 K1
y2' = y1 y2
1. Equilibriums:
K1 0 y12 K1
y1 0
0
0
0
y1
y12 y12 K1 = 0
y1 = K1 , y1 = 0 , y1 = 1
----y1 = K1
Ky22 = 0
Ky2 = 0
y2 = 0
----y1 = 0
Ky22 K1 = 0
0 =0
----y1 = 1
Ky22 = 0
y2 = 0
y2 = 0
----K1, 0 , 1, 0
2. Around [-1, 0]:
y1 = z1 K1
y2 = z2
z1' = z1 K1
2
Kz22 K1
z2' = z1 K1 z2
z1' = z12 Kz22 K2 z1
z2' = z1 z2 Kz2
z1' = K2 z1
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z2' = Kz2
z' =
K2
0
z
0 K1
λ = K2, v = 1, 0
λ = K1, v = 0, 1
stable
3. Around [1, 0]:
y1 = z1 C1
y2 = z2
z1' = z1 C1
2
Kz22 K1
z2' = z1 C1 z2
z1' = z12 Kz22 C2 z1
z2' = z1 z2 Cz2
z1' = 2 z1
z2' = z2
z' =
2 0
0 1
z
λ = 2, v = K1, 0
λ = 1, v = 0, K1
unstable
4. Sketch:
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Problem:
y1' = Ky22 Ky1 C2 y2 K3
y2' = y2 Cy1 C1
1. Equilibriums:
K1
2
Ky1 K3
1
y1 C1
0
0
1
y1 C1
Ky12 K5 y1 K6 = 0
y1 = K3 , y1 = K2
----y1 = K3
Ky22 C2 y2 = 0
y2 K2 = 0
y2 = 2
----y1 = K2
Ky22 C2 y2 K1 = 0
y2 K1 = 0
y2 = 1
----K3, 2 , K2, 1
2. Around [-3, 2]:
y1 = z1 K3
y2 = z2 C2
z1' = K z2 C2
2
Kz1 C4 C2 z2
z2' = z2 Cz1
2
z1' = Kz2 Kz1 K2 z2
z2' = z2 Cz1
z1' = Kz1 K2 z2
z2' = z2 Cz1
z' =
K1 K2
λ=I
λ = KI
1
1
z
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center
3. Around [-2, 1]:
y1 = z1 K2
y2 = z2 C1
z1' = K z2 C1
2
Kz1 C1 C2 z2
z2' = z2 Cz1
z1' = Kz22 Kz1
z2' = z2 Cz1
z1' = Kz1
z2' = z2 Cz1
z' =
K1 0
1 1
z
λ = K1, v = 2, K1
λ = 1, v = 0, 2
saddle
4. Sketch:
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Problem:
y1' = y12 Ky22 C2 y1
y2' = Ky1 y2 Ky2
1. Equilibriums:
K1
0
y12 C2 y1
Ky1 K1
0
0
0
Ky1 K1
0
Ky1 K1
2
y12 C2 y1 = 0
y1 = K2 , y1 = K1 , y1 = 0
----y1 = K2
Ky22 = 0
y2 = 0
y2 = 0
----y1 = K1
Ky22 K1 = 0
0 =0
----y1 = 0
Ky22 = 0
Ky2 = 0
y2 = 0
----K2, 0 , 0, 0
2. Jacobian:
2 y1 C2 K2 y2
J=
Ky2
Ky1 K1
3. Around [-2, 0]:
K2 0
z' =
z
0 1
λ = K2, v = 3, 0
λ = 1, v = 0, 3
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saddle
4. Around [0, 0]:
2 0
z' =
z
0 K1
λ = 2, v = K3, 0
λ = K1, v = 0, K3
saddle
5. Sketch:
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Problem:
y1' = y12 Cy22 K2 y1 K1
y2' = y1 Cy2 K1
1. Equilibriums:
1
0
y12 K2 y1 K1
1 y1 K1
0
1
0
y1 K1
2 y12 K4 y1 = 0
y1 = 0 , y1 = 2
----y1 = 0
y22 K1 = 0
y2 K1 = 0
y2 = 1
----y1 = 2
y22 K1 = 0
y2 C1 = 0
y2 = K1
----0, 1 , 2, K1
2. Jacobian:
2 y1 K2 2 y2
J=
1
1
3. Around [0, 1]:
K2 2
z' =
z
1 1
1
1
C
2
2
17 , v = 2,
3
1
C
2
2
17
1
1
K
2
2
17 , v = 2,
3
1
K
2
2
17
λ=K
λ=K
saddle
4. Around [2, -1]:
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z' =
2 K2
1
1
z
λ=
3
1
C
I 7
2
2
λ=
3
1
K
I 7
2
2
unstable
5. Sketch:
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Problem:
y1' = y1 Ky2 C1
y2' = y12 K2 y1 Ky2 C3
1. Equilibriums:
K1
y1 C1
K1 y12 K2 y1 C3
Ky12 C3 y1 K2 = 0
y1 = 1 , y1 = 2
----y1 = 1
2 Ky2 = 0
2 Ky2 = 0
y2 = 2
----y1 = 2
3 Ky2 = 0
3 Ky2 = 0
y2 = 3
----1, 2 , 2, 3
2. Jacobian:
K1
1
J=
2 y1 K2 K1
3. Around [1, 2]:
1 K1
z' =
z
0 K1
λ = 1, v = K1, 0
λ = K1, v = K1, K2
saddle
4. Around [2, 3]:
1 K1
z' =
z
2 K1
λ=I
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λ = KI
center
5. Sketch:
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1st-order, 1 variable (Exact)
Problem:
y' = 2 y K4
y K6
y(0) = 5
1. Find the general solution:
dy
= 2 dt
y K4 y K6
K
1
2 y K4
C
1
2 y K6
dy = 2 dt
1
1
ln y K4 C
ln y K6 = 2 t CC
2
2
K
Kln y K4 Cln y K6
ln
y K6
y K4
= 4 t CC
= 4 t CC
y K6
= e4 t C C
y K4
y K6
= C e4 t, C O 0
y K4
2. Find the particular solution:
C =1
y K6
= e4 t
y K4
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Problem:
y' = y K2
y K4 t
y(0) = 1
1. Find the general solution:
dy
= t dt
y K2 y K4
K
1
2 y K2
C
1
2 y K4
dy = t dt
1
1
1 2
ln y K2 C
ln y K4 =
t CC
2
2
2
K
Kln y K2 Cln y K4
ln
y K4
y K2
= t2 CC
= t2 CC
2
y K4
= et C C
y K2
2
y K4
= C et , C O 0
y K2
2. Find the particular solution:
C =3
2
y K4
= 3 et
y K2
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Problem:
y' = y C1
y K2 e3 t
y(0) = 3
1. Find the general solution:
dy
= e3 t dt
y C1 y K2
K
1
3 y C1
C
1
3 y K2
dy = e3 t dt
1
1
1 3t
ln y C1 C
ln y K2 =
e CC
3
3
3
K
Kln y C1 Cln y K2
ln
y K2
y C1
= e3 t CC
= e3 t CC
3t
y K2
= ee C C
y C1
3t
y K2
= C ee , C O 0
y C1
2. Find the particular solution:
1 K1
C=
e
4
y K2
1 K1 e3 t
=
e e
y C1
4
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Problem:
y' = 2 y K5
y K7
y(0) = 3
1. Find the general solution:
dy
= 2 dt
y K5 y K7
K
1
2 y K5
C
1
2 y K7
dy = 2 dt
1
1
ln y K5 C
ln y K7 = 2 t CC
2
2
K
Kln y K5 Cln y K7
ln
y K7
y K5
= 4 t CC
= 4 t CC
y K7
= e4 t C C
y K5
y K7
= C e4 t, C O 0
y K5
2. Find the particular solution:
C =2
y K7
= 2 e4 t
y K5
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Problem:
y' = y C2 y et
y(0) = 1
1. Find the general solution:
dy
= et dt
y C2 y
K
1
2 y C2
C
1
2y
dy = et dt
1
1
ln y C2 C
ln y = et CC
2
2
K
Kln y C2 Cln y = 2 et CC
ln
y
y C2
= 2 et CC
t
y
= e2 e C C
y C2
t
y
= C e2 e , C O 0
y C2
2. Find the particular solution:
1 K2
C=
e
3
y
1 K2 2 et
=
e e
y C2
3
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Problem:
y' = y C3
y C1 t2
y(0) = 3
1. Find the general solution:
dy
= t2 dt
y C3 y C1
K
1
2 y C3
C
1
2 y C1
dy = t2 dt
1
1
1 3
ln y C3 C
ln y C1 =
t CC
2
2
3
K
Kln y C3 Cln y C1
ln
y C1
y C3
=
=
2 3
t CC
3
2 3
t CC
3
2 3
t CC
y C1
3
=e
y C3
2 3
t
y C1
3
=C e , C O0
y C3
2. Find the particular solution:
2
C=
3
2 3
t
y C1
2 3
=
e
y C3
3
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1st-order, 2 variables (Exact)
Problem:
y1' =
y12 Ky2 C3
y2 K3 y2 K5
y2' =
y12 Ky2 C3
y1 K1 y1 K2
y1(0) = 0
y2(0) = 2
1. Find the general solution:
y1 K1 y1 K2
dy1/dy2 =
y2 K3 y2 K5
dy1
y1 K1 y1 K2
1
1
K
y1 K2
y1 K1
dy2
y2 K3 y2 K5
=
dy1 =
ln y1 K2 Kln y1 K1
=
2 ln y1 K2 K2 ln y1 K1
ln
y1 K2 2
y1 K1 2
= ln
y2 K5
y2 K3
1
1
K
2 y2 K5
2 y2 K3
1
1
ln y2 K5 K
ln y2 K3 CC
2
2
= ln y2 K5 Kln y2 K3 CC
CC
y1 K2 2
C y2 K5
=
, C O0
2
y2 K3
y1 K1
2. Find the particular solution:
4 =3 C
C=
4
3
y1 K2 2
4
=
2
3
y1 K1
y2 K5
y2 K3
dy2
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Problem:
y1' =
2 y2 K1 sin y1 cos y2
y2 C1 y2 K2
y2' =
sin y1 cos y2
y1 K1 y1 K3
y1(0) = 0
y2(0) = 0
1. Find the general solution:
2 y2 K1 y1 K1 y1 K3
dy1/dy2 =
y2 C1 y2 K2
dy1
y1 K1 y1 K3
1
2 y1 K3
K
=
2 y2 K1 dy2
y2 C1 y2 K2
1
2 y1 K1
dy1 =
1
1
ln y1 K3 K
ln y1 K1
2
2
ln y1 K3 Kln y1 K1
ln
y1 K3
y1 K1
1
1
C
y2 C1
y2 K2
= ln y2 C1 Cln y2 K2 CC
= 2 ln y2 C1 C2 ln y2 K2 CC
= ln y2 C1 2 y2 K2 2 CC
y1 K3
= C y2 C1 2 y2 K2 2, C O 0
y1 K1
2. Find the particular solution:
3 =4 C
C=
dy2
3
4
y1 K3
3
=
y2 C1 2 y2 K2 2
y1 K1
4
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Problem:
y1' =
y2 K4 ey1 K y2
y2 K1 y2 C2
y2' =
y1 K3 ey1 K y2
y1 K2 y1 K5
y1(0) = 0
y2(0) = K1
1. Find the general solution:
y2 K4 y1 K2 y1 K5
dy1/dy2 =
y2 K1 y2 C2 y1 K3
y1 K3 dy1
y1 K2 y1 K5
2
3 y1 K5
C
=
y2 K4 dy2
y2 K1 y2 C2
1
3 y1 K2
dy1 = K
2
1
ln y1 K5 C
ln y1 K2
3
3
2 ln y1 K5 Cln y1 K2
ln y1 K5
2
y1 K2
= ln
= K3 ln y2 K1 C6 ln y2 C2 CC
y2 C2 6
y2 K1 3
6
C = 400
y1 K5
2
y1 K2 =
dy2
= Kln y2 K1 C2 ln y2 C2 CC
CC
C y2 C2
y1 K5 y1 K2 =
, C O0
y2 K1 3
2. Find the particular solution:
1
50 =
C
8
2
1
2
C
y2 K1
y2 C2
400 y2 C2 6
y2 K1 3
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Problem:
1
y1' =
y2 K1
y2 K2
1
y2' =
y1 C1
y1 C3
y1(0) = 1
y2(0) = 0
1. Find the general solution:
y1 C1 y1 C3
dy1/dy2 =
y2 K1 y2 K2
dy1
y1 C1 y1 C3
K
1
2 y1 C3
C
=
dy2
y2 K1 y2 K2
1
2 y1 C1
1
1
C
y2 K1
y2 K2
dy1 = K
dy2
1
1
ln y1 C3 C
ln y1 C1 = ln y2 K2 Kln y2 K1 CC
2
2
K
Kln y1 C3 Cln y1 C1
ln
y1 C1
y1 C3
= ln
= 2 ln y2 K2 K2 ln y2 K1 CC
y2 K2 2
y2 K1 2
CC
2
y1 C1
C y2 K2
=
, C O0
y1 C3
y2 K1 2
2. Find the particular solution:
1
=4 C
2
C=
1
8
1
y1 C1
=
8
y1 C3
y2 K2 2
y2 K1 2