ORDINARY DIFFERENTIAL EQUATIONS
MINGFENG ZHAO
December 04, 2015
• Final Exam Review Session:
– Sections 101 and 104 Review: December 11, 2015 (Friday), 01:00PM–04:00PM, BUCH A104
• Final Exam Office Hours:
– December 14, 2015 (Monday), 2:00PM–4:00PM, LSK 300B
– December 15, 2015 (Tuesday), 2:00PM–4:00PM, LSK 300B
– December 16, 2015 (Wednesday), 2:00PM–4:00PM, LSK 300B
• PDF Files for final exam:
– Revised table of Laplace transform and variation of parameters.
– Two sample exams for Midterm exam 1, and Two Midterm exam 1.
– Two sample exams for Midterm exam 2, and Three Midterm exam 2.
– Four sample exams for Final exam.
– All sample exams for midterms and final, and midterm exams can be found on the common
webpage:
http://www.math.ubc.ca/~ttsai/courses/215-255-15F/.
• Contents for final exam: In addition to sections covered in MT1 and MT2, it will cover sections
3.3-3.5, 3.7, 3.9, and 8.1-8.3.
It will not cover 3.6, 3.8, 3.9.3.
It will not cover methods of
eigenvector decomposition and undetermined coefficients in 3.9.1, and global phase portrait in
8.3.
• Topoics for Chapter 1(The First Order Differential Equations): Page 2–Page 12.
• Topoics for Chapter 2(The Second Order Linear Differential Equations): Page 13–Page 36.
• Topoics for Chapter 6(The Laplace Transforms): Page 37–Page 44.
• Topoics for Chapter 3(The First Order Linear System): Page 45–Page 62.
• Topoics for Chapter 8(The First Order Nonlinear System): Page 63–Page 69.
1
2
MINGFENG ZHAO
Review of Chapter 1:
Integrals as solutions
Let’s consider the differential equation
y 0 = f (x).
By the definition of the antiderivative, we know that
Z
y=
f (x) dx.
In summary, we have
Z
0
y = f (x) =⇒ y =
f (x) dx .
Example 1. Find the general solution(i.e., all solution) of y 0 = 3x2 .
It’s easy to see that
Z
y=
3x2 dx = x3 + C .
Separable differential equations
A separable differential equation has the form of:
y0 =
dy
= f (x)g(y).
dx
There are two cases:
Case I: If g(a) = 0 for some constant a, then y(x) ≡ a is a solution.
dy
1
Case II: If g(y) 6= 0, since
= f (x)g(y), then
dy = f (x) dx, which implies that
dx
g(y)
Z
Z
1
dy = f (x) dx.
g(y)
In summary, we have



 1) y(x) ≡ a for some constant a such that g(a) = 0
0
Z
Z
y = f (x)g(y) =⇒
.
1


dy = f (x) dx.
 2)
g(y)
ORDINARY DIFFERENTIAL EQUATIONS
3
Example 2. Solve x2 y 0 = 1 − x2 − y 2 + x2 y 2 , y(1) = 0.
Fist, let’s find the general solution to x2 y 0 = 1 − x2 − y 2 + x2 y 2 . Since x2 y 0 = 1 − x2 − y 2 + x2 y 2 , then
y0 =
1 − x2 − y 2 + x2 y 2
.
x2
Since 1 − x2 − y 2 + x2 y 2 = (1 − x2 ) − y 2 (1 − x2 ) = (1 − x2 )(1 − y 2 ), then
1 − x2
dy
(1 − x2 )(1 − y 2 )
=
· (1 − y 2 ).
= y0 =
dx
x2
x2
Case I: If 1 − y 2 = 0, that is, y = ±1. So y(x) = ±1 are two solutions.
1 − x2
1
1 − x2
dy
(1 − x2 )(1 − y 2 )
2
=
·
(1
−
y
),
then
dy
=
dx, which implies that
Case II: If 1 − y 2 6= 0, since
=
x2
x2
1 − y2 x2
Z
Z dx 2
2
1
1
1
1
1
1
1−x
1
1−x
dy =
dx. Since
=− 2
=
−
and
= 2 − 1, then we get
1 − y2
x2
1 − y2
y −1
2 y+1 y−1
x2
x
1 y + 1 1
ln = − − x + C.
2
y−1
x
In summary, the general solution to x2 y 0 = 1 − x2 − y 2 + x2 y 2 is:
1 y + 1 1
ln = − − x + C.
2
y − 1
x
1 y + 1 1
1
Since y(1) = 0, then we only need look at the solution ln = − − x + C, we get 0 = ln 1 = −1 − 1 + C.
2
y − 1
x
2
So C = 2. Therefore, the solution to x2 y 0 = 1 − x2 − y 2 + x2 y 2 , y(1) = 0 is:
or y(x) = −1,
y(x) = 1,
or
1 y + 1 1
ln =− −x+2 .
2
y − 1
x
First order linear differential equations and the integrating factors
Let’s solve y 0 + p(x)y = f (x). Multiply r(x) on the both sides, we get r(x)y 0 + r(x)p(x)y = r(x)f (x). We are looking
for an integrating factor r(x), that is, r(x)y 0 + r(x)p(x)y = [r(x)y]0 = r(x)y 0 + r0 (x)y, then r0 (x) = r(x)p(x), which
implies that r(x) = e
R
p(x) dx
. So we have
[r(x)y]0 = r(x)f (x).
Z
Hence we have r(x)y =
r(x)f (x) dx, that is,
1
y=
r(x)
Z
r(x)f (x) dx = e
−
R
p(x) dx
Z R
e p(x)
dx
f (x) dx.
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MINGFENG ZHAO
1
y 0 + xy = 3, y(0) = 0.
x2 + 1
Rewrite the equation, we have y 0 + x(x2 + 1)y = 3(x2 + 1). Multiply r(x) on the both sides, we get r(x)y 0 + x(x2 +
Example 3. Solve
1)r(x)y = 3(x2 + 1)r(x). We are look for an integrating factor r(x), that is r(x)y 0 + x(x2 + 1)r(x)y = [r(x)y]0 =
r(x)y 0 + r0 (x)y, then r0 (x) = x(x2 + 1)r(x), which implies that r(x) = e
h
e
x4
4
2
+ x2
y
i0
= 3(x2 + 1)e
x4
4
2
x4
4
+ x2
. So we get
2
+ x2
.
So we get
e
x4
4
Z
2
+ x2
y=
2
3(x + 1)e
x4
4
Z
2
+ x2
dx = 3
x
t4
t2
e 4 + 2 3(t2 + 1) dt + C.
0
So we have
y = e−
x4
4
Z
2
− x2
·
x
t4
t2
e 4 + 2 3(t2 + 1) dt + Ce−
x4
4
2
− x2
.
0
Since y(0) = 0, then C = 0. Therefore, the solution to
y = e−
x4
4
Z
2
− x2
·
x2
x
1
y 0 + xy = 3, y(0) = 0 is:
+1
t4
t2
e 4 + 2 3(t2 + 1) dt .
0
Exact equations
Definition 1. Consider the differential equation M (x, y)+N (x, y)y 0 = 0, we say that the differential equation M (x, y)+
N (x, y)y 0 = 0 is exact if My (x, y) = Nx (x, y).
Remark 1. If the differential equation M (x, y) + N (x, y)y 0 = 0 is exact, that is, My (x, y) = Nx (x, y), then there exists
a function φ(x, y) such that
• φx (x, y) = M (x, y).
• φy (x, y) = N (x, y).
• The solution to M (x, y) + N (x, y)y 0 = 0 is given by φ(x, y) = C .
To solve an exact equation:
Let M (x, y) + N (x, y)y 0 = 0 be an exact equation, that is, My (x, y) = Nx (x, y), then we can find φ(x, y) such that
• φx (x, y) = M (x, y).
• φy (x, y) = N (x, y).
• The solution to M (x, y) + N (x, y)y 0 = 0 is given by φ(x, y) = C.
Now let’s find φ(x, y):
ORDINARY DIFFERENTIAL EQUATIONS
5
Step I: Since φx (x, y) = M (x, y), for any fixed y, we integrate with respect to x, we get
Z
φ(x, y) = M (x, y) dx + f (y), for some f (y).
Step II: Now we only need to compute f (y): Take the partial derivative with respect to y on both sides of φ(x, y) =
Z
M (x, y) dx + f (y), we get
Z
φy (x, y) = My (x, y) dx + f 0 (y).
Z
Since φy (x, y) = N (x, y), so we get f 0 (y) = N (x, y) − My (x, y) dx ( this is a function of y), which implies
that
f (y) =
Z Z
N (x, y) − My (x, y) dx dy.
In summary, we get
For exact equation: M (x, y) + N (x, y) y 0 = 0 =⇒
Z
M (x, y) dx +
Z Z
N (x, y) − My (x, y) dx dy = C .
Example 4. Solve 3y + ex + [3x + cos(y)]y 0 = 0.
Let M (x, y) = 3y + ex and N (x, y) = 3x + cos(y), then My (x, y) = 3 and Nx (x, y) = 3. So My (x, y) = Nx (x, y) = 3,
that is, [3y + ex ] + [3x + cos(y)] y 0 = 0 is exact. Then there exists some function φ(x, y) such that
• φx (x, y) = M (x, y) = 3y + ex .
• φy (x, y) = N (x, y) = 3x + cos(y).
• The solution to M (x, y) + N (x, y)y 0 = 0 is given by φ(x, y) = C.
Since φx (x, y) = 3y + ex then
Z
φ(x, y) =
(3y + ex ) dx + f (y) = 3xy + ex + f (y)
Take the partial derivative with respect to y on the both sides of φ(x, y) = 3xy + ex + f (y), then
φy (x, y) = 3y + f 0 (y).
Since φy (x, y) = 3x + cos(y), then 3x + cos(y) = N (x, y) = φy (x, y) = 3x + f 0 (y). So f 0 (y) = cos(y), which implies
that f (y) = sin(y). So we know that
φ(x, y) = 3xy + ex + sin(y).
Hence the general solution to 3y + ex + [3x + cos(y)]y 0 = 0 is:
3xy + ex y + sin(y) = C .
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MINGFENG ZHAO
Use the integrating factor r(x) to solve general equations
In general, M (x, y) + N (x, y)y 0 = 0 is not exact, that is, My (x, y) 6= Nx (x, y). Multiply r(x) on the both sides of
M (x, y) + N (x, y)y 0 = 0, then
r(x)M (x, y) + r(x)N (x, y)y 0 = 0.
We are looking for an integrating factor r(x), that is, r(x)M (x, y) + r(x)N (x, y)y 0 = 0 is exact, then
∂
∂
[r(x)M (x, y)] =
[r(x)N (x, y)].
∂y
∂x
So we get r(x)My (x, y) = r0 (x)N (x, y) + r(x)Nx (x, y), that is,
r0 =
In this case, we need
My (x, y) − Nx (x, y)
· r.
N (x, y)
My (x, y) − Nx (x, y)
is a function of only x. So since r(x)M (x, y) + r(x)N (x, y)y 0 = 0 is exact,
N (x, y)
then we can solve it.
Example 5. Solve y + (x2 y − x)y 0 = 0, x > 0.
Let M (x, y) = y and N (x, y) = x2 y − x, then
and Nx = 2xy − 1.
My = 1,
So y + (x2 y − x)y 0 = 0 is not exact. Multiply r(x) on the both sides of y + (x2 y − x)y 0 = 0, then
r(x)y + r(x)(x2 y − x)y 0 = 0.
We are looking for an integrating factor r(x), that is, r(x)y + r(x)(x2 y − x)y 0 = 0 is exact, then
∂
∂
[r(x)y] =
[r(x)(x2 y − x)].
∂y
∂x
So we get
r(x) = r0 (x)(x2 y − x) + r(x)(2xy − 1).
So we have
r(x)[2 − 2xy] = r0 (x)(x2 y − x) = r0 (x)x(xy − 1).
Hence 2r = −xr0 , that is, r0 = − x2 r, which implies that
r(x) = e−
R
2
x
dx
= e−2 ln x =
1
x2
ORDINARY DIFFERENTIAL EQUATIONS
That is, the equation
7
y
x2 y − x 0
+
y = 0 is exact. So there exist some φ(x, y) such that
x2
x2
y
.
x2
2
x y−x
1
• φy (x, y) =
=y− .
x2
x
• the solution to the differential equation is φ(x, y) = C.
• φx (x, y) =
Since φx (x, y) =
y
, then
x2
Z
φ(x, y) =
y
y
dx + f (y) = − + f (y).
x2
x
Take the partial derivative with respect to y on the both sides of the above identity, then
φy (x, y) = −
Since φy (x, y) = y −
1
+ f 0 (y).
x
1
1
y2
1
, then y − = − + f 0 (y). So f 0 (y) = y. We can take f (y) = . So we get
x
x
x
2
φ(x, y) = −
y
y2
+ .
x
2
So the solution is:
−
y
y2
+
=C .
x
2
Direction/slope fields
Definition 2. The direction/slope field of y 0 = f (x, y) is a picture on the xy-plane such that for each point (x, y) on
the plane, one draws a short line segment with the slope f (x, y) at the point (x, y).
To draw the slope field of y 0 = f (x, y):
1) select points in the xy-plane,
2) compute the numbers f (x, y) at the selected points (x, y),
3) at each selected point (x, y), draw a short tangent line whose slope is f (x, y).
Remark 2. For the differential equation y 0 = f (x, y), by the definition of the slope fields, we know that for any y = g(x)
which is a solution to y 0 = f (x, y), then at any point (x0 , y0 ) on the curve y = g(x), the line segment is the tangent line
of the curve y = g(x) at the point (x0 , y0 ).
Example 6. y 0 = xy. (It’s easy to see the general solution is: y = Ce
x2
2
.)
8
MINGFENG ZHAO
Figure 1. Slope field of y 0 = xy with graphs of solutions satisfying y(0) = ±0.2, and y(0) = ±1
Remark 3. For the directions of the short tangent line segment:
• “/” means positive slope
• “\” means a negative slope
• “−” means a zero slope
• “|” means infinity slope.
Theorem of the existence and uniqueness for the first order differential equations with
the initial value condition
Theorem 1 (Picard’s Theorem on Existence and Uniqueness). Let (x0 , y0 ) be a fixed point in RN , if f (x, y) is continuous
∂f
with respect to x and y near some (x0 , y0 ), and
(x, y) exists and is continuous with respect to x and y near the point
∂y
(x0 , y0 ), then a solution to

 y 0 = f (x, y),
 y(x ) = y ,
0
0
exists for some small interval containing x0 , and is unique.
Remark 4. For the uniqueness part in Theorem 1, if we have two solution curves y1 (x) and y2 (x) to the differential
equation y 0 = f (x, y), if these two curves y1 (x) and y2 (x) meet at one point, then y1 (x) = y2 (x) for all x belonging to
the domain.
ORDINARY DIFFERENTIAL EQUATIONS
9
Figure 2. This can NOT happen
p
Example 7. Is it possible to solve the equation y 0 = y |x| for y(0) = 0? Is the solution unique?
p
p
It’s easy to see that y(x) ≡ 0 is a solution to y 0 = y |x| for y(0) = 0. Let f (x, y) = y |x|, then f (x, y) is continuous
with respect to x and y, and
p
∂f
(x, y) = |x| is continuous near (0, 0).
∂y
By Theorem 1, we know that
y(x) ≡ 0 is the only solution to y 0 = y
p
|x| for y(0) = 0 .
Phase diagram of the autonomous equation
An autonomous equation is of the form y 0 = f (y), if f (a) = 0, y(x) ≡ a is called an equilibrium solution, and a is
called a critical point of y 0 = f (y).
To draw the phase diagram of y 0 = f (y):
1) Find all critical points of y 0 = f (y), that is, find all zeros of f (y) = 0.
2) Mark all critical points on a vertical line.
3) For any two neighboring critical points a and b, choose any point c which is between a and b, compute f (c):
– If f (c) > 0, draw an up arrow “↑” between a and b.
– If f (c) < 0, draw a down arrow “↓” between a and b.
Example 8. x0 = (x − 1)(x − 2)x2 .
• Critical Points: Solve (x − 1)(x − 2)x2 = 0, then x = 0, 1, 2 .
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MINGFENG ZHAO
• Phase Diagram:
Figure 3. Phase diagram of x0 = (x − 1)(x − 2)x2
• For the initial value problem: x(0) = 0.5. Since 0 < x(0) = 0.5 < 1, then x0 (0) = (0.5 − 1)(0.5 − 2)(0.5)2 > 0,
and
lim x(t) = 1 .
t→∞
Classification of critical points of the autonomous equation
Definition 3. Consider the autonomous equation y 0 = f (y), let a be a critical point of y 0 = f (y), that is, f (a) = 0.
Then
I. We say a is stable or a sink if any solution with initial condition close to a is asymptotic to a as x increases.
Figure 4. Stable or sink
ORDINARY DIFFERENTIAL EQUATIONS
11
II. We say a is a source if all solutions start close to a tend toward y as x decreases, and tend away from a as x
increases.
Figure 5. Source
III. We say a is a node if the equilibrium point a is neither a source nor a sink.
Figure 6. Node
IV. If a is either a source or a node, we say a is unstable.
Example 9. The phase diagram in Example 8 shows that 2 is a source(unstable), 1 is a sink(stable), and 0 is a
node(unstable).
Numerical methods: Euler’s method
Consider the initial value problem:
y 0 = f (x, y),
y(x0 ) = y0 .
Recall Taylor expansion:
y(x + ∆x) = y(x) + y 0 (x)∆x +
y 00 (x)
(∆x)2 + · · · .
2
Let’s take the linear approximation at x, then
(1)
Euler’s method:
y(x + ∆x) ≈ y(x) + y 0 (x)∆x.
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MINGFENG ZHAO
Given the initial condition y(x0 ) = y0 and the step size ∆x, for any k = 0, 1, 2, · · · , let yk = y(xk ) and
xk = xk−1 + ∆x = x0 + (k − 1)∆x.
Now we want to approximate yk : Replace x by xk in (1), then we get
yk+1 = yk + f (xk , yk )∆x.
Example 10. Use Euler’s method to approximate y(3) with step size ∆x = 1, where y 0 =
Recall the linear approximation, we get
y(x + ∆x) ≈ y(x) + y 0 (x)∆x.
Let yk = y(xk ), put x = xk and ∆x = 1 in the above formula, so we have
yk+1 = yk +
yk2
.
3
Then
k
xk
yk
0
0
1
1
1
y1 = y0 +
2
2
3
3
y02
12
4
=1+
=
3
3 3 2
4 1
4
y2
52
≈ 1.925
y2 = y1 + 1 = + ·
=
3
3 3
3
27
2
y2
52 1
52
5980
y3 = y2 + 2 =
+ ·
≈ 2.734
=
3
27 3
27
2187
y2
, y(0) = 1.
3
ORDINARY DIFFERENTIAL EQUATIONS
13
Review of Chapter 2:
Solving homogeneous differential equations
Theorem 2. Let p(x) and q(x) be continuous functions, y1 and y2 are two linearly independent solutions to a homogeneous equation y 00 + p(x)y 0 + q(x)y = 0. Then the general solution to y 00 + p(x)y 0 + q(x)y = 0 is:
y = C1 y1 (x) + C2 y2 (x).
Definition 4. Any two linearly independent solutions y1 and y2 to a homogeneous equation y 00 + p(x)y 0 + q(x)y = 0 is
called a fundamental set of solutions to this homogeneous equation y 00 + p(x)y 0 + q(x)y = 0.
Definition 5. Let a, b and c be three constants, the characteristic equation of the differential equation ay 00 +by 0 +cy = 0
is:
ar2 + br + c = 0.
Theorem 3. Let a, b and c be three constants, the roots of ar2 + br + c = 0 are:
√
√
−b − b2 − 4ac
−b + b2 − 4ac
, and r2 =
.
r1 =
2a
2a
Then
1) If b2 − 4ac > 0, then r1 and r2 are two different real numbers, and
√
√
−b
b2 − 4ac
−b
b2 − 4ac
r1 =
+
, and r2 =
−
.
2a
2a
2a
2a
The general solution to ay 00 + by 0 + cy = 0 is:
y = C1 er1 x + C2 er2 x .
2) If b2 − 4ac = 0, then r1 = r2 = −
b
is a real number. The general solution to ay 00 + by 0 + cy = 0 is:
2a
−b
−b
y = C1 e 2a ·x + C2 xe 2a ·x .
3) If b2 − 4ac < 0, then r1 and r2 are two different complex numbers, and
√
√
b
4ac − b2
b
4ac − b2
r1 = − + i
, and r2 = − − i
2a
2a
2a
2a
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MINGFENG ZHAO
The general solution to ay 00 + by 0 + cy = 0 is:
!
√
2
b
b
4ac
−
b
y = C1 e− 2a ·x cos
· x + C2 e− 2a ·x sin
2a
√
!
4ac − b2
·x .
2a
Remark 5. Let i be a number which is called the imaginary unit, that is, i2 = −1. A complex number is the form of:
a + ib,
for some real numbers a and b.
The arithmetic laws of complex numbers: Let a, b, c and d be four real numbers, then
• Addition:
(a + bi) + (c + di) = (a + c) + i(b + d).
• Subtraction:
(a + bi) − (c + di) = (a − c) + i(b − d).
• Multiplication:
(a + bi)(c + di) = ac + adi + bci + bdi2 = (ac − bd) + i(ad + bc).
• Division:
a + bi
(a + bi)(c − di)
ac − adi + bci − bdi2
(ac + bd) − i(ad + bc)
=
=
=
.
c + di
(c + di)(c − di)
c2 + d2
c2 + d2
• Conjugation:
a + bi = a − bi.
• Euler’s indentity:
ea+ib = ea · eib = ea [cos(b) + i sin(b)] = ea cos(b) + iea sin(b).
Example 11. Let a be a real constant, solve y 00 − 4y 0 − 5y = 0, y(0) = 6 and y 0 (0) = 6a. Then find the value of a such
that the solution approaches 0 as x → ∞.
Let y = erx be a solution to y 00 − 4y 0 − 5y = 0, then
y0
=
rerx
y 00
=
r2 erx
0
=
y 00 − 4y 0 − 5y
=
r2 erx − 4rerx − 5erx
=
erx (r2 − 4r − 5).
ORDINARY DIFFERENTIAL EQUATIONS
So r2 − 4r − 5 = 0, that is,
r1 = −1,
and r2 = 5.
So the general solution to y 00 − 4y 0 − 5y = 0 is:
y(x) = C1 e−x + C2 e5x .
Then
y 0 (x) = −C1 e−x + 5C2 e5x .
Since y(0) = 6 and y 0 (0) = 6a, then
C1 + C2 = 6,
and
− C1 + 5C2 = 6a.
Solve the above system, we get
C1 = 5 − a,
and C2 = 1 + a.
So we know that the solution to y 00 − 4y 0 − 5y = 0, y(0) = 6 and y 0 (0) = 6a is:
y(x) = (5 − a)e−x + (1 + a)e5x .
Moreover, if we assume y(x) → 0 as x → ∞, then we must have 1 + a = 0, that is, a = −1 .
Example 12. Solve y 00 − 8y 0 + 16y = 0, y(0) = 1 and y 0 (0) = 6.
Let y = erx be a solution to y 00 − 8y 0 + 16y = 0, then y 0 = rerx and y 00 = r2 erx . So we have
0
= y 00 − 8y 0 + 16y
= r2 erx − 8rerx + 16erx
= erx (r2 − 8r + 16).
So r2 − 8r + 16 = 0, that is,
r1 = r2 = 4.
So the general solution to y 00 − 8y 0 + 16y = 0 is:
y(x) = C1 e4x + C2 xe4x .
Since y(0) = 1, then C1 = 1 and
y(x) = e4x + C2 xe4x .
15
16
MINGFENG ZHAO
So
y 0 (x) = 4e4x + C2 e4x + 4C2 xe4x .
Since y 0 (0) = 6, then 6 = 4 + C2 , that is, C2 = 2. So the solution to y 00 − 8y 0 + 16y = 0, y(0) = 1 and y 0 (0) = 6 is:
y(x) = e4x + 2xe4x .
Example 13. Find the solution of y 00 − 6y 0 + 13y = 0, y(0) = 0 and y 0 (0) = 10.
Let y = erx be a solution to y 00 − 6y 0 + 13y = 0, then
y0
= rerx
y 00
= r2 erx
0
= y 00 − 6y 0 + 13y
= r2 erx − 6rerx + 13erx
= erx (r2 − 6r + 13).
So r2 − 6r + 13 = 0, that is,
r1 =
6+
√
√
6 + −13
36 − 4 · 13
=
= 3 + 2i,
2
2
and r2 = 3 − 2i.
Notice that
e r1 x
=
e(3+2i)x = e3x+2ix = e3x · ei(2x)
=
e3x [cos(2x) + i sin(2x)]
=
e3x cos(2x) + ie3x sin(2x).
By Euler’s identity
So the general solution to y 00 − 6y 0 + 13y = 0 is:
y(x) = C1 e3x cos(2x) + C2 e3x sin(2x).
Since y(0) = 0, then C1 = 0, that is, y(x) = C2 e3x sin(2x). So we get
y 0 (x) = 3C2 e3x sin(2x) + 2C2 e3x cos(2x).
Since y 0 (0) = 10, then 10 = y 0 (0) = 2C2 that is, C2 = 5. Therefore, the solution of y 00 − 6y 0 + 13y = 0, y(0) = 0 and
y 0 (0) = 10 is y(x) = 5e3x sin(2x) .
ORDINARY DIFFERENTIAL EQUATIONS
Free undamped motion [Simple harmonic motion]
Recall the Mass-Spring System:
Figure 7. Mass-Spring System
Let x(t) be the displacement of the mass, then
mx00 + cx0 + kx = F (t) .
The free undamped motion (that is, c = 0 and F (t) ≡ 0):
mx00 + kx = 0.
r
Let ω0 =
k
, then our equation becomes:
m
x00 + ω02 x = 0.
The general solution to x00 + ω02 x = 0 is:
x(t) = A cos(ω0 t) + B sin(ω0 t) .
By a trigonometric identity, we have
x(t) = A cos(ω0 t) + B sin(ω0 t) = C cos(ω0 t − γ) ,
where
C=
p
A2 + B 2 ,
Here are some terminology:
√
• C = A2 + B 2 is called the amplitude.
and γ = arctan
B
.
A
17
18
MINGFENG ZHAO
Figure 8. Free undamped motion
r
k
is called the frequency.
m
B
• γ = arctan
is called the phase shift.
A
2π
• T =
is called the period of the motion.
ω0
• ω0 =
Free damped motion
Now let’s focus on damped motion:
mx00 + cx0 + kx = 0.
Rewrite the equation as:
x00 + 2px0 + ω02 x = 0,
where
c
p=
(damping term),
2m
r
and
ω0 =
k
(frequency).
m
The characteristic equation of x00 + 2px0 + ω02 x = 0 is:
r2 + 2pr + ω02 = 0.
Solve r2 + 2pr + ω02 = 0, we get
r1,2 = −p ±
q
p2 − ω02 .
Hence the solutions to x00 + 2px0 + ω02 x = 0 depend on p2 − ω02 =
c2 − 4km
:
4m2
ORDINARY DIFFERENTIAL EQUATIONS
I. Overdamping: c2 − 4km > 0, that is, p2 − ω02 > 0. Then we have two different real roots:
r1 = −p −
q
p2 − ω02 < 0,
and r2 = −p +
q
p2 − ω02 < 0.
Then the solution is:
y(t) = Aer1 t + Ber2 t .
Since r1 < 0 and r2 < 0, then
y(t) → 0,
as t → ∞.
Figure 9. Overdamped motion
II. Critical damping: c2 − 4km = 0, that is, p2 − ω02 = 0. Then we have the same real roots:
r1 = r2 = −p < 0.
Then the solution is:
y(t) = Ae−pt + Bte−pt .
Since r1,2 = −p < 0, then
y(t) → 0,
as t → ∞.
III. Underdamping: c2 − 4km < 0, that is, p2 − ω02 < 0. Then we have two different complex roots:
q
r1 = −p − i ω02 − p2 ,
q
and r2 = −p + i ω02 − p2 .
Then the solution is:
y(t) = Ae
−pt
cos
q
ω02
−
p2 t
+ Be
−pt
sin
q
ω02
−
p2 t
= Ce−pt cos (ω1 t − γ) ,
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20
MINGFENG ZHAO
Figure 10. Critical damped motion
where
w1 =
q
ω02 − p2 < ω0 ,
and γ = arctan
B
A
Since p > 0, then
y(t) → 0,
as t → ∞.
Figure 11. Underdamped motion
In Figure 11, there are two envelop curves y1 (t) = Ce−pt and y2 (t) = −Ce−pt , the solution y(t) is oscillating
between these two envelop curves.
Example 14. Suppose that m = 2 kg and k = 8 N/m. The whole mass and spring setup is sitting on a truck that was
traveling at 1 m/s. The truck crashes and hence stops. The mass was held in place 0.5 meters forward from the rest
position. During the crash the mass gets loose. That is, the mass is now moving forward at 1 m/s, while the other end
of the spring is held in place. The mass therefore starts oscillating. What’s is the frequency of the resulting oscillation
and what is the amplitude.
ORDINARY DIFFERENTIAL EQUATIONS
21
Let x(t) be the displacement of the mass at time t (the moving forward is in the positive direction), them the
differential equation is:
mx00 + kx = 2x00 + 8x = 0.
By the assumption, we have
x(0) = 0.5,
and x0 (0) = 1.
Rewrite the equation, we have
x00 + 4x = 0.
The characteristic equation of x00 + 4x = 0 is r2 + 4 = 0, then r1 = 2i and r2 = −2i, which implies that the general
solution to 2x00 + 8x = 0 is:
x(t) = A cos(2x) + B sin(2x).
Then
x0 (t) = −2A sin(2x) + 2B cos(2x).
Since x(0) = 0.5 and x0 (0) = 1, then
A = 0.5,
and
2B = 1.
A = 0.5,
and B = 0.5
Then
Therefore, the solution to 2x00 + 8x = 0, x(0) = 0.5, x0 (0) = 1 is:
√
2
π
cos 2x +
.
2
4
√
p
2
2
2
2
Hence the frequency is 2 =
Hz , and the amplitude is
0.5 + 0.5 =
.
2π
2
x(t) = 0.5 cos(2x) + 0.5 sin(2x) =
Nohomogeneous equations
Theorem 4. Let yc (x) be the general solution to the homogeneous equation y 00 + p(x)y 0 + q(x)y = 0 ( which is call the
complementary solution), and yp (x) be any particular solution to the nonhomogeneous equation y 00 +p(x)y 0 +q(x)y = f (x),
then the general solution to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x) is:
y(x) = yp (x) + yc (x).
22
MINGFENG ZHAO
Undetermined Coefficients:
Let a, b and c be constants, consider the equation:
ay 00 + by 0 + cy = f (x).
Let pn (x) and p̃n (x) be polynomials with degree n, a particular solution yp (x) to ay 00 + by 0 + cy = f (x) can be taken as:
f (x)
yp (x)
pn (x)emx cos(kx) + p̃n (x)emx sin(kx)
xα [qn (x)emx cos(kx) + q̃n (x)emx sin(kx)]
where
• α is one of 0, 1 and 2 (α is the multiplicity of m+ki as the solutions to the characteristic equation ar2 +br +c = 0
of ay 00 + by 0 + cy = 0):
– If m + ki is not a root of ar2 + br + c = 0, then α = 0.
– If m + ki is a root of ar2 + br + c = 0 and ar2 + br + c = 0 has two different roots, then α = 1.
– If m + ki is a root of ar2 + br + c = 0 and ar2 + br + c = 0 only has one root m + ki, then α = 2.
• qn (x) and q̃n (x) are undetermined polynomials with degree n.
Example 15. Find the general solution to y 00 + 5y 0 + 6y = 2x + 1.
The characteristic equation to y 00 + 5y 0 + 6y = 0 is :
r2 + 5r + 6 = 0.
Solve r2 + 5r + 6 = 0, we get
r1 = −2,
and r2 = −3.
So the general solution to the homogeneous equation y 00 + 5y 0 + 6y = 0 is:
y(x) = C1 e−2x + C2 e−3x .
Let yp (x) = Ax + B be a particular solution to y 00 + 5y 0 + 6y = 2x + 1, then
yp00 + 5yp0 + 6yp = 0 + 5A + 6(Ax + B) = 2x + 1.
That is, we have
(6A − 2)x + 5A + 6B − 1 = 0.
ORDINARY DIFFERENTIAL EQUATIONS
23
So
6A − 2 = 0,
and
5A + 6B − 1 = 0.
So
A=
1
,
3
1
and B = − .
9
x 1
− is a particular solution to the nonhomogeneous equation y 00 + 5y 0 + 6y = 2x + 1. Therefore, he
3 9
general solution to the nonhomogeneous equation y 00 + 5y 0 + 6y = 2x + 1 is:
Hence yp (x) =
y(x) =
x 1
− + C1 e−2x + C2 e−3x .
3 9
Example 16. Find a particular solution to y 00 + 2y 0 + 3y = e5x .
The characteristic equation to y 00 + 2y 0 + 3y = 0 is :
r2 + 2r + 3 = 0.
Solve r2 + 5r + 6 = 0, we get
√
r1 = −1 + i 2,
√
and r2 = −1 − i 2.
Let yp (x) = Ae5x be a particular solution to y 00 + 2y 0 + 3y = e5x , then
So 38A = 1, that is, A =
1
38 .
yp0 (x)
=
5Ae5x
yp00 (x)
=
25Ae5x
yp00 + 2yp0 + 3yp
=
25Ae5x + 10Ae5x + 3Ae5x
=
38Ae5x
=
e5x
Hence a particular solution to y 00 + 2y 0 + 3y = e5x is yp (x) =
Example 17. Find a particular solution to y 00 − 6y 0 + 9y = e3x .
The characteristic equation of y 00 − 6y 0 + 9y = 0 is:
r2 − 6r + 9 = 0.
Solve r2 − 6r + 9 = 0, we get
r1 = r2 = 3.
1 5x
e .
38
24
MINGFENG ZHAO
Let yp (x) = Ax2 e3x be a particular solution to y 00 − 6y 0 + 9y = e3x , then
yp0 (x)
=
2Axe3x + 3Ax2 e3x
yp00 (x)
=
2Ae3x + 6Axe3x + 6Axe3x + 9Ax2 e3x
=
2Ae3x + 12Axe3x + 9Ax2 e3x
=
2Ae3x + 12Axe3x + 9Ax2 e3x − 6 2Axe3x + 3Ax2 e3x + 9Ax2 e3x
=
2Ae3x
=
e3x .
yp00 − 6yp0 + 9yp
So we have 2A = 1. That is, A =
1
1
. So a particular solution to y 00 − 9y = e3x is yp (x) = x2 e3x .
2
2
Example 18. Find a particular solution to y 00 + y = cos(2x).
The characteristic equation to y 00 + y = 0 is:
r2 + 1 = 0.
Solve r2 + 1 = 0, we have
r1 = i,
and r2 = −i
Let yp (x) = A cos(2x) + B sin(2x) be a particular solution to y 00 + 2y 0 + 2y = cos(2x), then
yp0 (x)
=
−2A sin(2x) + 2B cos(2x)
yp00 (x)
=
−4A cos(2x) − 4B sin(2x)
yp00 + yp
=
−4A cos(2x) − 4B sin(2x) + A cos(2x) + B sin(2x)
=
−3A cos(2x) − 3B sin(2x)
=
cos(2x).
So
−3A = 1,
and
− 3B = 0.
So we have
1
A=− ,
3
and B = 0.
1
So a particular solution to y 00 + y = cos(2x) is yp (x) = − cos(2x) .
3
ORDINARY DIFFERENTIAL EQUATIONS
25
Example 19. Find a particular solution to y 00 + y = cos(x).
The characteristic equation to y 00 + y = 0 is:
r2 + 1 = 0.
Solve r2 + 1 = 0, we have
r1 = i,
and r2 = −i
Let yp (x) = x[A cos(x) + B sin(x)] be a particular solution to y 00 + y = cos(x), then
yp0 (x)
= A cos(x) + B sin(x) + x[−A sin(x) + B cos(x)]
yp00 (x)
= −A sin(x) + B cos(x) − A sin(x) + B cos(x) + x[−A cos(x) − B sin(x)]
= −2A sin(x) + 2B cos(x) − x[A cos(x) + B sin(x)]
yp00 + yp
= −2A sin(x) + 2B cos(x) − x[A cos(x) + B sin(x)] + x[A cos(x) + B sin(x)]
= −2A sin(x) + 2B cos(x)
=
cos(x).
So
−2A = 0,
and
2B = 1.
So we have
A = 0,
So a particular solution to y 00 + y = cos(x) is yp (x) =
and B =
1
.
2
x sin(x)
.
2
Theorem 5. Let y1,p (x) be a particular solution to y 00 + p(x)y + q(x)y = f1 (x), and y2,p (x) be a particular solution to
y 00 + p(x)y + q(x)y = f2 (x), then yp (x) = y1,p (x) + y2,p (x) is a particular solution to y 00 + p(x)y + q(x)y = f (x), where
f (x) = f1 (x) + f2 (x).
Example 20. Find a particular solution to y 00 + y = x + ex .
The characteristic equation to y 00 + y = 0 is:
r2 + 1 = 0.
Solve r2 + 1 = 0, we have
r1 = i,
and r2 = −i
26
MINGFENG ZHAO
A particular solution to y 00 + y = x can be taken as y(x) = Ax + B, and a particular solution to y 00 + y = ex can be
y(x) = Cex , so let yp (x) = Ax + B + Cex be a particular solution to y 00 + y = ex , then
yp (x)0
= A + Cex
yp00 (x)
= Cex
yp00 + yp
= Cex + Ax + B + Cex
= Ax + 2Cex + B
= x + ex .
So we get
[2C − 1]ex + [A − 1]x + B = 0.
So
2C − 1 = 0,
2A − 1 = 0,
and B = 0.
That is,
A = 1,
B = 0,
and C =
1
.
2
Therefore, a particular solution to y 00 + y = x + ex can be:
yp (x) = x +
ex
.
2
Remark 6. Basically, you are looking for a particular solution for each term and then add those particular solutions
together.
Variation of Parameters
Definition 6. Let y1 (x) and y2 (x) be two solutions, the Wronskian of y1 and y2 is defined as:

W (y1 , y2 ) = det 
y1 (x) y2 (x)
y10 (x)
y20 (x)

 = y1 (x)y20 (x) − y10 (x)y2 (x).
To find a particular solution to the nonhomogeneous equation y 00 + p(x)y 0 + q(x)y = f (x):
ORDINARY DIFFERENTIAL EQUATIONS
27
I. Find a fundamental set of solutions y1 (x) and y2 (x) to the homogeneous equation y 00 + p(x)y 0 + q(x)y = 0, that
is, the general solution to y 00 + p(x)y 0 + q(x)y = 0 is:
yc (x) = C1 y1 (x) + C2 y2 (x).
II. Let yp (x) = C1 (x)y1 (x) + C2 (x)y2 (x) be a particular solution to y 00 + p(x)y 0 + q(x)y = f (x)
III. Compute yp0 (x), we get
yp0 (x) = C10 (x)y1 (x) + C20 (x)y2 (x) + C1 (x)y10 (x) + C2 (x)y20 (x)
Take
C10 (x)y1 (x) + C20 (x)y2 (x) = 0.
Then
yp0 (x)
= C1 (x)y10 (x) + C2 (x)y20 (x)
yp00 (x)
= C10 (x)y10 (x) + C1 (x)y100 (x) + C20 (x)y20 (x) + C2 (x)y200 (x).
IV. Plug yp (x), yp0 (x) and yp00 (x) into y 00 + p(x)y 0 + q(x)y = f (x), we get
C10 (x)y10 (x) + C20 (x)y20 (x) = f (x).
V. Solve C10 (x) and C20 (x) from the system:


 C10 (x)y1 (x) + C20 (x)y2 (x) = 0

 C 0 (x)y 0 (x) + C 0 (x)y 0 (x) = f (x).
1
1
2
2
Then
C10 (x) =
−y2 (x)f (x)
,
W (y1 , y2 )
and C20 (x) =
y1 (x)f (x)
.
W (y1 , y2 )
VI. Solve C1 (x) and C2 (x), then
Z
C1 (x) =
−y2 (x)f (x)
dx,
W (y1 , y2 )
Z
and C2 (x) =
y1 (x)f (x)
dx.
W (y1 , y2 )
VII. Write down the solution:
Z
yp (x) = −y1 (x)
y2 (x)f (x)
dx + y2 (x)
W (y1 , y2 )
Z
y1 (x)f (x)
dx .
W (y1 , y2 )
28
MINGFENG ZHAO
Example 21. Find the general solution of y 00 + y = csc(x) for 0 < x < π.
The characteristic equation of y 00 + y = 0 is:
r2 + 1 = 0.
Solve r2 + 1 = 0, we have
and r2 = −i.
r1 = i,
So the general solution to y 00 + y = 0 is:
y(x) = C1 cos(x) + C2 sin(x).
Let y(x) = C1 (x) cos(x) + C2 (x) sin(x) be particular solution to y 00 + y = csc(x), then
yp0 (x)
= C10 (x) cos(x) − C1 (x) sin(x) + C20 (x) sin(x) + C2 (x) cos(x).
Assume that
C10 (x) cos(x) + C20 (x) sin(x) = 0.
Then
yp0 (x)
= −C1 (x) sin(x) + C2 (x) cos(x)
yp00 (x)
= −C10 (x) sin(x) + C1 (x) cos(x) + C20 (x) cos(x) − C2 (x) sin(x)
yp00 (x) + yp (x)
= −C10 (x) sin(x) + C1 (x) cos(x) + C20 (x) cos(x) − u2 (x) sin(x) + C1 (x) cos(x) + C2 (x) sin(x)
= −C10 (x) sin(x) + C20 (x) cos(x)
=
csc(x).
So C10 and C20 satisfy


 C10 (x) cos(x) + C20 (x) sin(x) = 0

 −C 0 (x) sin(x) + C 0 (x) cos(x) = csc(x).
2
1
Solve C10 and C20 , we have
C10 (x) = − sin(x) · csc(x) = −1,
and C20 (x) = cos(x) · csc(x) = cot(x).
Then
Z
C1 (x) = −x,
and C2 (x) =
Z
cot(x) dx =
cos(x)
dx = ln sin(x).
sin(x)
ORDINARY DIFFERENTIAL EQUATIONS
Then a particular solution to y 00 + y = csc(x) can be:
yp (x) = −x cos(x) + sin(x) ln sin(x).
So the general solution to y 00 + y = csc(x) is:
yp (x) = C1 cos(x) + C2 sin(x) − x cos(x) + sin(x) ln sin(x) .
Forced oscillations and resonance
Mass-Spring System:
Figure 12. Mass-Spring System
Let x(t) be the displacement of the mass, then
mx00 + cx0 + kx = F (t) .
We are interested in periodic forcing, F (t) = F0 cos(ωt) with ω > 0. So we have
mx00 + cx0 + kx = F0 cos(ωt) .
Rewrite the equation, we have
x00 + 2px0 + ω02 x =
F0
cos(ωt) ,
m
where
c
p=
≥ 0,
2m
r
and ω0 =
k
> 0.
m
29
30
MINGFENG ZHAO
Undamped forced motion and resonance
The differential equation for the undamped forced motion (c = 0) is:
x00 + ω02 x =
F0
cos(ωt),
m
ω > 0.
The characteristic equation of x00 + ω02 x = 0 is:
r2 + ω02 = 0.
Solve r2 + ω02 = 0, we get
r1 = ω0 i,
and r2 = −ω0 i.
The general solution to x00 + ω02 x = 0 is:
xc (t) = A cos(ω0 t) + B sin(ω0 t) .
Since the nonhomogeneous function is
F0
cos(ωt), we have two cases:
m
I. If ω 6= ω0 , let xp (t) = D cos(ωt) + E sin(ωt) be a particular solution to x00 + ω02 x =
x0p (t)
= −Dω sin(ωt) + Eω cos(ωt)
x00p (t)
= −Dω 2 cos(ωt) − Eω 2 sin(ωt)
x00 + ω02 x
F0
cos(ωt), then
m
= −Dω 2 cos(ωt) − Eω 2 sin(ωt) + Dω02 cos(ωt)
= D[ω02 − ω 2 ] cos(ωt) − Eω 2 sin(ωt)
=
F0
cos(ωt).
m
So we get
F0
, and E = 0.
m[ω02 − ω 2 ]
F0
So a particular solution to x00 + ω02 x =
cos(ωt) can be:
m
D=
xp (t) =
F0
cos(ωt) .
− ω2 ]
m[ω02
II. If ω = ω0 . Let xp (t) = t[D cos(ω0 t) + E sin(ω0 t)] = Dt cos(ω0 t) + Et sin(ω0 t) be a particular solution to
F0
x00 + ω02 x =
cos(ω0 t), then
m
x0p (t)
= D cos(ω0 t) − Dω0 t sin(ω0 t) + E sin(ω0 t) + Eω0 t cos(ω0 t)
x00p (t)
= −Dω0 sin(ω0 t) − Dω0 sin(ω0 t) − Dω02 t cos(ω0 t) + Eω0 cos(ω0 t) + Eω0 cos(ω0 t) − Eω02 t sin(ω0 t)
ORDINARY DIFFERENTIAL EQUATIONS
31
= −2Dω0 sin(ω0 t) − Dω02 t cos(ω0 t) + 2Eω0 cos(ω0 t) − Eω02 t sin(ω0 t)
x00 + ω02 x
= −2Dω0 sin(ω0 t) − Dω02 t cos(ω0 t) + 2Eω0 cos(ω0 t) − Eω02 t sin(ω0 t)
+ω02 [Dt cos(ω0 t) + Et sin(ω0 t)]
= −2Dω0 sin(ω0 t) + 2Eω0 cos(ω0 t)
=
F0
cos(ω0 t).
m
Then we have
D = 0,
So a particular solution to x00 + ω02 x =
F0
.
2mω0
F0
cos(ω0 t) can be:
m
xp (t) =
In summary, the general solution to x00 + ω02 x =
x(t) =
and E =
F0
t sin(ω0 t) .
2mω0
F0
cos(ωt) is:
m



 A cos(ω0 t) + B sin(ω0 t) +
F0
cos(ωt), if ω0 6= ω,
m[ω02 − ω 2 ]


 A cos(ω0 t) + B sin(ω0 t) +
F0
t sin(ω0 t),
2mω0
.
if ω0 = ω
F0
F0
t and
t, in particular, the
2mω0
2mω0
amplitude of x(t) will go to infinity. This kind of behavior is called resonance or perhaps pure resonance.
In the case that ω0 = ω, when t → ∞, the graph of x(t) oscillates between −
Figure 13. Graph of
Damped forced motion and practical resonance
1
t sin(πt)
π
32
MINGFENG ZHAO
The differential equation for the damped forced motion (c > 0) is:
x00 + 2px0 + ω02 x =
F0
cos(ωt).
m
The characteristic equation of x00 + 2px0 + ω02 x = 0 is:
r2 + 2pr + ω02 = 0.
Solve r2 + 2pr + ω02 = 0, we have
r1,2 = −p ±
q
p2 − ω02 .
The general solution to x00 + 2px0 + ω02 x = 0 is:



Aer1 t + Ber2 t ,
overdamping, that is, p2 − ω02 > 0



Ae−pt + Bte−pt ,
critical damping, that is, p2 − ω02 = 0 .
xc (t) =

q
q



 Ae−pt cos
ω02 − p2 · t + Be−pt sin
ω02 − p2 · t , underdamping, that is, p2 − ω02 < 0
In any case, since p 6= 0, then xc (t) → 0 as t → ∞. Since ωi is not a solution to r2 + 2pr + ω02 = 0, then we can let
F0
cos(ωt), then
xp (t) = D cos(ωt) + E sin(ωt) be a particular solution to x00 + 2px0 + ω02 x =
m
x0p (t)
= −Dω sin(ωt) + Eω cos(ωt)
x00p (t)
= −Dω 2 cos(ωt) − Eω 2 sin(ωt)
x00p (t) + 2px0p (t) + ω02 xp (t)
= −Dω 2 cos(ωt) − Eω 2 sin(ωt) + 2p [−Dω sin(ωt) + Eω cos(ωt)]
+ω02 [D cos(ωt) + E sin(ωt)]
=
−Dω 2 + 2pEω + Dω02 cos(ωt) + −Eω 2 − 2pDω + Eω02 sin(ωt)
=
F0
cos(ωt).
m
Then we have
−Dω 2 + 2pEω + Dω02 =
F0
,
m
and
− Eω 2 − 2pDω + Eω02 = 0.
Solve D and E, we get
D=
F0 [ω02 − ω 2 ]
,
m [(2ωp)2 + (ω02 − ω 2 )2 ]
and E =
2ωpF0
.
m [(2ωp)2 + (ω02 − ω 2 )2 ]
ORDINARY DIFFERENTIAL EQUATIONS
So a particular solution to x00 + 2px0 + ω02 x =
xp (t) =
33
F0
cos(ωt) is:
m
F0 [ω02 − ω 2 ]
2ωpF0
· cos(ωt) +
sin(ωt) .
2
2
2
2
2
m [(2ωp) + (ω0 − ω ) ]
m [(2ωp) + (ω02 − ω 2 )2 ]
The general solution xc to x00 + 2px0 + ω02 x = 0 is called the transient solution, denoted by xtr , the particular solution
F0
xp found in above to x00 + 2px0 + ω02 x =
cos(ωt) is called the steady periodic solution, denoted by xsp , which is a
m
periodic function with frequency ω.
Since xtr (t) → 0 as t → ∞, so for any given initial data, at infinity, the solution will be close the the steady periodic
solution xsp (t).
Figure 14. Solutions with differential initial data for k = m = F0 = 1, c = 0.7 and ω = 1.1
For the steady periodic solution:
xsp (t) =
F0 [ω02 − ω 2 ]
2ωpF0
· cos(ωt) +
sin(ωt).
2
2
2
2
2
m [(2ωp) + (ω0 − ω ) ]
m [(2ωp) + (ω02 − ω 2 )2 ]
we know that
I. xsp (t) is a periodic function
II. xsp (t) has frequency ω, and the period
2π
.
ω
III. The amplitude of xsp is:
s
C(ω)
=
F0 [ω02 − ω 2 ]
m [(2ωp)2 + (ω02 − ω 2 )2 ]
=
F0
1
·p
.
m
(2ωp)2 + (ω02 − ω 2 )2
2
+
2ωpF0
m [(2ωp)2 + (ω02 − ω 2 )2 ]
2
34
MINGFENG ZHAO
Notice that
C 0 (ω)
=
F0
1
2(2ωp) · 2p + 2(ω02 − ω 2 ) · (−2ω)
· −
·
3
m
2
[(2ωp)2 + (ω02 − ω 2 )] 2
=
F0
2ω[ω02 − ω 2 − 2p2 ]
·
m [(2ωp)2 + (ω02 − ω 2 )] 23
=
2F0 −ω[ω 2 − (ω02 − 2p2 )]
·
.
m [(2ωp)2 + (ω02 − ω 2 )] 23
Then
– If ω02 − 2p2 ≤ 0, then C(ω) has the maximum value at ω = 0. But we assume that ω > 0, so C(ω) can not
F0
attain its maximum, that is, C(ω) < C(0) =
for all ω > 0.
mω02
p
– If ω02 − 2p2 > 0, then C(ω) has the maximum value at ω = ω02 − 2p2 , that is,
q
1
F0
2F0
max C(ω) = C( ω02 − 2p2 ) =
·p
=p
2
2
2
2
ω>0
m
c [4k 2 − c2 ]
4p [ω0 − p ]
In this case, ω =
1
p
4p2 [ω02
− p2 ]
p
ω02 − 2p2 is called the practical resonance frequency, and max C(ω) =
ω>0
F0
·
m
is called the practical resonance amplitude.
Figure 15. Graph of C(ω) for k = m = F0 = 1 with c = 0.4, 0.8 and 1.6
Example 22. A mass of 4 kg on a spring with k = 4 and a damping constant c = 1. Suppose F0 = 2. Using forcing
function F0 cos(ωt), find the ω that causes practical resonance and find the practical amplitude.
By the assumption, m = 4, c = 1, k = 4, and F0 = 2, we have
4x00 + x0 + 4x = 2 cos(ωt).
ORDINARY DIFFERENTIAL EQUATIONS
The characteristic equation of 4x00 + x0 + 4x = 0 is 4r2 + r + 4 = 0, then
√
−1 ± 1 − 4 · 4 · 4
r1,2 =
8
Let the stead periodic solution be:
xsp (t) = A cos(ωt) + B sin(ωt).
That is, xsp is a particular solution to 4x00 + x0 + 4x = 2 cos(ωt). Then
x0sp (t)
= −Aω sin(ωt) + Bω cos(ωt)
x00sp (t)
= −Aω 2 cos(ωt) − Bω 2 sin(ωt).
Then
4x00sp + x0sp + 4xsp
=
4 −Aω 2 cos(ωt) − Bω 2 sin(ωt) + [−Aω sin(ωt) + Bω cos(ωt)] + 4 [A cos(ωt) + B sin(ωt)]
=
=
2 cos(ωt).
−4Aω 2 + Bω + 4A cos(ωt) + −4Bω 2 − Aω + 4B sin(ωt)
Then
−4Aω 2 + Bω + 4A = 2,
and
− 4Bω 2 − Aω + 4B = 0.
Solve A and B, we get
A=
8 − 8ω 2
,
ω 2 + (4 − ω 2 )2
and B =
2ω
.
ω 2 + (4 − ω 2 )2
Then
xsp (t) = A cos(ωt) + B sin(ωt) =
2ω
8 − 8ω 2
cos(ωt) + 2
sin(ωt).
2
2
2
ω + (4 − 4ω )
ω + (4 − 4ω 2 )2
So the amplitude is:
C(ω)
=
=
=
=
=
p
A2 + B 2
s
2 2
8 − 8ω 2
2ω
+
ω 2 + (4 − 4ω 2 )2
ω 2 + (4 − 4ω 2 )2
p
(8 − 8ω 2 )2 + 4ω 2
ω 2 + (4 − 4ω 2 )2
p
2 (4 − 4ω 2 )2 + ω 2
ω 2 + (4 − 4ω 2 )2
2
p
ω 2 + (4 − 4ω 2 )2
35
36
MINGFENG ZHAO
=
=
2
+ 16 − 32ω 2 + 16ω 4
2
√
.
4
16ω − 31ω 2 + 16
√
ω2
So we get
C 0 (ω) = −
64ω 3 − 62ω
=
2ω(31 − 32ω 2 )
3
3 .
(16ω 4 − 31ω 2 + 16) 2
(16ω 4 − 31ω 2 + 16) 2
r
31
So C(ω) will achieve its maximum at ω =
, that is, the practical resonance frequency is:
32
r
31
ωpr =
≈ 0.984 .
32
The practical resonance amplitude is:
r
max C(ω) = C(ωpr ) = C
ω>0
31
32
!
2
=q
63
64
16
= √ ≈ 2.016 .
63
ORDINARY DIFFERENTIAL EQUATIONS
37
Review of Chapter 6:
The Laplace transform
Definition 7. Let f (t) be a function on [0, ∞), then
I. The Laplace transform of f , denoted by L[f ](s), is defined as:
Z
L[f (t)](s) =
∞
f (t)e−st dt,
for all s > 0.
0
II. If F (s) = L[f ](s), the inverse Laplace transform of F , denoted by L−1 [F ](t), is defined as:
L−1 [F (s)](t) = f (t),
for all t > 0.
Definition 8. Let a be a real constant, and the Heaviside function is defined as:

 1, if t ≥ a,
u(t − a) =
 0, if x < a.
Remark 7. Notice that when a = 0, we know that u(t − 0) = 1 for all t ≥ 0; when a = ∞, we define u(t − ∞) := 0 for
all t ≥ 0.
Definition 9. Let f (t) and g(t) be two functions on [0, ∞), the convolution of f and g is defined as:
Z
(f ∗ g)(t) =
t
f (τ )g(t − τ ) dτ.
0
Definition 10 (Dirac’s delta function). For any continuous function f (t) on (−∞, ∞), we have
Z
∞
δ(t)f (t) dt = f (0).
−∞
Proposition 1. The followings hold:
I. Transforms of derivatives:
L[f 0 ](s)
=
sL[f ](s) − f (0)
L[f 00 ](s)
=
s2 L[f ](s) − sf (0) − f 0 (0).
38
MINGFENG ZHAO
II. First Shifting Property:
L e−at f (t) (s)
= L[f ](s + a)
L−1 [F (s + a)](t)
= e−at L−1 [F (s)](t).
III. Second Shifting Property:
= e−as L[f (t)](s)
L[u(t − a)f (t − a)](s)
L−1 e−as F (s) = u(t − a)L−1 [F (s)](t − a).
IV. Transform of Integrals:
t
Z
L
f (τ ) dτ (s)
0
L−1
F (s)
(t)
s
L[f (t)](s)
s
Z t
=
L−1 [F (s)](τ ) dτ
=
0
V. Transform of Convolution:
L[(f ∗ g)(t)](s)
=
L[f (t)](s) · L[g(t)](s)
L−1 [F (s) · G(s)] (t)
=
L−1 [F (s)] ∗ L−1 [G(s)](t).
VI. Transform of Dirac delta:
L[δ(t − a)](s)
L−1 [e−as ](t)
= e−as
= δ(t − a).
Remark 8. In practice, we can use the second shifting property in the following way:
L[u(t − a)f (t)](s) = e−as L[f (t + a)](s).
Example 23. Find L[1].
In fact, we have
Z
L[1](s) =
∞
e
0
−st
∞
1 −st 1
dt = − e = ,
s
s
0
So
L[1] =
1
.
s
for all s > 0.
ORDINARY DIFFERENTIAL EQUATIONS
−1
Also we know that L
Example 24. Find L−1
1
=1.
s
1
.
s2 + 4s + 8
Notice that
1
1
=
.
s2 + 4s + 8
(s + 2)2 + 4
By looking up the table, we have
L[sin(2t)](s) =
2
.
s2 + 4
So we get
L
sin(2t)
1
.
(s) = 2
2
s +4
So we have
−1
L
1
2
s + 4s + 8
−1
1
(s + 2)2 + 4
=
L
=
1 −2t
e
sin(2t),
2
By the First Shifting Property.
Therefore, we get
L−1
1
1
= e−2t sin(2t) .
s2 + 4s + 8
2
Example 25. Solve x00 + x = f (t), x(0) = x0 (0) = 0, where

 1, if 1 ≤ t < 5
f (t) =
 0, otherwise.
Let X(s) = L[x] and F (s) = L[f (t)], apply the Laplace transform on the both sides of x00 + x = f (t), then
L[x00 ] + L[x] = L[f (t)] = F (s).
Since x(0) = x0 (0) = 0, then
L[x00 ] = s2 L[x] − sx(0) − x00 (0) = s2 X(s).
So we get
s2 X(s) + X(s) = F (s).
That is,
X(s) =
F (s)
.
s2 + 1
39
40
MINGFENG ZHAO
For F (s) = L[f (t)], by the definition of f (t), we know that
f (t) = u(t − 1) − u(t − 5),
∀t ≥ 0.
Then
F (s)
=
L[f (t)]
=
L[u(t − 1)] − L[u(t − 5)]
=
e−s
e−5s
−
.
s
s
So
X(s) =
e−s
e−5s
F (s)
=
−
.
2
2
s +1
s(s + 1) s(s2 + 1)
In order to compute L−1 [X(s)], by the Second Shifting Property, we need to compute L−1
of partial fractions, we have
1
A Bs + C
= + 2
.
s(s2 + 1)
s
s +1
Then
1 = A(s2 + 1) + (Bs + C)s = (A + B)s2 + Cs + A.
So
A + B = 0,
C = 0,
and A = 1.
Then
B = −1. and C = 0.
A = 1,
Then
1
1
s
= − 2
.
+ 1)
s s +1
s(s2
So we have
−1
L
1
s(s2 + 1)
−1
1
s
−1
−L
s
s2 + 1
=
L
=
1 − cos(t)
So we know that
x(t)
= L−1 [X(s)]
e−s
e−5s
−1
−1
= L
−L
s(s2 + 1)
s(s2 + 1)
By looking the table.
h
1
s(s2 +1)
i
. Use the method
ORDINARY DIFFERENTIAL EQUATIONS
= u1 (t)[1 − cos(t − 1)] − u5 (t)[1 − cos(t − 5)],
By the Second Shifting Property.
Therefore, the solution to x00 + x = f (t), x(0) = x0 (0) = 0 is:
x(t) = u(t − 1)[1 − cos(t − 1)] − u(t − 5)[1 − cos(t − 5)] .
Example 26. Find the Laplace transform of
f (t) =




1




t






0
if t < 1
if 1 ≤ t < 2
if t ≥ 2.
We can rewrite the piecewise definition of f using Heaviside functions:
f (t) = 1 · (u(t − 0) − u(t − 1)) + t · u(t − 1) − u(t − 2) + 0 · [u(t − 2) − u(t − ∞)]
= 1 − u(t − 1) + tu(t − 1) − tu(t − 2)
= 1 + (t − 1)u(t − 1) − tu(t − 2).
This last expression makes it easier to apply the second shifting formula:
F (s)
= L[1] + L[(t − 1)u(t − 1)] − L[tu(t − 2)]
=
=
=
1
+ e−s L[t + 1 − 1] − e−2s L[t + 2]
s
1
+ e−s L[t] − e−2s (L[t] + 2L[1])
s
1 e−s
2
1
−2s
+ 2 −e
+
s
s
s2
s
Remark 9. In general, let a0 = 0 < a1 < a2 < a3 < · · · < an < ∞, if



f1 (t),
if 0 ≤ t < a1 ,







f2 (t),
if a1 ≤ t < a2 ,





 f3 (t),
if a2 ≤ t < a3 ,
f (t) =
..
..


.
.







fn (t),
if an−1 ≤ t < an ,





 fn+1 (t), if t ≥ an ,
41
42
MINGFENG ZHAO
then,
f (t)
n
X
=
fk (t)[u(t − ak−1 ) − u(t − ak )] + fn+1 [u(t − an ) − u(t − ∞)]
k=1
f1 (t)[1 − u(t − a1 )] +
=
n
X
fk (t)[u(t − ak−1 ) − u(t − ak )] + fn+1 (t)u(t − an ).
k=2
1
.
s(s2 + 1)
By looking up the table, we have
Example 27. Find L−1
L−1
By the transform of integrals, we have
L−1
1
2
s(s + 1)
1
= sin(t).
s2 + 1
"
=
−1
1
s2 +1
t
s
L
Z
L
=
0
Z
=
−1
#
1
(τ ) dτ
s2 + 1
t
sin(τ ) dτ
0
=
1 − cos(t).
So
L−1
1
(t) = 1 − cos(t) .
s(s2 + 1)
Example 28. Let ω > 0, f (t) = sin(ωt) and g(t) = cos(ωt) for all t ≥ 0, find (f ∗ g)(t).
In fact, we have
t
Z
(f ∗ g)(t)
sin(ωτ ) cos(ω(t − τ )) dtτ
=
0
=
=
=
1
2
Z
t
[sin(ωt) − sin(ωt − 2ωτ )] dτ
0
t
1
1
sin(ωt)τ +
cos(2ωτ − ωt) 2
4ω
0
1
t sin(ωt).
2
Example 29. Let f (t) be any nice function on [0, ∞), find the solution to x00 + ω02 x = f (t), x(0) = x0 (0) = 0.
Let X(s) = L[x(t)](s) and F (s) = L[f (t)](s), apply the Laplace transform on the both sides of x00 + ω02 x = f (t), then
L[x00 ] + ω02 L[x] = L[f (t)].
ORDINARY DIFFERENTIAL EQUATIONS
43
By the transform of derivatives, we have
L[x00 ] = s2 L[x] − sx(0) − x00 (0) = s2 X(s).
Then we have
s2 X(s) + ω02 X(s) = F (s).
So we get
X(s) =
F (s)
1
= F (s) · 2
.
s2 + ω02
s + ω02
By the Laplace transform of convolution, we know that
x(t)
= L−1 [X(s)](t)
=
L−1 [F (s)] ∗ L−1
1
s2 + ω02
.
By looking up the table, we get
L
−1
1
sin(ω0 t)
=
.
2
2
s + ω0
ω0
So we have
Z
x(t)
t
f (t − τ ) ·
=
0
=
1
ω0
Z
sin(ω0 τ )
dτ
ω0
t
sin(ω0 τ )f (t − τ ) dτ.
0
Example 30. Solve y 00 + 4y 0 + 5y = δ(t − 3), y(0) = 1, y 0 (0) = 0.
Let Y (s) = L[y(t)], apply the Laplace transform on the both sides of y 00 + 4y 0 + 5y = δ(t − 3), then
L[y 00 ] + 4L[y 0 ] + 5L[y] = L[δ(t − 3)].
By the transform of derivatives, we have
L[y 0 ] = sL[y] − y(0) = sY (s) − 1,
and L[y 00 ] = s2 L[y] − sy(0) − y 0 (0) = s2 Y (s) − s.
By looking the table, we have
L[δ(t − 3)] = e−3s .
So we get
s2 Y (s) − s + 4[sY (s) − 1] + 5Y (s) = e−3s .
44
MINGFENG ZHAO
Then
Y (s)
=
=
=
=
e−3s + s + 4
s2 + 4s + 5
s+4
e−3s
+
s2 + 4s + 5 s2 + 4s + 5
s+4
e−3s
+
(s + 2)2 + 1 (s + 2)2 + 1
s+2
2
e−3s
+
+
.
2
2
(s + 2) + 1 (s + 2) + 1 (s + 2)2 + 1
By the first shifting property and looking up the table, we have
s+2
1
−1
−2t
−1
L
(t) = e
cos(t), and L
(t) = e−2t sin(t).
(s + 2)2 + 1
(s + 2)2 + 1
By the Second Shifting Property, we have
e−3s
−1
L
=
(s + 2)2 + 1
=
u(t − 3) · L
−1
1
(t − 3)
(s + 2)2 + 1
u(t − 3)e−2(t−3) sin(t − 3).
Then we get
y(t) = L−1 [Y (s)] = e−2t cos(t) + 2e−2t sin(t) + u(t − 3)e−2(t−3) sin(t − 3).
Therefore, the solution to y 00 + 4y 0 + 5y = δ(t − 3), y(0) = 1, y 0 (0) = 0 is:
y(t) = e−2t cos(t) + 2e−2t sin(t) + u(t − 3)e−2(t−3) sin(t − 3) .
ORDINARY DIFFERENTIAL EQUATIONS
45
Review of Chapter 3:
In this course, we only study the linear system in the following form:

 x0 = ax1 + bx2 + f1 (t)
1
 x0 = cx + dx + f (t)
1
2
2
2

Let ~x(t) = 
x1 (t)

,

A=
x2 (t)
a
b
c
d


and f~(t) = 
,
f1 (t)

, then we have ~x0 = A~x + f~(t) .
f2 (t)
Theorem 6. Let ~xc (t) be the general solution to the homogeneous system ~x0 = A~x, and ~xp (t) be a particular solution
to ~x0 = A~x + f~(t), then the general solution to ~x0 = A~x + f~(t) is:
~x(t) = ~xc (t) + ~xp (t).
Homogeneous system

Theorem 7. Let A be a 2 × 2 matrix, and ~y (t) = 
y1 (t)


 and ~z(t) = 
y2 (t)
z1 (t)

 be two linearly independent solutions
z2 (t)
to ~x0 = A~x, then the general solution to ~x0 = A~x is
~x(t) = C1 ~y (t) + C2 ~z(t).
~
In this case, the 2 × 2 matrix X(t)
= [~y (t) ~z(t)] is called a fundamental matrix of the system ~x0 = A~x. It’s easy to
~ 0 (t) = AX(t),
~
see that X
where

~ 0 (t) = 
X
y1 (t) z1 (t)
0

 := 
y2 (t) z2 (t)
y10 (t) z10 (t)
y20 (t)
z20 (t)

.
Theorem 8 (Eigenvalue Method). Let λ be an eigenvalue of A and ~v be an eigenvector corresponding to λ, then
~x(t) = eλt~v is a solution to ~x0 = A~x.
Remark 10. The non-zero eigenvectors corresponding to different eigenvalues are linearly independent.
Eigenvalue method with distinct real eigenvalues
46
MINGFENG ZHAO
Theorem 9. Let A be a 2 × 2 matrix with two distinct real eigenvalues λ1 and λ2 , and ~v1 and ~v2 are eigenvectors
corresponding to λ1 and λ2 , respectively, then the general solution to ~x0 = A~x is:
~x = C1 eλ1 t~v1 + C2 eλ2 t~v2 .
Example 31. Find the general solution to the system:

Let ~x(t) = 
x1 (t)


, A = 
x2 (t)

2
1
0
Since
1
A=
2
1
0
1
x01
=
2x1 + x2
x02
=
x2 .

, then we need to solve ~x0 = A~x. First, we need to find the eigenvalues of

 and corresponding eigenvectors.

det (A − λI2 ) = det 
2
1
0
1



 − λI2  = det 
2−λ
1
0
1−λ
then

I. When λ1 = 1, let’s solve 
2
0
λ1 = 1,
 
and λ2 = 2.

1
x1
x1

=
, that is,
1
x2
x2


1
1
0
0


x1



0
=
x2

.
0
Then we know that

x1



 = x1 
x2

So 
1
−1

1
−1
.

 is an eigenvector corresponding to λ1 = 1.

II. When λ1 = 2, let’s solve 
2
1

x1

0
1


 = 2
x2

0

0
x1

, that is,
x2
1
−1

x1

x2


=
0
0

.

 = (2 − λ)(1 − λ) = 0,
ORDINARY DIFFERENTIAL EQUATIONS
47
Then we know that


x1

 = x1 

x2

So 
1

1
.
0

 is an eigenvector corresponding to λ1 = 2.
0

In summary, the eigenvalues and corresponding eigenvectors of 

λ1 = 1 with 
2
1
0
1

 are:

1

1
 and λ2 = 2 with 
−1

.
0
Therefore, the general solution to ~x0 = A~x is:

x1 (t)



 = C 1 et 
x2 (t)

1

1
 + C2 e2t 
−1


=
C1 et + C2 e2t
−C1 et
0

.
Eigenvalue method with distinct complex eigenvalues
Theorem 10. Let A be a 2 × 2 matrix with two distinct complex eigenvalues λ1 and λ2 , and ~v1 and ~v2 are eigenvectors
corresponding to λ1 and λ2 , respectively, then λ2 = λ1 , ~v2 = ~v1 , and the general solution to ~x0 = A~x is:
~x = C1 Re eλ1 t~v1 + C2 Im eλ1 t~v1 .

Example 32. Find the general solution to the system ~x0 = 

First, let’s find the eigenvalues of A = 
1
1
−1
1
1
1
−1
1

 ~x.

, then

det (A − λI2 )
=
det 
1−λ
−1
=
(1 − λ)2 + 1
=
λ2 − 2λ + 2
=
0.
1
1−λ


48
MINGFENG ZHAO
Then
and λ2 = 1 − i.
λ1 = 1 + i,
For λ = 1 + i, let’s solve A~v = λ~v , that is,

1

−1
1

x1

1


x1
 = (1 + i) 
x2

.
x2
So we get

i
−1
1
i


x1



0
=
x2

.
0
So we have

x1



 = x1 
x2

That is, 
1
1

.
i


 is an eigenvalue corresponding to 1 + i, then e(1+i)t 
i
identity, we have
1

 is a solution to ~x0 = A~x. By Euler’s
i

e(1+i)t 
1


=
i
e(1+i)t
ie(1+i)t


=

et cos(t) + iet sin(t)
i [et cos(t) + iet sin(t)]

=
et cos(t)
−et sin(t)


 + i
et sin(t)
et cos(t)

.
Therefore, the general solution to ~x0 = A~x is:

~x(t) = C1 
et cos(t)
t
−e sin(t)


 + C2 
et sin(t)
t

.
e cos(t)
Eigenvalue method with the same real eigenvalue
Let A be a 2 × 2 matrix with the same real eigenvalue λ and ~v be the eigenvector corresponding to λ, then ~x(t) = eλt~v
is a solution to ~v 0 = A~x.

• If λI2 − A = 0, then 
1


0

 and   are two linearly independent eigenvectors corresponding to λ, that is, λ
0
1
is a complete eigenvalue of A. Hence the general solution to ~x0 = A~x is


 
1
0
 + eλt   .
~x(t) = eλt 
0
1
ORDINARY DIFFERENTIAL EQUATIONS
49
• If λI2 − A 6= 0, that is, λ is a defective eigenvalue of A, then another linearly independent solution ~y to ~x0 = A~x
has the form ~y (t) = eλt (t~v + w)
~ for some vector w
~ to be determined. That is, ~y (t) = t~x(t) + eλt w.
~ So we get
~y 0 (t) = ~x(t) + t~x0 (t) + λeλt w
~ = ~x(t) + tA~x(t) + λeλt w
~ = A~y (t) = tA~x(t) + eλt Aw.
~
Then ~x(t) + λeλt w
~ = eλt Aw,
~ that is, eλt~v + λeλt w
~ = eλt Aw,
~ so w
~ is a solution to ~v + λw
~ = Aw,
~ that is,
(A − λI2 )w
~ = ~v .
In this case, we have (A − λI2 )2 w
~ = (A − λI2 )~v = 0, we call w
~ a generalized eigenvector of A.

Example 33. Solve ~x0 = 

2
0
2
0
0
2

 ~x.

, it’s easy to see that A has only one eigenvalue λ = 2 and A − 2I2 = 0, that is, λ = 2 is a complete
0 2
eigenvalue of A. So the general solution to ~x0 = A~x is:
Let A = 

~x(t) = C1 e2t 
1


 + C2 e2t 
0
det (A − λI2 ) = det 
.
1

 x0 = 3x1 + x2
1
Example 34. Find the general solution to
.
 x0 = 3x
2
2


 x0 = 3x1 + x2
3
1
The coefficient matrix of the system
is A = 
 x0 = 3x
0
2
2


0
1

. First, let’s find eigenvalues of A, then
3

3−λ
1
0
3−λ
 = (3 − λ)2 = 0.
Then λ1 = λ2 = 3. Now let’s solve


3
1
0
3
0
−1
x1




 = 3
x2
x1

.
x2
Then



x1

0
0
x2


=
0
0

.
50
MINGFENG ZHAO

So 
x1


 = x1 
x2
1


. Hence we know that λ = 3 is the only one eigenvalue for A, and ~v = 
0
3t
1

 is the eigenvector.
0
0
Then ~y (t) = e ~v is a solution to ~x = A~x.
Let ~x = e3t (t~v + w)
~ = t~y (t) + e3t w
~ be another linearly independent solution to ~x0 = A~x, then
~x0 = ~y (t) + t~y 0 (t) + 3e3t w
~ = A(t~y (t) + e3t w)
~ = tA~y (t) + e3t Aw.
~
Since ~y (t) is a solution to ~x0 = A~x, then
~y (t) + 3e3t w
~ = e3t Aw.
~
Since ~y (t) = e3t~v , then e3t~v + 3e3t w
~ = e3t Aw,
~ that is, ~v + 3w
~ = Aw.
~ Hence, (A − 3I2 )w
~ = ~v , that is,

0
1


w1

0
0

So we can take w1 = 0 and w2 = 1, that is, w
~ =


=
w2

1

0

0
. So another solution is
1
 
e3t (t~v + w)
~ = e3t t 
1


+
0
0


 = e3t 
1
t

.
1
Therefore, the general solution to ~x0 = A~x is:

~x = C1 e3t 
1


 + C2 e3t 
0
t
−1

.
Undetermined coefficients
Theorem 11. Let ~xc (t) be the general solution to the homogeneous system ~x0 = A~x, and ~xp (t) be a particular solution
to ~x0 = A~x + f~(t), then the general solution to ~x0 = A~x + f~(t) is:
~x(t) = ~xc (t) + ~xp (t).
Let A be a 2 × 2 matrix,, consider the system:
~x0 = A~x + f~(t).
ORDINARY DIFFERENTIAL EQUATIONS
Let p~n (t) and ~qn (t) be “vector coefficients” polynomials with degree n. If f~(t) has the form:
p~n (t)emt cos(kt) + ~qn (t)emt sin(kt),
then a particular solution ~xp (t) to ~x0 = A~x + f~(t) can be take as:
~xp (t) = p~n+α (t)emt cos(kt) + ~qn+α (t)emt sin(kt),
where
• α is one of 0, 1 and 2 (α is the multiplicity of m + ki as the eigenvalues to A):
– If m + ki is not an eigenvalue of A, then α = 0.
– If m + ki is an eigenvalue of Aand A has two different eigenvalues, then α = 1.
– If m + ki is an eigenvalue of A and A has only one eigenvalue, then α = 2.
• p~n+α (t) and ~qn+α (t) are undetermined “vector coefficients” polynomials with degree n + α.

Example 35. Find the general solution to ~x0 = 

Let A = 
1
1
4
−2


 and f~(t) = − 
et
et

det (A − λI2 ) = det 

1
1
4
−2

 ~x − 
et
et

.

. First, let’s find the eigenvalues of A, that is,
1−λ
1
4
−2 − λ

 = (1 − λ)(−2 − λ) − 4 = λ2 + λ − 6 = 0.
Then
λ1 = −3,

Since − 
et
e
t


=
−1
−1
and λ2 = 2.

 et , let ~xp (t) = ~aet be a particular solution to ~x0 = A~x + f~(t), then

~x0p (t) = ~aet = A~aet + 
−1
−1
So we get

(I2 − A)~a = 
−1
−1

.

 et .
51
52
MINGFENG ZHAO
That is,


0
−1
−4
3


a1

−1
=

−1
a2

.
Then

1
~a = 

.
1

So ~xp (t) = 
1

 et is a particular solution to ~x0 = A~x + f~(t).
1

Example 36. Let A = 
−1
0
−2
1


. Find a particular solution of ~x0 = A~x + f~(t), where f~(t) = 
First, let’s find the eigenvalues of A, that is, det (A − λI2 ) = det 

f~(t) = 
et


1
=
1
0
−2
1−λ
 = (−λ − 1)(1 − λ) = 0, then


0
 et + 
0

 t.
1
So we can let ~xp (t) = (~at + ~b)et + ~ct + d~ be a particular solution to ~x0 = A~x = f~(t), then
= ~aet + ~atet + ~bet + ~c
~x0p (t)
= A~xp (t) + f~(t)

= A~atet + A~bet + A~ct + Ad~ + 
1


 et + 
0
0

 t.
1
Then we have

A~b + 
1


 = ~a + ~b,
A~c + 
0
0

 = 0,
~
and ~c = Ad.
1
Solve ~a, ~b, ~c, m and n, we can take
~a = 
0
−1
.

−1 − λ
λ1 = 1 and λ2 = −1. Notice that


t

A~a = ~a,
et

,

~b = 
1
2
0

,

~c = 
0
−1

,

and d~ = 
0
−1

.
ORDINARY DIFFERENTIAL EQUATIONS
53
So a particular solution to ~x0 = A~x + f~(t) is:

~xp (t) = 
0
−1


 tet + 
1
2


 et + 
0

0
−1

t + 
0
−1


=
1 t
2e
t
−te − t − 1

.
Variation of Parameters

Let A be a 2 × 2 matrix, and ~y (t) = 
y1 (t)


 and ~z(t) = 
y2 (t)

~x0 = A(t)~x, let X(t) = [~y (t) ~z(t)] = 
y1 (t) z1 (t)
z1 (t)

 be two linearly independent solutions to
z2 (t)

, then X(t) is a fundamental matrix, and X 0 (t) = A(t)X(t). Hence
y2 (t) z2 (t)
0
the general solution to ~x = A(t)~x is:

~x(t) = C1 ~y (t) + C2 ~z(t) = [~y (t) ~z(t)] 
C1
C2


=
y1 (t) z1 (t)

C1

y2 (t) z2 (t)

~
 = X(t)C.
C2
Let’s consider
~x0 = A(t)~x + f~(t).
Let ~xp (t) = X(t)~u(t) for some ~u(t) be a particular solution to ~x0 = A(t)~x + f~(t), then
~x0p = X 0 (t)~u(t) + X(t)~u0 (t) = A(t)~xp + f~(t) = A(t)X(t)~u(t) + f~(t).
Since X 0 (t) = A(t)X(t), then
A(t)X(t)~u(t) + X(t)~u0 (t) = A(t)X(t)~u(t) + f~(t).
That is, X(t)~u0 (t) = f~(t), which implies that
~u0 (t) = X(t)−1 f~(t).
Then
Z
~u(t) =
X(t)−1 f~(t) dt.
That is,
Z
~xp (t) = X(t)
X(t)−1 f~(t) dt.
54
MINGFENG ZHAO

Remark 11. Let A = 
a
b
c
d

, then
A−1

Example 37. Let A = 
−1
0
−2
1



d
1  d −b 
1

=
=
det A −c a
ad − bc −c
−b

.
a


. Find a particular solution of ~x0 = A~x + f~(t), where f~(t) = 
et

.
t

First, let’s find the eigenvalues of A, that is, det (A − λI2 ) = det 
−1 − λ
0
−2
1−λ

 = (−λ − 1)(1 − λ) = 0, then
and λ2 = −1.
λ1 = 1,
For λ1 = 1, let’s solve A~x = ~x, that is,



Then 
x1


 = x2 
x2
0


, that is, 
1
−2
0
−2
0
0


 et = 

Then 
x2
 = x1 
1


, tha is, 
1
1
0
e
For λ2 = −1, let’s solve A~x = −~x, that is,



=
x2
0

.
0
 is an eigenvalue corresponding to λ = 1, which implies that
1




0

x1
x1
1



t
0
0
−2
2

 is a solution to ~x0 = A~x.

x1



=
x2
0

.
0

 is an eigenvalue corresponding to λ = −2, which implies that
1

1

1


 e−t = 
e−t
e
−t

 is a solution to ~x0 = A~x.
Then we can take the fundamental matrix X(t) as:

X(t) = 
0
e−t
t
−t
e
e

.
ORDINARY DIFFERENTIAL EQUATIONS
55
Then det X(t) = −1, and

X(t)−1 = − 
e−t
−e−t
−et
0


=
−e−t
e−t
et
0

.
Let ~xp (t) = X(t)~u(t) for some ~u(t) be a particular solution to ~x0 = A(t)~x + f~(t), then
~x0p = X 0 (t)~u(t) + X(t)~u0 (t) = A(t)~xp + f~(t) = A(t)X(t)~u(t) + f~(t).
Since X 0 (t) = AX(t), then X(t)~u0 (t) = f~(t), that is, ~u0 (t) + X(t)−1 f~(t). So we get
Z
~u(t) =
X(t)−1 f~(t) dt

Z
=


Z
=
−e−t
e−t
et
0
−1 + te−t
e
=
2t

 dt
t

 dt
−t − te−t − e−t

et




1 2t
2e

.
Then
~x(t)
=
=
=
X(t)~u(t)



0 e−t
−t − te−t − e−t



1 2t
t
−t
e e
2e


1 t
e
2

.
t
−te − t − 1 + 12 et
So a particular solution to ~x0 = A~x + f~(t) is:

~xp (t) = 
t
−te − t − 1 +
Two dimensional systems
and their vector fields

To draw the vector field of
I. Plot the x1 x2 -plane.
 x0 = f1 (x1 , x2 )
1
:
 x0 = f (x , x )
2
1
2
2

1 t
2e
1 t
2e
.
56
MINGFENG ZHAO
II. Select points as many as possible in the plane, say P1 , P2 , · · · , Pn .
III. At each point Pi , draw a short arrow with direction (f1 (Pi ), f2 (Pi )).

 x0 = xy
Example 38. Draw the vector field of the system
 y0 = x + y
Remark 12. For any given initial data ~x(t0 ) = P0 , the solution ~x(t) to ~x0 = f~(~x) and ~x(t0 ) = P0 is a curve ~x(t) =
(x1 (t), x2 (t) which is parametrized by t such that this curve passes through the point P0 and has tangent vector f~(~x(t))
at each point on the curve t 7→ ~x(t). This curve in the x1 x2 -plane is called a trajectory or orbit. A representative set of
trajectories is referred to as a phase portrait.
Notice that for the trajectories, since the trajectory is a curve in the x1 x2 plane, so if we think the trajectory locally,
than x2 (or x1 ) will be a function of x1 (or x2 ), that is, x2 (t) = x2 (x1 (t)), take the derivative with respect to t, by the
chain rule, we have
dx2
dx2 dx1
=
·
.
dt
dx1 dt
So (x1 , x2 ) will satisfy
(2)
dx2
=
dx1
dx2
dt
dx1
dt
=
x02 (t)
f2 (x1 , x2 )
=
.
0
x1 (t)
f1 (x1 , x2 )
So to draw the trajectories by plotting all points (x1 (t), x2 (t)) for a certain range of t, we should solve the first order
differential equation (2).
ORDINARY DIFFERENTIAL EQUATIONS
57
Definition 11. Consider the autonomous system ~x0 = f~(~x), let ~a be a constant vector, we say that ~a is a critical point
of the autonomous system ~x0 = f~(~x) if f~(~a) = 0. In this case, we say that ~x(t) ≡ ~a is an equilibrium solution to the
system ~x0 = f~(~x).

 x0 = sin(x)
Example 39. For the system
. Find all critical points.
 y 0 = sin2 (x) + y 2
Solve sin(x) = 0 and sin2 (x) + y 2 = 0, then we get sin(x) = 0 and y = 0. So we have x = kπ for k ∈ Z and y = 0. So

 x0 = sin(x)
all critical points of the system
are (kπ, 0) with k ∈ Z.
 y 0 = sin2 (x) + y 2
Definition 12. Let ~a be a critical point of the system ~x0 = f~(~x), that is, f~(~a) = 0. Then
I. We say that the critical point ~a is stable if for any given > 0, there exists some δ > 0 such that for any
solutions ~x(t) to the initial value problem ~x0 = f~(~x), ~x(0) = ~x0 with |~x0 − ~a| < δ, we have
|~x(t) − ~x0 | < ,
for all t ≥ 0.
II. We say that the critical point ~a is unstable if ~a is not stable.
III. We say that the critical point ~a is asymptotically stable if ~a is stable, and there exists some δ > 0 such that for
any solutions ~x(t) to the initial value problem ~x0 = f~(~x), ~x(0) = ~x0 with |~x0 − ~a| < δ, we have
lim ~x(t) = ~a.
t→∞

In the following, let A = 
a
b
c
d

 be a 2 × 2 matrix, λ1 and λ2 be two nonzero different eigenvalues of A. Notice
that det(A − λI2 ) = λ2 − (a + d)λ + (ad − bc). Since λ1 6= 0 and λ2 6= 0, then det A 6= 0, which implies that the origin
(0, 0) is the only critical point to the system ~x0 = A~x.
For the behaviors of solutions to ~x0 = A~x near the origin (0, 0), we have the following six cases:
Eigenvalues
Type of Critical Point (0,0)
Stability
I.
λ1 , λ2 are real and both positive
source
unstable
II.
λ1 , λ2 are real and both negative
sink
asymptotically stable
III. λ1 , λ2 are real and opposite signs
saddle point
unstable
IV.
λ1 , λ2 are complex with positive real part
spiral source
unstable
V.
λ1 , λ2 are complex with negative real part
spiral sink
asymptotically stable
VI.
λ1 , λ2 are complex with zero real part
center point
stable
58
MINGFENG ZHAO

Remark 13. Let λ be an eigenvalue of A and ~v = 

method solution to ~x0 = A~x is ~x(t) = eλt~v = eλt 
v1
v1
v2


 be an eigenvector corresponding to λ, then the eigenvalue

=
v1 eλt
λt

, that is, x1 (t) = v1 eλt and x2 (t) = v2 eλt . So in
v2
v2 e
x1
v1
the x1 x2 -plane, we have
= , that is, v2 x1 − v1 x2 = 0 which is a straight line and passing (0, 0) in the x1 x2 -plane.
x2
v2
Now let’s look at each case separately.
I. Both eigenvalues λ1 and λ2 are real and positive, then the vector field of ~x0 = A~x is like a source with arrows
coming out from the origin, we say the critical point (0, 0) is a source and unstable.

Example 40. Let A = 

~v1 = 
1


 and ~v2 = 
0
solution to ~x0 = A~x is:
1
1
1

, then it’s easy to know that λ1 = 1 and λ2 = 2 are two eigenvalues of A, and
0 2

 are eigenvectors corresponding to λ1 = 1 and λ2 = 2, respectively. So the general
1

~x(t) = C1 et 
1
0


 + C2 e2t 
1

.
1
Figure 16. Example source vector field with eigenvectors and solutions
II. Both eigenvalues λ1 and λ2 are real and negative, then the vector field of ~x0 = A~x is like a sink with arrows
pointing to the origin, we say the critical point (0, 0) is a sinkand asymptotically stable.
ORDINARY DIFFERENTIAL EQUATIONS

Example 41. Let A = 
−1
A, and ~v1 = 
1


−1
, then it’s easy to know that λ1 = −1 and λ2 = −2 are two eigenvalues of
−2
0


 and ~v2 = 
1
59

 are eigenvectors corresponding to λ1 = −1 and λ2 = −2, respectively. So
0
1
the general solution to ~x0 = A~x is:

~x(t) = C1 e−t 
1


 + C2 e−2t 
0
1

.
1
Figure 17. Example sink vector field with eigenvectors and solutions
III. Both eigenvalues λ1 and λ2 are real, one is positive, and the other is negative, then we reverse the arrows
on the line corresponding to the negative eigenvalue, we say the critical point (0, 0) is a saddle point and unstable

1
1

, then it’s easy to know that λ1 = 1 and λ2 = −2 are two eigenvalues of A,
0 −2




1
1
 and ~v2 = 
 are eigenvectors corresponding to λ1 = 1 and λ2 = −2, respectively. So the
and ~v1 = 
0
−3
general solution to ~x0 = A~x is:
Example 42. Let A = 

~x(t) = C1 et 
1
0


 + C2 e−2t 
1
−3

.
IV. Both eigenvalues λ1 and λ2 are complex, and have positive real parts, then the solutions grow in magnitude while spinning around the origin, we say the critical point (0, 0) is a spiral source and unstable.
60
MINGFENG ZHAO
Figure 18. Example saddle vector field with eigenvectors and solutions

Example 43. Let A = 

of A, and ~v1 = 
1

1
1
−4
1

, then it’s easy to know that λ1 = 1 + 2i and λ2 = 1 − 2i are two eigenvalues

 and ~v2 = 
1

 are eigenvectors corresponding to λ1 = 1 + 2i and λ2 = 1 − 2i,
2i
−2i
respectively. Notice that




1
cos(2t)
 e(1+2i)t = et 
,
Re 
2i
−2 sin(2t)
So the general solution to ~x0 = A~x is:

~x(t) = C1 et 

and
cos(2t)
−2 sin(2t)
Im 
1


 e(1+2i)t = et 
2i


 + C2 et 
sin(2t)
2 cos(2t)
sin(2t)

.
2 cos(2t)
Figure 19. Example spiral source vector field

.
ORDINARY DIFFERENTIAL EQUATIONS
61
V. Both eigenvalues λ1 and λ2 are complex, and have negative real parts, then the solutions shrink in magnitude while spinning around the origin, we say the critical point (0, 0) is a spiral sink and asymptotically
stable.

Example 44. Let A = 
−1
−1
4

−1

1

, then it’s easy to know that λ1 = −1 − 2i and λ2 = −1 + 2i are two

1

 and ~v2 = 
 are eigenvectors corresponding to λ1 = −1 − 2i and
2i
−2i
λ2 = −1 + 2i, respectively. Notice that
eigenvalues of A, and ~v1 = 

Re 
1
2i


 e(−1−2i)t = e−t 
cos(2t)

,

and
Im 
2 sin(2t)
1


 e(−1−2i)t = e−t 
2i
− sin(2t)

.
2 cos(2t)
So the general solution to ~x0 = A~x is:

~x(t) = C1 e−t 
cos(2t)
2 sin(2t)


 + C2 e−t 
− sin(2t)

.
2 cos(2t)
Figure 20. Example spiral sink vector field
VI. Both eigenvalues λ1 and λ2 are purely imaginary, that is, the eigenvalues are ±ik, then we get ellipses of
solutions, we say the critical point (0, 0) is a center point and stable
Let ~v be an eigenvector corresponding to eik , then the general solution is:
~x(t) = C1 Re(eik~v ) + C2 Im(eik~v ).
62
MINGFENG ZHAO

Example 45. Let A = 

A, and ~v1 = 
1

0
−4
, then it’s easy to know that λ1 = 2i and λ2 = −2i are two eigenvalues of
0

 and ~v2 = 
2i

1

1
−2i
 are eigenvectors corresponding to λ1 = 2i and λ2 = −2i, respectively.
Notice that

Re 
1
2i


 ei2t = 
cos(2t)
−2 sin(2t)


,
Im 
1


sin(2t)
 ei2t = 
2i

.
2 cos(2t)
Figure 21. Example center vector field
Remark 14. Let λ = m + ik be a complex eigenvalue of the real 2 × 2 matrix A, then we can take an eigenvector of


1
, so we know that
the form ~v = 
a + ib

Re(eλt~v ) = emt 
cos(kt)

.
A cos(kt) + B sin(kt)
To draw the correct direction of the above spiral or ellipse, you just need to know the positions of five points at t = 0,
π π 3π
2π
, ,
and
.
2k k 2k
k
ORDINARY DIFFERENTIAL EQUATIONS
63
Review of Chapter 8:
Linearization of a system

 x0 = f (x, y)
, that is, f (x0 , y0 ) =
Definition 13. Let (x0 , y0 ) be a critical point to the autonomous system
 y 0 = g(x, y)

 x0 = f (x, y)
g(x0 , y0 ) = 0. Let u = x − x0 and v = y − y0 , the linearization at (x0 , y0 ) of the autonomous system
is:
 y 0 = g(x, y)

in which the matrix 

u0

v0

=
fx (x0 , y0 ) fy (x0 , y0 )

u

gx (x0 , y0 )
fx (x0 , y0 ) fy (x0 , y0 )
gx (x0 , y0 )

gy (x0 , y0 )

,
v


 is called the Jacobian matrix of the vector function 
gy (x0 , y0 )
f (x, y)

 at the
g(x, y)
point (x0 , y0 ).

 x0 = f (x, y)
Remark 15. The Linearization of an autonomous system
 y 0 = g(x, y)
at a critical point (x0 , y0 ) is the just the
linearizations of each component f (x, y) and g(x, y) at this critical point (x0 , y0 ).

 x0 = y
.
Example 46. Find all critical points and linearizations at these critical points for the system
 y 0 = −x + x2
It’s easy to see that there are two critical points: (0, 0) and (1, 0). Let f (x, y) = y and g(x, y) = −x + x2 , then the


f (x, y)
 is:
Jacobian matrix of 
g(x, y)

fx (x, y) fy (x, y)

gx (x, y)
u0

v
0


=
0
−1
1
0
1
−1 + 2x 0

.
is:

u

0

=
gy (x, y)

 x0 = y
• At (0, 0) the linearization of
 y 0 = −x + x2


v

,
with u = x and v = y.
64
MINGFENG ZHAO

 x0 = y
• At (1, 0) the linearization of
 y 0 = −x + x2

u0

v0


=
0
1
1
0

is:

u
with u = x − 1 and y = y.
,

v
Example 47. Find all critical points and linearizations at these critical points for the system x0 = (1 − y)(2x − y),
y 0 = (2 + x)(x − 2y).
Let’s solve
(1 − y)(2x − y) = 0,
and
(2 + x)(x − 2y) = 0.
Then there are four critical points:
(−2, 1),
(−2, −4)
(2, 1),
and
(0, 0).
Let f (x, y) = (1 − y)(2x − y) = 2x − y − 2xy + y 2 and g(x, y) = (2 + x)(x − 2y) = 2x − 4y + x2 − 2xy, then the


f (x, y)
 is:
Jacobian matrix of 
g(x, y)

fx (x, y) fy (x, y)

gx (x, y)


=
gy (x, y)
2 − 2y
−1 − 2x + 2y
2 + 2x − 2y
−4 − 2x

.
Therefore, we know that

 x0 = (1 − y)(2x − y)
I. At (−2, 1) the linearization of
is:
 y 0 = (2 + x)(x − 2y)

u0

v0


=
0
5
−4
0

u



with u = x + 2 and v = y − 1.
v

 x0 = (1 − y)(2x − y)
II. At (2, 1) the linearization of
is:
 y 0 = (2 + x)(x − 2y)

u0

v
0


=
0
4
−3
−4

u



v
with u = x − 2 and v = y − 1.
ORDINARY DIFFERENTIAL EQUATIONS
65

 x0 = (1 − y)(2x − y)
III. At (−2, −4) the linearization of
is:
 y 0 = (2 + x)(x − 2y)

u0

v
0


=
10 −5

u

6
0


with u = x + 2 and v = y + 4.
v

 x0 = (1 − y)(2x − y)
IV. At (0, 0) the linearization of
is:
 y 0 = (2 + x)(x − 2y)

u0

v0


=
2
−1
2
−4

u



with u = x and v = y.
v
Isolated critical points and almost linear systems

 x0 = f (x, y)
Definition 14. Let (x0 , y0 ) be a critical point of the system
, that is, f (x0 , y0 ) = g(x0 , y0 ) = 0, then
 y 0 = g(x, y)
I. We say that the critical point (x0 , y0 ) is isolated if it is the only critical point in a small “neighborhood” of
(x0 , y0 ).

 x0 = f (x, y)
II. We say that the system
is almost linear at critical point (x0 , y0 ) if the critical point ~a is isolated,
 y 0 = g(x, y)


f (x, y)
 at (x0 , y0 ) is invertible.
and the Jacobian matrix of 
g(x, y)
Remark 16. If the Jacobian matrix at ~a is invertible, then ~a is isolated. Therefore, it suffices to verify that the Jacobian
at ~a is invertible.
NonExample 1. It’s easy to see (0, 0) is a critical point of the system x0 = x2 , y 0 = y 2 , but the system is


0 0
 which is not invertible.
not almost linear at (0, 0), because the Jacobian matrix at (0, 0) is 
0 0
Stability and classification of isolated critical points

 x0 = f (x, y)
In the following, we assume that (x0 , y0 ) is a critical point of the system
, that is, f (x0 , y0 ) =
 y 0 = g(x, y)


f (x, y)
 at (x0 , y0 ) is invertible. Also we assume that the Jacobian matrix
g(x0 , y0 ) = 0, and the Jacobian matrix of 
g(x, y)
66
MINGFENG ZHAO

of 
f (x, y)

 at (x0 , y0 ) has two different nonzero eigenvalues λ1 and λ2 . Notice that the linearization of the system
g(x, y)

 x0 = f (x, y)
at (x0 , y0 ) is:
 y 0 = g(x, y)

u0

v0


=
fx (x0 , y0 ) fy (x0 , y0 )

u

gx (x0 , y0 )
gy (x0 , y0 )

with u = x − x0 and v = y − y0 .

v

 x0 = f (x, y)
The following table shows the behavior of the system
near this isolated critical point (x0 , y0 ):
 y 0 = g(x, y)
Eigenvalues of the Jacobian matrix at (x0 , y0 )
Type of (x0 , y0 )
Stability of (x0 , y0 )
I.
λ1 and λ2 are real and both positive
source
unstable
II.
λ1 and λ2 are real and both negative
sink
asymptotically stable
saddle point
unstable
III. λ1 and λ2 are real and opposite signs
IV.
λ1 and λ2 are complex with positive real part
spiral source
unstable
V.
λ1 and λ2 are complex with negative real part
spiral sink
asymptotically stable
VI.
λ1 and λ2 are complex with zero real part
center
Example 48. Find all critical points and classify the stabilities of these critical points for the system x0 = −y − x2 ,
y 0 = −x + y 2 .
Let’s solve −y − x2 = 0 and −x + y 2 = 0, then
(x, y) = (0, 0),
and
(x, y) = (1, −1).
That is, all critical points of the system x0 = −y − x2 , y 0 = −x + y 2 are:
(0, 0) and (1, −1) .

Let f (x, y) = −y − x2 and g(x, y) = −x + y 2 , then Jacobian matrix of 
f (x, y)
g(x, y)

fx (x, y) fy (x, y)

gx (x, y)
Therefore, we know that
gy (x, y)


=
−2x
−1
−1
2y

.

 is
ORDINARY DIFFERENTIAL EQUATIONS

I. At (0, 0) the linearization is 
u0
v
0


=
−1
0
−1

u

0
67


 with u = x and v = u. Notice that 
v

is invertible, then the system is almost linear at (0, 0). Since the eigenvalues of 
0
−1
−1
0
0
−1
−1


0

 are λ1 = −1 and
λ2 = 1, then (0, 0) is an unstable critical point of the system, solutions behaves like a saddle point near (0, 0).





u0
2 −1
u
 = 

 with u = x − 1 and v = y + 1. Notice that
II. at (1, −1) the linearization is 
v0
−1 −2
v




2 −1
2 −1

 is invertible, then the system is almost linear at (1, −1) . Since the eigenvalues of 

−1 −2
−1 −2
are λ1 = −1 and λ2 = −3, then (1, −1) is an asymptotically stable critical point of the system, solutions
behaves like a sink near (0, 0).
Conservative equations
Definition 15. An equation of the form x00 + f (x) = 0 is called a conservative equation.
Let y = x0 , then we have the system:
x0 = y,
and y 0 = −f (x).
Then the trajectories will satisfy
f (x)
dy
=−
.
dx
y
Solve the above equation, we get
1 2
y +
2
Z
f (x) dx = C,
which is called the Hamiltonian or energy of the system.

 

p
y
0
1
 is 
 which has eigenvalues λ1,2 = ± −f 0 (x).
It’s easy to see that the Jacobian matrix of 
−f 0 (x) 0
−f (x)
Therefore, there are only two possibilities for critical points, either an unstable saddle point, or a stable center;
but never any asymptotically stable points for the system x0 = y, y 0 = −f (x).
Example 49. Find the trajectories for the equation x00 + x − x2 = 0.
Let y = x0 , then
x0 = y,
y 0 = x00 = x2 − x.
68
MINGFENG ZHAO
So we get
dy
x2 − x
=
, so the trajectories satisfy
dx
y
1 2
1
1
y = x3 − x2 + C .
2
3
2
Figure 22. Phase portrait with some trajectories of x0 = y, y 0 = −x + x2

 x0 = y
Remark 17. For the system
, it’s easy to see that there are two critical points: (0, 0) and (1, 0). Let
 y 0 = −x + x2


f
(x,
y)
 is:
f (x, y) = y and g(x, y) = −x + x2 , then the Jacobian matrix of 
g(x, y)

fx (x, y) fy (x, y)

gx (x, y)
u0

v0


=

Notice that eigenvalues of 
0
1
−1
0
0
1
−1
0

=
gy (x, y)

 x0 = y
• At (0, 0) the linearization of
 y 0 = −x + x2


0
1
−1 + 2x 0

.
is:

u


,
with u = x and v = y.
v

 are λ1,2 = ±i, so (0, 0) is a center .
ORDINARY DIFFERENTIAL EQUATIONS

 x0 = y
• At (1, 0) the linearization of
 y 0 = −x + x2

u0

v
0


=
0
1
Notice that eigenvalues of 
is:
u

1


0
1
1
0
0
69

,
with u = x − 1 and y = y.
v

 are λ1,2 = ±1, so (1, 0) is a unstable saddle point .
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca