ZacharyTseng more notes
Math 251: Ordinary Di↵erential Equations
and Partial Di↵erential Equations
Pichkitti Bannangkoon
July 13, 2015
To my students
Preface
This note is an outcome of Math 251 course given at Penn state from 2010-2015. I’ve tried
to make this note clear as much as possible to students. When new terminologies or concepts
are introduced, examples with complete solutions are immediately followed. Most examples are
carefully selected from past Math 251 exams at Penn state university. So students will feel wellprepared for the exam when they read this note. It is important to emphasize that this note is not
a substitution for textbook but the supplement one. Also this note does not cover every material,
insights, shortcuts I give in class. . On the other hand, occasionally I don’t have enough time to
provide complete solutions or various examples in class but you may find it in this note. That’s
why both class attendance and note reading are strongly recommended for students’ achievement.
This note is far from perfect. Please excuse my poor English since it is not my first language. If
you find some typos or have comments or suggestions, please don’t hesitate to let me know. It will
benefit all of us. Finally, I would like to thank all of my students who have inspired me to write
this note. This note does not happen without their encouragement, support and love. I hope you
enjoy this note and find it is helpful. Don’t forget to be 1.01.
Fall 2014
Penn state university
With Love,
Pichkitti Bannangkoon
ii
Contents
1 Introduction
1.1 Classification of Di↵erential Equations . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Solutions of Di↵erential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Direction Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
8
11
2 First Order Di↵erential Equations
2.1 Solving First Order Di↵erential Equations . . . . . . . . . . . . . . . . . .
2.1.1 Separable Equation . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.2 Homogeneous Polar Equation* . . . . . . . . . . . . . . . . . . . .
2.1.3 Method of Integrating Factor . . . . . . . . . . . . . . . . . . . . .
2.1.4 Bernoulli Equation* . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.5 Riccati Equation* . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.6 Exact Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.7 Recognizable Exact Equation and Integrating Factor* . . . . . . .
2.1.8 More Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 The Existence and Uniqueness Theorem . . . . . . . . . . . . . . . . . . .
2.3 Stability of Equilibrium Solutions . . . . . . . . . . . . . . . . . . . . . . .
2.4 Modeling with First Order Di↵erential Equation . . . . . . . . . . . . . .
2.4.1 Mixing Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.2 Falling Object/Motion of an Object in a Resistive Fluid Medium*
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15
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49
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74
3 Second Order Di↵erential Equations
3.1 Fundamental Set of Solutions & Wronskian . . . . . . . . . . . . . . . . .
3.2 Linear Homogeneous Equation with Constant Coefficients . . . . . . . . .
3.2.1 Two Distinct Real roots . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 Complex Conjugate Roots . . . . . . . . . . . . . . . . . . . . . . .
3.2.3 Repeated Real Roots . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.4 Summary & More Examples . . . . . . . . . . . . . . . . . . . . . .
3.3 Linear Homogeneous Equation . . . . . . . . . . . . . . . . . . . . . . . .
3.3.1 Euler Equation* . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.2 Reduction of Order . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 The Existence and Uniqueness Theorem . . . . . . . . . . . . . . . . . . .
3.5 More on Wronskian & Abel’s Theorem . . . . . . . . . . . . . . . . . . . .
3.6 Linear Nonhomogeneous Equation . . . . . . . . . . . . . . . . . . . . . .
3.6.1 Method of Undetermined Coefficients . . . . . . . . . . . . . . . .
3.6.2 Variation of Parameters* . . . . . . . . . . . . . . . . . . . . . . .
3.7 Modeling with Second Order Di↵erential Equation: Mechanical Vibrations
3.7.1 Undamped Free Vibrations . . . . . . . . . . . . . . . . . . . . . .
3.7.2 Damped Free Vibrations . . . . . . . . . . . . . . . . . . . . . . . .
3.7.3 Undamped Forced Vibrations . . . . . . . . . . . . . . . . . . . . .
3.8 Nonlinear Equation* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.8.1 Equations of the Type F (t, y 0 , y 00 ) = 0* . . . . . . . . . . . . . . .
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81
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108
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133
136
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142
150
157
157
iii
CONTENTS
iv
3.8.2
3.8.3
Equations of the Type F (y, y 0 , y 00 ) = 0* . . . . . . . . . . . . . . . . . . . . 158
Equations of the Type F (t, y, y 0 , y 00 ) = 0 with F homogeneous* . . . . . . . 160
4 Higher Order Linear Equation
162
4.1 General Theory of nth Order linear Equations . . . . . . . . . . . . . . . . . . . . . 162
4.2 Homogeneous Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . 168
5 Laplace Transform
5.1 Definition of Laplace Transform . . . . . . . . . . . . . .
5.2 The Gamma Function* . . . . . . . . . . . . . . . . . . .
5.3 Inverse Laplace Transform . . . . . . . . . . . . . . . . .
5.4 Solving Initial Value Problems with Laplace Transform .
5.4.1 Equations with Constant Coefficients . . . . . . .
5.4.2 Equations with Variable Coefficients* . . . . . .
5.5 Step Functions . . . . . . . . . . . . . . . . . . . . . . .
5.6 Impulse Functions . . . . . . . . . . . . . . . . . . . . .
5.7 The Convolution Integral* . . . . . . . . . . . . . . . . .
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175
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183
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216
226
6 Systems of First Order Linear Equations
6.1 Introduction . . . . . . . . . . . . . . . . .
6.2 Review of Matrices . . . . . . . . . . . . .
6.3 Solutions to Systems of First Order Linear
6.3.1 Real and Distinct Eigenvalues . . .
6.3.2 Complex Conjugate Eigenvalues .
6.3.3 Repeated Real Eigenvalues . . . .
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229
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Equations .
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7 Nonlinear Di↵erential Equations and Stability
263
7.1 Autonomous Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
7.2 Locally Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
8 Partial Di↵erential Equation
8.1 Boundary Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1.1 BVP vs IVP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1.2 Eigenvalues Problems . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.1 The Fourier Convergence Theorem . . . . . . . . . . . . . . . . . . . .
8.2.2 Even and Odd Functions and their Fourier Series . . . . . . . . . . . .
8.2.3 The Cosine and Sine Series Extensions . . . . . . . . . . . . . . . . . .
8.3 Heat Equations & Separation of Variables . . . . . . . . . . . . . . . . . . . .
8.3.1 Type I: Homogeneous Boundary Conditions & Separation of Variables
8.3.2 Type II: Nonhomogeneous Boundary Conditions . . . . . . . . . . . .
8.3.3 Type III: Bar with Both Ends Insulated . . . . . . . . . . . . . . . . .
8.3.4 More Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3.5 Summary of Heat Conduction Problems . . . . . . . . . . . . . . . . .
8.4 Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4.1 Solutions to Wave Equation Problem . . . . . . . . . . . . . . . . . . .
8.4.2 Special Cases of Wave Equation Problems . . . . . . . . . . . . . . . .
8.4.3 Summary of Wave Equation Problems . . . . . . . . . . . . . . . . . .
8.5 Laplace’s Equation* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.5.1 Dirichlet Problem for a Rectangle* . . . . . . . . . . . . . . . . . . . .
8.5.2 Neumann Problem for a Rectangle* . . . . . . . . . . . . . . . . . . .
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275
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337
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Bibliography
346
Index
347
Chapter 1
Introduction
A di↵erential equation is any equation which contains derivatives, either ordinary derivatives
or partial derivatives.
Example 1 (Newton’s Second Law of motion). If an object of mass m is moving with
acceleration a and being acted on with force F then we have a relation F = ma. Indeed it is
a di↵erential equation by using
a=
dv
dt
or
a=
d2 u
,
dt2
where v is the velocity of the object and u is the position function of the object at time t. So
Newton’s Second Law can now be written as
F (t, v) = m
1.1
dv
dt
or
F (t, u) = m
d2 u
.
dt2
Classification of Di↵erential Equations
Order:
The order of a di↵erential equation is the “largest” derivative present in the di↵erential equation.
• We write first, second and third derivatives of y as y 0 , y 00 and y 000 respectively.
• For fourth and higher derivatives of y we write y (n) as nth derivative of y.
• Make sure you see the di↵erence between y n (means y · · · y, n times) and y (n) (means nth
derivative of y).
ODE vs PDE:
• A di↵erential equation is called an ordinary di↵erential equation, abbreviated by ODE,
di y
if it has ordinary derivatives in it. (when you see y (i) or i for natural number i)
dt
• Similarly, an equation is called a partial di↵erential equation, abbreviated by PDE, if it
@2U
or Utx for examples)
has partial derivatives in it. (when you see
@x@t
1
CHAPTER 1. INTRODUCTION
2
Linear vs Nonlinear:
• A linear ordinary di↵erential equation is any di↵erential equation that can be written
in the form
an (t)y (n) (t) + an
1 (t)y
(n 1)
(t) + . . . + a1 (t)y 0 (t) + a0 (t)y(t) = g(t),
where all ai (t), g(t) are functions of t and y (i) is i-th derivative of y.
• If an ordinary di↵erential equation can’t be written in the above form then it is called
non-linear or nonlinear.
To consider whether the equation linear or nonlinear, we determine only on y and its derivatives since all ai (t) and g(t) can be any linear or nonlinear functions. Also, according to the
definitions, if the followings appear in the di↵erential equation, it is nonlinear:
• Products of y and its derivative (for examples: yy 0 , y 0 y (5) ).
• Either functionpy or its derivative raises to any power other than first power (for examples: y 2 , (y 0 )4 , y (3) ).
• Nonlinear function of y and its derivative (for examples: sin y,
y
1
, ey , ).
y+2
y
Example 2. Consider the following di↵erential equations.
Di↵erential equations
y 00
3y 0 + 5 = 0
y 000 + 2 sin(t)e3t y 00 + cos(t)y = 3
d
d5 y
= 5
dt
dt
d9 y d12 y
2
(y + 2y 1) 9
= 18
dt
dt12
t2 3 0 5
y (4) 9 sin(t)e t y 00 +
(y ) = 1
18 7t
@2U
@U
9 2 =
@x
@t
(
d3 y 2
)
dt3
ety
2Uxx = 3Utx
Utt
@3U
@U
=1+
@x2 @t
@y
Order
ODE/PDE
Linear/Nonlinear
2
ODE
linear
3
ODE
linear
5
ODE
nonlinear
12
ODE
nonlinear
4
ODE
nonlinear
2
PDE
2
PDE
3
PDE
CHAPTER 1. INTRODUCTION
3
Example 3 (Summer 2002 Midterm Exam I).
Which of the following is a linear second order di↵erential equation?
(a) y 0 + ty = 1
(b) y 00 = t2 y + et
(c) (y 0 )2 = (y + 2)(y
3)
(d) y 00 + 3y 0 + 2y = sin y
Solution In a), the di↵erential equation is linear but first order. In c), the equation is
nonlinear because of the terms (y 0 )2 appearing on the left-hand side and y 2 on the right-hand
side after multiplication. Similarly, the one in d) is also nonlinear because of the term sin y.
So b) is the right answer.
Homogeneous vs Nonhomogeneous:
• A linear di↵erential equation
an (t)y (n) (t) + an
1 (t)y
(n 1)
(t) + . . . + a1 (t)y 0 (t) + a0 (t)y(t) = g(t)
is said to be homogeneous if g(t) = 0 for all t.
• Otherwise, the linear equation is called nonhomogeneous.
Example 4 (Spring 2002 Midterm Exam I).
Which of the following is TRUE?
a) y 0 =
t
is a first order linear di↵erential equation.
y
b) sin(t)y 00 + (1
t2 )y 0 + cos(t)y = 0 is a second order linear di↵erential equation.
c) y 00 + (y 0 )3 + y = 0 is a nonlinear di↵erential equation of order 3.
d) y 00 + y 0 + y = t is a second order homogeneous di↵erential equation.
e)
@y
@t
+ ty = 0 is an ordinary di↵erential equation.
1
. The
y
one in c) has order 2. d) is a nonhomomgeneous equation since the right hand side is not
a zero function. Finally the last equation is partial di↵erential equation because of the term
@y
. So b) is the correct answer.
@t
Solution The di↵erential equation in a) is not a linear equation because of the term
CHAPTER 1. INTRODUCTION
4
Example 5 (Spring 2008, Spring 2011, Summer 2010 Midterm Exam I).
For each of the di↵erential equations below, state its order and whether it is linear or nonlinear.
Di↵erential equations
Order
Linear/Nonlinear
3
linear
2
nonlinear
5t
1
linear
y 000 + ty 0 + (cos2 y)t = t3
3
nonlinear
y 0 + t2 = y
1
linear
(1 + y 3 )y 00 + ty 0 + y = et
2
nonlinear
y 0 (t) = ty 2 (t)
1
nonlinear
y 00 (t) = ln(t)y(t)
2
linear
2
nonlinear
2y 000 + ty 00
y = y2
y0
y 0 + et y = 0
y 00
cos(2t)y = t2
ty(t)y 0 (t)y 00 (t)
e
1=0
Example 6 (Spring 2004 Midterm Exam I).
Indicate whether each equation below is linear or non-linear.
a) sin(x) + ex
dy
= y4 .
dx
b) y 00 + p(t)y + et = 0.
c) y 0 = y 2 cos(t).
d) y 00 = cos(y).
e) y 000 + t4 y 0 + t3 y = 0.
Solution Di↵erential equations in b) and e) are linear while the ones in a), c) and d) are
non-linear because of the term y 4 , y 2 and cos(y), respectively.
CHAPTER 1. INTRODUCTION
5
Exercises 1.1
1. Classify the following equations as linear or non-linear, and state their order (Fall 2000
Midterm Exam I)
d2 y
dy
+ t2
+ t3 y = cos t
dt2
dt
d3 y
dy
(b) t 3 + t2
+ t3 y = cos y
dt
dt
dy
2y 3
(c)
=
dx
2x + 2
(a) t
2. Which of the following is a linear second order di↵erential equation? (Summer 2002 Midterm
Exam I)
(a) y 0 + ty = 1
(b) y 00 = t2 y + et
(c) (y 0 )2 = (y + 2)(y
00
3)
0
(d) y + 3y + 2y = sin y
3. Classify the following di↵erential equations as linear or non-linear and state their order.
(Spring 2003 Midterm Exam I)
(a) ln(t)
(b) 2y 0
d2 y
dy
+ 3et
2
dt
dt
y 2 = et
(c) y 000 + (t2
(d) y
00
y sin t = 0
1)y + cos t = 0
sin(t + y)y 0 + (t2 + 1)y = 0
4. Which of the following is a second order, linear, homogeneous di↵erential equation? (Summer
2003 Midterm Exam I)
(a) y 00 + y 2 = 0
(b) y 00 + y 0 + 2y = t ln y
(c) ty 00 + y = 0
(d) (y 0 )2
t2 y = 1
5. For each of the di↵erential equations below, state its order and whether it is linear or
nonlinear. (Fall 2003 Midterm Exam I)
(a) y 0 + t2 y = et
(b) 2y 00 + 3y 0
y = te
(c) ty 0 = y(y + 1)(y
(d) y
000
0
2y + ty
t
1)
2
y =0
6. Identify each of the following equations as linear or non-linear and also determine their order.
(Fall 2004 Midterm Exam I)
dy
=t
dt
d2 y
dy
(b) ( 2 )3 + ( )3 + y 3 = t3
dt
dt
dy
d2 y
(c) sin t
+ t5 y = (1 t2 ) 2
dt
dt
(a) y
CHAPTER 1. INTRODUCTION
(d) (1 + y) sin2 t + (
6
d3 y
+ y) cos2 t = 1
dt3
7. Which of the equations below is a second order nonlinear di↵erential equation? (Fall 2013
Midterm Exam I)
(a) t2 y
y 0 )2 = e2t
(1
t 00
sin(t)y 0 = tan(t)
(b) e y
(c) ty 00
5 sin(2t)y = ey
(d) t2 y 0
e3 cos(t) y = t4
8. Which of the equations below is a second order nonlinear di↵erential equation? (Fall 2012
Midterm Exam I)
1
(a) y 0 + e t y = cos( )
t
(b) y 00 = et sin 2y
(c) (y 0 )2 + y = 0
(d) y 00
t ln(t)y 0 + e5t y = 6
9. Which of the equations below is a second order linear ordinary di↵erential equation? (Fall
2011 Midterm Exam I)
(a) y 0 + yy 0 = et sin(t)
(b) (y 0 )2 + sin y = t2
(c) y 00 + yy 0 = 0
(d) y 00 + t3 sin(t)y 0
sec(t)et y =
3t
t2
9
10. Which of the equations below is a second order linear di↵erential equation? (Fall 2010
Midterm Exam I)
(a) t2 y 00 = 10y + 7t4
(b) y 00
2y 0 = y 2 + 3
(c) ty 0 = t2 e99t
0
2y
2
(d) y = 4y + 4y + 1
11. Which of the following di↵erential equations is linear? (Fall 2009 Midterm Exam I)
(a) yy 0 + ty = t + 1
3
(b) ln(t2 )y 00 + t 2 y 0 + et+3
(c) ty 0 + cos y = 1
(d) 2y 0
y = y2
12. For part (a) through (e) below, a list of di↵erential equations is given. For each part, write
down the letter corresponding to the equation on the list with the specified properties. There
is only one correct answer to each part. (Summer 2009 Midterm Exam I)
A. y 0 = 2y + t
B. y 0 = e2y
C. y 0 = ey
1
00
0
00
t 0
D. y + 4y
t
5y = 2
E. y + e y + t2 y = 0
CHAPTER 1. INTRODUCTION
F. y 00
4y =
7
t
y
G. y 00 + 3y 00 + 3y 0 + y = t5 + ln t
H. y 000 + y 0 y = e
2t
sin 5t
(a) Second order homogeneous linear equation.
(b) Third order nonlinear equation.
(c) Second order nonhomogeneous linear equation.
(d) First order linear equation.
Answers
1. (a) second order and linear, (b) third order and non-linear, (c) first order and linear
2. (b)
3. (a) second order and linear, (b) first order and non-linear, (c) third order and linear, (d)
second order and non-linear
4. (c)
5. (a) first order and linear, (b) second order and linear, (c) first order and non-linear, (d) third
order and non-linear
6. (a) first order and non-linear, (b) second order and non-linear, (c) second order and linear,
(d) third order and linear
7. (c)
8. (a)
9. (d)
10. (a)
11. (b)
12. (a) E, (b) H, (c) D, (d) A
CHAPTER 1. INTRODUCTION
1.2
8
Solutions of Di↵erential Equations
A solution to a di↵erential equation is any function which satisfies the di↵erential equation.
• Therefore, to check whether the certain function is a solution or not, we simply substitute that function into y.
• If that function satisfies the equation, it is a solution. If not, it is not a solution.
• Di↵erential equation does not necessary have the unique solution.
Example 7 (Fall 2009 Midterm Exam I).
Which function is a solution of the di↵erential equation
(y 0 )2
(a) y(t) = t2 (b)
y(t) = e5t (c)
5ty = 5t2 + 1
y(t) =
?
t (d)
1
5t
y(t) =
Solution Let’s consider each choice separately,
(a) Since ((t2 )0 )2
5t(t2 ) = 4t2
5t3 6= 5t2 + 1
t2 is not a solution.
)
(b) Since ((e5t )0 )2
5t(e5t ) = 25e10t
5te5t 6= 5t2 + 1
(c) Since (( t)0 )2
5t( t) = 5t2 + 1
)
(d) Since ((
1 0 2
5t ) )
5t(
1
5t )
1
(5t)4
=
e5t is not a solution.
)
t is a solution.
+ 1 6= 5t2 + 1
1
5t
)
is not a solution.
Example 8 (Summer 2003 Midterm Exam I).
Which equation has y1 = e
(a)
2y 0 +3y = 0
2t
and y2 = e3t as two solutions?
y 00 +y 0 6y = 0 (c)
(b)
y 00 +y 0 +6y = 0 (d)
2y 00 +10y 0 12y = 0
Solution Again let’s take a look each choice one by one,
(a) For y = e
2t
)
(b) For y = e
2t
)
(c) For y = e 2t
For y = e3t
)
)
2t
)
(d) For y = e
solution.
2(e
(e
2t 0
) + 3(e
2t 00
) + (e
2t
2t 0
)
) = 7e
6e
2t
2t
=
6= 0
)
2t
6= 0
e
)
2t
is not a solution.
e
2t
is not a solution.
(e 2t )00 + (e 2t )0 + 6(e 2t ) = 0 ) e 2t is a solution.
(e3t )00 + (e3t )0 + 6(e3t ) = 0 ) e3t is a solution.
2(e
2t 00
) + 10(e
2t 0
)
12e
2t
=
30e
Example 9 (Spring 2007 Midterm Exam I).
Consider the equation
(t
1)y 00
ty 0 + y = 0.
Verify that the functions y1 = t and y2 = et are its solutions.
Solution Consider
e
2t
6= 0
)
e
2t
is not a
CHAPTER 1. INTRODUCTION
9
• Check y1 = t: We substitute y by y1 = t into the equation. We have
(t
1)y 00
ty 0 + y = (t
1)(t)00
t(t)0 + t = (t
1)(0)
t(1) + t =
t + t = 0.
• Check y2 = et : We substitute y by y2 = et into the equation. Then
(t 1)y 00
ty 0 +y = (t 1)(et )00
t(et )0 +t = (t 1)(et ) t(et )+t = tet
Since both functions satisfy the equation (t
1)y 00
et
tet +et = 0.
ty 0 + y = 0, they are the solutions.
Example 10 (Summer 2007 Midterm Exam I).
All of the equations below have y(t) = 5e6t as a particular solution, EXCEPT
a) y 00
12y 0 + 36y = 0
b) y 00
5y 0
c) y 00 + 4y 0
d) y 00
6y = 0
12y = 0
36y = 0
Solution We plug in y(t) by 5e6t into each equations to check whether it is a solution. For
a), we have
y 00
12y 0 + 36y = (5e6t )00
= 5(36e6t )
= 180e6t
12(5e6t )0 + 36(5e6t )
12(30e6t ) + 36(5e6t )
360e6t + 180e6t = 0.
For b), we have
y 00
5y 0
6y = (5e6t )00
= 5(36e6t )
= 180e6t
5(5e6t )0
6(5e6t )
5(30e6t )
150e6t
30e6t
30e6t = 0.
For c), we have
y 00 + 4y 0
12y = (5e6t )00 + 4(5e6t )0
12(5e6t )
= 5(36e6t ) + 4(30e6t )
= 180e6t + 120e6t
60e6t
60e6t = 240e6t .
For d), we have
y 00
36y = (5e6t )00
= 5(36e6t )
36(5e6t )
36(5e6t ) = 0.
Since 5e6t satisfies the di↵erential equations in a), b) and d), it is the solution to these
equations. On the other hand, 5e6t fails to satisfy the di↵erential equation in c), it is not a
solution. Therefore c) is the correct answer.
CHAPTER 1. INTRODUCTION
10
Exercises 1.2
1. Show that the functions y1 (t) = t2 , y2 (t) = t 2 are solutions of the di↵erential equation (Fall
2002 Midterm Exam I)
t2 y 00 + ty 0 4y = 0
2. Are the functions y1 (t) = t3 and y2 (t) =
2003 Midterm Exam I)
t2 y 00
1
solutions to the di↵erential equation (Spring
t
ty 0
3. Which of the equations below has y(t) = 5e
Midterm Exam I)
3y = 0?
3t
as a particular solution? (Spring 2009
(a) y 00 + 9y = 0
(b) y 00
6y 0 + 9y = 0
(c) y 00 + y 0
(d) y
00
2y
6y = 0
0
3y = 0
4. Which of the equations below has y(t) = 19e4t as a particular solution? (Fall 2011 Midterm
Exam I)
(a) y 00 + 16y = 0
(b) y 00
8y 0 + 16y = 0
(c) y 00 + 4y 0 = 0
(d) y 00 + 5y 0 + 4y = 0
5. Which function below is a solution of the linear di↵erential equation (Spring 2014 Midterm
Exam I)
y (4) 4y 00 = 0.
p
I y = 3t 2 + 7e 2t
II y = 3 + t2
2e2t
III y = e2t+4 + e
e
2t
2t+9
(a) I only.
(b) I and II.
(c) I and III.
(d) I, II, and III.
Answers
1. 2. yes
3. (c)
4. (b)
5. (c)
CHAPTER 1. INTRODUCTION
1.3
11
Direction Fields
• Direction field is a tool to study the behavior of the solutions of the 1st order di↵erential
equation
y 0 = f (t, y)
without solving it. To draw the direction field, do the following steps:
1. Draw ty-lane.
2. For each point (t0 , y0 ), evaluate the value of f (t0 , y0 ) = y 0 |t=t0 ,y=y0 .
3. Draw a short line at (t0 , y0 ) representing the slope of line tangent to the solution passing
(t0 , y0 ).
4. Repeat steps 2. and 3. with other points on the plane.
• Curves passing through the arrows are called integral curves.
• Direction field and integral curves is used to find information about the long term behavior
of the solution ( lim y(t)) without knowing solution of the di↵erential equation.
t!1
• The first order di↵erential equation in the form
dy
= f (y),
dt
(f is free from independent variable t)
is called autonomous.
• Equilibrium solutions are constant functions y satisfying y 0 = 0. (for autonomous equation it means the function y for which f (y) = 0).
Example 11. Draw the direction field of y 0 = y
t.
Next let’s consider the autonomous equation, in which its direction filed is easier to be drawn since
f is free from independent variable t.
CHAPTER 1. INTRODUCTION
12
Example 12. Find the equilibrium solution(s), draw the direction field and determine the
behavior of y as t ! 1.
1. y 0 = 3
2y
2. y 0 = 1 + 2y
Solution
1. For y 0 = 3
2y, the equilibrium solution is y such that y 0 = 3
2y = 0
All solutions appear to converge to the equilibrium solution y(t) =
3
words, lim y(t) = .
t!1
2
2. For y 0 = 1 + 2y, set y 0 = 0
equilibrium solution.
)
1 + 2y = 0
)
y =
3
2
1
2.
)
y = 32 .
as t ! 1. In other
So y =
1
2
is the
CHAPTER 1. INTRODUCTION
In this case, all solutions diverge away from the equilibrium solution y(t) =
t ! 1. So lim y(t) does not exist.
t!1
Can you answer the following questions?
Example 13 (Fall 2001, Summer 2013, Summer 2014 Midterm Exam I).
Give an example of the following:
a) A first order, nonlinear, autonomous, ordinary di↵erential equation.
b) A second order, linear, homogeneous, ordinary di↵erential equation.
c) A first order, linear, autonomous, ordinary di↵erential equation.
d) A third order partial di↵erential equation.
e) A second order, linear, nonhomogeneous, ordinary di↵erential equation.
13
1
2
as
CHAPTER 1. INTRODUCTION
14
Exercises 1.3
1. Determine the di↵erential equation whose direction field is given below.
Midterm Exam I)
(Spring 2011
(a) y 0 = y 2
(b) y 0 = t
(c) y 0 = y 2 + t
(d) y 0 = y 2
t
2. Determine the di↵erential equation whose direction field is give below. (Spring 2013 Midterm
Exam I)
(a) y 0 = y
x
(b) y 0 = x + y
(c) y 0 = y(y
0
(d) y = x(x
2)
2)
Answers
1. c)
2. a)
Chapter 2
First Order Di↵erential Equations
2.1
Solving First Order Di↵erential Equations
In this chapter, we will take a look closely at solving first order di↵erential equations ,
dy
= f (y, t).
dt
Unfortunately, there is no general formula for the solution to above equation. What we can do
instead is to consider at special cases. We will study three methods to solve first order di↵erential
equation in this course.:
• Separable equation
• Integrating factors
• Exact equation
2.1.1
Separable Equation
A separable equation is any di↵erential equation that we can write in the following form
P (y) dy = Q(t) dt.
(separate y’s and dy to t’s and dt)
In other words, it is the equation that we can separate one variable to one side of the equation
and another to the opposite side of the equation. If we fail to do this, we can’t use this method
to solve the di↵erential equation. Note here that we always place dy and dt as the last terms for
each side of the equation. Solving separable equation is very easy, you just integrate both sides,
Z
Z
P (y) dy = Q(t) dt.
Before we see the examples, let’s learn some new terminologies here.
• The general solution is the solution with undetermined constant c.
• An explicit solution is any solution that is given in the form y = y(t).
• The initial value problem, abbreviated IVP, is a di↵erential equation along with initial condition(s). The initial condition is used to determine the value of constant c in the
general solution. The solution with determined value of c obtained by applying the initial
condition(s) is called the particular solution .
15
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
16
Example 14 (Fall 2005 Midterm Exam I).
Find the general solution to the following di↵erential equation
3y 0 = 4y 4 t3 .
dy
Solution Replace y 0 by
and then move y’s and dy to the left and t’s and dt to the right
dt
of the equation,
3y 0 = 4y 4 t3
)
3
dy
= 4y 4 t3
dt
Integrating on both sides of the equation,
Z
Z
3
dy
=
4t3 dt
y4
3
dy = 4t3 dt.
y4
)
1
= t4 + c.
y3
)
So the general solution (in implicit form) to the above equation is
1
= t4 + c.
y3
From the above example, if we simplify it further, we have
1
= t4 + c
y3
)
3
y =
1
4
t +c
)
y=
r
3
t4
1
+c
as the general solution in explicit form.
Example 15 (Summer 2006 Midterm Exam I).
Find the solution of the initial value problem
x
,
y
y0 =
y(0) =
2
in explicit form.
dy
(notice here that, unlike the
dx
previous example, the independent variable is x) and then move y’s and dy to the left and x’s
and dx to the right of the equation,
Solution We use the same tactic here by first replacing y 0 to
y0 =
x
y
)
dy
=
dx
x
y
Z
x dx
)
)
y dy =
x dx.
Integrating on both sides,
Z
y dy =
Then we impose the initial condition y(0) =
( 2)2
=
2
y2
=
2
x2
+ c.
2
2, which means, if x=0 then y =
02
+c
2
)
c = 2.
2,
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
17
Therefore the particular solution (in implicit form) to the above equation is
y2
=
2
x2
+ 2.
2
Simplify a few more steps to get the solution in explicit form,
y2
=
2
x2
+2
2
)
y2 =
x2 + 4
)
y=±
p
x2 + 4.
p
Unfortunately, one of the solution, namely y =
x2 + 4 fails to agree with the initial condition y(0) = 2, so
p
x2 + 4
y=
is the particular solution to the above IVP in the explicit form.
After a few examples, we get one useful observation here:
We often use the method of separable equation to solve first order di↵erential equation which
contains fraction expressions. Also don’t forget to check the solutions to the IVP with the
initial condition to get the final particular solution in the explicit form when you take the
square root.
Example 16 (Fall 2002 Midterm Exam I).
Find the explicit solution of the initial value problems
y0 =
cos t + 1
,
y
y(0) =
3.
Solution Rewrite the di↵erential equation,
y0 =
cos t + 1
y
Integrating both sides,
Z
y dy =
dy
cos t + 1
=
dt
y
)
Z
(cos t + 1) dt
Now we apply the given initial condition y(0) =
( 3)2
=
2
)
)
y dy = (cos t + 1) dt.
y2
= sin t + t + c.
2
3, to determine the value of c,
sin(0) + 0 + c
)
c=
9
.
2
Therefore the particular solution is
y2
9
= sin t + t + .
2
2
Then we rewrite it in the explicit form,
y 2 = 2 sin t + 2t + 9
)
p
y = ± 2 sin t + 2t + 9.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
18
Since the negative sign is chosen to satisfy the initial condition y(0) =
of the above IVP is
p
y=
2 sin t + 2t + 9.
3, the explicit solution
Example 17 (Summer 2002 Final Exam ).
Solve explicitly the initial value problem
y0 =
3x2 + 4x + 2
,
2y 6
y(0) = 4.
dy
Solution The first step to solve the separable equation is to change y 0 to
and separate all
dx
expressions of y to the left-hand side and all of x to the right-hand side.
dy
3x2 + 4x + 2
=
dx
2y 6
)
(2y
6) dy = (3x2 + 4x + 2) dx.
Then we integrate both sides and then simplify,
Z
Z
(2y 6) dy = (3x2 + 4x + 2) dx )
y2
6y = x3 + 2x2 + 2x + c.
16
24 = c
Now we impose the initial condition, y(0) = 4,
42
6(4) = 03 + 2(02 ) + 2(0) + c
)
)
c=
8.
So the particular solution in implicit form is
y2
6y = x3 + 2x2 + 2x
8.
To write it in explicit form, we have to complete the square on the left-hand side of the
equation by adding 32 .
y2
6y + 32 = x3 + 2x2 + 2x
8 + 32
)
(y
3)2 = x3 + 2x2 + 2x + 1.
After we take the square root and add 3 to both sides of the equation, we have
p
y = 3 ± x3 + 2x2 + 2x + 1.
p
Notice that y = 3
x3 + 2x2 + 2x + 1 fails to satisfy the initial condition y(0) = 4. Instead
it gives y(0) = 2 So in the end the explicit particular solution of the above IVP is
p
y = 3 + x3 + 2x2 + 2x + 1.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
19
Exercises 2.1.1
1. Solve explicitly for y(t) in the following initial value problem
et
yy 0 = 0,
y0 =
4x sin x
,
y 2
y(0) = 1.
2. Solve the initial value problem
(a) y = 2 +
(b) y = 2
(c) y = 2 +
(d) y = 2
p
p
p
p
4x2 + 2 cos x
1
4x2
1
+ 2 cos x
2x2
cos x + 1
2x2
cos x + 1
y(0) = 1.
3. Find the explicit solution of the initial value problem
2
y0 =
4tet
, y(0) =
y
4.
4. Solve, explicitly for y(t), the initial value problem
y 0 = e2y (4t
cos t), y(0) = 1.
5. Solve the initial value problem
y0 =
3x2 + 8x3
,
2 + 2y
2
y(1) =
Give your solution in the explicit form.
6. Solve the initial value problem explicitly
y0 =
xe
2y
2x
4
,
Answers
1. y(t) =
p
2et
1
2. (b)
3. y =
p
4et2 + 12
1
ln( 4t2 + 2 sin t + e 2 )
2
p
5. y = 1
2x x3 + 2x4 2
1p
6. 2
2xe 2x e 2x + 5
2
4. y =
y(0) = 1.
2.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.1.2
20
Homogeneous Polar Equation*
A homogeneous polar equation is one of the form
y
y 0 = f ( ).
x
(2.1)
To solve this equation, we make the substitution
y(x) = xv(x)
By the product rule,
,
v=
y
.
x
y 0 = xv 0 + v.
Substitute (2.2) into (2.1), it gives
(2.2)
xv 0 + v = f (v).
This is a separable di↵erential equation since we can rewrite it to
x
dv
+ v = f (v)
dx
dv
f (v)
=
dx
x
)
v
)
dv
dx
=
.
f (v) v
x
y
After we integrate to both sides of the equation and then replace v by , the general solution
x
follows.
Example 18. Solve the following IVP
(x2 + 2xy)y 0 = 2(xy + y 2 ),
y(1) = 2.
Solution Note that
2(xy+y 2 )
x2
x2 +2xy
x2
2(xy + y 2 )
y = 2
=
x + 2xy
0
Then
f (v)
Therefore
Z
Since v =
dv
=
f (v) v
v=
Z
2v + 2v 2
1 + 2v
dx
x
)
Z
=
v=
2 · xy + 2( xy )2
2v + 2v 2
=
= f (v).
y
1+2· x
1 + 2v
2v + 2v 2 v(1 + 2v)
v
=
.
1 + 2v
1 + 2v
1 + 2v
dv =
v
Z
dx
x
)
ln v + 2v = ln x + c.
y
, the general solution is
x
ln
y
2y
+
= ln x + c.
x
x
Now let’s use the initial condition y(1) = 2,
ln 2 + 2(2) = ln 1 + c
)
c = 4 + ln 2.
So the particular solution is
ln
y
2y
+
= ln x + 4 + ln 2
x
x
)
ln
y
x
ln x
ln 2 = 4
2y
x
)
ln
y
=4
2x2
2y
.
x
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
21
Example 19. Find the general solution of the di↵erential equation
y0 =
x y
.
x+y
Solution This is actually homogeneous polar di↵erential equation. We can see it by diving
both numerator and denominator of the RHS term by x.
y0 =
Then
f (v)
Therefore
Z
Note that
v=
x y
=
x+y
1 v
1+v
dv
=
f (v) v
1
Z
x y
x
x+y
x
v=
dx
x
=
1
1+
1
v
)
Z
y
x
y
x
1 v
= f (v).
1+v
v(1 + v)
1 2v + v 2
=
.
1+v
1+v
1+v
dv =
1 2v + v 2
Z
dx
.
x
1+v
1+v
1
2
=
=
+
,
2v + v 2
(1 v)2
1 v (1 v)2
by partial fraction decomposition. Then we have
Z
Z
1
2
dx
+
dv
=
)
2
1 v (1 v)
x
Since v =
=
ln(1
v) +
2
1
v
= ln x + c.
y
, the general solution is
x
ln(1
y
2
)+
x
1
y
x
= ln x + c
)
ln
✓
x
y
x
◆
+
2x
x
y
= ln x + c.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.1.3
22
Method of Integrating Factor
In order to use this method, the di↵erential equation must be rewritten in the following form
dy
+ p(t)y = g(t),
dt
(2.3)
where p and g are given functions of t. This form is called standard form or canonical form of
1st order linear equation. To solve it, we assume that there is some magical function µ(t) satisfying
µ(t)p(t) = µ0 (t).
(2.4)
We called µ(t) an integrating factor. Now we change the form of the equation by multiplying
Equation (2.3) both sides by µ(t):
µ(t)
dy
+ µ(t)p(t)y = µ(t)g(t).
dt
(2.5)
Then change the second term of the left hand side of Equation (2.5) by using Equation (2.4),
µ(t)
dy
+ µ0 (t)y = µ(t)g(t).
dt
Now the left hand side is simply (µ(t)y(t))0 , by product rule of di↵erentiation, and so we get
(µ(t)y(t))0 = µ(t)g(t)
d(µ(t)y(t))
= µ(t)g(t)
dt
)
)
d(µ(t)y(t)) = µ(t)g(t) dt.
Integrating both sides of the equation,
Z
Z
d(µ(t)y(t)) = µ(t)g(t) dt
Z
µ(t)y(t) = µ(t)g(t) dt + c
R
µ(t)g(t) dt + c
y(t) =
,
µ(t)
which is the general solution of Equation (2.3). Now let’s find out what the formula of the magical
function µ(t) is. From Equation (2.4),
µ(t)p(t) = µ0 (t)
Notice here that
)
µ0 (t)
= p(t).
µ(t)
µ0 (t)
= (ln µ(t))0
µ(t)
by chain rule. Therefore
)
d(ln µ(t))
= p(t)
dt
Integrating both sides,
Z
Z
d(ln µ(t)) = p(t) dt
Z
(ln µ(t))0 = p(t)
)
ln µ(t) =
)
p(t) dt
Hence the formula for the integrating factor µ(t) is
µ(t) = e
R
p(t) dt
.
d(ln µ(t)) = p(t) dt.
)
µ(t) = e
R
p(t) dt
.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
23
In conclusion, we have the following:
The process of using the method of integrating factor,
1. Rewrite the di↵erential equation into the standard form to identify p(t),
dy
+ p(t)y = g(t).
dt
2. Find the integrating factor, µ(t), by using
µ(t) = e
R
p(t) dt
.
(omit constant from this integration to make our life easier and still give the same
result)
3. Multiply both sides to the equation in step 1 (not the original equation) by µ(t).
4. Change the left hand side of the equation to (µ(t)y(t))0 by product rule of di↵erentiation.
5. Integrate both sides of the equation.
6. Find y(t) as the general solution.
Example 20 (Summer 2006 Midterm Exam I).
Solve the following initial value problem
ty 0
2y = t3 cos t,
y(⇡) = 2.
Solution Rewrite the di↵erential equation in canonical form,
2
y = t2 cos t.
t
y0
So we have p(t) =
(2.6)
2
and hence the integrating factor is
t
µ(t) = e
R
p(t) dt
=e
R
2
t
dt
=e
2 ln t
= eln t
2
=t
2
.
Multiply both sides of Equation (2.6) (not the original one) by µ(t),
t
2 0
y
2t
Rewrite the left hand side of the equation to (t
(t
2
y)0 = cos t
)
Integrating both sides,
Z
Z
2
d(t y) = cos t dt
d
(t
dt
)
2
t
3
y = cos t.
2
y)0 and then simplify,
y) = cos t
2
)
y = sin t + c
d(t
)
2
y) = cos t dt
y = t2 sin t + ct2 .
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
24
Then we apply the given initial condition y(⇡) = 2,
2 = ⇡ 2 sin ⇡ + c⇡ 2
)
c=
2
⇡2
Therefore the particular solution is
y = t2 sin t +
R
2 2
t .
⇡2
2
From the above example, we notice that e t dt = t 2 . The number
sides is not a coincidence. Let’s consider the general case,
e
R
k
t
dt
2 which appears on both
k
= ek ln t = eln t = tk .
So we have the shortcut here,
e
R
k
t
dt
= tk
for any natural number k. Let’s practice on this method with a few more examples .
Example 21 (Summer 2010 Midterm Exam I).
Use the integrating factor technique to solve the following equation
2y 0 (t) + y(t) = 3t.
(a) What is the integrating factor µ(t)?
Solution Rewrite the given equation to,
1
3
y 0 + y = t.
2
2
(2.7)
Therefore the integrating factor is
µ(t) = e
R
p(t) dt
=e
R
1
2
dt
t
= e2 .
(b) What is the general solution of the above di↵erential equation?
t
Solution Multiply both sides of Equation (2.7) by µ(t) = e 2 ,
t
1 t
3 t
e 2 y 0 + e 2 y = te 2
2
2
Now change 0 to
)
t
(e 2 y)0 =
3 t
te 2 .
2
d
dt ,
3 t
d t
(e 2 y) = te 2
dt
2
)
t
d(e 2 y) =
3 t
te 2 dt.
2
Integrating both sides and using integration-by-parts for the right hand side of the equation,
Z
Z
t
t
t
t
t
t
3 t
3
2
d(e y) =
te 2 dt ) e 2 y = (2te 2 4e 2 ) + c = 3te 2 6e 2 + c.
2
2
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
25
t
Divide both sides by e 2 to get the general solution,
y = 3t
6 + ce
t
2
.
Example 22 (Fall 2000 Midterm Exam I).
Solve the initial value problem
sin(t)y 0 + cos(t)y = 3t sin(t),
⇡
y( ) = 0.
2
Solution The standard form of the given equation is,
y 0 + cot(t)y = 3t.
(2.8)
So the integrating factor is
µ(t) = e
R
p(t) dt
=e
R
cot t dt
= eln sin t = sin t.
Multiply both sides of Equation (2.8) by sin t,
sin(t)y 0 + cos(t)y = 3t sin(t),
which gives us the original di↵erential equation in the question (this is just a coincidence).
Rewrite the left hand side of the equation to (sin(t)y)0 and then simplify,
(sin(t)y)0 = 3t sin(t)
d
(sin(t)y) = 3t sin t
dt
)
)
d(sin(t)y) = 3t sin t dt.
Integrating both sides (using integration-by-parts for the left hand side term),
Z
Z
d(sin(t)y) = 3t sin t dt ) sin(t)y = 3t cos t + 3 sin t + c.
Then apply the given initial condition y( ⇡2 ) = 0,
⇡
sin( )0 =
2
3
⇡
⇡
⇡
cos + 3 sin + c
2
2
2
)
c=
Hence the particular solution is
y=
1
( 3t cos t + 3 sin t
sin t
3).
Example 23 (Fall 2009 Midterm Exam I).
Solve the following initial value problem:
ty 0 + (t + 1)y = 1,
y(1) = 1.
Solution As usual, we rewrite the equation into the canonical form,
y0 +
t+1
1
y= .
t
t
3.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
Thus p(t) =
26
t+1
. So the integrating factor is
t
µ=e
R
p(t) dt
=e
R
t+1
t
dt
=e
R
1+ 1t dt
= et+ln t = et · eln t = tet .
Now let’s multiply both sides to the equation
y0 +
t+1
1
y= ,
t
t
by µ(t) = tet ,
tet y 0 + (t + 1)et y = et
)
(tet y)0 = et
Then we integrate both sides,
Z
Z
t
d(te y) = et dt
)
d(tet y)
= et
dt
)
tet y = et + c.
To get the particular solution, we use the initial condition y(1) = 1,
1e1 (1) = e1 + c
)
c = 0.
Therefore the particular solution for the above IVP is
tet y = et
)
y=
1
.
t
)
d(tet y) = et dt.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
27
Exercises 2.1.3
1. Solve the following initial value problem. You may leave your answers in implicit form.
sin(t)y 0 + cos(t)y = 3t sin(t),
2. Find the general solution of
t2
dy
+ 3ty = et ,
dt
⇡
y( ) = 0
2
t > 0.
3. What is a suitable integrating factor that can be used to solve the equation
(1 + t2 )y 0
(a) µ(t) =
2ty = 4e
⇡t
?
1
1 + t2
(b) µ(t) = et
2
(c) µ(t) = (1 + t2 )2
2 arctan t
(d) µ(t) = e
4. True or false: The function µ(t) = t2 e 3t is a suitable integrating factor that can be used
to solve the equation below. Justify your answer by finding the correct µ(t).
t2 y 0
(3t2
2t)y = e4t
5. Which of the following is an integrating factor for the di↵erential equation
(x2
2xy + x,
x2
(a) µ(t) = e
(b) µ(t) = x2
(c) µ(t) = e
1)y 0 =
1
x2
(d) µ(t) = ln(x2
1)
Answers
1. y =
3t cot t + 3
2. y(t) = t
3. (a)
4. True
5. (b)
2 t
e
t
3 csc t
3 t
e + Ct
3
x > 1?
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.1.4
28
Bernoulli Equation*
Bernoulli equation is a type of first order di↵erential equation that can be written as
y 0 + p(x)y = q(x)y n ,
(2.9)
where n is an integer. If n = 0, the equation is simply
y 0 + p(x)y = q(x)
which can be solved by the method of integrating factor. When n = 1, the above equation can be
rewritten as
dy
dy
y 0 = (q(x) p(x))y )
= (q(x) p(x))y )
= (q(x) p(x)) dx,
dx
y
which is solvable by the method of separable equation. Hence we assume here that n =
6 0, 1. So
the Bernoulli equation is always nonlinear because of that. If we multiply (2.9) by (1 n)y n , we
get
(1 n)y n y 0 + (1 n)p(x)y 1 n = (1 n)q(x).
Note that, by chain rule,
(1
n 0
y = (y 1
n)y
n 0
).
Now the equation becomes
(y 1
If we replace y 1
n
n 0
) + (1
n)p(x)y 1
n
= (1
n)q(x).
by u, we have
u0 + (1
n)p(x)u = (1
n)q(x).
This can be solved by the method of an integrating factor as we learned before. Note that this
method of solution was found by Leibniz in 1696. Let’s consider the examples below.
Example 24. Solve
y 0 + xy =
x
,
y3
y 6= 0.
Solution Notice that this is a Bernoulli equation with n = 3. As we discussed before, we
must multiply the original equation by (1 n)y n = 4y 3 . Then we have
4y 3 y 0 + 4xy 4 = 4x.
Then the first term of the equation above can be written as (y 4 )0 . Now the equation becomes
(y 4 )0 + 4xy 4 = 4x.
Substitute y 4 by u, the result is
u0 + 4xu = 4x.
An integrating factor is
µ=e
Multiply (2.10) by µ,
R
4x dx
(2.10)
2
= e2x .
2
2
2
e2x u0 + 4xe2x u = 4xe2x
Therefore
Z
Z
2
2
d(e2x u) = 4xe2x dx
)
2
)
2
2
(e2x u)0 = 4xe2x
e2x u =
Z
4xe2x
2
2
)
2
d(e2x u)
= 4xe2x .
dx
2
d(2x2 )
= e2x + c
4x
)
u = 1 + ce
2x2
.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
Replace u back to y 4 ,
29
2x2
y 4 = 1 + ce
is the particular solution in implicit form.
Example 25. Solve
t2 y 0 + 2ty
y 3 = 0,
t > 0.
Solution First, we rewrite the original equation to
t2 y 0 + 2ty = y 3
y 0 + 2t
)
1
y=t
2 3
y .
This is a Bernoulli equation with n = 3. Then we multiply the rewritten equation by
(1
Now the equation becomes
3 0
2y
n
n)y
y
=
1
4t
2y
2
y
3
=
.
2
2t
.
Notice that the first term of the equation above can be written as (y
(y
Substitute y
2
2 0
)
1
4t
2
y
=
2t
2
2 0
) . So we have
.
by u, the result is
u0
1
4t
u=
2t
2
.
(2.11)
Then an integrating factor is
µ(t) = e
Multiply (2.11) by µ = t
t
4 0
u
4t
5
4
R
4
t
dx
=e
R
4 ln t dt
=t
4
.
,
u=
2t
Z
2t
6
)
4
(t
u)0 =
6
2t
d(t 4 u)
=
dt
)
2t
6
.
Therefore
Z
d(t
4
u) =
Finally let’s change u back to y
6
dt
2
)
t
4
u=
2
t
5
5
+c
, the particular solution is
y
2
=
2
t
5
1
+ ct4 .
)
u=
2
t
5
1
+ ct4 .
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.1.5
30
Riccati Equation*
Riccati equation is a first order di↵erential equation in the form
y 0 = q0 (t) + q1 (t)y + q2 (t)y 2 ,
(2.12)
where q0 , q1 and q2 are given functions, with q0 , q2 6= 0. Suppose that some particular solution y1
of this equation is known. Then the general solution of this equation can be obtained through the
substitution
1
y = y1 (t) +
.
(2.13)
v(t)
This process will change the equation to a linear first order di↵erential equation which is solvable
by the method of integrating factor. To see this, let’s compute
1
y 0 = (y1 + )0 = y10
v
v0
.
v2
(2.14)
Combine (2.12) and (2.14) together,
q0 + q1 y + q2 y 2 = y10
v0
.
v2
Then we apply (2.13),
✓
◆
✓
◆2
1
1
q 0 + q 1 y1 +
+ q 2 y1 +
= y10
v
v
v0
v2
(2.15)
Since y1 is the particular solution of y 0 = q0 (t) + q1 (t)y + q2 (t)y 2 ,
y10 = q0 + q1 y1 + q2 y12 .
(2.16)
Put (2.15) and (2.16) together,
✓
◆
✓
◆2
1
1
q 0 + q 1 y1 +
+ q 2 y1 +
= q0 + q1 y1 + q2 y12
v
v
q1
2q2 y1
q2
q 0 + q1 y1 +
+ q2 y12 +
+ 2 = q0 + q1 y1 + q2 y12
v
v
v
q1
2q2 y1
q2
v0
+
+ 2 =
.
v
v
v
v2
v0
v2
v0
v2
Multiply the last equation by v 2 and rearrange terms,
v0
(q1 + 2q2 y1 )v + q2 =
)
v 0 + (q1 + 2q2 y1 )v =
q2 .
Now we can solve this equation by an integrating factor method as we discussed earlier.
Example 26. Solve the following IVP
y0 =
cos t + (2
tan t)y
(sec t)y 2 ,
y(0) = 0,
where y1 (t) = cos t is the particular solution.
Solution Note that
q0 (t) =
cos t,
q1 (t) = 2
Now we have
y = y1 (t) +
tan t,
q2 (t) =
1
1
= cos t +
v(t)
v
sec t.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
31
and after substitution the equation becomes
v 0 + (q1 + 2q2 y1 )v =
v 0 + ((2
)
q2
)
v
0
tan t)
2 sec t cos t) v = sec t
(tan t)v = sec t.
After we solve the above equation by the method of integrating factor,
t+c
.
cos t
v(t) =
Therefore
✓
1+
)
c=
1.
◆
cos t =
1
cos t
= cos t +
=
v
t+c
y = cos t +
1
t+c
◆
cos t.
Then we invoke the initial condition y(0) = 0,
1
=0
c
1+
So the particular solution of the IVP is
✓
y(t) = 1 +
1
t
1
t cos t
.
t 1
Example 27. Solve the following IVP
y0 = t
2
1
+ 3t
(4t
1
+ 3)y + 2y 2 ,
y(1) =
5
,
2
1
is the particular solution.
t
Solution Note that
where y1 (t) =
q0 (t) = t
2
1
+ 3t
,
q1 (t) =
Now we have
y = y1 (t) +
(4t
1
+ 3),
q2 (t) = 2.
1
1 1
= +
v(t)
t
v
and after substitution the equation becomes
0
v + (q1 + 2q2 y1 )v =
q2
0
)
v +
)
v0
✓
(4t
3v =
1
2(2)
+ 3) +
t
2.
Using the method of integrating factor, we obtain
v(t) =
Therefore
y=
2
+ ce3t .
3
1 1
1
+ = +
t
v
t
2
3
1
.
+ ce3t
◆
v=
2
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
32
Then we plug in the initial condition y(1) = 52 ,
5
=1+
2
2
3
1
+ ce3
)
2
2
+ ce3 =
3
3
)
Hence the particular solution of the IVP is
y=
1
1
1 3
3t + 2
+ 2 = + =
.
t
t
2
2t
3
c = 0.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.1.6
33
Exact Equation
We call the di↵erential equation exact if we can rewrite the di↵erential equation in the form
M (x, y) + N (x, y)
and there is a function
dy
=0
dx
(2.17)
(x, y) which has continuous second partial derivatives such that
x
= M (x, y) and
= N (x, y).
y
(2.18)
You might ask yourself already that why we need these complicated conditions. Here’s why.
Combine Equations (2.17) and (2.18) together,
x
+
y
dy
= 0.
dx
Using the chain rule, the left hand side is simply
d
dx
(x, y(x)). Therefore,
d
(x, y(x)) = 0.
dx
After we integrate both sides, we get
(x, y(x)) = c,
as an implicit solution to our di↵erential equation. Since second partials of
Schwarz’s theorem, we get the symmetry of second derivatives which means
xy
=
are continuous, by
yx .
However, we also have the following
xy
=(
x )y
= My
yx
=(
y )x
= Nx .
Therefore, if a di↵erential equation is exact and
(x, y) must satisfy
My = N x .
(2.19)
This condition is called exactness. Likewise if Equation(2.19) is not true there is no way for the
di↵erential equation to be exact. Therefore we will use this equation as a test for exactness of the
equation. In other words, an equation of the form
M (x, y) + N (x, y)
In conclusion, we have
dy
= 0 is an exact equation if and only if My = Nx .
dx
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
34
The process of exact equation method
1. Rewrite the di↵erential equation into the form
M (x, y) + N (x, y)
dy
= 0,
dx
to identify M and N .
2. Check exactness of the equation by checking My = Nx . (if it is not, we can’t use this
method.)
3. After using these two conditions
x
y
=M
=N
)
=
)
=
Z
Z
M dx;
N dy,
compare these two equations to get
4. Then the general solution in implicit form is
(x, y) = c.
Example 28 (Spring 2005 Midterm Exam I).
a) Verify that the following ODE is exact:
2x + ey + (xey
cos y)
dy
= 0.
dx
Solution First we identify functions M and N ,
M (x, y) = 2x + ey ,
N (x, y) = xey
cos y.
To check exactness, let’s compute My and Nx .
@
@
@ y
(2x + ey ) =
2x +
e = 0 + ey = ey ,
@y
@y
@y
@
@
@
Nx =
(xey cos y) = ey x
cos y = ey (1)
@x
@x
@x
My =
0 = ey .
Since My = Nx , the equation is exact.
b) Find the general solution to the ODE.
Solution Recall that
Z
Z
Z
Z
y
y
)
= M dx = (2x + e ) dx = 2x dx + e
dx = x2 + xey + c1 (y),
x =M
Z
Z
Z
Z
y
y
=
N
)
=
N
dy
=
(xe
cos
y)
dy
=
x
e
dy
cos y dy = xey sin y + c2 (x).
y
Comparing two results,
x2 + xey + c1 (y) = xey
sin y + c2 (x).
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
Therefore c1 (y) =
35
sin y and c2 (x) = x2 . And so
(x, y) = x2 + xey
sin y.
Therefore the general solution is
x2 + xey
sin y = c.
( (x, y) = c)
c) Find the solution to the ODE in part a) which satisfies y(1) =
Solution Apply the initial condition, y(1) =
⇡
12 + (1)e 2
sin
⇡
2.
⇡
2,
⇡
=c
2
⇡
)
c = e2.
⇡
Hence the particular solution is x2 + xey
sin y = e 2 .
Example 29 (Summer 2006 Midterm Exam I).
Consider the initial value problem
6x2
2xy + ex+y + (ex+y
x2 )
dy
= 0,
dx
y(1) =
1.
a) Verify that the equation is exact.
Solution In this example, we have
M (x, y) = 6x2
2xy + ex+y ,
N (x, y) = ex+y
x2 .
Compute My and Nx .
@
@
@
(6x2 2xy + ex+y ) = 0 2x y + ex ey =
@y
@y
@y
@ x+y
@
@ 2
Nx =
(e
x2 ) = ey ex
x = ex+y 2x.
@x
@x
@x
My =
2x + ex+y ,
So the equation is exact because My = Nx .
b) Solve this IVP. You may leave your answer in implicit form.
Solution Since
x
=M
)
y
=N
)
Z
M dx = (6x2
Z
Z
= N dy = (ex+y
=
Z
2xy + ex+y ) dx = 2x3
x2 ) dy = ex+y
x2 y + ex+y + c1 (y),
x2 y + c2 (x).
Then we compare these two equations,
2x3
Hx+y
x2⇢
y +H
ex+y
H
H + c1 (y) = e H
H
⇢
x2⇢
y + c2 (x).
⇢
Thus c1 (y) = 0 and c2 (x) = 2x3 . And so
(x, y) = 2x3
x2 y + ex+y .
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
36
Hence the general solution is
2x3
x2 y + ex+y = c.
Now we impose the initial condition, y(1) =
2(1)3
(1)2 ( 1) + e1
1,
1
=c
)
c = 4.
Thus the particular solution is
2x3
x2 y + ex+y = 4.
Example 30 (Summer 2002 Final Exam ).
Solve the initial value problem;
(2x + 4xy +
y
) dx + (2x2 + ln x + 3y 2 ) dy = 0,
x
y(1) =
1.
Solution From the form of the equation, it suggests us to solve it with the exact equation
method. Note that we have
M (x, y) = 2x + 4xy +
y
and N (x, y) = 2x2 + ln x + 3y 2 .
x
Then
@
y
@
1 @
1
(2x + 4xy + ) = 0 + 4x y +
y = 4x +
@y
x
@y
x @y
x
@
@
@
1
Nx =
(2x2 + ln x + 3y 2 ) =
2x2 +
ln x + 0 = 4x + .
@x
@x
@x
x
My =
Since My = Nx , this equation is exact. To find the general solution, we use the following
relations:
Z
Z
y
)
= M dx = (2x + 4xy + ) dx = x2 + 2x2 y + y ln x + c1 (y)
x =M
x
Z
Z
)
= N dy = (2x2 + ln x + 3y 2 ) dy = 2x2 y + y ln x + y 3 + c2 (x).
y =N
By comparison of above two equations, we then have
X
X
x2 + 2x2 y + X
y ln
x + c1 (y) = 2x2 y + X
y ln
x + y 3 + c2 (x)
X
X
)
c1 (y) = y 3 , c2 (x) = x2 .
So the general solution is
(x, y) = c
)
x2 + 2x2 y + y ln x + y 3 = c.
Now let’s apply the given initial condition, y(1) =
12 + 2(1)2 ( 1) + ( 1) ln 1 + ( 1)3 = c
1,
)
c=1
2
1=
2.
Substitute the obtained value of c back into the general solution. So the particular solution to
the IVP is
x2 + 2x2 y + y ln x + y 3 = 2.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
37
Example 31 (Fall 2001 Midterm Exam I).
Find the general solution of the following equation:
xexy )y 0 = 2 + yexy .
(2y
You may leave your answer in implicit form.
Solution First move every terms to the left-hand side of the equation,
So M (x, y) =
equation,
2
xexy )y 0 = 0.
yexy + (2y
2
yexy and N (x, y) = 2y
@
( 2
@y
@
Nx =
(2y
@x
@ xy
ye = (xyexy + exy ).
@y
@ xy
xe = (xyexy + exy ).
@x
yexy ) = 0
My =
xexy . Now we check the exactness of this
xexy ) = 0
Since My = Nx , the equation above is indeed an exact equation. So we can proceed as usual
Z
Z
)
= M dx = ( 2 yexy ) dx = 2x exy + c1 (y)
x =M
Z
Z
)
= N dy = (2y xexy ) dy = y 2 exy + c2 (x).
y =N
We compare both equations above,
2x
exy + c1 (y) = y 2
Therefore c1 (y) = y 2 and c2 (x) =
solution is
2x and so
2x
exy + c2 (x).
(x, y) =
2x
exy + y 2 . Thus the general
exy + y 2 = c.
Example 32 (Spring 2004 Midterm Exam I).
Given that
2
2
y 2 exy + 4x3 + (2xyexy + 2)
dy
= 0,
dx
y(0) = 2.
a) Verify that the equation is exact.
Solution Here, we have
2
2
M (x, y) = y 2 exy and N (x, y) = 2xyexy + 2.
To check an exactness of the equation by considering
2
2
2
@ 2 xy2
@
@ 2 xy2
(y e ) = y 2 exy +
(y )e
= 2xy 3 exy + 2yexy
@y
@y
@y

2
2
2
2
@
@
@
@
Nx =
(2xyexy + 2) = 2y (xexy ) + 0 = 2y x exy + ( x)exy
@x
@x
@x
@x
My =
2
2
2
2
= 2y(xy 2 exy + exy ) = 2xy 3 exy + 2yexy .
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
Therefore the equation is exact since My = Nx .
b) Solve the initial value problem. Leave your answer in implicit form.
Solution Consider the following:
Z
Z
2
2
=
M
)
=
M
dx
=
(y 2 exy + 4x3 ) dx = exy + x4 + c1 (y)
x
Z
Z
2
2
)
= N dy = (2xyexy + 2) dy = exy + 2y + c2 (x).
y =N
Hence c1 (y) = 2y and c2 (x) = x4 . And so the general solution is
)
(x, y) = c
2
exy + x4 + 2y = c.
Apply the initial condition, y(0) = 2. We get
2
e0(2) + 04 + 2(2) = c
)
c = 1 + 4 = 5.
Therefore the particular solution is
2
exy + x4 + 2y = 5.
38
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
Exercises 2.1.6
1. Find the value for the constant b, for which given equation is exact.
(ex sin y + bx2 y 2 ) dx + (ex cos y + x3 y) dy = 0
(a) b = 0
1
(b) b =
3
(c) b = 3
3
(d) b =
2
(e) b = 1
2. Find the general solution of
(2xy
3x2 ) dx + (x2 + 2y) dy = 0
(you may keep your solution in implicit form).
3. Consider the equation
3x5 y 4
2x + (2x6 y 3 + 6e3y )y 0 = 0.
(a) Verify that it is an exact equation.
(b) Find its general solution. You may leave your answer in implicit form.
(c) Find the particular solution satisfying the initial condition y(3) = 0.
4. What is the general solution of the equation
4x3 + 2y + y cos x + (2x + sin x + 2y)y 0 = 0?
(a) 2 + cos x = C
(b) 12x2
y sin x = C
4
(c) x + 2xy + y sin x = C
(d) x4 + 2xy + y sin x + y 2 = C
5. Consider the equation
(2 x5 y 3
(a) Find the value of
1
) + (3x6 y 2
x2
4 )y 0 = 0.
such that the equation becomes an exact equation.
(b) Find its general solution. You may leave your answer in implicit form.
6. Consider the equation
1
(x3 y 2 + xey + ln x) + ( x4 y
2
Find the value
(a)
=
(b)
=1
(c)
=
(d)
=2
1
2
6y sin y 2 + x2 ey )y 0 = 0.
such that the above equation is exact.
39
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
40
7. What is the general solution of the following equation
cos(x + y) + 2x + (cos(x + y) + 4y)y 0 = 0?
Answers
1. (d)
2. x2 y
x3 + y 2 = C
3. (a) My = 12x5 y 3 = Nx , (b)
1 6 4
x y
2
4. (d)
5. (a)
= 3, b) x6 y 3 +
1
x
6. (d)
7. sin(x + y) + x2 + 2y 2 = c
12y = C
1
x2 + 2e3y = C, (c) x6 y 4
2
x2 + 2e3y =
7
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.1.7
41
Recognizable Exact Equation and Integrating Factor*
It is sometimes possible to convert a di↵erential equation that is not exact into an exact equation
by multiplying the equation by a suitable factor. This factor is called an integrating factor. To
explore the possibility of this idea, let us multiply the equation
M (x, y) + N (x, y)y 0 = 0
(2.20)
by a function µ = µ(x, y) and then try to choose µ so that the resulting equation
µM (x, y) + µN (x, y)y 0 = 0
is exact. This means
(µM )y = (µN )x
)
µMy + µy M = µNx + µx N.
As a result,
µy M
µx N + (My
Nx )µ = 0
(2.21)
Unfortunately, Equation (2.20) which determines the integrating factor µ is at least as hard to
solve as the original equation Equation (2.21). Therefore we need to consider them in special cases
only. We consider six possibilities.
1. µ is a function only of x, i.e., µ = µ(x). Then µy = 0, so the Equation (2.21) becomes
µx N + (My
)
Nx )µ = 0
µx N = (My
)
Nx )µ
dµ
My N x µ
=
.
dx
N
My N x
Then, if
is a function of x only, then we can solve the above di↵erential equation
N
as a separable equation,
Z
Z
My N x
dµ
My N x
dµ
=
µ )
=
dx.
dx
N
µ
N
Therefore
ln µ =
✓Z
My
Nx
N
◆
dx + c0
)
µ = c exp
✓Z
My
Nx
N
dx
◆
For simplicity, we can omit the constant c. So in then end, the integrating factor is
✓Z
◆
My N x
µ(x) = exp
dx .
N
Example 33. Find an integrating factor for the equation
(3xy + y 2 ) + (x2 + xy)y 0 = 0
and then solve the equation.
Solution We have M (x, y) = 3xy+y 2 and N (x, y) = x2 +xy. Note that this di↵erential
equation is not exact since My 6= Nx :
@
(3xy + y 2 ) = 3x + 2y
@y
@ 2
Nx =
(x + xy) = 2x + y.
@x
My =
Consider
My
Nx
N
=
(3x + 2y) (2x + y)
x+y
1
=
= .
x2 + xy
x(x + y)
x
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
Since
My
Nx
N
42
is a function of x only, then the integrating factor is
µ(x) = exp
✓Z
My
Nx
N
dx
◆
= exp
✓Z
1
dx
x
◆
= exp(ln x) = x.
Multiplying the given equation by this integrating factor, we obtain
(3x2 y + xy 2 ) + (x3 + x2 y)y 0 = 0.
Now this equation is exact since Py = Qx ,
@
(3x2 y + xy 2 ) = 3x2 + 2xy
@y
@ 3
Qx =
(x + x2 y) = 3x2 + 2xy.
@x
Py =
Then we can find a function by
Z
Z
x2 y 2
)
= P dx = (3x2 y + xy 2 ) dx = x3 y +
+ c1 (y)
x =P
2
Z
Z
x2 y 2
)
= Q dy = (x3 + x2 y) dy = x3 y + +
+ c2 (x).
y =Q
2
Compare these two equations, we have c1 (y) = c2 (x) = 0 and so we have
(x, y) = x3 y +
x2 y 2
.
2
Therefore the general solution is
x3 y +
x2 y 2
= c.
2
2. µ is a function only of y, i.e., µ = µ(y). Then µx = 0, so the Equation (2.21) becomes
µy M + (My
Similar to first case if
Nx
My
M
)
Nx )µ = 0
µy =
Nx
My
M
µ.
is a function of y only, then the integrating factor is
µ(y) = exp
✓Z
Nx
My
M
◆
dy .
Example 34. Find an integrating factor for the equation
y + (2xy
e
2y
)y 0 = 0
and then solve the equation.
Solution We have M (x, y) = y and N (x, y) = 2xy
e
2y
. This equation is not exact
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
43
since My 6= Nx :
@
y=1
@y
@
Nx =
(2xy e
@x
My =
Consider
My
Nx
N
=
1
2xy
2y
) = 2y.
2y
.
e 2y
It is clearly not a function of x only. So let’s consider
Nx
My
M
=
2y
1
y
=2
1
.
y
Since it is a function of y only, then the integrating factor is
✓Z
◆
✓Z
◆
N x My
1
µ(y) = exp
dy = exp
(2
) dy = e2y
M
y
ln y
=
e2y
.
y
Multiplying the original equation by this integrating factor,
1 0
)y = 0.
y
e2y + (2xe2y
Now the obtained equation is actually exact since Py = Qx ,
@ 2y
(e ) = 2e2y
@y
@
1
Qx =
(2xe2y
) = 2e2y .
@x
y
Py =
Let’s find the expression for
x
=P
)
y
=Q
)
(x, y) from
Z
Z
= P dx = (e2y ) dx = xe2y + c1 (y)
Z
Z
1
= Q dy = (2xe2y
) dy = xe2y ln |y| + c2 (x).
y
Compare these two equations, we have c1 (y) =
(x, y) = xe2y
ln y and c2 (x) = 0 and hence
ln |y|.
Therefore the general solution is
xe2y
ln |y| = c.
3. µ is a function of xy, i.e., µ = µ(u), where u = xy. Since u = xy, we have
@u
= x,
@y
@u
= y.
@x
Therefore, by chain rule,
@µ(u)
dµ(u) @u
dµ(u)
=
·
=x
,
@y
du
@y
du
@µ(u)
dµ(u) @u
dµ(u)
µx =
=
·
=y
.
@x
du
@x
du
µy =
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
44
Substituting everything back in (2.21) and simplify the result, we have
µy M
µx N + (My
Nx )µ = 0
)
x
dµ(u)
M
du
) (xM
)
)
dµ(u)
µ(u)
dµ(u)
µ(u)
dµ(u)
N + (My Nx )µ(u) = 0
du
dµ(u)
yN )
= (My Nx )µ(u)
du
My N x
=
du
xM yN
My N x
=
du
yN xM
y
If the coefficient of du in the equation above also a function of u = xy, say F (u) = F (xy) so
that
My N x
F (u) =
,
yN xM
then the equation above will change to
dµ(u)
= F (u) du.
µ(u)
This is a separable equation and so it is easily to find the solution as we show below
Z
Z
Z
dµ(u)
= F (u) du ) ln(µ(u)) = F (u) du.
µ(u)
Hence,
µ(u) = e
R
F (u) du
is an integrating factor of this case where u = xy. Let’s study the following example for
clarification.
Example 35. First show that the di↵erential equation
(y 3 + xy 2 + y) + (x3 + x2 y + x)y 0 = 0
is not exact and then find an integrating factor.
Solution We have M (x, y) = y 3 + xy 2 + y and N (x, y) = x3 + x2 y + x. Note that
@ 3
(y + xy 2 + y) = 3y 2 + 2xy + 1
@y
@ 3
Nx =
(x + x2 y + x) = 3x2 + 2xy + 1.
@x
My =
Therefore the equation is not exact since My is not the same as Nx . Consider
My
yN
Nx
(3y 2 + 2xy + 1)
=
xM
y(x3 + x2 y + x)
(3x2 + 2xy + 1)
3y 2
=
x(y 3 + xy 2 + y)
yx3
Simplify further, we obtain
My
yN
Therefore F (u) =
Nx
3(y 2 x2 )
=
=
xM
yx(x2 y 2 )
3
=
xy
3
.
u
3
and then the integrating factor is
u
µ(u) = e
R
F (u) du
=e
R
3
u
du
=u
3
= (xy)
3
.
3x2
.
xy 3
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
45
Multiplying the original equation by this integrating factor, we get
3
(xy)
(x
3
2
+x
y
1
(y 3 + xy 2 + y) + (xy)
+x
3
y
2
) + (y
3
+x
3
1
(x3 + x2 y + x)y 0 = 0
y
2
+x
2
y
3
)y 0 = 0.
Now let’s check the exactness of new equation,
@
(x
@y
@
Qx =
(y
@x
Py =
3
+x
2
y
1
+x
3
y
2
)=
x
2
y
2
2x
3
y
3
,
3
+x
1
y
2
+x
2
y
3
)=
x
2
y
2
2x
3
y
3
.
Since Py = Qx , the new equation is exact and we can proceed as usual to get the solution.
4. µ is a function of
x
x
x
, i.e., µ = µ(u), where u = . Since u = , we have
y
y
y
@u
=
@y
x
,
y2
@u
1
= .
@x
y
Therefore, by chain rule,
@µ(u)
dµ(u) @u
x dµ(u)
=
·
=
,
@y
du
@y
y 2 du
@µ(u)
dµ(u) @u
1 dµ(u)
µx =
=
·
=
.
@x
du
@x
y du
µy =
Substituting everything back in (2.21) and simplify the result, we have
µy M
µx N + (My
Nx )µ = 0
x dµ(u)
1 dµ(u)
M
N + (My Nx )µ(u) = 0
2
y du
y du
x
1
dµ(u)
) ( 2M
N)
= (My Nx )µ(u)
y
y
du
dµ(u)
My N x
)
=
du
x
1
µ(u)
y2 M
yN
)
)
dµ(u)
My N x
= x
du
1
µ(u)
y2 M + y N
)
dµ(u)
y 2 (My Nx )
=
du
µ(u)
xM + yN
If the coefficient of du in the equation also simplifies to a function of u =
so that
G(u) =
x
x
, say G(u) = G( )
y
y
y 2 (My Nx )
,
xM + yN
then the equation above will change to
dµ(u)
= G(u) du.
µ(u)
This is a separable equation and so it is easily to find the solution as we show below
Z
Z
Z
dµ(u)
= G(u) du ) ln(µ(u)) = G(u) du.
µ(u)
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
46
Hence an integrating factor for this case is
µ(u) = e
where u =
x
.
y
R
G(u) du
,
Example 36. First show that the di↵erential equation
xy 0 = 0
3y
is not exact and then find an integrating factor.
Solution We have M (x, y) = 3y and N (x, y) =
My =
@
(3y) = 3,
@y
x. Note that
Nx =
@
( x) =
@x
1.
Since My 6= Nx , the equation is not exact. Let’s consider
y 2 (My Nx )
y 2 (3 ( 1))
4y 2
yx
=
=
=2
xM + yN
3xy xy
2xy
.
Therefore G(u) =
2
and then the integrating factor is
u
µ(u) = e
R
F (u) du
=e
R
2
u
du
= u2 =
x2
.
y2
Multiplying the original equation by this integrating factor, we get
3y
xy 0 = 0
)
x2
(3y)
y2
x2
(xy 0 ) = 0
y2
)
3x2 y
1
x3 y
2 0
y = 0.
Let’s check the exactness of new equation,
Py =
@
(3x2 y
@y
1
)=
3x2 y
2
,
Qx =
@
( x3 y
@x
2
)=
3x2 y
2
.
Since Py = Qx , the new equation is exact and we can follow usual procedure to obtain
solution.
y
y
5. µ is a function of , i.e., µ = µ(u), where u = . Similar to the previous case, an
x
x
integrating factor is
R
µ(u) = e H(u) du ,
y
whrere u = and
x
x2 (Nx My )
H(u) =
.
xM + yN
Example 37. First show that the di↵erential equation
y
3xy 0 = 0
is not exact and then find an integrating factor.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
Solution We have M (x, y) = y and N (x, y) =
My =
@
(y) = 1,
@y
47
3x. Note that
Nx =
@
( 3x) =
@x
3.
Since My 6= Nx , the equation is not exact. Then we consider
x2 (Nx My )
x2 ( 3 1))
=
=
xM + yN
xy 3xy
Therefore H(u) =
4x2
xy
=2
2xy
.
2
and then the integrating factor is
u
µ(u) = e
R
H(u) du
=e
R
2
u
du
= u2 =
y2
.
x2
Multiplying the original equation by this integrating factor, we get
y
3xy 0 = 0
)
y2
(y)
x2
y2
(3xy 0 ) = 0
x2
)
x
2 3
y
3x
1 2 0
y y = 0.
Let’s check the exactness of new equation,
Py =
@
(x
@y
2 3
y ) = 3x
2 2
y ,
Qx =
@
( 3x
@x
1 2
y ) = 3x
2 2
y .
Then the new equation is exact since Py = Qx . To obtain solution, we simply follow
usual steps as before.
6. If a di↵erential equation can be written in the form
y(Axp y q + Bxr y s ) + x(Cxp y q + Dxr y s )y 0 = 0,
where A, B, C, D are constants. Then it can be shown that an integrating factor of the above
equation is in the form xa y b where a and b are suitably chosen constants. We illustrate details
below.
Example 38. First show that the di↵erential equation
y(2x2 y 3 + 3) + x(x2 y 3
1)y 0 = 0
is not exact and then find an integrating factor.
Solution We have M (x, y) = 2x2 y 4 + 3y and N (x, y) = x3 y 3
My =
@
(2x2 y 4 + 3y) = 8x2 y 3 + 3,
@y
Nx =
@ 3 3
(x y
@x
x. Note that
x) = 3x2 y 3
1.
Since My 6= Nx , the equation is not exact. Since we expect that an integrating factor
will have the form xa y b , multiply both sides of the equation by xa y b , we get
xa y b (2x2 y 4 + 3y) + xa y b (x3 y 3
2xa+2 y b+4 + 3xa y b+1 + (xa+3 y b+3
x)y 0 = 0
xa+1 y b )y 0 = 0.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
New equation will be exact if
@
@ a+3 b+3
(2xa+2 y b+4 + 3xa y b+1 ) =
(x y
xa+1 y b )
@y
@x
2(b + 4)xa+2 y b+3 + 3(b + 1)xa y b = (a + 3)xa+2 y b+3 (a + 1)xa y b .
Therefore a and b must satisfy
2(b + 4) = a + 3,
Two equations above give a =
7
and b =
5
3(b + 1) =
(a + 1).
9
. Hence an integrating factor is
5
7
xa y b = x 5 y
9
5
.
48
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.1.8
49
More Examples
Separable Equation vs Exact Equation
Example 39 (Spring 2012 Final Exam ).
True or false.
a) Every exact equation is also separable.
Solution Recall that the exact equation is in the form
M (x, y) + N (x, y)y 0 = 0 and it satisfies My = Nx .
Now let’s try to separate all expressions of y from expressions of x,
M (x, y) + N (x, y)
dy
=0
dx
)
M (x, y) dx =
N (x, y) dy.
As you can see, we fail to completely separate variable y from x. So it’s not necessary
that exact equation is separable. Thus the statement is false.
*b) Every separable equation is also exact.
Solution So now let’s start with the separable equation which we can write it in the
form
P (x) dx = Q(y) dy.
Now rewrite it in the form of exact equation and then check its exactness.
P (x) dx + Q(y) dy = 0
)
P (x) + Q(y)y 0 = 0.
So M (x, y) = P (x) and N (x, y) = Q(y). Notice that My = Nx = 0. So every
separable equation is an exact equation as well. Therefore this statement is true.
Example 40 (Fall 2000 Midterm Exam I).
Solve the following equations. You may leave your answers in implicit form.
a) y 0 =
x2
2y
dy
Solution We first change y 0 =
and then we solve it by using the separable equation
dx
method.
Z
Z 2
dy
x2
x2
x
y2
x3
=
) y dy =
dx )
y dy =
dx )
=
+ c.
dx
2y
2
2
2
6
So the general solution in implicit form is
y2 =
b) x
y + (y
x3
+ c.
3
x)y 0 = 0
Solution We can use two methods to solve this equation. In the end, we’ll check that
both methods give us the same solution.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
50
(a) Exact equation method: We have M (x, y) = x y and N (x, y) = y
that My = Nx = 1. So this is an exact equation. Now we consider
x
=M
)
y
=N
)
Z
Z
x2
M dx = (x y) dx =
xy + c1 (y)
2
Z
Z
y2
= N dy = (y x) dy =
xy + c2 (x).
2
=
Now we compare both equations to get the right form of
x2
2
So c1 (y) =
x. Notice
y2
2
xy + c1 (y) =
,
xy + c2 (x).
y2
x2
and c2 (x) =
. Therefore,
2
2
x2
2
xy +
xy +
y2
= c.
2
(x, y) =
So the general solution is
x2
2
y2
.
2
(b) Separable equation method: Rewrite the equation,
x
y + (y
x)y 0 = 0
)
(y
x)y 0 = y
)
x
y0 =
y
y
x
= 1.
x
This is the separable equation and so
0
y =1
dy
=1
dx
)
)
Z
dy =
Z
1 dx
)
y = x + c.
Even though it seems like both methods give di↵erent solutions, but indeed they are the
same. Consider the solution from the first method again,
x2
2
xy +
y2
=c
2
x2
)
2xy + y 2 = c
)
(y
Example 41 (Summer 2007 Midterm Exam I).
Consider the initial value problem
y0 =
4x3 + 1
,
2y 4
y(0) = 0.
Which of the following statements is false?
(a) The equation is separable.
(b) The equation is exact.
(c) The implicit form of its solution is y 2
4y = x4 + x.
p
(d) The explicit form of its solution is y = 2 + x4 + x + 4.
x)2 = c
)
y = x + c.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
51
Solution The statement in a) is true since we can rewrite it as follows:
dy
4x3 + 1
=
dt
2y 4
)
4) dy = (4x3 + 1) dx.
(2y
To check the validity of the statement in b), we rewrite the equation to
y0 =
4x3 + 1
2y 4
)
(4x3 + 1) + (2y
4)y 0 = 0.
Hence M (x, y) = (4x3 + 1) and N (x, y) = 2y 4. Notice that My = Nx = 0. So it is exact.
Now let’s find the solution for the above initial value problem by using the method of separable
equation. We continue from what we did earlier,
Z
Z
(2y 4) dy = (4x3 + 1) dx ) y 2 4y = x4 + x + c.
Then we use the initial condition, y(0) = 0, which gives c = 0. So the particular solution in
implicit form is
y 2 4y = x4 + x.
Therefore the statement in c) is valid as well. Finally let’s consider part d) by finding the
explicit form of the solution. As before, we need to complete the square of the expression in
the LHS of the equation and solve for y.
p
y 2 4y + 22 = x4 + x + 22 ) (y 2)2 = x4 + x + 4 ) y = 2 ± x4 + x + 4.
p
Observe that y = 2 + x4 + x + 4 disagrees with the initial condition y(0) = 0 since it gives
y(0) = 4. Therefore the implicit form of the particular solution is
p
y=2
x4 + x + 4.
This is why, d) is the correct answer since its statement is invalid.
Example 42 (Spring 2011 Midterm Exam I).
Let y(x) be the solution to the initial value problem
y0
2y 2 = xy 2 ,
y(0) = 1.
Which of the following values is y(1)?
Solution Be careful here since we can’t use the method of an integrating factor here because
it’s not in the form of
y 0 + p(x)y = g(t).
Indeed, it is a separable equation since
y 0 = xy 2 + 2y 2
)
y 0 = (x + 2)y 2
)
dy
= (x + 2)y 2
dx
We integrate both sides to obtain the general solution,
1
x2
=
+ 2x + c.
y
2
)
1
dy = (x + 2) dx.
y2
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
52
Now we impose the initial condition, y(0) = 1,
1=0+c
So the particular solution is
)
1
x2
=
+ 2x
y
2
c=
1.
1.
Then we can find y(1) by simply plugging in x = 1,
1
12
=
+ 2(1)
y(1)
2
1
)
1
3
=
y(1)
2
)
y(1) =
2
.
3
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
Exercises 2.1.8
1. Consider the di↵erential equation (Summer 2012 Midterm Exam I)
(t2
9)y 0
t
t
1
y=
9 sin t
.
t
Answer the following questions. Briefly explain your answers.
(a) Is the equation linear?
(b) Is the equation separable?
2. Consider the di↵erential equation (Spring 2012 Midterm Exam I)
yy 0 = 0.
t2 + t3 y + sin(t)y
Answer the following questions. Briefly explain your answers.
(a) What is the equation’s order?
(b) Is the equation linear?
(c) Is the equation separable?
(d) Is the equation exact?
(e) Is the equation autonomous?
3. Consider the di↵erential equation
yy 0 = 1
x2 .
Answer the following questions. You must give a reason to justify each of your answers.
(a) Is the equation first order linear?
(b) Is the equation separable?
(c) Is the equation exact?
(d) Is the equation autonomous?
Answers
1. (a) yes, (b) no
2. (a) first order, (b) nonlinear, (c) no, (d) no, (e) no
3. (a) no, (b) yes, (c) yes, (d) no
53
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.2
54
The Existence and Uniqueness Theorem
So far, we have seen a number of initial value problems, each of which has the unique solution.
This observation raises a natural question:
Does every initial value problem has exactly one solution or at least has a solution?
This is important question. Because you might want to know that initial value problem has a
solution before spending time and e↵ort in trying to find it. Luckily, for linear first order equations
with mild conditions, the answer to this question is positive and it is given by the next theorem.
Theorem 1 (The Existence and Uniqueness Theorem). If the functions p and g are continuous on an open interval I : ↵ < t < containing the point t = t0 , then there exists a unique
function y = (t) that satisfies the di↵erential equation
y 0 + p(t)y = g(t)
for each t in I, and that also satisfies the initial condition y(t0 ) = y0 . In other words, neither
existence nor uniqueness of a solution is guaranteed at a discontinuity of either the coefficient
functions p(t) or g(t).
Before we see the examples, let’s learn a new terminology here.
• The interval of validity for an IVP of n th-order ODE with initial condition(s)
y(t0 ) = y0
and/or
y (k) (t0 ) = yk , where k = 1, · · · , n
1,
is the largest possible interval on which the solution is valid and contains t0 .
Example 43 (Summer 2003 Midterm Exam I).
What is the largest interval on which the solution of
ty 0 + 2y = 3t,
y(2) = 5,
is guaranteed to exist?
Solution Firstly, rewrite the given equation in the canonical form,
2
y 0 + y = 3,
t
y(2) = 5,
2
with p(t) =
and g(t) = 3. So p has one discontinuity at t = 0 and g is continuous
t
everywhere. By the Existence and Uniqueness Theorem , the solution of this initial value
problem guaranteed to exist uniquely on any interval containing t0 = 2 but not containing
discontinuity t = 0.
Hence the interval of validity is (0, 1).
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
55
Example 44 (Summer 2008 Midterm Exam I).
Consider the initial value problem
(t2
t
t+1
16)y 0 + (sin )y =
,
5
t 1
y(⇡) =
1
.
2
Without solving the equation, what is the largest interval in which a unique solution is guaranteed to exist?
Solution As usual, write down the equation in the canonical form together with the fact that
t2 16 = (t 4)(t + 4), we have
y0 +
sin 5t
(t + 4)(t
4)
y=
t+1
1)(t + 4)(t
(t
4)
,
y(⇡) =
sin 5t
t+1
and g(t) =
.
(t + 4)(t 4)
(t 1)(t + 4)(t 4)
So p has discontinuities at t = 4, 4 and g has discontinuities at t =
1
,
2
with p(t) =
4, 1, 4.
By the Existence and Uniqueness Theorem , the largest interval in which a unique solution is
guaranteed to exist is (1, 4).
Before we move on, let’s remind ourselves of the useful trig facts here:
sin(t) = 0
)
t = n⇡ for all integer n
cos(t) = 0
)
t=
(2n
1)⇡
2
for all integer n
Example 45 (Fall 2000 Midterm Exam I). Consider the initial value problem
sin(t)y 0 + cos(t)y = 3t sin(t),
⇡
y( ) = 0.
2
On what interval is the solution guaranteed to exist? Justify your answers!
Solution First, rewrite the equation in the standard form,
y0 +
cos(t)
y = 3t,
sin(t)
cos(t)
and g(t) = 3t.
sin(t)
So p has discontinuities at t = n⇡, where n is an integer and g is continuous everywhere.
with p(t) =
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
56
By the Existence and Uniqueness Theorem , the interval of validity is (0, ⇡).
It’s worth noting that many functions we deal with so far have a domain of all real numbers. But
the natural logarithm ln t has di↵erent domain. This function is defined only when t > 0, in other
words, its domain is the set of positive real numbers or (0, 1). Roughly speaking, number that
you are allowed to put after “ln” is only the positive ones. Base on this fundamental fact, we
have to be careful when we talk about the domain of other types of function involving the natural
logarithm. For example, ln(t a) is defined only when t a > 0 or t > a. We list others below:
ln t is defined when t > 0
ln(t
ln |t
a) is defined when t > a
ln |t| is defined when t 6= 0
a| is defined when t 6= a
Example 46 (Spring 2010 Midterm Exam I). Consider the initial value problem
(t + 1)(t
⇡)y 0
ln(t)y = cos(3t),
y(2) = 5.
Without solving the equation, what is the largest interval in which a unique solution is guaranteed to exist?
Solution The standard form of the given equation is,
y0
ln(t)
cos(3t)
y=
,
(t + 1)(t ⇡)
(t + 1)(t ⇡)
ln(t)
cos(3t)
and g(t) =
.
(t + 1)(t ⇡)
(t + 1)(t ⇡)
So p has discontinuities at t = 1, ⇡ (since t = 1 is not in the domain of ln) and g has
discontinuities at t = 1, ⇡. Since the natural logarithm appears in p, we draw the the line
starting from 0 to infinity (not the whole real numbers) reflecting the fact that its domain is
only the set of all positive real numbers.
with p(t) =
Therefore the largest such interval is (0, ⇡).
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
57
Example 47 (Summer 2010 Midterm Exam I). Consider the initial value problem
(t2
16)y 0
ln(t
1)y = sin(3t),
y(⇡) = 5.
Without solving the equation, what is the largest interval in which a unique solution is
guaranteed to exist?
Solution Again write the equation in standard form,
y0
ln(t 1)
sin(3t)
y=
,
(t + 4)(t + 4)
(t + 4)(t + 4)
ln(t 1)
sin(3t)
and g(t) =
.
(t + 4)(t + 4)
(t + 4)(t + 4)
Note that the domain of ln(t 1) is t > 1. So p has discontinuities at t =
is not in the domain of ln(t 1)) and g has discontinuities at t = 4, 4.
with p(t) =
4, 4 (since t =
4
By the Existence and Uniqueness Theorem , the largest interval in which a unique solution is
guaranteed to exist is (1, 4).
The Existence and Uniqueness Theorem requires a linear di↵erential equation. There is a also
similar theorem for non-linear first order di↵erential equations.
Theorem 2. Let the functions f and @f
<
@y be continuous in some rectangle ↵ < t < ,
y < containing the point (t0 , y0 ). Then, in some interval t0 h < t < t0 + h contained in
↵ < t < , there is a unique solution y = (t) of the initial value problem
y 0 = f (t, y)
y(t0 ) = y0 .
Example 48. Determine all possible solutions to the following IVP,
1
y0 = y 3 ,
y(0) = 0.
Solution First, notice that the given nonlinear di↵erential equation does not meet the requirement of the theorem above. We have
1
f (t, y) = y 3
)
@f
1
=
2 .
@y
3y 3
Even though f is continuous everywhere but its derivative with respect to y is not continuous
at y = 0 and so will not be continuous at any interval containing y = 0.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
58
Now let’s solve this problem by using the method of separable equation,
Z
Z
1
1
dy
3 2
= y3 )
y 3 dy =
dt )
y 3 = t + c.
dt
2
Apply initial condition, y(0) = 0 gives c = 0 and so the particular solution is
3 2
y 3 = t.
2
Write it in explicit form,
3 2
y3 = t
2
)
2
y = t
3
2
3
)
2
y = ( t)3
3
2
)
y=±
r
2
( t)3 .
3
Both of them also satisfy the initial condition. Also there is a third solution to this IVP,
namely y = 0, which satisfy both the given equation and the initial condition.
The previous example shows that the IVP does not always give the unique solution.
Example 49. Which of the following initial value problems has more than one solution?
(a) y 0 = 2y,
1
(b) y 0 = y 3 ,
y(0) = 0
y(0) = 0
(c) y 00 + ty 0 + 2y = 0,
(d) (1 + t2 )y 0 + ty = 0,
y(0) = 0,
y 0 (0) = 0
y(0) = 0
Solution Every equation has a unique solution by the Existence and Uniqueness Theorem
except the one in (b). It is nonlinear.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
59
Exercises 2.2
1. Consider the initial value problem
(t2 + t)y 0 +
1
(t
4)
y = e2t ,
y(2) = 3⇡.
According to the Existence and Uniqueness Theorem, what is the largest interval in which
a unique solution is guaranteed to exist?
(a)
(b)
(c)
(d)
(4, 1)
( 1, 4)
(0, 4)
(1, 4)
2. Consider the initial value problem
(t2
4)y 0 + 3y = ln |5
t|,
y(3) = 0.
According to the Existence and Uniqueness Theorem, what is the largest interval in which
a unique solution is guaranteed to exist?
(a)
(b)
(c)
(d)
(2, 1)
( 2, 2)
(2, 5)
( 1, 5)
3. The initial value problem
(4
t2 )y 0 + ln(t)y = sin(t),
y(1.33) = 3.14159
is certain to have a unique continuous solution on the interval.
(a)
(b)
(c)
(d)
( 4, 4)
( 2, 2)
(0, 2)
(0, 3.14159)
4. Consider the initial value problem
(t2
9)y 0 + (2t sin t)y = ln(1
t),
y(0) = 4.
According to the Existence and Uniqueness Theorem, what is the largest interval in which
a unique solution is guaranteed to exist?
(a)
(b)
(c)
(d)
( 3, 3)
(3, 1)
(1, 3)
( 3, 1)
Answers
1. (c)
2. (c)
3. (c)
4. (d)
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.3
60
Stability of Equilibrium Solutions
Recall that
• The first order di↵erential equation in the form
dy
= f (y)
dt
(f is free from independent variable t),
is called autonomous.
• Equilibrium solutions or critical points are constant functions y satisfying y 0 = f (y) = 0.
What we are about to study will tell us about the long term behavior of y without solving the
equation like direction filed. But it’s quicker and more e↵ective.
Given an autonomous equation,
dy
= f (y).
dt
• First, let’s draw the graph of f on yf (y) plane.
• The y-axis along with the arrows (which we’ll talk about it shortly) is called phase line.
• It is more customary to draw it in vertical orientation.
• Here’s how you draw the arrow. On any interval where f (y) > 0, y is increasing and we
denote this fact by drawing the right arrow (upward arrow for vertical version). Similarly,
on interval where f (y) < 0, y is decreasing we draw left arrow (downward arrow for vertical
version).
Take a look at this example below.
Figure 2.1: Left: Graph of f (y) and phase line, Right: Phase line in vertical orientation.
As you notice from the right figure, the number 1 and 3 appear on the axis reflecting the fact
that y = 1 and y = 3 are equilibrium solutions of the given autonomous equation. In general,
there are three types of stability of each equilibrium solutions based on the picture of phase line.
Let y = a be an equilibrium solution.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
In our example, y = 3 is (asymptotically) stable while y =
61
1 is unstable.
Example 50 (Spring 2004 Midterm Exam I).
For the autonomous equation y 0 = y 3 + 2y 2
3y.
a) Find all equilibrium solutions.
Solution To find equilibrium solutions, we set y 0 = 0. Hence
0 = y 3 + 2y 2
So y =
3y = y(y 2 + 2y
3) = y(y + 3)(y
1).
3, 0, 1 are equilibrium solutions.
b) Determine the stability of each equilibrium solution you find in part a).
Solution
• Method 1: Phase line
Now f (y) = y(y + 3)(y 1). To consider the sign for each interval, we simply pick
the sample point and evaluate f at that point.
So y = 1 is unstable, y = 0 is (asymptotically) stable and y =
• Method 2: Checking f 0 . Since f (y) = y 3 + 2y 2
f 0 (1) = 3 + 4
0
f (0) = 0 + 0
f 0 ( 3) = 27
3>0
3<0
12
3>0
3y, f 0 (y) = 3y 2 + 4y
)
y = 1 is unstable
)
y=
)
3 is unstable.
3.
y = 0 is stable
3 is unstable
c) Let y(t) be the solution whose initial condition is y(0) = 4. What is the behavior of y(t)
as t ! 1?
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
62
Solution Since 4 lies between the interval (1, 1),
lim y(t) = 1,
t!1
from phase line above.
d) Let y(t) be the solution whose initial condition is y(0) =
y(t) as t ! 1?
Solution Since
1. What is the behavior of
1 lies between the interval ( 3, 0),
lim y(t) = 0,
t!1
from phase line above.
e)* Let y(t) be the solution whose initial condition is y(100) =
y(t) as t ! 1?
Solution Since
⇡. What is the behavior of
⇡ lies between the interval ( 1, 3),
lim y(t) =
t!1
1,
from the phase line.
f )* If y(⇡) =
3, then find y(338). What is the behavior of y(t) as t ! 1?
Solution Since 3 is an equilibrium solution, the value of y is never change. In other
words, y(t) = 3 for all t. In particular y(338) = 3 and so lim y(t) = 3.
t!1
Example 51 (Spring 2002 Midterm Exam I).
Let y(t) be the solution of the initial value problem
dy
= y3
dt
y,
y(0) =
9
.
10
Then lim y(t) is equal to
t!1
(a) 1
(b)
9
10
(c) 1
(d) 0
(e)
1
Solution Notice that the question ask not for y but for the limit of y when t approaches
infinity. This is exactly when phase line can be used. The di↵erential equation is autonomous
with
f (y) = y 3 y = y(y 2 1) = y(y 1)(y + 1).
So we get the phaseline as follows.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
63
9
Since the initial condition gives the value of y0 =
which lies between 0 and 1, by the
10
phaseline,
lim y(t) = 0.
t!1
So (d) is the correct answer.
Example 52 (Summer 2006 Midterm Exam I).
Consider the first order autonomous equation
dy
= (y
dt
6)2 (y 2
25).
a) Find all equilibrium solutions.
Solution We can find equilibrium solutions by letting y 0 = 0 which is
0 = (y
So y =
6)2 (y 2
25) = (y
6)2 (y
5)(y + 5).
5, 5, 6 are equilibrium solutions.
b) For each equilibrium solution, classify its stability. Justify your answer.
Solution We gonna use the phase line to determine the stability of each equilibrium
solution. In this example, we have
f (y) = (y
6)2 (y
5)(y + 5).
Therefore y = 6 is semistable, y = 5 is unstable and y =
5 is (asymptotically) stable.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
64
c) If y(1) = 0, what is lim y(t)?
t!1
Solution Since 0 lies between the interval ( 5, 5),
lim y(t) =
t!1
5.
d) Suppose y(0) = y0 and lim y(t) = 5. Find the value(s) of y0 .
t!1
Solution If y(0) = 5, since 5 is an equilibrium solution, lim y(t) = 5. If the value of
t!1
y(0) is not 5, then from the phase line we can see that lim y(t) is never equal to 5. So
t!1
y0 = 5 only.
e)* Suppose y(0) = y0 and lim y(t) =
t!1
Solution If y(0) =
5, since
5. Find the value(s) of y0 .
5 is an equilibrium solution, lim y(t) =
t!1
5. If the
value of y(0) lies between either ( 1, 5) or ( 5, 5), then from the phase line we can
see that lim y(t) is also 5. So y0 < 5 or ( 1, 5).
t!1
f )* Suppose y(0) = y0 and lim y(t) = 6. Find the value(s) of y0 .
t!1
Solution Of course if y(0) = 6, lim y(t) = 6 since 6 is an equilibrium solution. If the
t!1
value of y(0) is from 5 to 6, then from the phase line we can see that lim y(t) is also
t!1
6. On the other hand if y(0) lies from 6 to 1, from the phase line, lim y(t) = 1.
t!1
Therefore 5 < y0  6 or (5, 6].
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
65
Exercises 2.3
1. For the following equation,
dy
= (y + 1)(y 2)(1 y).
dt
Determine and classify the critical points (equilibrium solutions).
2. Consider the autonomous di↵erential equation
y 0 = (y
1)(y
2)(y + 3)2 .
(a) Find all of its equilibrium solutions.
(b) Classify the stability of each equilibrium solution. Clearly explain how you have obtained your answer.
(c) If y( 3) =
1, then what is lim y(t)?
t!1
(d) If y(500) = 2, then what is y(2500)? Without solving the equation, briefly explain your
conclusion.
3. Consider the autonomous di↵erential equation
y 0 = 3y(y + 2)2 (4
y).
(a) Find all of its equilibrium solutions.
(b) Classify the stability of each equilibrium solution. Justify your answer.
(c) If y( 2) =
1, then what is lim y(t)?
t!1
(d) If y(10) = 0, then what is y(1000)? Without solving the equation, briefly explain your
conclusion.
4. Consider the solution y(t) of the initial value problem
y0
2y = 2,
y(0) = 2.
What is lim y(t)?
t!1
(a)
(b)
1
1
(c) 2
(d) 1
5. Consider the following autonomous di↵erential equation
y 0 = cos(y).
The function y(t) =
5⇡
is
2
(a) a semi-stable equilibrium solution.
(b) an unstable equilibrium solution.
(c) an asymptotically stable equilibrium solution.
(d) not an equilibrium solution.
6. Consider the autonomous di↵erential equation
y0 =
y 4 + 16y 2 .
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
66
(a) Find all of its equilibrium solutions.
(b) Classify the stability of each equilibrium solution. Justify your answer.
(c) If y(t) is a solution that satisfies y( 1) = 4, what is y(0)? Without solving the
equation, briefly explain your conclusion.
(d) If y(t) is a solution that satisfies y(3) = 3, then what is lim y(t)?
t!1
7. Consider the autonomous equation
y 0 = 9y 2
y 4 = y 2 (3
y)(3 + y)
Answer the following questions without solving the equation.
(a) Find all equilibrium solutions.
(b) Classify the stability of each equilibrium solution. Justify your answer.
22
(c) If y( ) = ⇡, what is lim y(t)?
t!1
7
(d) If y(2⇡) = 3, find y(t).
(e) If y(4) = . For what value (or range of values) of would lim y(t) = 0?
t!1
8. Consider the autonomous equation
y0 = y3
12y 2 + 20y = y(y
2)(y
10).
⇡
Suppose y( ) = 10. What is lim y(t)?
t!1
2
(a)
(b)
(c)
(d)
0
2
10
1
9. Consider the autonomous di↵erential equation
y 0 = (y
1)2 (y 2
4).
(a) Find all of its equilibrium solutions.
(b) Classify the stability of each equilibrium solution. Justify your answer.
(c) If y(1325) = 1.5, then what is lim y(t)?
t!1
(d) If y( 1325) = 2, then what is lim y(t)?
t!1
Answers
1. The equilibrium solutions are: y =
1(stable), y = 1(unstable), and y = 2(stable).
2. (a) y =
3, 1, 2, (b) semistable, stable, unstable, (c) lim y(t) = 1, (d) y(2500) = 2
3. (a) y =
2, 0, 4, (b) semistable, unstable, stable, (c) lim y(t) =
t!1
t!1
2, (d) y(1000) = 0
4. (d)
5. (c)
6. (a) y =
4, 0, 4, (b) unstable, semistable, stable, (c) y(0) =
7. (a) y =
3, 0, 3, (b) unstable, semistable, stable, (c) 3, (d)
4, (d) lim y(t) = 0
t!1
3, (e)
3<
0
8. (c)
9. (a) y =
2, 1, 2, (b) stable, semistable, unstable, (c) lim y(t) = 1, (d) lim y(t) = 2
t!1
t!1
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.4
67
Modeling with First Order Di↵erential Equation
Modeling is the process of writing a di↵erential equation to describe a physical situation. One
of the modeling with first order di↵erential equation is mixing problems. Here is the set up:
2.4.1
Mixing Problems
Figure 2.2: Mixing problems
• A holding tank initially contains amount of substance dissolved in certain amount of liquid.
• The same substance is also dissolved in a liquid which will be entering the tank.
• Assume that the content of the mixing tank is stirred very quickly such that the solution
within is always of uniform concentration.
• The mixed content is then leaving the tank.
• We want to develop a di↵erential equation that can predict the amount of the substance
dissolved in the tank at any time.
Let Q(t) be the amount of the substance dissolved in the liquid in the tank at any time t. Then
we have,
Rate of change of Q(t) = Rate at which Q(t) enters the tank
Rate at which Q(t) exits the tank
Note here that
• Rate of change of Q(t) is
dQ
or Q0 (t).
dt
• Rate at which Q(t) enters the tank
=(Flow rate of liquid entering)⇥(concentration of substance in liquid entering)
= ri c i
• Rate at which Q(t) exits the tank
=(Flow rate of liquid exiting)⇥(concentration of substance in liquid exiting)
Amount of substance in the tank at time t
=(Flow rate of liquid exiting)⇥
Volume of liquid in the tank at time t
Q(t)
= ro
,
V0 + (ri ro )t
where V0 is the initial volume of liquid in the tank.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
Combine everything altogether, we have
Modeling for Mixing Problems
Q0 (t) = ri ci
ro
Q(t)
,
V0 + (ri ro )t
Q(0) = Q0
as the initial value problem for mixing problems where
• Q(t) = the amount of the substance dissolved in the liquid in the tank at time t.
• ri = flow rate of liquid entering (flow rate in).
• ci = concentration of substance in liquid entering (conc. in).
• ro = flow rate of liquid exiting (flow rate out).
• V0 = the initial volume of liquid in the tank.
• Q(0) = the initial amount of the substance dissolved in the liquid in the tank.
Example 53 (Spring 2011 Midterm Exam I).
A tank initially contains 200 gallons of water with a salt concentration of 0.5 oz/gal. Water
containing a salt concentration of 2 + sin t oz/gal flows into the tank at a rate of 5 gal/min
and the mixture in the tank flows out at the same rate. Let Q(t) be the amount of salt in the
tank at time t. Find the initial value problem which accurately describes the situation.
Solution First, let Q(t) be the amount of salt dissolved in the water in the tank at time t
(oz). From the information above, we have
• V0 = 200 gal.
• Q(0) = 0.5 oz/gal ⇥200 gal = 100 oz. Here we use the fact that
amount of substance
volume of the liquid
) amount of substance = concentration of substance ⇥ volume of the liquid
concentration of substance =
• ci = 2 + sin t oz/gal.
• ri = 5 gal/min.
• ro = 5 gal/min.
Substitute everything back into modeling for mixing problem. Then we have the initial value
problem describing the situation above as
Q0 (t) = 5(2 + sin t)
5
Q(t)
,
200 + (5 5)t
which is
Q0 (t) = 5(2 + sin t)
Q(t)
,
40
Q(0) = 100
Q(0) = 100.
68
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
69
Example 54 (Spring 2008 Midterm Exam I).
A 1000-gallon above-ground swimming pool is initially filled with 800 gallons of rain water.
Water containing 5 g/gal of Chlorine flows into the pool at a rate of 6 gallons per minute.
The well-mixed Chlorine solution is pumped out at a rate of 3 gallons per minute. Let Q(t)
denote the amount of Chlorine in the pool at any time t > 0 and before the pool eventually
overflows. Write down an initial value problem that Q(t) must satisfy.
Solution From the problem, we get
• V0 = 800 gal (not 1000 which is the maximum capacity).
• Q(0) = 0 g (rain water).
• ci = 5 g/gal.
• ri = 6 gal/min.
• ro = 3 gal/min.
• t is at most
200
min. (After that, the pool will overflow the swimming pool).
3
Therefore we have the initial value problem describing the situation above as follow
Q0 (t) = 6(5)
3
Q(t)
,
800 + (6 3)t
which is
Q0 (t) = 30
3
Q(t)
,
800 + 3t
where 0 < t 
Q(0) = 0
Q(0) = 0,
where 0 < t 
200
3
200
.
3
Example 55 (Spring 2009 Midterm Exam I).
A 100-liter vat initially contains 80 liters of 2 grams/liter sodium hydroxide solution. At
t = 0, sodium hydroxide solution with a concentration of 5 grams/liter begins to flow into the
vat at the rate of 2 liters/min. The thoroughly mixed content of the vat is drawn o↵ at the
rate of 3 liters/min. Find the initial value problem which best describe the quantity of sodium
hydroxide, Q(t), in the vat at time t, 0 < t < 80.
Solution We have
• V0 = 80 l (not 100 which is the maximum capacity).
• Q(0) = 2 g/l ⇥80 l = 160.
• ci = 5 g/gal.
• ri = 2 gal/min.
• ro = 3 gal/min.
• t < 80 min (After that, the vat will be drained).
Then we get the IVP describe the situation as follows,
Q0 (t) = 2(5)
3
Q(t)
,
80 + (2 3)t
Q(0) = 160
where 0 < t < 80
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
which is
Q0 (t) = 10
3
Q(t)
,
80 t
Q(0) = 160,
70
where 0 < t < 80.
Example 56 (Fall 2004 Midterm Exam I).
A tank has 100 gal of water and 100 lb of salt mixed in it. Water enters the tank at the rate of
3 gal/min with concentration of salt in it, at time t given by e t lb/gal. A well mixed solution
leaves the tank at the same rate of 3 gal/min.
i) Find a formula for the amount of salt in the tank at any time t.
Solution Let Q(t) be the amount of salt dissolved in the water in the tank at time t
(lb). From the information, we have
• V0 = 100 gal.
• Q(0) = 100 lb.
• ci = e
t
lb/gal.
• ri = 3 gal/min.
• ro = 3 gal/min.
Thus, we have the IVP for Q,
Q0 (t) = 3(e t )
3
Q(t)
,
100 + (3 3)t
Q(0) = 100
which is
Q(t)
,
Q(0) = 100.
100
Now we’re about to solve this IVP by using the method of integrating factor. So let’s
write the equation in the standard form,
Q0 (t) = 3e
t
3
Q0 (t) + 3
Q(t)
= 3e t .
100
(2.22)
And the integrating factor is
µ(t) = e
R
p(t) dt
=e
R
3
100
dt
3t
= e 100 .
Multiply both sides of Equation 2.22 by µ and change the left hand side of the equation
to one term,
3t
3t
Q(t)
= 3e t e 100
100
3t
97t
(e 100 Q(t))0 = 3e 100
3t
e 100 Q0 (t) + 3e 100
d 3t
(e 100 Q(t)) = 3e
dt
3t
d(e 100 Q(t)) = 3e
97t
100
97t
100
dt
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
71
Now integrate both sides of the equation and simplify it to get Q,
Z
Z
3t
97t
d(e 100 Q(t)) = 3 e 100 dt
3t
e 100 Q(t) =
300
e
97
3t
100
Q(t) = (e
) Q(t) =
97t
100
300
e
97
)(
300
e
97
+c
t
97t
100
3t
100
+ ce
+ c)
.
Imposing the initial condition Q(0) = 100,
100 =
300
+c
97
10, 000
.
97
)
c=
+
10, 000
e
97
Hence the formula for Q(t) is
Q(t) =
300
e
97
t
3t
100
.
ii) Find also the eventual concentration of salt in the tank.
Solution Again we use the following relation,
amount of salt in the tank at time t
volume of the water in the tank at time t
Q(t)
=
V0 + (ri ro )t
Q(t)
=
100 + (3 3)(t)
Q(t)
=
100
concentration of salt in the tank at time t =
Then the eventual concentration of salt in the tank is
Q(t)
1
=
lim Q(t)
t!1 100
100 t!1
1
300
=
lim
e
100 t!1
97
= 0.
lim
t
+
10, 000
e
97
3t
100
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
72
Exercises 2.4.1
1. A tank is filled with 200 liters of a solution containing 100 grams of salt. A solution containing
a concentration of 2 g/liter salt enters the tank at the rate 4 liters/minute and the well-stirred
mixture leaves the tank at the same rate. Set up the initial value problem for the amount of
salt in the tank at time t, find the particular solution and find the limiting amount of salt
in the tank as t ! 1.
2. A tank originally contains 100 liters of pure water. Salt water with a concentration of 2
kg/L is pumped into the tank at 3 L/min, and the well-mixed solution is drained at the
same rate.
(a) Set up an initial value problem describing the situation. Be sure to explain all of your
variables.
(b) Solve the initial value problem to find the amount of salt in the tank at any time t.
(c) What is the limiting concentration of salt in the tank?
3. A 800-gallon tank initial contains 500 gallons of water and 30 pounds of salt dissolved in
it. Water enters the tank at the rate of 3 gal/min with concentration 4 lb/gal of salt in it.
The well mixed solution leaves the tank at the rate of 1 gal/min. Which of the initial value
problems below models the change of the amount of salt Q(t) inside the tank during the
time interval 0  t  150?
Q(t)
, Q(0) = 30
800
Q(t)
(b) Q0 (t) = 12
, Q(0) = 800
500 2t
Q(t)
(c) Q0 (t) = 12
, Q(0) = 30
500 + 2t
Q(t)
(d) Q0 (t) = 12
, Q(0) = 500
500 + t
(a) Q0 (t) = 3
4. A jar contains a sugar solution. Initially it had 2 liters of water and 15 grams of dissolved
sugar. Water containing 5 grams of sugar per liter enters the jar at a rate of 2 liters/min.
The well stirred mixture flows out at the same rate. How many grams of dissolved sugar is
present in the jar after ln 5 minutes?
5. A 400-liter tank is initially filled with 100 liters of dye solution with a dye concentration 5
g/l. Pure water flows into the tank at a rate 3 liters per minute. The well-stirred solution
is drained at a rate of 2 liters per minute. Find the concentration of dye in the tank at the
time that the tank is completely filled.
6. A 200 m3 room initially contains fresh air. At t = 0, a faulty heating system causes gas
containing 20% carbon monoxide to be pumped into the room at a rate of 3 m3 per minute.
The well-mixed air is vented out at the same rate.
(a) Write a di↵erential equation, and give the initial condition, that describe this event.
(b) Solve the initial value problem.
(c) A carbon monoxide detector in the room is triggered when the carbon monoxide reaches
1%. Find the time when the detector will sound the alarm.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
73
7. A swimming pool is initially filled with 500 m3 of water with a chlorine concentration of 10
g/m3 . Water containing 50 g/m3 of chlorine flows into the pool at a rate of 5 m3 per minute.
The well-mixed water in the pool is drained away at the same rate.
(a) Find the amount of chlorine in the pool at anytime t > 0.
(b) What is the concentration of chlorine in the pool as t ! 1
8. A tank initially contains 120 liters of pure water. A salt solution with a concentration of
grams/liter of salt enters the tank at a rate of 2 liters/min and the well-stirred mixture
leaves the tank at the same rate. Find (in terms of ) an expression for the amount of salt
in the tank at any time t and the limiting amount of salt in the tank as t approaches infinity.
9. A swimming pool is initially filled with 400 m3 of fresh water. At t = 0 water containing
50 g/m3 of chlorine starts to flow into the swimming pool at a rate of 2 m3 per minute.
Well-mixed water is drained from the pool at the same rate.
(a)
(b)
(c)
(d)
Set up an initial value problem modeling this process.
Solve the initial value problem.
Find the time when the chlorine concentration within the pool reaches 25 g/m3 .
(As t ! 1) what is the limiting amount of chlorine that will be in the swimming pool?
10. A culinary experiment that went horribly awry has filled a 60 m3 kitchen with air that
contains 2 g/m3 of smoke and soot. At t = 0, the ventilation system is switched on so that 3
m3 /min of fresh air is pumped in. The well-mixed smokey air is drawn o↵ at the same rate.
(a) Let Q(t) denote the amount of smoke and soot in the air at any time t > 0. Write down
an initial value problem that Q(t) must satisfy.
(b) Solve the initial value problem to find Q(t).
(c) How much time would it take for the concentration of smoke and soot in the air to go
1
down to 10
of its original level?
Answers
0
1. The initial value problem is Q = 8
And lim Q(t) = 400.
Q
, Q(0) = 100. The solution is Q(t) = 400
50
t
300e 50 .
t!1
3t
200e 100 , iii) The limiting concentration is
3Q
, Q(0) = 0, ii) Q(t) = 200
100
Q(t)
200
lim
=
= 2.
t!1 100
100
3. c)
2. i) Q0 = 6
4. 11
5
64
3t
200
20
40e 200 , c) t =
(ln )
3
19
5. The tank is filled at t = 300 then the concentration at that moment is
6. a) Q0 = 0.6
3
Q,
200
7. a) Q = 25, 000
20, 000e
8. a) Q = 120
120 e
9. a) Q0 = 100
1
Q,
200
10. a) Q0 =
Q
,
20
Q(0) = 0, b) Q(t) = 40
1
60 t
1
100 t
, b) 50
, b) 120
Q(0) = 0, b) Q(t) = 20, 000
Q(0) = 120, b) Q(t) = 120e
t
20
20, 000e
, c) 20 ln 10
t
200
, c) t = 200 ln 2, d) 20, 000
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
2.4.2
74
Falling Object/Motion of an Object in a Resistive Fluid Medium*
Suppose that an object is falling in the atmosphere near sea level. Our goal is to formulate a
di↵erential equation that describes the motion of this object. We begin by reminding ourself
about Newton’s second law. It states that the mass of object times its acceleration is equal to the
net force on the object. Therefore this laws is expressed mathematically by the equation
F = ma,
where m is the mass of the object, a is its acceleration, and F is the net force exerted on the
object. Also a is related to v by a = v 0 , so we can rewrite the equation above in the form
F = mv 0 .
(2.23)
Next, let’s consider the forces that act on the object as it falls. Gravity exerts a force equal to
the weight of the object, or mg, where g is the acceleration due to gravity. There is also a force
due to air resistance, or drag. It is often assumed that the drag is proportional to the velocity
and we make this assumption here. So the drag force is v, where is a constant called the drag
coefficient. This constant varies widely from one object to another.
Figure 2.3: Free-body diagram of the forces on a falling object.
Remember that gravity acts in the downward (positive) direction, whereas drag acts in the upward
(negative) direction, as show in figure above. Thus
F = mg
v.
(2.24)
Now we combine equations (2.23) and (2.24) together then becomes
mv 0 = mg
v.
This equation is a mathematical model of an object falling in the atmosphere near sea level where
m, g and are all constants and it can be solved by using the method of integrating factor.
In conclusion, we have
Modeling for Falling Object
mv 0 = mg
v,
v(0) = v0 ,
where
• v(t) = the velocity of the falling object at time t. (m/s)
• m = the mass of the object. (kg)
• g = the acceleration due to gravity. (9.8 m/s2 or 10 m/s2 )
•
= drag coefficent. (kg/s)
• v0 = the initial velocity of the object.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
75
Example 57 (Spring 2002 Midterm Exam I).
A ball of mass 2 kg is dropped from rest in a viscous liquid. As a result the ball experiences
a drag force which is twice the magnitude of its velocity. Find the distance the ball travels in
the first 2 seconds of its motion. Assume that g = 10 m/sec2 .
Solution We have
With m = 2 kg,
mv 0 = mg
2v 0 = 2g
2v
2v,
v(0) = 0.
v0 = g
)
)
v
v 0 + v = 10.
Then an integrating factor is
µ(t) = e
R
p(t) dt
=e
R
1 dt
= et .
Multiply both sides of the equation by µ(t) = et ,
et v 0 + et v = 10et
)
Z
Z
Then
t
d(e v) =
(et v)0 = 10et
10et dt
)
)
d(et v)
= 10et .
dt
et v = 10et + c.
Now we divide both sides of the equation by et , then the general solution is
v(t) = ce
Impose the initial condition v(0) = 0, c =
v(t) =
t
+ 10.
10. Therefore we have
10e
t
+ 10.
So the distance the ball travels in the first 2 seconds is
Z 2
Z 2
2
v(t) dt =
( 10e t + 10) dt = 10e t + 10t 0 = (10e
0
2
+ 20)
(10) = 10e
2
+ 10.
0
Limiting Velocity As the name suggests, it is the eventual velocity of the object. So its definition
is
vL = lim v(t).
t!1
In order to find it, we don’t actually need to solve the di↵erential equation to find v(t) and then
take the limit.
Notice that vL is also the maximum velocity achievable. This is true since we give infinite amount
of time to the object to accelerate. So, since vL is the maximum velocity, it must occurs at a
critical point of v(t). Therefore in order to find vL , we simply
set v 0 = 0
)
solve for vL .
Example 58. Find the limiting velocity of the a ball described in the previous example.
Method 1: Use the definition. We showed in the above example that
v(t) =
10e
t
+ 10.
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
76
Therefore the limiting velocity vL is
vL = lim v(t) = lim ( 10e
t!1
t
t!1
+ 10) = 10.
Method 2: Set v 0 = 0. From the above example, we derive the equation as
v 0 + v = 10.
By setting v 0 = 0, we have vL = 10.
In general, by setting v 0 = 0, we easily find the limiting velocity as follows:
mv 0 = mg
v
)
0 = mg
)
v
vL =
mg
.
In conclusion, we have
Modeling for Falling Object
mv 0 = mg
v,
v(0) = v0 ,
where
• v(t) = the velocity of the falling object at time t. (m/s)
• m = the mass of the object. (kg)
• g = the acceleration due to gravity. (9.8 m/s2 or 10 m/s2 )
•
= drag coefficent. (kg/s)
• v0 = the initial velocity of the object.
The limiting velocity is
vL =
mg
.
Example 59 (Spring 2013 Midterm Exam I).
An object with mass m = 0.5 kg is thrown upward with initial velocity v0 = 20 meters per
second from the roof of a building which is 30 meters high. Assume there is a force due to
air resistance that is proportional to the velociy v of the object with a positive constant of
1
proportionality (coefficient of drag) = 10
. You may use g = 10 meters per second squared
as the gravitational constant.
a) Construct an initial value problem that models for velocity of the object in motion at
any time t.
Solution We have m = 0.5 kg, v0 = 20 m/s,
describing the motion is
mv 0 = mg
v
)
0.5v 0 = (0.5)(10)
with the initial condition v(0) =
upward.)
=
1
10
and g = 10 m/s2 . So the IVP
1
v
10
)
5v 0 = 50
v,
20. (It is negative because the object is thrown
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
77
b) Solve the initial value problem to find v(t).
Solution We’re gonna solve this di↵erential equation by using the method of integrating
factor. Firstly, we rewrite it in the standard form.
5v 0 = 50
v
5v 0 + v = 50
)
v0 +
)
v
= 10.
5
So the integrating factor is
µ(t) = e
R
p(t) dt
1
t
= e 5 dt = e 5 .
Then we multiply both sides of the equation by µ(t),
t
t
v t
e 5 v 0 + e 5 = 10e 5
5
Hence
Z
t
5
d(e v) =
Z
)
t
10e 5 dt
)
t
t
(e 5 v)0 = 10e 5 .
t
t
e 5 v = 50e 5 + c.
To obtain the particular solution, we impose the initial condition v(0) =
t
t
e 5 v = 50e 5 + c
Therefore
t
)
t
e 5 v = 50e 5
70
20 = 50 + c
)
)
v(t) = 50
c=
70e
t
5
20,
70.
.
c) Suppose the motion could indefinite. To what value would the object’s velocity approach
eventually.
Solution We can either find it directly from the definition
vL = lim v(t) = lim (50
t!1
t!1
70e
t
5
) = 50
or use the formula we derived above
vL =
mg
=
(0.5)(10)
1
10
= 50.
Note that for small, slowly falling objects, the assumption that the drag force is proportional to
the velocity is a good one. For larger, more rapidly falling objects, it is more accurate to assume
that the drag force is proportional to the square of the velocity.
Example 60 (Fall 2013 Midterm Exam I).
The velocity, given in meters per second, of a certain particle is given by the initial value
problem
dv
1 2
= 1000
v , v 0, v(0) = 5.
dt
40
Approximately how fast will the particle be moving after a very long time?
Hint: the answer can be deduced without having to solve the initial value problem.
(a) 25 m/s
(b) 200 m/s
(c) 40000 m/s
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
78
(d) 1
Solution To find the limiting velocity, we set v 0 = 0. Therefore
1000
1 2
v
40
)
v2
)
40000 = 0
(v
200)(v + 200) = 0
)
⇠
⇠
vL = 200,⇠200.
Example 61 (Spring 2010 Midterm Exam I).
Suppose the velocity v(t) of a speedboat is given by the equation
v 0 = 200
1 2
v ,
2
v
0.
What is its limiting velocity, lim v(t)?
t!1
(a) 20
(b) 100
(c) 400
(d) 1
Solution To find the limiting velocity, we set v 0 = 0. Therefore
200
1 2
v =0
2
)
v2
400 = 0
)
(v
20)(v + 20) = 0
)
vL = 20, 20.
Next we consider the motion of an object in a resistive fluid medium. According to fluid dynamics,
the drag force exerted on an object moving in fluid medium is proportional to the square of its
speed. Therefore the equation that models the motion of an object in a resistive fluid medium is
mv 0 = p
v2 ,
v
0,
p > 0,
where p is a propulsive force (gravity, thrust of an engine, for examples.) Here m, p and are
all constants. This is a nonlinear 1st order di↵erential equation which can be solved by using the
separable equation method:
mv 0 = p
v2
)
m
dv
=p
dt
v2
)
m
p
v2
dv = 1 dt.
Then we integrate on both sides of the equations and solve for v. Also, in this modeling, we can
find the limit velocity by again setting v 0 = 0.
mv 0 = p
In conclusion, we have
v2
)
0=p
v2
)
v2 =
p
)
vL =
r
p
,
r
p
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
Modeling for Motion of an Object in a Resistive Fluid Medium
mv 0 = p
v2 ,
v(0) = v0 ,
where
• v(t) = the velocity of the falling object at time t. (m/s)
• m = the mass of the object. (kg)
• p = the propulsive force. (N)
•
= drag coefficent. (kg/s)
• v0 = the initial velocity of the object.
The limiting velocity is
vL =
r
p
.
Example 62. A 100 kg Unmanned Aerial Vehicle (UAV) possesses propulsive force of 10, 000
N and has drag coefficient k = 4. Find the velocity function of its flight in explicit form.
Solution Here we have
• m = 100 kg
•
=4
• p = 10, 000 N
• v(0) = 0 as the initial condition.
So we have the initial value problem,
100v 0 = 10000
Dividing out by 4, we have
4v 2 ,
25v 0 = 2500
v(0) = 0.
v2 .
Note that the limiting velocity is 50. This follows by setting v 0 = 0,
0 = 2500
v 2 = (50
v)(50 + v)
)
vL = 50, 50.
So the range of the object’s velocity is from 0 to 50 or 0  v  50. To solve it, we use the
method of separable equation,
Z
Z
dv
25
25
25
= 2500 v 2 )
dv = dt )
dv =
dt.
dt
(50 v)(50 + v)
(50 v)(50 + v)
By using the partial fraction decomposition,
(50
1
1
25
4
4
=
+
.
v)(50 + v)
50 v 50 + v
79
CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS
80
Therefore
Z
1
4
50
v
+
1
4
50 + v
dv =
Z
dt
1
ln |50
4
)
v| +
1
ln |50 + v| = t + c.
4
We apply the initial condition, v(0) = 0, here
1
1
ln 50 + ln 50 = 0 + c
4
4
)
c = 0.
Hence the particular solution is
1
ln |50
4
v| +
1
ln |50 + v| = t.
4
Now let’s write it explicitly:
ln |50
v| + ln |50 + v| = 4t
Since 0  v  50,
ln
Consequently,
ln
✓
50 + v
= ln
50 v
50 + v
50 v
◆
= 4t
✓
)
)
50 + v
50 v
50 + v
= 4t.
50 v
ln
◆
.
50 + v
= e4t .
50 v
Simplify further to get the expression of y,
50 + v = 50e4t
ve4t
)
ve4t + v = 50e4t
Hence
v(t) =
50
)
(e4t + 1)v = 50(e4t
50(e4t 1)
.
e4t + 1
(Notice also that
50(e4t 1)
= 50.)
t!1
e4t + 1
vL = lim v(t) = lim
t!1
1).
Chapter 3
Second Order Di↵erential
Equations
In this chapter we will focus only on second order linear di↵erential equations. Its most general
form is
p(t)y 00 + q(t)y 0 + r(t)y = g(t),
where the coefficients p, q, r and g are functions of t.
Homogeneous vs Nonhomogeneous:
• A second order linear equation is said to be homogeneous if g(t) = 0 for all t.
• Otherwise, the equation is called nonhomogeneous.
Example 63 (Summer 2003 Midterm Exam I).
Which of the following is a second order, linear, homogeneous di↵erential equation?
a) y 00 + y 2 = 0
b) y 00 + y 0 + 2y = t ln y
c) ty 00 + y = 0
d) (y 0 )2
t2 y = 1
Solution The equation in c) is the right answer since di↵erential equations in a), b) and d)
are nonlinear by the terms y 2 , ln y and (y 0 )2 respectively. Notice that the equation in c) is
homogeneous since the right-hand side of the equation is zero.
For the nonhomogeneous equation, we will consider only when all of the coefficients p, q and r are
real constant. So we will study the following di↵erential equation,
ay 00 + by 0 + cy = g(t),
where a 6= 0, b, c are arbitrary real numbers.
81
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.1
82
Fundamental Set of Solutions & Wronskian
Let’s consider the following example.
Example 64. Determine some solutions to
y 00
y = 0.
Solution It is easy to see that et is a solution and so is c1 et . Another solution is e
is c2 e t . We can check further that any function in the form
y = c 1 et + c 2 e
t
and so
t
is a solution of the above di↵erential equation. In fact, it is a general solution which we will
show later.
This example guides us to this fact:
Theorem 3 (Principle of superposition).
equation
If y1 and y2 are solutions of the homogeneous
y 00 + p(t)y 0 + q(t)y = 0,
then so is c1 y1 + c2 y2 for arbitrary constant c1 and c2 .
Also it is worth noting that
• Unlike 1st order di↵erential equation, the general solution of 2nd order equation has two
arbitrary coefficients.
• To find a particular solution, we require two initial conditions y(t0 ) = y0 and y 0 (t0 ) = y00 .
As in the example above, it suggests us that the solution of ay 00 + by 0 + cy = 0 should be in the
form y = ert . Plug this function into the the equation,
a(ert )00 + b(ert )0 + c(ert ) = 0
)
a(r2 ert ) + b(rer t) + cert = 0
Since ert 6= 0,
)
ert (ar2 + br + c) = 0.
ar2 + br + c = 0.
• This equation is typically called the characteristic equation for ay 00 + by 0 + cy = 0.
• This is a quadratic equation and so we should expect two roots r1 and r2 . Once we have
these two roots, we have two solutions to ay 00 + by 0 + cy = 0
y1 (t) = er1 t and y2 (t) = er2 t .
• Apply the principle of superposition,
y = c 1 e r1 t + c 2 e r2 t
is a solution to ay 00 + by 0 + cy = 0.
So the only remaining question is that
Does it contain every possible solutions to ay 00 + by 0 + cy = 0?
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
83
It can be answered by the theorem below involving Wronskian determinant1 .
Theorem 4 (Fundamental Set of Solutions, Wronskian).
solutions of
Suppose that y1 and y2 are two
y 00 + p(t)y 0 + q(t)y = 0.
Then
y = c1 y1 + c2 y2 is the general solution of
(or the solutions y1 and y2 are said to form a fundamental set of solution of )
y 00 + p(t)y 0 + q(t)y = 0
if and only if the Wronskian of y1 and y2 , written as W (y1 , y2 ), is not a zero function where
W (y1 , y2 ) :=
y1
y10
y2
= y1 y20
y20
y10 y2 .
Example 65 (Fall 2000 Midterm Exam I).
Show y1 (t) = t2 and y2 (t) = t3 , t > 0 form a fundamental set of solutions for
t2 y 00
4ty 0 + 6y = 0.
Solution To be a fundamental set of solutions, not only they are solutions but also their
Wronskian is not a zero function. Let’s check both conditions down below:
• Verify that y1 (t) = t2 and y2 (t) = t3 are solutions to t2 y 00
t2 y100
4ty10 + 6y1 = t2 (t2 )00
4t(t2 )0 + 6(t2 ) = t2 (2)
t2 y200
4ty20 + 6y2 = t2 (t3 )00
4t(t3 )0 + 6(t3 ) = t2 (6t)
4ty 0 + 6y = 0.
4t(2t) + 6t2 = 2t2
4t(3t2 ) + 6t3 = 6t3
8t2 + 6t2 = 0
12t3 + 6t3 = 0.
• Verify a nonzero-function Wronskian.
W (t2 , t3 ) =
t2
2t
t3
= t2 (3t2 )
3t2
2t(t3 ) = 3t4
2t4 = t4 6= 0.
Therefore y1 (t) = t2 and y2 (t) = t3 form a fundamental set of solutions. Consequently, the
general solution to t2 y 00 4ty 0 + 6y = 0 is
y = c 1 y 1 + c 2 y 2 = c 1 t2 + c 2 t 3 .
1 Wronskian determinants are named for Josef Maria Hoene-Wronski (1776
1853), who was born in Poland
but spent most of his life in France. Wronski was a gifted but troubled man, and his life was marked by frequent
heated disputes with other individuals and institutions.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.2
84
Linear Homogeneous Equation with Constant Coefficients
Recall that the characteristic equation of a di↵erential equation ay 00 + by 0 + cy = 0 is
ar2 + br + c = 0,
which will gives us two roots:
b+
r1 =
p
b2
2a
4ac
,
b
r2 =
p
b2
2a
4ac
.
We’ll take a closer look on each 3 possible cases depend on the type of roots as follow:
1. Two distinct real roots (it happens when b2
4ac > 0).
2. Two complex conjugate roots (it happens when b2
3. Repeated real roots (it happens when b2
3.2.1
4ac < 0).
4ac = 0).
Two Distinct Real roots
Assume that r1 and r2 are two distinct real roots of the characteristic equation ar2 + br + c = 0.
This gives two solutions of ay 00 + by 0 + cy = 0,
y1 = e r1 t
and
y2 (t) = er2 t .
Compute their Wronskian,
W (y1 , y2 ) =
e r1 t
r1 e r 1 t
e r2 t
= r2 e(r1 +r2 )t
r2 e r 2 t
r1 e(r1 +r2 )t = (r2
r1 )e(r1 +r2 )t .
It is nonzero since r1 6= r2 and so is exponential function. By Theorem 4, y1 and y2 form the
fundamental set of solutions of ay 00 + by 0 + cy = 0. It means that the general solution is
y = c 1 e r1 t + c 2 e r2 t .
)
y = c 1 y1 + c 2 y 2
If r1 , r2 are distinct real roots of characteristic equation, then the general solution is
y = c 1 e r 1 t + c 2 e r2 t .
Example 66 (Spring 2002 Midterm Exam 1). Solve the IVP
y 00
4y = 0,
y(0) = 4,
Solution The characteristic equation of y 00
r2
4=0
)
(r
0
y (0) = 4
4y = 0 is
2)(r + 2) = 0
So the general solution is y(t) = c1 e2t + c2 e
Impose the initial conditions,
y(0) = 4
y 0 (0) = 4.
2t
)
)
)
r = 2, 2.
. Then y 0 (t) = 2c1 e2t
4 = c1 + c2
4 = 2c1
2c2 .
2c2 e
2t
.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
85
Then we have c1 = 3 and c2 = 1. So the particular solution is y = 3e2t + e
2t
.
Example 67 (Summer 2009 Midterm Exam 1).
Find the particular solution to
y 00
5y 0 + 4y = 0,
y(0) = 2,
y 0 (0) =
1.
What is the behavior of the solutions when t ! 1?
Solution The characteristic equation of y 00
r2
)
5r + 4 = 0
5y 0 + 4y = 0 is
(r
1)(r
)
4) = 0
r = 1, 4.
Hence the general solution is y(t) = c1 et + c2 e4t . Note that y 0 (t) = c1 et + 4c2 e4t .
Plug in the initial conditions,
)
y(0) = 2
0
y (0) =
Then we have c1 = 3 and c2 =
2 = c1 + c2
)
1
1 = c1 + 4c2 .
1. So the particular solution is
y(t) = 3et
e4t .
Moreover,
lim y(t) = lim (3et
t!1
t!1
e4t ) = lim et (3
e3t ) =
t!1
1.
Example 68 (Summer 2002 Midterm Exam I).
a) Solve the initial value problem:
y 00
y0
6y = 0,
y(0) = ↵,
y 0 (0) = 4.
Solution First, consider the roots of its characteristic equation as follow:
r2
r
6=0
)
(r + 2)(r
3) = 0
)
So the general solution is
2t
y(t) = c1 e
+ c2 e3t .
Note here that its first derivative is
y 0 (t) =
2c1 e
2t
+ 3c2 e3t .
Now we apply initial conditions,
)
y(0) = ↵
0
)
y (0) = 4
↵ = c1 + c2
4=
2c1 + 3c2 .
After solving the above equations, we find that
c1 =
3↵
4
5
and c2 =
2↵ + 4
.
5
r=
2, 3.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
86
Therefore the particular solution for the above initial value problem is
3↵
y(t) =
4
5
e
2t
+
2↵ + 4 3t
e .
5
b) If lim y(t) = 0, what is the value of ↵?
t!1
2↵ + 4 3t
e . In order to
5
3t
get a zero limit for y, its constant coefficient must be zero since lim e is infinity. So
Solution Since lim e
t!1
2t
is always zero, then lim y(t) = lim
t!1
t!1
t!1
2↵ + 4
=0
5
)
)
2↵ + 4 = 0
↵=
2.
c)* If lim y(t) = 0, what is the value of ↵?
t! 1
Solution We use the same tactic we just did in part b). Since
2t
lim e
t! 1
=1
lim e3t = 0.
and
t! 1
To get lim y(t) = 0, the constant coefficient in front of e 2t must be zero. In other
t!1
words,
3↵ 4
4
= 0 ) 3↵ 4 = 0 ) ↵ = .
5
3
Example 69 (Fall 2004 Midterm Exam 1).
Find ↵ so that the solution to the IVP
y 00 + 3y 0
4y = 0,
converge to 0 as t ! 1.
Solution The characteristic equation of y 00 + 3y 0
r2 + 3r
)
4=0
(r + 4)(r
Therefore the general solution is y(t) = c1 e
Plug in the initial conditions,
y(0) = ↵
0
y (0) = 1
Then we get c1 =
↵
1
5
and c2 =
y 0 (0) = 1,
y(0) = ↵,
4t
4y = 0 is
1) = 0
)
r=
+ c2 et . Then y 0 (t) =
)
4, 1.
4c1 e
4t
+ c2 et .
↵ = c1 + c2
)
1=
4c1 + c2 .
4↵ + 1
. So the particular solution is
5
y=
↵
1
5
e
4t
+
4↵ + 1 t
e.
5
4↵ + 1 t
e.
t!1
t!1
t!1
5
t
Since lim e = 1, the only way to get lim y(t) = 0 is to have the zero coefficient. Hence
Note that since lim e
4t
= 0, lim y(t) = lim
t!1
t!1
4↵ + 1
=0
5
)
4↵ + 1 = 0
)
↵=
1
.
4
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
87
Example 70 (Fall 2000 Midterm Exam I).
a) Find the solution to the initial value problem
y 00 + 3y 0 + 2y = 0,
y 0 (0) = ↵.
y(0) = 0,
Solution The corresponding characteristic equation of y 00 + 3y 0 + 2y = 0 is
r2 + 3r + 2 = 0
)
)
(r + 1)(r + 2) = 0
t
So the general solution is y(t) = c1 e
we apply the initial conditions,
)
y(0) = 0
y 0 (0) = ↵
2t
+ c2 e
r=
. Also y 0 (t) =
1, 2.
c1 e
t
2t
2c2 e
0 = c1 + c2
)
↵=
c1
2c2
After solving the above linear system, we find c1 = ↵ and c2 =
solution for the IVP is
y(t) = ↵e t ↵e 2t .
↵. So the particular
b) For what value(s) of ↵ is the lim y(t) = 0?
t!1
Solution Since lim e
t!1
t
t!1
t
lim y(t) = lim ↵e
t!1
2t
= 0 and lim e
t!1
↵e
2t
= 0, we have
= lim ↵(e
t
t!1
e
2t
) = ↵(0) = 0.
So all values of ↵ would make the limit 0.
c) For what value(s) of ↵ is the lim y(t) = 0?
t! 1
Solution Note that
y(t) = ↵e
Note that lim e
t
t! 1
t
↵e
2t
= 1 and lim e
t! 1
lim y(t) = lim ↵
t! 1
Since lim
t! 1
= ↵(e
t
t! 1
✓
et 1
e2t
e
2t
◆
2t
1
) = ↵( t
e
1
)=↵
e2t
✓
et 1
e2t
◆
.
= 1. By L’Hospital’s rule, we have
= lim ↵
t! 1
✓
et
2e2t
◆
= lim ↵
t! 1
✓
1
2et
◆
1
= 1, its constant coefficient ↵ must be zero. So ↵ = 0 only.
2et
Example 71 (Summer 2007 Midterm Exam I).
Consider all the nonzero solutions of
y 00 + 4y 0 + 3y = 0.
How will the solutions behave as t ! 1?
(a) They all go to 0.
(b) They all go to
1.
. Then
.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
88
(c) They all go to 1.
(d) Some go to 1, the others go to
1.
Solution As usual, we consider the corresponding characteristic equation and find its roots.
0 = r2 + 4r + 3 = (r + 1)(r + 3)
)
r=
1, 3.
So the general solution is
y(t) = c1 e
t
+ c2 e
3t
,
t
where c1 and c2 are arbitrary. Notice that both e and e 3t converge to zero as t approaches
infinity. So all solution eventually go to 0. Thus a) is the correct answer.
3.2.2
Complex Conjugate Roots
Assume that r1,2 = ± µi are the roots of the characteristic equation ar2 + br + c = 0. Then we
get two solutions to the di↵erential equation ay 00 + by 0 + cy = 0 which are
y1 = e (
+µi)t
y2 = e (
and
µi)t
.
Let’s consider their Wronskian,
W (y1 , y2 ) =
e( +µi)t
( + µi)e( +µi)t
(
e( µi)t
µi)e(
µi)t
= [(
( + µi)]e2
µi)
t
=
2iµe2
t
6= 0,
since exponential function is nonzero and µ 6= 0.
So y1 and y2 form a fundamental set of solutions. Hence the general solution is
y(t) = c1 e(
+µi)t
+ c 2 e(
µi)t
by Theorem 4. But we would like to express the solution as the real valued functions (without i).
In order to achieve this, we need the Euler’s formula,
ei✓ = cos ✓ + i sin ✓.
From this we have
e
i✓
= cos( ✓) + i sin( ✓) = cos(✓)
i sin ✓.
And now we are ready to rewrite the general solution,
y(t) = c1 e(
= c1 e
+µi)t
t
+ c2 e(
· eiµt + c2 e
µi)t
t
·e
iµt
= c1 e t (cos µt + i sin µt) + c2 e t (cos µt
t
= (c1 + c2 )e cos µt + (c1
i sin µt)
t
c2 )ie sin µt.
After we rename the constant, we get the new form of the general solution,
y(t) = c1 e t cos µt + c2 e t sin µt
or
y(t) = e t (c1 cos µt + c2 sin µt)
In conclusion,
If r1,2 =
± µi are complex roots of the characteristic equation, then the general solution is
y(t) = e t (c1 cos µt + c2 sin µt).
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
89
Example 72 (Spring 2004 Midterm Exam 1).
Find the general solutions to the y 00 + 2y 0 + 3y = 0.
Solution The characteristic equation of y 00 + 2y 0 + 3y = 0 is
p
p
2 ± 4 4(3)
2±
8
2
r + 2r + 3 = 0 ) r =
=
=
2
2
p
p
So the general solution is y(t) = e t (c1 cos 2t + c2 sin 2t).
1±
p
2i.
Example 73 (Summer 2006 Midterm Exam 1).
Solve the following IVP
y 00 + 4y 0 + 13y = 0,
y 0 (0) =
y(0) = 3,
9.
Be sure to express your solution in terms of real-valued functions only.
Solution The characteristic equation of y 00 + 4y 0 + 13y = 0 is
p
4 ± 16 52
4 ± 6i
2
r + 4r + 13 = 0 ) r =
=
=
2
2
2 ± 3i.
Therefore the general solution is
y(t) = e
2t
(c1 cos 3t + c2 sin 3t).
Note that
y 0 (t) = e
2t
( 3c1 sin 3t + 3c2 cos 3t)
2e
2t
(c1 cos 3t + c2 sin 3t).
Plug in the initial conditions,
)
y(0) = 3
0
y (0) =
Thus c1 = 3 and c2 =
)
9
3 = c1
9 = 3c2
2c1 .
1. So the particular solution is
y(t) = e
2t
(3 cos 3t
sin 3t).
Example 74 (Summer 2003 Midterm Exam 1).
Solve the IVP:
y 00
2y 0 + 5y = 0,
y(0) = 2,
y 0 (0) =
Solution The characteristic equation of y 00
r2
2r + 5 = 0
)
4.
2y 0 + 5y = 0 is
p
2 ± 4 20
2 ± 4i
r=
=
= 1 ± 2i.
2
2
Hence the general solution is
y(t) = et (c1 cos 2t + c2 sin 2t).
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
Then
90
y 0 (t) = et ( 2c1 sin 2t + 2c2 cos 2t) + et (c1 cos 2t + c2 sin 2t).
Plug in the initial conditions,
y(0) = 2
y 0 (0) =
So c1 = 2 and c2 =
4
)
)
4 = 2c2 + c1 .
3. So the particular solution is
y(t) = et (2 cos 3t
3.2.3
2 = c1
3 sin 3t).
Repeated Real Roots
Now assume that r1 = r2 = r are repeated real roots of characteristic equation ar2 + br + c = 0.
So two solutions of ay 00 + by 0 + cy = 0 are
y1 (t) = er1 t = ert
and
y2 (t) = er2 t = ert .
Compute their Wronskian,
W (y1 , y2 ) =
ert
rert
ert
= re2rt
rert
re2rt = 0.
So y1 and y2 don’t form the fundamental set of solutions. We will use y1 (t) = ert as the first
solution, but we’re going to need a new second solution. To get the second solution, we need a new
tool namely the method of reduction of order which we will study about it shortly. After using
this method, we will receive y2 (t) = tert as the second solution. Let’s check their Wronkskian,
W (y1 , y2 ) =
ert
rert
tert
= ert (rtert + ert )
rte + ert
rt
rte2rt = rte2rt + e2rt
rte2rt = e2rt 6= 0.
Consequently, y1 and new y2 form the fundamental set of solutions which means the general
solution is
y = c1 ert + c2 tert
by Theorem 4. In conclusion,
If r1 = r2 = r are repeated real roots of the characteristic equation, the general solution is
y(t) = c1 ert + c2 tert .
Example 75 (Spring 2004 Midterm Exam 1).
Find the general solutions to
y 00 + 6y 0 + 9y = 0.
Solution The characteristic equation of y 00 + 6y 0 + 9y = 0 is
r2 + 6r + 9 = 0
Thus y(t) = c1 e
3t
+ c2 te
3t
)
(r + 3)2 = 0
is the general solution.
)
r=
3, 3.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
91
Example 76 (Summer 2008 Midterm Exam 1).
Consider the IVP
y 00
8y 0 + 16y = 0,
y(0) = 3,
y 0 (0) = 10.
1. Find the solution, y(t), of this IVP.
Solution The characteristic equation of y 00
r2
)
8r + 16 = 0
(r
8y 0 + 16y = 0 is
4)2 = 0
)
r = 4, 4.
So the general solution is y(t) = c1 e4t + c2 te4t .
Note that
y 0 (t) = 4c1 e4t + c2 (4te4t + e4t ).
Plug in the initial conditions,
y(0) = 3
0
y (0) = 10
Thus c1 = 3 and c2 =
)
)
3 = c1
10 = 4c1 + c2 .
2. Therefore the particular solution is
y(t) = 3e4t
2te4t .
2. What is lim y(t)?
t!1
Solution lim y(t) = lim (3e4t
t!1
2te4t ) = lim e4t (3
t!1
t!1
1.
2t) =
Example 77 (Fall 2007 Midterm Exam I).
What is the general solution of
1
9y 00 + 6y 0 + y = 0?
1
(a) c1 e 3 t + c2 e 3 t
3t
(b) c1 e
t
3
(c) c1 e
+ c2 e3t
+ c2 te
t
3
t
(d) c1 e 3 + c2 e3t
Solution The characteristic equation of 9y 00 + 6y 0 + y = 0 is
9r2 + 6r + 1 = 0
)
Hence the general solution is y(t) = c1 e
(3r + 1)2 = 0
1
3t
+ c2 te
1
3t
)
1
,
3
r=
. So (c) is the correct answer.
Example 78 (Spring 2003 Midterm Exam I).
Solve the initial value problem
y 00 + 6y 0 + 9y = 0,
1
.
3
y(0) = 2,
y 0 (0) =
1.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
92
Solution Let’s find out the roots of its characteristic equation,
r2 + 6r + 9 = 0
)
(r + 3)2 = 0
)
r=
3, 3.
Then the general solution is
3t
+ c2 te
3t
.
+ c2 ( 3te
3t
+e
y(t) = c1 e
Note here that,
y 0 (t) =
3c1 e
3t
3t
).
Now we impose the given initial conditions,
y(0) = 2
0
y (0) =
1
)
2 = c1
)
1=
)
3c1 + c2
c2 = 5.
Hence we obtain the following particular solution,
3t
y(t) = 2e
3.2.4
+ 5te
3t
.
Summary & More Examples
We summary all we’ve done here:
Let r1 and r2 be roots of characteristic equation ar2 + br + c = 0 of the di↵erential equation
ay 00 + by 0 + cy = 0.
1. Two distinct real roots: r1 6= r2
The general solution is y = c1 er1 t + c2 er2 t .
2. Complex conjugate roots: r1,2 =
t
± µi
The general solution is y(t) = e (c1 cos µt + c2 sin µt).
3. Repeated real roots: r1 = r2 = r
The general solution is y(t) = c1 ert + c2 tert .
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
93
Example 79 (Spring 2003 Midterm Exam I).
What of the equations below has y(t) = c1 et + c2 e
(a) y 0
2y = 0
(b) y 00
y0
as its general solution?
2y = 0
(c) 2y 00 + 2y 0
(d) y 00
2t
4y = 0
3y 0
2y = 0
Solution According to the general solution, we have r1 = 1 and r2 =
the roots of the characteristic equation. So
(r
r2 + r
)
1)(r + 2) = 0
)
2=0
2. Both of them are
y 00 + y 0
2y = 0.
Multiply both sides of the equation by 2, we have (c) as the correct answer.
Example 80 (Fall 2010 Midterm Exam I).
Consider the equation
y 00 + 4y 0 + 4y = 0.
a) Find its general solution.
Solution The characteristic equation of y 00 + 4y 0 + 4y = 0 is
r2 + 4r + 4 = 0
Thus y(t) = c1 e
2t
+ c2 te
2t
)
(r + 2)2 = 0
)
r=
2, 2.
is the general solution.
b) Find the particular solution satisfying
y 0 (365) =
y(365) = 1,
6.
Solution
• Method 1: Usual way. By using the general solution from part a) ,we have
y 0 (t) =
2c1 e
2t
+ c2 ( 2te
2t
+e
2t
).
Then impose the initial conditions,
y(365) = 1
y 0 (365) =
6
)
)
1 = c1 e
6=
2(365)
2c1 e
+ c2 (365)e
2(365)
2(365)
+ c2 ( 2(365)e
2(365)
+e
After a long and messy calculation, we have c1 = 1461e730 and c2 =
the particular solution for the given IVP is
y = 1461e730 e
2t
4e730 te
2t
= (1461
4t)e730
2t
.
2(365)
).
4e730 . Hence
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
94
• Method 2: Translation. We introduce new variable T by letting T = t
the initial conditions change as follow:
)
y(t = 365) = 1
0
y (t = 365) =
y(T = 0) = 1
y 0 (T = 0) =
)
6
365. Then
6.
Now apply new initial conditions to the general solution we have in part a),
)
y(0) = 1
0
y (0) =
Then we get c1 = 1 and c2 =
1 = c1
)
6
6=
2c1 + c2 .
4. Hence the particular solution is
2T
y(T ) = e
4T e
2T
.
The final step is to change the variable from T back to t using T = t
y(t) = e
2(t 365)
4(t
365)e
2(t 365)
365,
.
• Comparison Method 1 to Method 2. After we simplify the solution from Method 2,
we can see that it’s actually the same as the solution we get from Method 1. Here
is the detail:
2(t 365)
4(t
= (1
4(t
365))e
= (1
4t + 4(365))e
y(t) = e
= (1461
4t)e730
365)e
2t
2(t 365)
2(t 365)
2t+2(365)
.
c) For the solution found in b), determine lim y(t).
t!1
Solution Using L’Hospital’s rule, we have
lim y(t) = lim (1461
t!1
t!1
4t)e730
2t
= lim
t!1
4t)e730
(1461
e2t
= lim
t!1
4e730
= 0.
2e2t
Example 81 (Fall 2005 Midterm Exam I).
a) Find the general solution of the following di↵erential equation,
y 00 + 6y 0 + 13y = 0.
Solution The characteristic equation of y 00 + 6y 0 + 13y = 0 is
p
6 ± 36 52
6 ± 4i
2
r + 6r + 13 = 0 ) r =
=
=
2
2
So the general solution is y(t) = e
3t
(c1 cos 2t + c2 sin 2t).
b) Solve the following IVP for the above ODE:
y(0) = 1,
y 0 (0) = 3.
3 ± 2i.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
95
Solution From part a), we have
y 0 (t) = e
3t
( 2c1 sin 2t + 2c2 cos 2t)
3e
3t
(c1 cos 2t + c2 sin 2t).
Then apply initial conditions,
)
y(0) = 1
0
1 = c1
)
y (0) = 3
3 = 2c2
3c1 .
So c1 = 1 and c2 = 3. Then the particular solution is
y(t) = e
3t
(cos2t + 3 sin 2t).
c) Solve the following IVP for the above ODE:
y 0 (999) = 3.
y(999) = 1,
Solution Use the technique of the translation by introducing new variable T = t
Then the initial conditions change:
y(t = 999) = 1
y 0 (t = 999) = 3
)
)
y(T = 0) = 1
y 0 (T = 0) = 3.
By part b), we then have the solution
y(T ) = e
3T
(cos2T + 3 sin 2T ).
Now change it back to variable t by using T = t
y(t) = e
3(t 999)
(cos2(t
as the particular solution of the given IVP.
999. So we have
999) + 3 sin 2(t
999)),
999.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
96
Exercises 3.2
1. Consider the second order linear equation
y 00
8y 0 + 16y = 0.
(a) Find the general solution of the equation.
(b) Find the solution satisfying the initial conditions y(735) = 2, y 0 (735) =
1.
(c) Find lim y(t).
t!1
2. Construct a second order homogeneous linear equation with constant coefficients, such that
1
y1 = e t and y2 = ⇡e3t are two of its solutions.
⇡
3. Consider the second order linear equation
y 00 + 6y 0 + 10y = 0.
(a) Find its general solution.
(b) Find the solution satisfying the initial conditions y(5031) = 6 and y 0 (5031) =
(c) Let y(t) be the solution found in ii). Evaluate lim y(t).
t!1
4. What is the solution of the initial value problem
y 00
(a) y =
e2t
6
(b) y =
13e2t
(c) y =
e2t+6
+ 4te2t
6
4y 0 + 4y = 0,
y(3) =
y 0 (3) = 2?
1,
6
+ 4te2t
6
4te2t+6
(d) y = 11e2t+6 + 4te2t+6
5. Let y(t) be the solution of the initial value problem
y 00 + 2y 0
8y = 0,
y(0) = ↵,
y 0 (0) =
Suppose lim y(t) = 0, find the value of ↵.
t!1
(a) ↵ =
1
2
(b) ↵ =
1
(c) ↵ =
4
(d) ↵ = 8
6. Consider all nonzero solutions of the linear equation
y 00
2y 0 + 5y = 0.
As t ! 1, they will
(a) oscillate between 1 and
(b) all approach 0.
1, and not approach any limit.
(c) some approach 1, all the rest approach
1.
(d) some approach 0, some approach 1, some approach
1.
2.
1.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
97
7. Which equation below has the function y = 19e4t as a particular solution?
(a) y 00 + 16y = 0
(b) y 00
8y 0 + 16y = 0
(c) y 00 + 4y 0 = 0
(d) y 00 + 5y 0 + 4y = 0
8. Let y(t) be the solution of the initial value problem
y 00 + 3y 0
10y = 0,
y(0) = ↵,
y 0 (0) =
2.
Suppose lim y(t) = 0, find the value of ↵.
t!1
(a)
1
(b) 10
2
(c)
5
4
(d)
5
9. Which of the following second order linear equations below has y = 6e
(a) y 00 + 3y 0 = 0
(b) y 00 + 2y 0
(c) 2y
(d)
00
4y
y 00
3t
3y = e
0
sin t
6y = 0
2y 0 + 3y = 0
10. Find the general solution of
t
16y 00 + 8y 0 + y = 0.
t
(a) y = c1 e 4 + c2 e 4
t
4
(b) y = c1 e
4t
(c) y = c1 e
t
4
(d) y = c1 e
t
4
+ c2 te
+ c2 e4t
+ c2 e
4t
11. Consider the initial value problem
y 00
7y 0 + 12y = 0,
y(0) = 4,
y 0 (0) = 11.
(a) Find the solution y(t) of this initial value problem.
(b) Determine lim y(t).
t!1
12. (a) Find the solution to the initial value problem
y 00 + 3y 0 + 2y = 0,
y(0) = 0,
y 0 (0) = ↵
(b) For what value(s) of ↵ is the lim y(t) = 0?
t!1
(c) For what value(s) of ↵ is the lim y(t) = 0?
t! 1
13. Find the general solution for each of the following equations.
(a) y 00
00
y0
2y = 0
0
(b) y + 4y + 4y = 0
3t
3et as a solution?
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
98
(c) y 00 + 4y 0 + 5y = 0
14. Find the particular solution to
y 00
5y 0 + 4y = 0,
y(0) = 2,
y 0 (0) =
1.
What is the behavior of the solution as t approaches positive infinity?
15. In each of the parts (a), (b), and (c), find the real-valued general solution of the second order
linear equation given.
(a) y 00 + 12y 0 + 36y = 0
(b) y 00
7y 0
00
0
(c) y
8y = 0
4y + 13y = 0
For each of the next parts, consider the solutions of the three equations above and write
down the correct equation whose solutions behave as stated.
(a) Every solution of this equation approaches 0 as t ! 1
(b) Every nonzero solution of this equation does not approach a finite limit, nor does it
have a limit of 1 or 1, as t ! 1.
Answers
1. (a) y = c1 e4t + c2 te4t , (b) y=2e4(t
2. y 00
2y 0
735)
9(t
735)e4(t
735)
sin t, (b) y = 6e
3(t 5031)
cos(t
, (c)
1
3y = 0
3t
3. (a) y = c1 e
5031), (c) 0
cos t + c2 e
3t
5031) + 17e
3(t 5031)
sin(t
4. (b)
5. (a)
6. (a)
7. (b)
8. (c)
9. (d)
10. (b)
11. (a) y = 5e3t
12. (a) y = ↵e
e4t , ii)
t
↵e
2t
1
, (b) All real numbers, (c) ↵ = 0 only
13. (a) y = c1 e2t + c2 e t , (b) y = c1 e
14. y = 3et
e4t ,
15. (a) y = c1 e
6t
2t
+ c2 te
2t
, (c) y = c1 e
2t
cos t + c2 e
2t
sin t
1
+ c2 te
6t
, (b) y = c1 e8t + c2 e t , (c) y = c1 e2t cos 3t + c2 e2t sin 3t, (d) i, e) iii
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.3
99
Linear Homogeneous Equation
3.3.1
Euler Equation*
An equation of the form
at2 y 00 + bty 0 + cy = 0,
t > 0,
or
d2 y
dy
+ bt
+ cy = 0, t > 0,
dt2
dt
is called a (homogeneous) Euler equation. To solve it, we make the substitution t = es (or
s = ln t). By doing this, we change the Euler equation to one with constant coefficients which is
solvable by the technique we’ve just learned from previous section. Note first that
at2
dy
dy ds
dy 1
dy 1
=
=
=
.
dt
ds dt
ds t
ds es
Therefore
dy
t
= es
dt
Moreover
✓
dy 1
ds es
◆
=
(3.1)
dy
.
ds
(3.2)
⇣ ⌘
⇣ ⌘
✓ ◆
d dy
d dy
dt
dt ds
d2 y
d dy
ds
=
=
=
· .
2
dt
dt
ds dt
ds dt
dt
Now we use (3.1) and follow by the product rule,
✓
◆
✓
◆✓ ◆
d2 y
d dy 1
ds
dy 1
d2 y 1
1
=
·
=
+
dt2
ds ds es
dt
ds es
ds2 es
t
✓
◆✓ ◆
d2 y
dy
2
dy 1
d y 1
1
2
=
+ 2 s
= ds 2s ds .
s
ds e
ds e
es
e
Hence
d2 y
2
t2 2 = (es )
dt
d2 y
ds2
dy
ds
e2s
!
=
d2 y
ds2
dy
.
ds
◆
✓
(3.3)
Substitute (3.2) and (3.3) into Euler equation,
at2
d2 y
dy
+ bt
+ cy = 0
dt2
dt
)
a
✓
d2 y
ds2
dy
ds
+b
dy
ds
◆
+ cy = 0.
Therefore the transformed homogeneous equation with constant coefficients is
or
where y is a function of s.
d2 y
+ (b
ds2
a)
ay 00 + (b
a)y 0 + cy = 0,
dy
+ cy = 0,
ds
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
Example 82. Solve
2t2 y 00 + ty 0
3y = 0,
t > 0.
Solution This is an Euler equation with a = 2, b = 1 and c =
above, the equation becomes
ay 00 + (b
a)y 0 + cy = 0
2y 00 + (1
)
100
2)y 0
3. By using the substitution
)
3y = 0
2y 00
y0
3y = 0.
Write down its characteristic equation and find roots,
2r2
r
)
3=0
(r + 1)(2r
So the general solution is
y(s) = c1 e
)
3) = 0
3
1, .
2
r=
3
s
+ c2 e 2 s .
The last step is to substitute s = ln t, we then have
y(t) = c1 e
Example 83. Solve
t2 y 00
ln t
3
+ c2 e 2 ln t =
ty 0 + 5y = 0,
Solution This is an Euler equation with a = 1, b =
the equation becomes
ay 00 + (b
a)y 0 + cy = 0
)
y 00 + ( 1
3
c1
+ c2 t 2 .
t
t > 0.
1 and c = 5. By using t = es or s = ln t,
1)y 0 + 5y = 0
Write down its characteristic equation and find roots,
p
2± 4
r2 2r + 5 = 0 ) r =
2
20
)
y 00
2y 0 + 5y = 0.
= 1 ± 2i.
Then the general solution in terms of s is
y(s) = c1 es cos 2s + c2 es sin 2s.
Finally, we substitute s = ln t,
y(t) = c1 eln t cos 2 ln t + c2 eln t sin 2 ln t = c1 t cos(2 ln t) + c2 t sin(2 ln t).
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.3.2
101
Reduction of Order
Now we look at solutions to non constant coefficient homogeneous second order di↵erential equation
of the form
y 00 + p(t)y 0 + q(t)y = 0.
In general, finding solutions to this di↵erential equation is difficult. However, if we already knew
one of its solution, we can use this method to find a second solution. Together their linear
combination forms a fundamental set of solutions of the above equation.
The basic idea is to assume that the second solution has similar form as the first solution by
assume that
y2 (t) = v(t)y1 (t),
where v is the function of t. Our job is to find this certain function. Along the way, we will be
forced to solve the second order di↵erential equation which we can’t if we try to solve it directly.
Instead we introduce new variable u by setting u = v 0 to reduce the order of the equation to first
order in which we have learned how to solve it from previous chapter. This procedure is then
called the method of reduction of order. Since y2 (t) = v(t)y1 (t),
y20 (t) = vy10 + v 0 y1 ,
y200 (t) = (vy10 + v 0 y1 )0 = vy100 + v 0 y10 + v 0 y10 + v 00 y1 = vy100 + 2v 0 y10 + v 00 y1 .
Since y2 is a solution to y 00 + p(t)y 0 + q(t)y = 0, we find that
y200 + p(t)y20 + q(t)y2 = 0
)
)
(vy100 + 2v 0 y10 + v 00 y1 ) + p(t)(vy10 + v 0 y1 ) + q(t)vy1 = 0
y1 v 00 + (2y10 + p(t)y1 )v 0 + (y100 + p(t)y10 + q(t)y1 )v = 0.
Since y1 is also a solution to y 00 + p(t)y 0 + q(t)y = 0, we have y100 + p(t)y10 + q(t)y1 = 0 so that the
equation above becomes
y1 v 00 + (2y10 + p(t)y1 )v 0 = 0.
Even though its appearance is a second order equation, but actually it is a first order equation for
the function u = v 0 . It can be solved as a separable equation as we show below:
y1 u0 + (2y10 + p(t)y1 )u = 0
)
y1 u 0 =
(2y10 + p(t)y1 )u
)
u0
=
u
(2y10 + p(t)y1 )
.
y1
Once v 0 has been found, then v is obtained by an integration. After we get v, we can find y2 from
y2 = vy1 . Then the general solution to y 00 + p(t)y 0 + q(t)y = 0 is
y = c 1 y1 + c 2 y2 .
The last step is true by Theorem 4 (as long as v is not a constant function) since their Wronskian
is nonzero,
y
vy1
W (y1 , y2 ) = 10
= y1 (vy10 + y1 v 0 ) vy1 y10 = y12 v 0 6= 0.
y1 vy10 + y1 v 0
(y1 is nonzero and so is v 0 if v is not a constant function.)
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
102
The process of the reduction of order method
1. Assume that the second solution of p(t)y 00 + q(t)y 0 + r(t)y = 0 is
y2 (t) = v(t)y1 (t)
where y1 (t) is the given first solution.
2. Using y2 from the first step to compute y20 and y200 . Then substitute them back in the
given di↵erential equation.
3. Reduce the order of di↵erential equation from 2nd order to 1st order by introducing
new variable u = v 0 .
4. Solve for u.
5. Using u = v 0 and then solve for v.
6. Substitute v back into y2 (t) = v(t)y1 (t) to get the second solution y2 .
7. Choose suitable constant to get y2 with no repeated term in y1 .
8. Then the general solution to p(t)y 00 + q(t)y 0 + r(t)y = 0 is
y(t) = c1 y1 + c2 y2 .
Example 84 (Summer 2010 Midterm Exam I).
The linear homogeneous di↵erential equation
t2 y 00 + ty 0
4y = 0,
t>0
has a solution y1 (t) = t2 . Use the reduction of order to find the second solution y2 of this
di↵erential equation.
Solution Suppose that y2 (t) = v(t)y1 (t) = vt2 is the second solution. Then the first and
second derivative of y2 are
y20 = (vt2 )0 = 2tv + t2 v 0
y200 = (y20 )0 = (2tv + t2 v 0 )0 = 2tv 0 + 2v + t2 v 00 + 2tv 0 = 4tv 0 + 2v + t2 v 00 .
Since y2 is a solution to t2 y 00 + ty 0
4y = 0,
0 = t2 y200 + ty20
4y2
= t (4tv + 2v + t2 v 00 ) + t(2tv + t2 v 0 )
2
0
= 4t3 v 0 + 2t2 v + t4 v 00 + 2t2 v 0 + t3 v 0
4(vt2 )
4vt2
= t4 v 00 + 5t3 v 0 .
Reducing the order of di↵erential equation by letting u = v 0 , the di↵erential equation becomes
0 = t4 u0 + 5t3 u.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
103
This is a separable first order equation.
t4
du
=
dt
5t3 u
)
du
=
u
5
dt
t
)
Z
du
=
u
+c0
= eln t
Z
5
dt
t
)
ln u =
5 ln t + c0 .
(Here we use the fact that t > 0)
Hence
ln u = ln t
5
+ c0
u = eln t
)
But u = v 0 , so we have
5
v 0 = ct
5
5
0
· ec = ct
y2 (t) = v(t)y1 (t) = (
Choose c =
c
t
4
4
0
where c = ec .
.
Again we use the method of separable equation to solve it,
Z
Z
dv
= ct 5 )
dv = ct 5 dt ) v(t) =
dt
Consequently,
5
c
t
4
+ k)t2 =
2
c
t
4
4
+ k.
+ kt2 .
4 and k = 0, we have
y2 (t) = t
2
.
(The general solution is y(t) = c1 y1 + c2 y2 = c1 t2 + c2 t
2
)
Example 85 (Spring 2009 Midterm Exam I).
Given that y1 (t) = t3 is a known solution of the second order linear di↵erential equation
t2 y 00
5ty 0 + 9y = 0,
t > 0.
Find the general solution of the equation.
Solution Let y2 (t) = v(t)y1 (t) = vt3 be the second solution. Then the first and second
derivative of y2 are
y20 = (vt3 )0 = 3t2 v + t3 v 0
y200 = (y20 )0 = (3t2 v + t3 v 0 )0 = 3t2 v 0 + 6tv + t3 v 00 + 3t2 v 0 = 6t2 v 0 + 6tv + t3 v 00 .
Since y2 is a solution to t2 y 00
0 = t2 y200
5ty 0 + 9y = 0,
5ty20 + 9y2
= t2 (6t2 v 0 + 6tv + t3 v 00 )
= 6t4 v 0 + 6t3 v + t5 v 00
5t(3t2 v + t3 v 0 ) + 9(vt3 )
3⇠
⇠
15t⇠
v 5t4 v 0 + 9vt3
= t4 v 0 + t5 v 00 .
Divide out both sides of the equation by t5 (it is allowed since t > 0),
0=t
1 0
v + v 00 .
Reducing the order of di↵erential equation by letting u = v 0 , the di↵erential equation becomes
0=t
1
u + u0 .
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
This is a separable first order equation.
Z
Z
du
du
t 1u =
)
=
dt
u
dt
t
)
ln t + c0
ln u =
104
)
c
u= .
t
But u = v 0 ,
c
v0 = .
t
Again we use the method of separable equation to solve it,
Z
Z
dv
c
c
=
)
dv =
dt ) v = c ln t + k.
dt
t
t
Hence
y2 (t) = v(t)y1 (t) = (c ln t + k)t3 = ct3 ln t + kt3 .
Choose c = 1 and k = 0, we have y2 (t) = t3 ln t. Therefore the general solution is
y(t) = c1 y1 + c2 y2 = c1 t3 + c2 t3 ln t.
Example 86 (Fall 2010 Midterm Exam I).
Given that y1 (t) = t4 is a known solution of the second order linear di↵erential equation
t2 y 00
6ty 0 + 12y = 0,
t > 0.
Find the general solution of the equation.
Solution Let y2 (t) = v(t)y1 (t) = vt4 be the second solution. Then the first and second
derivative of y2 are
y20 = (vt4 )0 = 4t3 v + t4 v 0
y200 = (y20 )0 = (4t3 v + t4 v 0 )0 = 4t3 v 0 + 12t2 v + t4 v 00 + 4t3 v 0 = 8t3 v 0 + 12t2 v + t4 v 00 .
Since y2 is a solution to t2 y 00
0 = t2 y200
6ty 0 + 12y = 0,
6ty20 + 12y2
= t2 (8t3 v 0 + 12t2 v + t4 v 00 ) 6t(4t3 v + t4 v 0 ) + 12(vt4 )
4⇠
4⇠
4
⇠⇠
= 8t5 v 0 + ⇠
12t⇠
v + t6 v 00 ⇠
24t⇠
v 6t5 v 0 + ⇠
12vt
= 2t5 v 0 + t6 v 00 .
Divide out both sides of the equation by t6 (it is allowed since t > 0),
0 = 2t
1 0
v + v 00 .
Use u = v 0 to reduce the order of the di↵erential equation. So
0 = 2t
This is a separable first order equation.
Z
Z
du
du
2
2t 1 u =
)
=
dt
dt
u
t
1
u + u0 .
)
ln u =
2 ln t + c0
)
u = ct
2
.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
But u = v 0 ,
v 0 = ct
2
105
.
Using the method of separable equation to solve it,
Z
Z
dv
2
= ct
)
dv = ct 2 dt
dt
)
v(t) =
ct
1
+ k.
Therefore
y2 (t) = v(t)y1 (t) = ( ct
Choose c =
1
+ k)t4 =
ct3 + kt4 .
1 and k = 0, we have y2 (t) = t3 . Therefore the general solution is
y(t) = c1 y1 + c2 y2 = c1 t4 + c2 t3 .
Example 87 (Summer 2002 Final Exam ).
Given that y1 (t) = t3 is a solution, find the general solution of
t2 y 00 + 2ty 0
12y = 0,
t > 0.
Solution Suppose that the second solution is y2 (t) = v(t)y1 (t) = vt3 . Note that
y20 (t) = (vt3 )0 = 3t2 v + t3 v 0
y200 (t) = (vt3 )00 = (3t2 v + t3 v 0 )0 = 3t2 v 0 + 6tv + t3 v 00 + 3t2 v 0 = 6t2 v 0 + 6tv + t3 v 00 .
Since y2 is a solution to t2 y 00 + 2ty 0
t2 y200 + 2ty20
12y2 = 0
12y = 0, we have
)
t2 (6t2 v 0 + 6tv + t3 v 00 ) + 2t(3t2 v + t3 v 0 ) 12vt3 = 0
3
⇠⇠
6t4 v 0 + 6t3 v + t5 v 00 + 6t3 v + 2t4 v 0 ⇠
12vt
=0
)
t5 v 00 + 8t4 v 0 = 0.
)
Since t 6= 0, we divide both sides by t5 . So we get
8
v 00 + v 0 = 0.
t
We then reduce the order of the above equation from second order to first order by using
u = v 0 . Then we use the method of separable equation to solve it.
8
u0 + u = 0
t
)
du
=
dt
8
u
t
)
We then integrate both sides and simplify the expression,
Z
Z
1
8
du =
dt ) ln u =
u
t
du
=
u
8
dt.
t
8 ln t + c.
Use the exponential function on both sides,
eln u = e
8 ln t+c
= eln t
8
· ec
)
u = ct
8
.
Now we change variable u back to variable v by again using u = v 0 , then solve it:
Z
Z
dv
c 7
v 0 = ct 8 )
= ct 8 )
dv = ct 8 dt ) v(t) =
t + k.
dt
7
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
106
We simplify the form of v by choosing c = 7 and k = 0 to obtain v(t) = t
solution is
y2 (t) = vt3 = (t 7 )t3 = t 4 .
7
. So the second
Therefore the general solution is
y(t) = c1 y1 + c2 y2 = c1 t3 + c2 t
4
.
In a certain type of di↵erential equation, we can apply the technique that we have learned from
the method of reduction of order to solve it.
Example 88 (Summer 2007 Midterm Exam I).
Solve the following initial value problem
ty 00 + y 0 = 0,
t > 0,
y 0 (1) = 1.
y(1) = 2,
Hint: Use the substitutions u = y 0 and u0 = y 00 to convert the equation into a first order
linear equation in terms of u. Then integrate your answer to find y and lastly apply the
initial conditions.
Solution We follow the hint by writing u = y 0 and so u0 = y 00 . Then the equation becomes
ty 00 + y 0 = 0
)
tu0 + u = 0.
Notice that it is the separable equation and so
tu0 + u = 0
)
t
du
+u=0
dt
)
du
=
u
dt
.
t
Now we integrate both sides then take the natural log of both sides of an equation,
Z
Z
1
1
du =
dt ) ln u = ln t + c ) u = ct 1 .
u
t
Now let’s substitute u by y 0 and again solve it by the technique of separable equations,
Z
Z
dy
0
1
1
y = ct
)
= ct
)
dy = ct 1 dt ) y = c ln t + k.
dt
c
So y(t) = c ln t + k. (Also y 0 (t) = .) Then we apply both initial conditions here,
t
y(1) = 2
0
y (1) = 1
)
)
2 = c(0) + k
c
1=
1
)
k=2
)
c = 1.
Consequently, the particular solution of the above IVP is
y = ln t + 2.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
107
Exercises 3.3.2
1. Given that y1 (t) = t2 is a known solution of the second order linear equation
t2 y 00
4ty 0 + 6y = 0,
t > 0.
Use reduction of order to find the general solution of the equation.
2. Given that y1 (t) =
1
is a known solution of the second order linear equation
t2
t2 y 00
ty 0
8y = 0,
t > 0.
3ty 0 + 4y = 0,
t > 0.
Find the general solution of the equation.
3. Consider the second order linear equation
t2 y 00
(a) Verify that y1 (t) = t2 is a solution of this equation.
(b) Find the equation’s general solution using the method of reduction of order.
4. Given that y1 (t) = t is a known solution of the second order linear di↵erential equation
t2 y 00
4ty 0 + 4y = 0,
t > 0.
Find the general solution of the equation.
5. Given that y1 (t) = t3 is a solution, find the general solution of
t2 y 00 + 2ty 0
6. Given that y1 (t) = t
1
12y = 0,
t > 0.
is a solution to the equation,
t2 y 00 + 3ty 0 + y = 0,
t > 0.
(a) Use the method of reduction of order to find another solution y2 which is not a scalar
multiple of y1 .
(b) Find the general solution of the equation.
(c) Find a solution satisfying the following initial conditions: y(e) = 1e , y 0 (e) =
7. Given that y1 (t) = t4 is a known solution of the linear di↵erential equation
t2 y 00
7ty 0 + 16y = 0,
t > 0.
Use reduction of order to find the general solution of the equation.
Answers
1. y = c1 t2 + c2 t3
2. y = c1 t
2
+ c 2 t4
3. (a) -, (b) y = c1 t2 + c2 t2 ln t
4. y = c1 t + c2 t4
5. y = c1 t3 + c2 t
6. (a) t
1
4
ln t, (b) y(t) = c1 t
7. y = c1 t4 ln t + c2 t4
1
+ c2 t
1
ln t, (c) y(t) =
t
1
+ 2t
1
ln t
1
e2 .
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.4
108
The Existence and Uniqueness Theorem
This theorem is analogous to the one for first order linear equation.
Theorem 5 (The Existence and Uniqueness Theorem (for second order linear equation)).
Consider the initial value problem
y 00 + p(t)y 0 + q(t)y = g(t),
y(t0 ) = y0 ,
y 0 (t0 ) = y00 ,
where p, q and g are continuous on an open interval I that contains the point t0 . Then there
is exactly one solution y = (t) of this problem, and the solution exists throughout the interval
I.
Example 89 (Fall 2003 Midterm Exam I).
The existence and uniqueness theorem guarantees that the solution to
sin(t)y 00 +
1
t
3
y 0 + e t y = t3 ,
y(1) = 0,
y 0 (1) = 1
is valid on
(a) (0, 3) (b)
(0, ⇡) (c)
( 1, 3) (d)
( 1, 1)
Solution Rewrite the given equation in suitable form,
y 00 +
(t
1
et
t3
y0 +
y=
.
3) sin(t)
sin(t)
sin(t)
Then the discontinuities of p are t = n⇡ for all integer n and t = 3 while the discontinuities
of q and r are t = n⇡ for all integer n
Since t0 = 1 lies in (0, 3), the interval of validity is (0, 3). So (a) is the correct answer.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
109
Example 90 (Spring 2009 Midterm Exam I).
Consider the IVP
sin(t)y 00 + tan(t)y 0 + ty = et ,
4⇡
⇡
y( ) =
,
4
3
⇡
y0 ( ) =
4
⇡
.
4
Without solving the equation, what is the largest interval in which a unique solution is guaranteed to exist?
(a) ( 1, 1) (b)
( ⇡2 , ⇡2 ) (c)
(0, ⇡2 ) (d)
( ⇡2 , 3⇡
2 )
Solution Rewrite the given equation,
⇠
sin⇠t
t
et
0
y
+
y
=
.
⇠
sin⇠t cos t
sin t
sin t
y 00 +
So the discontinuities of p are t =
and r are t = n⇡ for all integer n.
Since t0 =
(2n
1)
2
⇡ for all integer n while the discontinuities of q
⇡
⇡
⇡
lies in (0, ), the interval of validity is (0, ). Then (c) is the final answer.
4
2
2
Example 91 (Spring 2011 Midterm Exam I).
Consider the IVP
(t + 5)y 00 +
t+2 0
y + ln(t)y = 0,
t ⇡
y(1) =
10,
y 0 (1) = 3.
Without solving the equation, what is the largest interval in which a unique solution is guaranteed to exist?
(a) ( 1, 1) (b)
( 2, 1) (d)
( 5, ⇡) (c)
(0, ⇡)
Solution Change the form of the equation by dividing out by t + 5,
y 00 +
(t
t+2
ln(t)
y0 +
y=0
⇡)(t + 5)
t+5
Then p has discontinuities at t = ⇡, 5, q has no discontinuity since t =
domain of ln t while r is continuous everywhere.
5 is not in the
Since t0 = 1 lies in (0, ⇡), the largest such interval is (0, ⇡) and so (d) is the correct answer.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
110
Example 92 (Fall 2002 Midterm Exam I).
What is the largest interval on which a solution of the IVP is guaranteed to exist? Do not
solve the di↵erential equation.
y 00 + ln |t
10|y 0 + y =
1
t2
t
,
y 0 (8) = 0.
y(8) = 8,
Without solving the equation, what is the largest interval in which a unique solution is guaranteed to exist?
Solution Since t2
t = t(t
1), we have
y 00 + ln |t
10|y 0 + y =
1
t(t
1)
.
Then p is is not defined at t = 10, q is continuous while r has discontinuities at t = 0, 1.
Since t0 = 8 lies between (1, 10), the interval of validity is (1, 10).
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
111
Exercises 3.4
1. Consider the initial value problem
(t2
16)y 00 +
t 1 0
t
y +
y = 0,
t+⇡
2 t
y(3) = 10,
y 0 (3) =
2.
What is the largest interval in which a unique solution is guaranteed to exist?
(a) (2, 1)
(b) ( ⇡, 2)
(c) ( 4, 4)
(d) (2, 4)
2. The Existence and Uniqueness Theorem guarantees that the solution to
(t + 2)y 00
sin ty 0 +
ty
t
4
=
e2t
,
t
y( 1) = 0
(a) is valid on ( 1, 1).
(b) is valid on ( ⇡, 0).
(c) is valid on ( 2, 0).
(d) does not exist.
3. Consider the initial value problem
(t + 5)y 00 +
t+2 0
y + ln(t)y = 0,
t ⇡
y(1) =
10,
y 0 (1) = 2.
What is the largest interval in which a unique solution is guaranteed to exist?
(a) ( 1, 5)
(b) ( 5, ⇡)
(c) ( 2, 1)
(d) (0, ⇡)
4. The largest interval on which the di↵erential equation
(t2
1)y 00 + sin(t)y 0 + cos(t)y = 0,
y(5) = 0,
is certain to have a unique twice di↵erentiable solution is
(a) ( 1, 5)
(b) ( 1, 1)
(c) (5, 1)
(d) (1, 1)
(e) ( 1, 1)
Answers
1. (d)
2. (c)
3. (d)
4. (d)
y 0 (5) = 1.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.5
112
More on Wronskian & Abel’s Theorem
Let y1 and y2 be solutions of y 00 + p(t)y 0 + q(t)y = g(t). Then y1 and y2 form a fundamental set
of solution if their Wronskian is not a zero function. And the Wronskian can be found by
y1
y10
W (y1 , y2 ) :=
y2
= y1 y20
y20
y10 y2 .
Let’s consider a few more examples about it here.
Example 93 (Fall 2002 Midterm Exam I).
Are y1 (t) = t2 , y2 (t) = t
a fundamental set of solutions of t2 y 00 + ty 0
2
4y = 0?
Solution We completely answer this question in 2 steps as follow:
• Check y1 (t) = t2 , y2 (t) = t
2
are solutions to t2 y 00 + ty 0
t2 y100 + ty10
4y1 = t2 (t2 )00 + t(t2 )0
t2 y200 + ty20
4y2 = t2 (t
2 00
) + t(t
4y = 0.
4(t2 ) = 2t2 + 2t2
2 0
)
4(t
2
) = t2 (6t
4t2 = 0
4
) + t( 2t
3
)
• Compute their Wronskian:
W (y1 , y2 ) =
t2
2t
t 2
=
2t 3
2t
1
2t
1
=
4t
Therefore y1 and y2 form a fundamental set of solutions of t2 y 00 + ty 0
1
6= 0.
4y = 0.
Example 94 (Spring 2011 Midterm Exam I).
Which pair of functions below can be a fundamental set of solutions?
a) 1
2t,
b) 3 cos t,
e2t ,
c)
d) 0,
4t
2
2 sin t
e2t+5
3 sin 2t
Solution Let’s consider each choice one by one,
(a) W (1
2t, 4t
1
2) =
2t
2
4t
2
= (4
4
8t)
( 8t + 4) = 0.
3 cos t
2 sin t
= 6 cos2 t
3 sin t
2 cos t
(Here we use the identity: sin2 t + cos2 t = 1.)
(b) W (3 cos t, 2 sin t) =
(c) W ( e2t , e2t+5 ) =
(d) W (0, 3 sin 2t) =
0
0
e2t
2e2t
e2t+5
=
2e2t+5
3 sin 2t
=0
6 cos 2t
2e4t+5
0 = 0.
6 sin2 t =
( 2e4t+5 ) = 0.
6 6= 0.
4t
2
= 0.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
113
So b) is the correct answer since it it the only one which gives us a nonzero function Wronskian.
Besides the definition, there is another way to compute Wronkskian without knowing solutions y1
and y2 by using Abel’s theorem2 .
Theorem 6 (Abel’s Theorem).
If y1 and y2 are solutions of
y 00 + p(t)y 0 + q(t)y = 0,
where p and q are continuous on an open interval I, then the Wronskian W (y1 , y2 )(t) is given
by
R
W (y1 , y2 )(t) = ce p(t) dt ,
where c is a certain constant that depends on y1 and y2 but not on t. Further, W (y1 , y2 )(t)
either is zero for all t in I (if c = 0) or else is never zero in I (if c 6= 0).
Proof. Since y1 and y2 are solution of y 00 + p(t)y 0 + q(t)y = 0, they satisfy
y100 + p(t)y10 + q(t)y1 = 0
y200 + p(t)y20 + q(t)y2 = 0
Then we multiply the first equation by
y2 and the second equation by y1 ,
y100 y2
p(t)y10 y2
q(t)y1 y2 = 0
y1 y200
p(t)y1 y20
+ q(t)y1 y2 = 0.
+
Then we add these two equations together,
(y1 y200
Recall that W (y1 , y2 )(t) = y1 y20
W 0 (t) = (y1 y20
y100 y2 ) + p(t)(y1 y20
y10 y2 ) = 0.
(3.4)
y10 y2 and we let W (t) = W (y1 , y2 )(t). Observe that
y10 y2 )0 = y1 y200 + y10 y20
y10 y20
y100 y2 = y1 y200
y100 y2 .
Then we can write Equation (3.4) in the form,
W 0 + p(t)W = 0.
This is a separable first order equation. Let’s solve it here.
W0 =
p(t)W
)
dW
=
dt
p(t)W
)
Z
dW
=
W
Z
p(t) dt
2 The result in Abel’s Theorem was derived by the Norwegian mathematician Niels Henrik Abel (1802
1829)
in 1827 and is known as Abel’s formula.
Abel also showed that there is no general formula for solving a quintic, or fifth degree, polynomial equation in
terms of explicit algebraic operations on the coefficients. Ever since the cubic and quartic equations had been solved
in the sixteenth century, men had studied the quintic. Abel at first thought he had hit on a solution. He believed
that he had solved the quintic in 1821 and submitted a paper to the Danish mathematician Ferdinand Degen, for
publication by the Royal Society of Copenhagen. Degen asked Abel to give a numerical example of his method
and, while trying to provide an example, Abel discovered the mistake in his paper.
Finally, in 1824, he published a memoir, On the Algebraic Resolution of Equations in which he reached the
opposite conclusion. He gave the first proof that no solution is possible, thus putting an end to the long search.
An earlier proof, less satisfactory and generally overlooked, of the insolvability of the quintic had been published
in 1799 by Paolo Ruffini (1765 1822), and hence the result is now referred to as the Abel-Ruffini theorem.
His greatest contributions, however, were in analysis, particularly in the study of elliptic functions. Unfortunately,
his work was not widely noticed until after his death. The distinguished French mathematician Legendre called it
a monument more lasting than bronze.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
Thus
ln W =
✓Z
◆
p(t) dt + c
0
)
W = c exp
✓ Z
114
◆
p(t) dt ,
where c is a constant. The value of c depends on y1 and y2 . However, since the exponential
function is never zero, W (t) is not zero unless c = 0, in which case W (t) is zero for all t. This
completes the proof.
Example 95 (Spring 2011 Midterm Exam I).
a) Let y1 , y2 be two solutions to the equation
ty 00
2y 0
y = 0.
Determine the Wronskian W (y1 , y2 ) of y1 , y2 .
Solution Rewrite the di↵erential equation into the standard form,
2 0
y
t
y 00
So p(t) =
2
t.
y
= 0.
t
By Abel’s theorem,
W (y1 , y2 )(t) = ce
R
p(t) dt
R
= ce
2
t
dt
= ce2 ln t = ct2 .
b) If W (y1 , y2 )(2) = 1, then determine W (y1 , y2 )(3)?
Solution If W (y1 , y2 )(2) = 1, then by part a) we have
1 = c(2)2
Hence W (y1 , y2 )(t) =
)
c=
1
.
4
t2
and so
4
W (y1 , y2 )(3) =
32
9
= .
4
4
Example 96 (Spring 2007 Midterm Exam I).
Consider the second order linear di↵erential equation
ty 00
2y 0 + y = 0.
Suppose y1 (t) and y2 (t) are two fundamental sets of solutions of the equation satisfying
y1 (1) = 2,
y10 (1) = 0,
y2 (1) = 2,
y20 (1) = 2.
Compute their Wronskian W (y1 , y2 )(t) as a function of t.
Solution Again write down the di↵erential equation in the standard form,
y 00
2 0 y
y + = 0.
t
t
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
So p(t) =
2
t.
115
By Abel’s theorem,
W (y1 , y2 )(t) = ce
R
p(t) dt
= ce
R
2
t
dt
= ce2 ln t = ct2 .
(3.5)
Note here that
W (y1 , y2 )(1) = c(1)2 = c
(3.6)
On the other hand, by definition, we know
W (y1 , y2 )(t) =
y1
y10
y2
= y1 (t)y20 (t)
y20
) W (y1 , y2 )(1) = y1 (1)y20 (1)
y10 (1)y2 (1) = 2(2)
Comparing Equations (3.6) and (3.7), c = 4. Hence
W (y1 , y2 )(t) = 4t2 ,
according to Equation (3.5).
y10 (t)y2 (t).
0(2) = 4.
(3.7)
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
116
Exercises 3.5
1. Let y1 (t) and y2 (t) be any two solutions of the second order linear equation
(t + 2)y 00 + 4y 0 + (t2 + 9)e5t y = 0.
What is the general form of their Wronskian, W (y1 , y2 )(t)?
(a) ce
4t
(b) ce4t
(c) c(t + 2)4
c
(d)
(t + 2)4
2. Suppose y1 (t) = et and y2 (t) = cos t are both solutions of the second order linear equation
y 00 + p(t)y 0 + q(t)y = 0.
All of functions below must also be solutions of the same equation, EXCEPT
(a) y = 100⇡ cos t
(b) y = 2et cos t
(c) y = 7et+1
3 cos t
(d) y = 0
3. Suppose y1 (t) = 4et and y2 (t) =
linear equation
2t are two solutions of a certain second order homogeneous
y 00 + p(t)y 0 + q(t)y = 0.
(a) Find the Wronskian W (y1 , y2 )(t).
(b) True or false: y1 and y2 form a set of fundamental solutions of this equation. Why
or why not?
(c) Write down a general solution of the di↵erential equation.
(d) Find the particular solution satisfying the initial conditions y(0) = 8 and y 0 (0) =
1.
(e) True or false: y3 = 0 is also a solution of this equation. Why or why not?
(f) True or false: y4 =
8tet is also a solution of this equation. Why or why not?
4. Suppose y1 (t) = 15t3 and y2 (t) = 4t3 ln(t) are two solutions of a certain second order
homogeneous linear equation
y 00 + p(t)y 0 + q(t)y = 0.
(a) Find the Wronskian W (y1 , y2 )(t).
(b) True or false: y1 and y2 form a set of fundamental solutions of this equation. Why
or why not?
(c) Write down a general solution of the di↵erential equation.
(d) True or false: y3 = 9t6 ln(t) is also a solution of this equation. Why or why not?
(e) True or false: y4 = 0 is also a solution of this equation. Why or why not?
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
117
5. Let y1 (t) and y2 (t) be any two solutions of the second order linear equation
y 00 + 2 tan(t)y 0 + t3 y = 0.
What is the general form of their Wronskian, W (y1 , y2 )(t)?
(a) c cos2 t
(b) cesec
2
t
(c) ce2 sin t
(d) c sec2 t
6. Which pair of functions below can be a fundamental set of solutions?
5te2t
(a) 0,
(b) sin 3t,
t
(c) 6e ,
(d) 4t
sin 3t
6et
cos t,
3 cos t
7. Suppose y1 (t) =
equation
12t
t2 and y2 (t) = 3t2 ln(t) are both solutions of the second order linear
y 00 + p(t)y 0 + q(t)y = 0,
t > 0.
Which statement below is FALSE?
(a) y = 2t2
10t2 ln(t) is also a solution of the equation.
(b) W (y1 , y2 )(t) = 0
(c) y = 0 is also a solution of the equation.
(d) y = 1 can never be a solution of the equation.
8. Show y1 (t) = t2 and y2 (t) = t3 , t > 0 form a fundamental set of solutions for
t2 y 00
4ty 0 + 6y = 0.
9. Given the linear ordinary di↵erential equation
ty 00
4y 0 +
4et
y=0
t
and two fundamental solutions y1 (t), y2 (t) such that
y1 (1) = 1,
y10 (1) = 0,
y2 (1) = 2,
and y20 (1) = 3,
compute their Wronskian W (y1 (t), y2 (t)) as a function of time, using Abel’s Theorem. Use
the initial condition to determine the constant of the Wronskian.
10. Let y1 and y2 be two solutions to the linear equation
2t2 y 00
ty 0
y = 0.
Then the Wronskian of y1 and y2 must be a constant multiple of
p
(a) t
(b) t
(c) 1
(d) t2
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
118
11. Let y1 (t) and y2 (t) be any two solutions of the second order linear equation
(t2 + 4)y 00 + 2ty 0
t3 y = 0.
In what general form must their Wronskian, W (y1 , y2 )(t), appear?
(a) c(t2 + 4)
p
(b) c t2 + 4
c
(c) 2
(t + 4)
(d) c(t2 + 4)2
12. Given that y1 (t) = 1 and y2 (t) = arctan(t) are both solutions of the second order homogeneous linear equation
(t2 + 1)y 00 + 2ty 0 = 0.
Determine whether each of the following statements is true or false. State a brief reason
that justifies each answer.
(a) Wronskian W (y1 , y2 )(t) = 0.
(b) y1 and y2 from a set of fundamental solutions of this equation.
(c) y3 (t) = 7 + 5 arctan(t) is also a solution of the equation.
(d) There are additional solutions that cannot expressed as a linear combination of y1 and
y2 .
(e) Each solution is unique to its corresponding initial conditions only on the interval
1 < t < 1.
13. y1 (t) = tet and y2 (t) = t2 are both solutions of the second order linear di↵erential equation
y 00 + p(t)y 0 + q(t)y = 0.
(a) Compute W (y1 , y2 )(t) = 0.
(b) (TRUE or FALSE) y1 and y2 is not a fundamental set of solutions.
(c) (TRUE or FALSE) y = 0 is a solution of the equation.
(d) (TRUE or FALSE) y = (10t
4et )t is not a solution of the equation.
Answers
1. (d)
2. (b)
3. (a) 8et + 8tet , (b) True since W (y1 , y2 ) 6= 0, (c) y(t) = c1 et + c2 t, (d) y(t) = 8et
True, (f) False
4. (a)
9t, (e)
60t5 , (b) True since W (y1 , y2 ) 6= 0, (c) y(t) = c1 t3 + c2 t3 ln(t), (d) False, (e) True
5. (a)
6. (c)
7. (b)
8. First substitute y1 and y2 into the equation to verify that they both satisfy it. Then calculate
their Wronskian, W (y1 , y2 )(t) = t4 6= 0 when t > 0.
9. 3t4
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
10. (a)
11. (c)
12. (a) F, (b) T, (c) T, (d) F, (e) F
13. (a) t2 et
t3 et , (b) F, (c) T, (d) F
119
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.6
120
Linear Nonhomogeneous Equation
For the nonhomogeneous equation,
p(t)y 00 + q(t)y 0 + r(t)y = g(t).
(3.8)
The general solution to Equation (3.8) can be written as
y(t) = yc (t) + Y (t),
where yc (t) (which is called complementary solution )
= the general solution of the corresponding homogeneous equation p(t)y 00 + q(t)y 0 + r(t)y = 0.
Y (t) or yp (t) (which is called particular solution)
= some specific solution of the nonhomogeneous equation (3.8).
3.6.1
Method of Undetermined Coefficients
In this course we focus on a simpler case where the coefficients are constant:
ay 00 + by 0 + cy = g(t),
where a 6= 0, g(t) 6= 0 and a, b, c are constants. The complementary solution yc (t) (the general
solution to ay 00 +by 0 +cy = 0) can be easily found from the roots of characteristic equation that we
studied before. So the only task remaining is to find the particular solution Y (t). The method
of undetermined coefficients will give us the answer.
Let
g(t) = g1 (t) + · · · + gn (t).
This method gives us the form of
Y (t) = Y1 (t) + · · · + Yn (t),
with undetermined coefficients where the form of Yi (t) depends on the form of gi (t) by the table
below.
gi (t)
1. Pn (t) = a0 tn + a1 tn
Yi (t)
1
ts (A0 tn + A1 tn
1
+ · · · + An )
2. Pn (t)e↵t
ts (A0 tn + A1 tn
1
+ · · · + An )e↵t
3. Pn (t)e↵t sin t, or Pn (t)e↵t cos t, or
ts (A0 tn + A1 tn
1
+ · · · + An )e↵t cos t
Pn (t)e↵t sin t + Pn (t)e↵t cos t
+ts (B0 tn + B1 tn
+ · · · + an
1
+ · · · + Bn )e↵t sin t
Here s is the smallest nonnegative integer (s = 0, 1, or 2) that will ensure that no term in Yi (t) is
a solution of the corresponding homogeneous equation.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
Example 97. Assume that the complementary solution is
yc (t) = c1 e3t cos t + c2 e3t sin t.
gi (t)
Yi (t)
3
A
2t
At + B
5t4
3t2 + 1
At4 + Bt3 + Ct2 + Dt + E
tet
(At + B)et
6 cos 3t
A cos 3t + B sin 3t
3 sin 2t
A cos 2t + B sin 2t
2t sin 5t
(At + B) cos 5t + (Ct + D) sin 5t
e3t cos t
t(Ae3t cos t + Be3t sin t)
te3t sin t
t{(At + B)e3t cos t + (Ct + D)e3t sin t}
t2 e2t sin 3t
(At2 + Bt + C)e2t cos 3t + (Dt2 + Et + F )e2t sin 3t
Example 98. Suppose that the complementary solution is
yc (t) = c1 e2t + c2 te2t .
gi (t)
t2
Yi (t)
5t
At2 + Bt + C
2te2t
t2 (At + B)e2t
sin 2t
A cos 2t + B sin 2t
e3t cos 2t
Ae3t cos 2t + Be3t sin 2t
3e
t
Ae
4t2 sin t
5t
2e2t
3
t
(At2 + Bt + C) cos t + (Dt2 + Et + F ) sin t
At + B
t2 (Ae2t )
121
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
122
Example 99 (Spring 2003 Midterm Exam I).
What is the form of the particular solution of
y 00 + 4y 0 + 4y = te
2t
+ 2t2 cos 2t
3?
Solution First consider the corresponding homogeneous equation,
y 00 + 4y 0 + 4y = 0.
Its characteristic equation is r2 + 4r + 4 = 0 which is (r + 2)2 = 0. So r =
complementary solution is
yc (t) = c1 e 2t + c2 te 2t .
2, 2. Then the
Now let’s consider the form of the particular solution Y ,
g1 (t) = te
2t
g2 (t) = 2t2 cos 2t
g3 (t) =
)
Y1 (t) = t2 (At + B)e
)
Y3 (t) = I.
Y2 (t) = (Ct2 + Dt + E) cos 2t + (F t2 + Gt + H) sin 2t
)
3
2t
So the form of the particular solution is
Y (t) = t2 (At + B)e
2t
+ (Ct2 + Dt + E) cos 2t + (F t2 + Gt + H) sin 2t + I.
Example 100 (Fall 2007 Midterm Exam I).
Consider the nonhomogeneous second order linear equation of the form
y 00
4y 0 + 8y = g(t).
a) Find its complementary solution yc (t).
Solution Consider y 00
4y 0 + 8y = 0. Its characteristic equation is r2
p
4 ± 16 32
4 ± 4i
r=
=
= 2 ± 2i.
2
2
4r + 8 = 0.
So yc (t) = c1 e2t cos 2t + c2 e2t sin 2t.
For each of parts b) through d) write down the correct choice of the form of particular solution
that you would use to solve the given equation using the Method of Undetermined Coefficients.
b) y 00
4y 0 + 8y = 2e2t
5t2 + sin 2t.
Solution From g(t) = 2e2t
5t2 + sin 2t, then
g1 (t) = 2e2t
g2 (t) =
5t2
g3 (t) = sin 2t
)
Y1 (t) = Ae2t
)
Y3 (t) = E cos 2t + F sin 2t.
)
Y2 (t) = Bt2 + Ct + D
So Y (t) = (Ae2t ) + (Bt2 + Ct + D) + (E cos 2t + F sin 2t).
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
c) y 00
4y 0 + 8y =
123
e2t sin 2t + 1.
e2t sin 2t + 1, we have
Solution From g(t) =
e2t sin 2t
g1 (t) =
Y1 (t) = t(Ae2t cos 2t + Be2t sin 2t)
)
)
g2 (t) = 1
Y2 (t) = C.
So Y (t) = t(Ae2t cos 2t + Be2t sin 2t) + C.
d) y 00
4y 0 + 8y = t2 e
t
cos 5t.
Solution Since g(t) = t2 e
t
cos 5t,
Y (t) = (At2 + Bt + C)e
t
cos 5t + (Dt2 + Et + F )e
t
sin 5t.
Example 101 (Fall 2008 Midterm Exam I).
Consider the nonhomogeneous second order linear equation of the form
y 00 + 5y 0 + 4y = g(t).
a) Find its complementary solution yc (t).
Solution Consider the corresponding homogeneous equation y 00 + 5y 0 + 4y = 0. Hence
Its characteristic equation is
r2 + 5r + 4 = 0
4t
Therefore yc (t) = c1 e
)
(r + 4)(r + 1) = 0
)
r=
4, 1.
+ c2 e t .
For each of parts b) through d) write down the correct choice of the form of particular solution
that you would use to solve the given equation using the Method of Undetermined Coefficients.
b) y 00 + 5y 0 + 4y = 6e
4t
+ 2et .
Solution From g(t) = 6e
4t
+ 2et , then
4t
g1 (t) = 6e
g2 (t) = 2e
So Y (t) = Ate
4t
t
)
)
Y1 (t) = t(Ae
Y2 (t) = Be
4t
)
t
+ Bet .
c) y 00 + 5y 0 + 4y = 2t3 e t .
Solution Since g(t) = 2t3 e t ,
Y (t) = t(At3 + Bt2 + Ct + D)e t .
d) y 00 + 5y 0 + 4y = t2 e
4t
sin t.
Solution Since g(t) = t2 e
4t
sin t,
Y (t) = (At2 + Bt + C)e
4t
cos t + (Dt2 + Et + F )e
4t
sin t.
After we learned how to find an appropriate form of the particular solution Y (t) by using the
method of undetermined coefficients, we are ready to solve nonhomogeneous equation
ay 00 + by 0 + cy = g(t).
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
124
The process of solving nonhomogeneous equation ay 00 + by 0 + cy = g(t)
1. Find the complementary solution yc (t) by solving the corresponding homogeneous equation ay 00 + by 0 + cy = 0.
2. Find appropriate form of the particular solution Y (t) by using the method of undetermined coefficients.
3. Substitute Y (t) back into ay 00 + by 0 + cy = g(t) to determine the coefficients of Y (t).
4. Then the general solution of ay 00 + by 0 + cy = g(t) is
y(t) = yc (t) + Y (t).
Example 102 (Spring 2011 Midterm Exam I).
Find the general solution of nonhomogeneous second order linear equation,
y 00
4y 0 + 5y = e2t
10t.
Solution First, find yc (t) by considering y 00
r2
)
4r + 5 = 0
4y 0 + 5y = 0. Its characteristic equation is
p
4 ± 16 20
4 ± 2i
r=
=
= 2 ± i.
2
2
Therefore yc (t) = c1 e2t cos t + c2 e2t sin t. Here we have g(t) = e2t
form of the particular solution Y as
10t. Then it gives the
Y (t) = Ae2t + (Bt + C).
Then Y 0 (t) = 2Ae2t + B and Y 00 = 4Ae2t . Since Y satisfies the given equation,
e2t
10t = Y 00
4Y 0 + 5Y
= (4Ae2t )
= (4Ae2t
= (4A
4(2Ae2t + B) + 5(Ae2t + Bt + C)
8Ae2t + 5Ae2t ) + 5Bt + ( 4B + 5C)
8A + 5A)e2t + 5Bt + (5C
2t
= Ae + 5Bt + (5C
4B)
4B)
Comparing the coefficients from both sides of the equation,
A = 1,
So A = 1, B =
2 and C =
5B =
10,
8
. So Y (t) = e2t
5
5C
2t
4B = 0.
8
. Thus the general solution is
5
y(t) = yc (t) + Y (t) = c1 e2t cos t + c2 e2t sin t + e2t
2t
8
.
5
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
125
Example 103 (Summer 2008 Midterm Exam I).
Consider the nonhomogeneous second order linear equation of the form
y 00
y0
2y = g(t).
a) Find its complementary solution, yc (t).
Solution The characteristic equation of y 00
r2
r
2=0
)
(r
y0
2y = 0 is
2)(r + 1) = 0
)
r = 2, 1.
So the complementary solution is
yc (t) = c1 e2t + c2 e t .
b) Find a particular solution Y (t) satisfying y 00
y0
2y = 3 sin 2t.
Solution Since
g(t) = 3 sin 2t
)
Y (t) = A cos 2t + B sin 2t.
Then Y 0 (t) = 2A sin 2t + 2B cos 2t and Y 00 (t) = 4A cos 2t
particular solution to y 00 y 0 2y = 3 sin 2t, we have
3 sin 2t = Y 00
Y0
4B sin 2t. Since Y is the
2Y
= ( 4A cos 2t
4B sin 2t)
= ( 4A
2B
2A) cos 2t + ( 4B + 2A
( 2A sin 2t + 2B cos 2t)
= ( 6A
2B) cos 2t + (2A
2(A cos 2t + B sin 2t)
2B) sin 2t
6B) sin 2t.
By the comparison of the coefficients, we have
6A
Then A =
3
and B =
20
2B = 0,
and
2A
6B = 3.
9
. So
20
Y (t) =
3
cos 2t
20
9
sin 2t.
20
c) Write down the correct choice of the form of particular solution that you would use to
solve the equation below using the method of undetermined coefficients.
y 00
Solution Since g(t) = t3 e
t
y0
2y = t3 e
t
7e2t cos 6t.
7e2t cos 6t, we have
Y (t) = t(At3 + Bt2 + Ct + D)e
t
+ (Ee2t cos 6t + F e2t sin 6t).
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
126
Example 104 (Spring 2003 Midterm Exam I).
a) Find the solution to the homogeneous di↵erential equation
y 00 + 3y 0
4y = 0.
Solution The characteristic equation is
r2 + 3r
)
4=0
(r + 4)(r
1) = 0
)
r=
4, 1.
So the general solution is
y = c1 e
b) Find the general solution to y 00 + 3y 0
Solution From part a), yc (t) = c1 e
g(t) = 50 sin 2t
4t
+ c2 et .
4y = 50 sin 2t.
4t
+ c2 et . Since
)
Y (t) = A cos 2t + B sin 2t.
Then Y 0 (t) = 2A sin 2t + 2B cos 2t and Y 00 (t) = 4A cos 2t
particular solution to y 00 + 3y 0 4y = 50 sin 2t, we have
50 sin 2t = Y 00 + 3Y 0
4B sin 2t. Since Y is the
4Y
= ( 4A cos 2t
4B sin 2t) + 3( 2A sin 2t + 2B cos 2t)
= ( 4A + 6B
4A) cos 2t + ( 4B
= ( 8A + 6B) cos 2t + ( 6A
6A
4(A cos 2t + B sin 2t)
4B) sin 2t
8B) sin 2t.
Comparing the coefficients, we have
8A + 6B = 0,
They give A = 3 and B =
to y 00 + 3y 0 4y = 0 is
and
4. So Y (t) =
y(t) = yc (t) + Y (t) = c1 e
6A
8B = 50.
3 cos 2t 4 sin 2t. Thus the general solution
4t
+ c2 et
3 cos 2t
4 sin 2t.
Example 105 (Summer 2007 Midterm Exam I).
Find the general solution of the nonhomogeneous linear equation,
y 00
2y 0 + 5y = 5t2 + 6t
Solution The characteristic equation of y 00
r2
2r + 5 = 0
)
12.
2y 0
5y = 0 is
p
2 ± 4 20
r=
= 1 ± 2i.
2
Thus yc (t) = c1 et cos 2t + c2 et sin 2t. Since g(t) = 5t2 + 6t
Y (t) = At2 + Bt + C.
12, then
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
127
Then Y 0 (t) = 2At + B and Y 00 (t) = 2A. Since Y satisfies the given equation,
5t2 + 6t
12 = Y 00
2Y 0 + 5Y
2(2At + B) + 5(At2 + Bt + C)
= (2A)
= 5At2 + ( 4A + 5B)t + (2A
2B + 5C)
Comparing the coefficients,
5A = 5,
Then A = 1, B = 2 and C =
4A + 5B = 6,
2A
2B + 5C =
2. Therefore Y (t) = t2 + 2t
12.
2 and so the general solution is
y(t) = yc (t) + Y (t) = c1 et cos 2t + c2 et sin 2t + t2 + 2t
2.
Example 106 (Summer 2011 Final Exam ).
Suppose it is known that y = 5t2 is a solution of y 00 + 16y = g(t). What is the general solution
of this equation?
Solution Let’s first consider the solution to y 00 + 16y = 0. To do this, we write down its
characteristic equation and find its roots.
r2 + 16 = 0
)
r = ±4i
)
y = c1 cos 4t + c2 sin 4t.
Notice that 5t2 is not in this form so g(t) 6= 0. So we have yc (t) = c1 cos 4t + c2 sin 4t and
Y (t) = 5t2 . Therefore the general solution for y 00 + 16y = g(t) is
y(t) = yc (t) + Y (t) = c1 cos 4t + c2 sin 4t + 5t2 .
Example 107 (Fall 2011 Final Exam ).
p
Suppose it is known that y1 = 6e3t + 5t2 et and y2 =
2e
second order linear equation
y 00 + by 0 + cy = g(t),
2t
+ 5t2 et are two solutions of a
where b and c are constants. Then what is the general solution of this equation?
(a) y = c1 e3t + c2 e
2t
+ 10t2 et
(b) y = c1 (6e3t + 5t2 et ) + c2 (
(c) y = c1 e3t + c2 e
2t
+ c 3 t2 e t
(d) y = c1 e3t + c2 e
2t
+ 5t2 et
p
2e
2t
+ 5t2 et )
Solution The first tactic is to consider whether a second order equation is homogeneous or
nonhomogeneous. Notice that the term 5t2 et which appears in both y1 and y2 is not in the
form of the solution of homogeneous equation in all 3 cases. This suggests that this second
order equation we deal with is nonhomogeneous with the particular solution
Y (t) = 5t2 et .
p
and
2e 2t are special case of the complementary
From the expression of y1 and y2 , 6e3t
solution. So
yc (t) = c1 e3t + c2 e
2t
.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
128
Therefore the general solution to y 00 + by 0 + cy = g(t) is
y = c1 e3t + c2 e
2t
+ 5t2 et
which gives (d) the right answer.
Example 108 (Summer 2012 Final Exam ).
Suppose it is known that y1 = 3e6t + 3t cos(5t) and y2 = 10te6t + 3t cos(5t) are two solutions
of a second order linear equation
y 00 + by 0 + cy = g(t),
where b and c are constants. Then what is the general solution of this equation?
Solution Observe that 3t cos(5t) is not the solution of any second order linear equation with
constant coefficients. This suggests that g(t) is not a zero function and
Y (t) = 3t cos(5t).
Additionally, 3e6t and 10te6t give us some hint that the complementary solution of the
nonhomogeneous equation is
yc (t) = c1 e6t + c2 te6t .
Hence, the general solution to y 00 + by 0 + cy = g(t) is
y(t) = yc (t) + Y (t) = c1 e6t + c2 te6t + 3t cos(5t).
Example 109 (Summer 2014 Final Exam ).
Find a second order linear equation which has
y(t) = c1 et + c2 e
2t
+ 3t
1
as its general solution.
Solution This is a solution for nonhomogeneous second linear equation where
yc (t) = c1 et + c2 e
2t
,
Y (t) = 3t
1.
In order to find the corresponding homogeneous equation, we take a look at the form of yc (t).
Clearly r1 = 1 and r2 = 2. And so
(r
1)(r + 2) = 0
)
r2 + r
2=0
)
y 00 + y 0
2y = 0.
So now we know that the equation we’re seeking is in the form of
y 00 + y 0
2y = g(t),
where g(t) is a non-zero function. To find g(t), we simply plug in Y (t) = 3t
equation above. Note that Y 0 (t) = 3 and Y 00 (t) = 0. So
Y 00 + Y 0
2Y = g(t)
)
0+3
2(3t
1) = g(t)
)
g(t) =
1 back into the
6t + 5.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
129
So the equation we’re looking for is
y 00 + y 0
2y =
6t + 5.
Example 110 (Fall 2009 Final Exam ).
Y1 (t) is a solution to the equation y 00 + p(t)y 0 + q(t)y = et . Y2 (t) is a solution to the equation
y 00 + p(t)y 0 + q(t)y = 3et . Which of the following is a solution to the equation
y 00 + p(t)y 0 + q(t)y = 2et ?
(a) y = Y1 (t) + Y2 (t).
(b) y = Y1 (t)
Y2 (t).
(c) y = 2Y1 (t).
(d) y = 2Y2 (t).
Solution Since Y1 (t) is a solution to the equation y 00 + p(t)y 0 + q(t)y = et , we have
Y100 + p(t)Y10 + q(t)Y1 = et .
Also Y2 (t) is a solution to the equation y 00 + p(t)y 0 + q(t)y = 3et , we get
Y200 + p(t)Y20 + q(t)Y2 = 3et .
Consider the following
• If we add both equations together, we then have
(Y1 + Y2 )00 + p(t)(Y1 + Y2 )0 + q(t)(Y1 + Y2 ) = 4et .
So Y1 (t) + Y2 (t) is a solution to y 00 + p(t)y 0 + q(t)y = 4et .
• If we subtract second equation to first equation, we have
(Y1
Hence Y1 (t)
Y2 )00 + p(t)(Y1
Y2 )0 + q(t)(Y1
Y2 ) =
Y2 (t) is a solution to y 00 + p(t)y 0 + q(t)y =
2et .
• If we multiply both sides of the first equation by 2, then
(2Y1 )00 + p(t)(2Y1 )0 + q(t)(2Y1 ) = 2et .
So 2Y1 (t) is a solution to y 00 + p(t)y 0 + q(t)y = 2et .
• If we multiply both sides of the second equation by 2, then
(2Y2 )00 + p(t)(2Y2 )0 + q(t)(2Y2 ) = 6et .
So 2Y2 (t) is a solution to y 00 + p(t)y 0 + q(t)y = 6et .
Therefore the correct answer is (c).
2et .
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
130
Exercises 3.6.1
1. Consider the nonhomogeneous second order linear equation of the form
y 00
8y 0 + 16y = g(t).
(a) Find its complementary solution, yc (t).
(b) Find the general solution of y 00
8y 0 + 16y = 10e2t .
(c) What is the correct form of particular solution that you would use to solve the equation
below using the Method of Undetermined coefficients? DO NOT ATTEMPT TO
SOLVE THE COEFFICIENTS.
y 00
8y 0 + 16y = 9 cos 4t
5t2 e4t
2. Consider the second order nonhomogeneous linear equation
y 00
8y 0
9y = 18t + 20.
(a) Find yc (t), the solution of its corresponding homogeneous equation.
(b) Find its general solution.
(c) What is the form of particular solution Y that you would use to solve the following
equation using the Method of Undetermined coefficients? DO NOT ATTEMPT TO
SOLVE THE COEFFICIENTS.
y 00
8y 0
9y = te
t
+ 2e9t sin 4t.
3. Consider the second order nonhomogeneous linear equation
y 00
10y 0 + 25y = 6e2t .
(a) Find yc (t), the solution of its corresponding homogeneous equation.
(b) Use the Method of Undetermined Coefficients to find its general solution.
(c) What is the form of particular solution Y that you would use to solve the following
equation using the Method of Undetermined coefficients? DO NOT ATTEMPT TO
SOLVE THE COEFFICIENTS.
y 00
10y 0 + 25y = 9te5t
3e
5t
cos t.
4. Consider the second order nonhomogeneous linear equation
y 00 + 4y 0 = 2et + t.
(a) Find yc (t), the solution of its corresponding homogeneous equation.
(b) Find the general solution of the equation.
(c) What is the form of particular solution Y that you would use to solve the following
equation using the Method of Undetermined coefficients? DO NOT ATTEMPT TO
SOLVE THE COEFFICIENTS.
y 00 + 4y 0 = 11te
4t
(sin 6t
3 cos 6t) + 7e
5. What is the form of a particular solution of the equation:
y 00
Do not solve for the constants.
6y 0 + 9y = t2 + e3t
4 cos t?
4t
.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
131
6. What is the form of the particular solution of
y 00 + 4y 0 + 4y = te
(a) y(x) = (At3 + Bt2 )e
2t
(b) y(x) = (At2 + Bt)e
(c) y(x) = (At + B)e
2t
(d) y(x) = (At3 + Bt2 )e
7. Given that y(t) =
2t
2t
+ 2t2 cos 2t
3?
+ (Ct2 + Dt + E) cos 2t
+ (Ct2 + Dt + E) cos 2t + F
+ (Ct2 + Dt + E) cos 2t + (F t2 + Gt + H) sin 2t + I
2t
+ (Ct2 + Dt + E) cos 2t + (F t2 + Gt + H) sin 2t + I
2
1
+ 2 is a solution to the di↵erential equation
t
t
y 00 + 2y 0 + 4y =
8
6
+ 4,
t
t
find its general solution.
c1
c2
2
1
+ 4 + + 2
t
t
t
t
p
(b) y = c1 e t cos 3t + c2 e
(a) y =
p
2
1
+ 2
t
t
p
p
8
6
t
t
(c) y = c1 e cos 3t + c2 e sin 3t + + 4
t
t
c1
c2
(d) y =
+ 2
t
t
t
sin
3t +
8. In Parts a) and b) determine the form of a particular solution yp = yp (t) having the least
number of unknown constants. DO NOT DETERMINE the unknown constants appearing in your answers in Parts i) and ii).
(a) y 00
14y 0 + 49y = 2t2 e7t
(b) y 00
50y 0 + 49y = 3tet
9. Of what form will the particular solution to the following di↵erential equation be? Do not
solve the equation.
y 00 4y 0 + 4y = e2t + t2 e3t sin(2⇡t)
(a) Ae2t + Bt2 e3t + Cte3t + De3t + E sin(2⇡t) + F cos(2⇡t)
(b) Ate2t + Bt2 e3t + Cte3t + De3t + E sin(2⇡t) + F cos(2⇡t)
(c) At2 e2t + Bt2 e3t + Cte3t + De3t + E sin(2⇡t) + F cos(2⇡t)
(d) Ae2t + Bt2 e3t + E sin(2⇡t) + F cos(2⇡t)
10. Solve the following initial value problem:
y 00
5y 0
14y =
14t2
10t
26,
y(0) = 0,
y 0 (0) = 13.
11. When using the method of undetermined coefficients to solve the following equation, what
is the form of the particular solution? Do not solve for the constants.
y 00
y0
2y = t2 e3t + 3e
t
+ te2t cos(2t)
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
132
12. Consider the nonhomogeneous second order linear equation of the form
y 00
4y 0 + 8y = g(t).
(a) Find its complementary solution, yc (t).
For each of parts b) through d), write down the correct choice of the form of particular
solution that you would use to solve the given equation using the Method of Undetermined
Coefficients? DO NOT ATTEMPT TO SOLVE THE COEFFICIENTS.
(a) y 00
(b) y 00
(c) y 00
4y 0 + 8y = 2e2t 5t2 + sin 2t
4y + 8y 0 = e2t sin 2t + 1
4y + 8y 0 = t2 e t cos 5t
13. Consider the second order nonhomogeneous linear equation
y 00
2y 0
3y = 3te2t .
(a) Find yc (t), the solution of its corresponding homogeneous equation.
(b) Find its general solution.
(c) Find the solution satisfying the conditions y(0) = 1 and y 0 (0) = 1.
14. Consider the second order nonhomogeneous linear equation
y 00 + 3y 0 + 2y = 20 sin(2t) + 4t.
(a) Find yc (t), the solution of its corresponding homogeneous equation.
(b) Find its general solution.
(c) Find the solution satisfying the conditions y(0) = 0 and y 0 (0) = 2.
Answers
1. (a) yc (t) = c1 e4t + c2 te4t , (b) y = c1 e4t + c2 te4t + 25 e2t , (c) Y = A cos 4t + B sin 4t + (Ct4 +
Dt3 + Et2 )e4t
2. (a) yc = c1 e9t + c2 e t , (b) y = c1 e9t + c2 e
De9t sin(4t)
t
4
9,
2t
(c) Y = (At2 + Bt)e
t
+ Ce9t cos(4t) +
3. (a) yc = c1 e5t + c2 te5t , (b) y = c1 e5t + c2 te5t + 23 e2t , (c) Y = (At3 + Bt2 )e5t + Ce
De 5t sin(t)
4. (a) yc = c1 + c2 e 4t , (b) y = c1 + c2 e
(Ct + D)e 4t sin 6t + Ete 4t
4t
+ 25 et + 18 t2
1
16 t,
(c) Y = (At + B)e
5t
cos(t) +
4t
cos 6t +
5. Y = At2 + Bt + C + Dt2 e3t + E cos t + F sin t
6. (d)
7. (b)
8. (a) (At2 + Bt + c)e7t t2 , (b) (At + B)et t
9. (c)
10. y(t) = e7t
3e
2t
+ t2 + 2
11. Y (t) = (At2 + Bt + C)e3t + Dte
t
+ (Et + F )e2t cos 2t + (Gt + H)e2t sin 2t
12. (a) yc = c1 e2t cos 2t+c2 e2t sin 2t, (b) Ae2t +Bt2 +Ct+D+E cos 2t+F sin 2t (c) Ate2t cos 2t+
Bte2t sin 2t + C, (d) (At2 + Bt + C)e t cos 5t + (Dt2 + Et + F )e t sin 5t
13. (a) yc = c1 e
t
+ c2 e3t , (b) Y (t) = c1 e
t
+ c2 e3t
14. (a) yc = c1 e 2t + c2 e t , (b) Y (t) = c1 e 2t + c2 e
4e 2t + 10e t sin(2t) 3 cos(2t) + 2t 3
(t + 23 )e2t , (c) Y (t) = 23 e
t
sin(2t)
t
+ e3t
3 cos(2t) + 2t
(t + 23 )e2t
3, (c) Y (t) =
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.6.2
133
Variation of Parameters*
We use this method for finding a particular solution Y (t) of the nonhomogeneous equation
y 00 + p(t)y 0 + q(t)y = g(t),
(3.9)
once the solutions of the corresponding homogeneous equation
y 00 + p(t)y 0 + q(t)y = 0
(3.10)
are known. Let y1 (t) and y2 (t) be a known fundamental set of solutions of the homogeneous
equation (3.9). We’re looking for a particular solution Y (t) of the nonhomogeneous equation
(3.10) of the form
Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t).
(3.11)
In other words, we’re trying to find functions u1 (t) and u2 (t) so that u1 (t)y1 (t) + u2 (t)y2 (t) is a
solution of (3.9). Let’s use (3.11) to compute Y 0 ,
Y 0 (t) = (u1 y1 + u2 y2 )0
= (u1 y10 + u01 y1 ) + (u2 y20 + u02 y2 )
= (u1 y10 + u2 y20 ) + (u01 y1 + u02 y2 ).
Notice that Y 00 will contain no second-order derivatives of u1 and u2 if we impose the condition
u01 y1 + u02 y2 = 0.
on the functions u1 (t) and u2 (t). Next let’s consider
0
Y 00 + p(t)Y 0 + q(t)Y = (u1 y10 + u2 y20 ) + p(t) (u1 y10 + u2 y20 ) + q(t) (u1 y1 + u2 y2 )
= u1 y100 + u01 y10 + u2 y200 + u02 y20 + u1 p(t)y10 + u2 p(t)u2 y20 + u1 q(t)y1 + u2 q(t)y2
= (u01 y10 + u02 y20 ) + u1 (y100 + p(t)y10 + q(t)y1 ) + u2 (y200 + p(t)y20 + q(t)y2 )
= u01 y10 + u02 y20 ,
since y1 and y2 are solutions of homogeneous equation (3.10). In summary, Y (t) is a particular
solution of the nonhomogeneous equation (3.9) if u1 and u2 satisfy the following two equations,
u01 y1 + u02 y2 = 0
(3.12)
u01 y10
(3.13)
+
u02 y20
= g(t).
Multiplying (3.12) by y20 , and (3.13) by y2 , we get
u01 y1 y20 + u02 y2 y20 = 0
u01 y10 y2 + u02 y20 y2 = g(t)y2 .
Subtraction gives
u01 (y1 y20
y10 y2 ) =
g(t)y2
)
u01 (t) =
g(t)y2
=
y1 y20 y10 y2
g(t)y2
.
W (y1 , y2 )(t)
By similar fashion, we multiply (3.12) by y10 , and (3.13) by y1 , we get
u01 y1 y10 + u02 y2 y10 = 0
u01 y10 y1 + u02 y20 y1 = g(t)y1 .
After subtraction, we get
u02 (y1 y20
y10 y2 ) = g(t)y1
)
u02 (t) =
g(t)y1
g(t)y1
=
.
0
0
y1 y2 y 1 y 2
W (y1 , y2 )(t)
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
134
Now we have
g(t)y2
g(t)y1
, u02 (t) =
.
W (y1 , y2 )(t)
W (y1 , y2 )(t)
Therefore by integrating, we obtain the formula for u1 and u2 ,
Z
Z
g(t)y2
g(t)y1
u1 (t) =
dt, u2 (t) =
dt.
W (y1 , y2 )(t)
W (y1 , y2 )(t)
u01 (t) =
(3.14)
Then a particular solution can be found by (3.11), which is,
Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t).
In the end, the solution to nonhomogeneous equation (3.9) is
y(t) = yc (t) + Y (t)
= (c1 y1 + c2 y2 ) + (u1 (t)y1 (t) + u2 (t)y2 (t))
✓
◆
Z
Z
g(t)y2
g(t)y1
) y(t) = (c1 y1 + c2 y2 ) +
y1
dt + y2
dt .
W (y1 , y2 )(t)
W (y1 , y2 )(t)
Since c1 and c2 are arbitrary, this method is known as the method of variation of parameters.
Example 111. Find the general solution of the equation
y 00 + y = tan t,
⇡
⇡
<t< .
2
2
Solution Let’s consider first its corresponding homogeneous equation,
y 00 + y = 0.
Its characteristic equation is r2 +1 = 0 which gives r = ±i as its roots. So the complementary
solution is
yc (t) = c1 cos t + c2 sin t,
where y1 (t) = cos t and y2 (t) = sin t. Moreover its Wronskian is
W (y1 , y2 )(t) =
cos t
sin t
sin t
= sin2 t + cos2 t = 1.
cos t
Now we use the formulas in (3.14), with g(t) = tan t,
Z
Z
g(t)y2
u1 (t) =
dt =
tan t sin t dt
W (y1 , y2 )(t)
Z
Z
sin2 t
cos2 t 1
=
dt =
dt
cos t
cos t
Z
= (cos t sec t) dt = sin t ln(sec t + tan t)
and
u2 (t) =
Z
g(t)y1
dt =
W (y1 , y2 )(t)
Therefore a particular solution is
Z
tan t cos t dt =
Z
sin t dt =
Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t)
= (sin t
=
ln(sec t + tan t)) cos t
cos t ln(sec t + tan t).
cos t sin t
cos t.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
135
Hence the solution to y 00 + y = tan t is
y(t) = yc (t) + Y (t) = c1 cos t + c2 sin t
cos t ln(sec t + tan t).
Example 112. Find the solution of the equation
y 00
3y 0 + 2y = sin e t .
Solution Let’s first consider the corresponding homogeneous equation,
y 00
3y 0 + 2y = 0.
Its characteristic equation is r2 3r + 2 = (r
Then the complementary solution is
1)(r
2) = 0 which gives r = 1, 2 as its roots.
yc (t) = c1 et + c2 e2t ,
where y1 (t) = et and y2 (t) = e2t. Moreover its Wronskian is
W (y1 , y2 )(t) =
et
et
e2t
= 2e3t
2e2t
e3t = e3t .
Now we use the formulas in (3.14), with g(t) = sin e t ,
Z
Z 2t
g(t)y2
e sin e t
u1 (t) =
dt =
dt
W (y1 , y2 )(t)
e3t
Z
Z
d(e t )
=
e t sin e t dt =
e t sin e t
e t
t
= cos e
and
Z t
g(t)y1
e sin e t
dt =
dt
W (y1 , y2 )(t)
e3t
Z
Z
d(e t )
= e 2t sin e t dt = e 2t sin e t
e t
Z
=
e t sin e t d(e t ) = e t cos e t sin e t .
u2 (t) =
Z
Therefore the particular solution is
Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t)
=
et cos e
t
+ e2t e
t
=
et cos e
t
+ et cos e
=
e2t sin e t .
cos e
t
t
sin e
e2t sin e
t
t
Hence the general solution is
y(t) = yc (t) + Y (t) = c1 et + c2 e2t
e2t sin e t .
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
1
Example 113. If y1 (t) = t and y2 (t) = t
136
are two solutions of the di↵erential equation
t2 y 00 + ty 0
y = 0,
find the general solution of the equation
t2 y 00 + ty 0
t 6= 0..
y = t,
Solution The Wronskian of y1 and y2 is
W (y1 , y2 )(t) =
t
1
1
t
t
2
=
1
t
t
1
=
2t
1
.
Since their Wronskian is nonzero, the complementary solution is
yc (t) = c1 t + c2 t
1
.
To find g(t), we need to rewrite the original equation to the form y 00 + p(t)y 0 + q(t)y = g(t)
by diving by t2 ,
y 00 + t 1 y 0 t 2 y = t 1 .
So g(t) = t
1
. To find u1 and u2 , we apply formulas in (3.14),
Z
Z
Z
g(t)y2
t 1·t 1
1
1
ln t
u1 (t) =
dt =
dt =
dt =
W (y1 , y2 )(t)
2t 1
2
t
2
and
u2 (t) =
Z
g(t)y1
dt =
W (y1 , y2 )(t)
Therefore the particular solution is
Z
1
t
Y (t) = u1 (t)y1 (t) + u2 (t)y2 (t) =
2t
·t
1
1
2
dt =
t2
·t
4
ln t
·t
2
Z
1
t dt =
t2
.
4
t ln t
2
t
.
4
=
Hence the general solution is
y(t) = yc (t) + Y (t) = c1 t + c2 t
3.7
1
+
t ln t
2
t
.
4
Modeling with Second Order Di↵erential Equation: Mechanical Vibrations
A mass m is hooked at the end of the spring. Its weight stretches the spring by a length L to
reach the equilibrium position. After that the mass is then set in the motion. We want to find
the displacement of the mass from its equilibrium position at any time.
Let u(t), measured positive downward, denote the displacement of the mass from its equilibrium
position at time t. The di↵erential equation governing the movement of the mass is
mu00 (t) + u0 (t) + ku(t) = F (t),
where constant m, k > 0,
u(t0 ) = u0 ,
0 and
• m = mass (m > 0) (kg or slug) which can be found from
m=
w
.
g
u0 (t0 ) = u00 ,
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
Figure 3.1: A spring mass system
• w = weight of the object. (N or lb)
• g = gravitational constant (10 m/s2 or 32 ft/s2 ).
•
= damping constant (
0) which can be found from
=
Fd
.
u0
• Fd = damping force or resistive force.
• u0 = velocity.
• k = spring constant (k > 0) which can be found from
k=
mg
w
= .
L
L
• L = elongation of the spring caused by the weight of mass. (metre or ft)
• F (t) = external force.
• u(t0 ) = initial displacement of the mass.
• u0 (t0 ) = initial velocity of the mass.
In conclusion, we have
137
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
138
Modeling for Mechanical Vibrations
mu00 (t) + u0 (t) + ku(t) = F (t),
u(t0 ) = u0 ,
u0 (t0 ) = u00 ,
is the IVP for mechanical vibrations in which its coefficients can be found from
m=
w
,
g
=
Fd
,
u0
k=
mg
w
= ,
L
L
where
• m = mass (m > 0) (kg or slug).
• w = weight of the object (N or lb).
• g = gravitational constant. (10 m/s2 or 32 ft/s2 )
•
= damping constant (
0).
• Fd = damping force or resistive force.
• u0 = velocity.
• k = spring constant (k > 0).
• L = elongation of the spring caused by the weight of mass. (metre or ft)
• F (t) = external force.
• u(t0 ) = initial displacement of the mass.
• u0 (t0 ) = initial velocity of the mass.
Example 114.
A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is displaced in additional
6 in the positive direction and then released. The mass is in a medium that exerts a vicious
resistance of 6 lb when the mass has a velocity of 3 ft/s. Under the assumption, formulate
the IVP that governs the motion of the mass.
Solution We will measure t in second and displacement u in feet. The initial conditions are
u(0) =
6
1
= ,
12
2
u0 (0) = 0.
We also have
m=
w
4 lb
1
=
2 = 8,
g
32 ft/s
=
Fd
6 lb
=
= 2,
u0
3 ft/s
k=
w
4 lb
= 2
= 24.
L
12 ft
Nothing is said about the external force, F (t) = 0. So the IVP describes the u(t) is
1 00
u + 2u0 + 24u = 0,
8
u(0) =
1
,
2
u0 (0) = 0.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
139
Now we ’re taking a closer look to the solution of the modeling of mechanical vibrations
mu00 (t) + u0 (t) + ku(t) = F (t)
in a few special cases.
3.7.1
Undamped Free Vibrations
Suppose that the system has neither damping ( = 0) nor external force (F (t) = 0). Then the
equation becomes
mu00 + ku = 0.
q
k
Its characteristic equation is mr2 + k = 0 which gives roots r1,2 = ± m
i.
So the general solution is
u(t) = c1 cos w0 t + c2 sin w0 t,
(3.15)
q
k
where w0 = m
.
• w0 is called the natural frequency of the system.
• The (natural) period of the motion T is given by
r
2⇡
m
T =
= 2⇡
.
w0
k
To get a clear picture of what’s going on, we rewrite the solution of u(t) to the form
u = R cos(w0 t
),
(3.16)
where R > 0 is chosen. Note that Equation (3.16) can be written as
u = R cos(w0 t) cos( ) + R sin(w0 t) sin( ).
(3.17)
Comparing Equations (3.15) and (3.17), we have
c1 = R cos ,
These two relations give,
c2
R cos
=
c1
R sin
c2 = R sin .
= tan .
Also
c21 + c22 = (R cos )2 + (R sin )2 = R2 (cos2 + sin2 ) = R2
• The angle
)
R=
is called the phase of displacement.
• R is the amplitude of the motion.
Using the new form of the solution, it is easier to graph as it is shown below.
We summarize all the details here:
q
c21 + c22 .
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
140
Figure 3.2: Undamped free vibration
Undamped Free Vibrations ( = 0, F (t) = 0)
If we assume that the system has neither damping ( = 0) nor external force (F (t) = 0),
mu00 + ku = 0
gives the general solution as
u(t) = c1 cos w0 t + c2 sin w0 t = R cos(w0 t
where
tan =
c2
,
c1
R=
)
)
q
c21 + c22 ,
• w0 = the natural frequency of the system, w0 =
r
k
.
m
2⇡
• T = the (natural) period of the motion, T =
= 2⇡
w0
•
lim u(t) does not exist.
t!1
r
m
.
k
= the phase of displacement.
• R = amplitude of the motion.
Example 115 (Fall 2007 Math 250 Midterm Exam I).
A mass of 2 kg stretches a spring 10 cm. The mass is pulled down 20 cm from the equilibrium
position and then released with downward initial velocity 2 m/s. Ignore air resistance and
take g=10 m/s2 .
a) Write down the di↵erential equation governing the motion of the mass.
Solution From the information above, we have
u(0) =
20
1
= m,
100
5
u0 (0) = 2 m/s2 ,
m = 2 kg,
L=
10
1
=
m
100
10
Since we ignore air resistance, = 0. There is no external force mentioned so F (t) = 0.
Also we can find the spring constant from,
k=
mg
2(10)
= 1 = 200.
L
10
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
141
So the IVP governing the motion of the mass is
2u00 + 200u = 0,
u(0) =
1
,
5
u0 (0) = 2.
b) Determine the position of the mass at any time t.
Solution From part a), we have u00 + 100u = 0. Its characteristic equation is
r2 + 100 = 0
)
r = ±10i.
So we have u(t) = c1 cos 10t+c2 sin 10t. Hence u0 (t) =
apply initial conditions,
1
5
u0 (0) = 2
u(0) =
Therefore c1 = c2 =
)
)
10c1 sin 10t+10c2 cos 10t. Then
1
5
10c2 = 2.
c1 =
1
and then
5
u(t) =
1
1
cos 10t + sin 10t.
5
5
c) Find an amplitude, frequency, period and phase of the motion.
Solution
r
r
k
200
• Frequency: w0 =
=
= 10.
m
2
2⇡
⇡
2⇡
• Period: T =
=
= .
w0
10
5
r
p
q
1 2
1 2
2
2
2
• Amplitude: R = c1 + c2 = ( ) + ( ) =
.
5
5
5
• Phase: Since tan =
c2
=
c1
1
5
1
5
=1
)
= tan
1
1=
⇡
.
4
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.7.2
142
Damped Free Vibrations
If we include the e↵ect of damping to the system ( > 0) without the external force (F (t) = 0),
then the di↵erential equation of u(t) is
mu00 + u0 + ku = 0.
The roots of corresponding characteristic equation mr2 + r + k = 0 is
p
2
±
4mk
r1,2 =
.
2m
Depending on the sign of
1.
2
2
4mk, the solution u is one of the following form:
4mk > 0. This mass-spring system is called overdamped.
In this case the roots of characteristic equation are two distinct real roots, both actually are
negative. Hence
u(t) = c1 er1 t + c2 er2 t ,
where both r1 6= r2 are negative. Then lim u(t) = 0. This system does not oscillate and it
t!1
could cross the equilibrium position at most once.
2.
2
4mk = 0. We call this mass-spring system critically damped.
For this case, r = r1 = r2 =
2m
which are repeated negative real roots. So
u(t) = c1 ert + c2 tert and hence lim u(t) = 0.
t!1
Figure 3.3: Critically damped motions: u00 + u0 + 0.25u = 0; u = (A + Bt)e
t
2
Like the first case, the mass could cross its equilibrium position at most one time.
3.
2
4mk < 0 (the most interesting case). A system is called underdamped. This case we
p
2
4mk
r1,2 =
±
i.
2m
2m
So the solution is
u(t) = c1 e t cos µt + c2 e t sin µt
p
2
4mk
where =
< 0 and µ =
> 0. Since < 0, we have lim u(t) = 0. Also we
t!1
2m
2m
can use the same tactic that we used in undamped free vibration case to rewrite the solution
to
u(t) = Re t cos(µt
),
p
c2
where R = c21 + c22 and tan = . Here
c1
have
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
143
• µ is called the quasi frequency of the system.
• Td is called the quasi period which can be found from Td =
2⇡
µ .
Figure 3.4: Underdamped motion
In this case the mass crosses its equilibrium position infinitely many times. The displacement
function u(t) is oscillating (not periodic) but the amplitude is decaying exponentially.
Note here that in all 3 cases, the displacement function u(t) tends to zero as t ! 1 reflecting the
fact that we we include the e↵ect of damping to the system.
Example 116 (Spring 2008 Midterm Exam I).
A mass-spring system is described by the initial value problem,
u00 + 4u0 + 20u = 0,
u(0) = 2,
u0 (0) = 0.
a) Find the real-valued particular solution of the initial value problem.
Solution Consider its characteristic equation,
p
4 ± 16
0 = r2 + 4r + 20 ) r =
2
80
=
4 ± 8i
=
2
2 ± 4i.
Therefore the general solution is
y(t) = c1 e
2t
cos 4t + c2 e
2t
sin 4t.
Then apply the initial conditions to receive c1 = 2 and c2 = 1. (check!) So the particular
solution is
y(t) = 2e 2t cos 4t + e 2t sin 4t.
b) What are the quasi-frequency and quasi-period of the mass-spring system?
Solution The quasi frequency can be found from
p
p
2
4(1)(20)
4mk
µ=
=
2m
2
42
Therefore the quasi period of the system is
Td =
2⇡
2⇡
⇡
=
= .
µ
4
2
=
p
64
= 4.
2
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
144
c) True or False: Every nonzero solution of this mass-spring system will cross the equilibrium position more than once.
Solution True since this system is underdamped ( 2 4mk = 42 4(1)(20) =
(Indeed, it crosses the equilibrium position infinitely many times.)
64 < 0).
Example 117 (Summer 2010 Midterm Exam I).
A mass-spring system is described by the following equation,
3u00 + u0 + 12u = 0.
a) If
= 6, what is the quasi-period of this mass-spring problem?
Solution The system is underdamped since
2
4mk = 62
Then its quasi frequency µ is
p
µ=
So the quasi period is Td =
4(3)(12) = 36
108 < 0.
p
p
108
6 3 p
=
=
= 3.
2(3)
6
2
4mk
2m
144 =
2⇡
2⇡
=p .
µ
3
b) For which values of , the system is overdamped.
Solution The system is over damped when
2
4mk > 0
2
)
)
2
> 12 or
<
4(3)(12) > 0
144 > 0.
And so
(
12)( + 12) > 0
But for the mass-spring system,
c) For which values of
times.
)
0. Hence
12.
> 12 is the final answer.
, there is a solution that crosses equilibrium position exactly 5
Solution None, it could not happen since a solution either crosses equilibrium position
at most once (for overdamped or critically damped system) or infinitely many times (for
underdamped system).
Example 118 (Summer 2011 Midterm Exam I).
A mass weighing 0.25 kg stretches a spring 1.25 m. The mass-spring system has a damping
constant = 12 kg/s. At t = 0 the mass is displaced an additional 50 cm downward from
its equilibrium position and set in motion with an upward velocity of 2 m/s. (You may use
g = 10 m/s2 as the gravitational constant.)
a) Write an initial value problem that describes the motion of the mass.
Solution We have m = 0.25 =
1
4
kg, L = 1.25 =
5
4
m,
=
1
2
kg/s. Since there is no
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
145
external force mentioned here, F (t) = 0. Note that the spring constant is
k=
1
4
mg
=
L
· 10
5
4
= 2.
So the equation describing the motion is
mu00 + u0 + ku = F (t)
1 00 1 0
u + u + 2u = 0
4
2
)
For the initial condition, we have
1
m (positively downward),
2
2 m/s (negatively upward)
u(0) = 50 cm =
u0 (0) =
So the IVP governing the motion of the mass is
1 00 1 0
u + u + 2u = 0,
4
2
u(0) =
1
,
2
u0 (0) =
2.
b) Determine the system’s quasi frequency and quasi period.
Solution The quasi frequency is
q
q
p
1
1 2
4(
)(2)
(
)
2
2
4mk
4
2
µ=
=
=
1
1
2m
2( 4 )
2
1
4
=
q
7
4
1
2
=
p
7 2 p
· = 7.
2 1
Hence the quasi period of the system is
2⇡
2⇡
=p .
µ
7
Td =
Example 119 (Fall 2012 Midterm Exam I).
A mass-spring system is described by the equation
5u00 + u0 + ku = F (t).
a) Suppose the mass originally stretched the spring 2 m to reach its equilibrium position.
What is the spring constant k? (Assume g = 10 m/s2 to be the gravitational constant.)
Solution From the equation above, m = 5 kg. Also L = 2 m. Therefore the spring
constant is
mg
5(10)
k=
=
= 25.
L
2
b) Suppose k = 45. For what value(s) of
would this system be critically damped?
Solution The system is critically damped when
2
4mk = 0
)
2
)
4(5)(45) = 0
2
900 = 0.
And so
(
30)( + 30) = 0
But for the mechanical vibration system,
)
= 30, 30.
0, hence
= 30 only.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
c) Suppose
146
= 0 and k = 400. What is the natural frequency of this system?
Solution The natural frequency of this system is
r
r
p
k
400 p
w0 =
=
= 80 = 4 5.
m
5
A second mass-spring system is described by the initial value problem
2u00 + 12u0 + 20u = 0,
u0 (0) = 2.
u(0) = 0,
3⇡
.
2
Solution Consider the corresponding characteristic equation of the equation above,
d) Find the position of the mass when t =
2r2 + 12r + 20 = 0
r2 + 6r + 10 = 0.
)
Now we use the quadratic formula to find its roots,
p
p
6 ± 62 4(1)(10)
6±
4
r=
=
=
2
2
3 ± i.
So we have
u(t) = c1 e
3t
cos t + c2 e
3t
3t
cos t + c2 e
sin t.
Note here that its first derivative is
u0 (t) = c1
e
3t
sin t
3e
3t
cos t
3e
3t
sin t .
Apply initial conditions,
u(0) = 0
u0 (0) = 2
)
)
0 = c1 ,
2=
3c1 + c2 .
Then c1 = 0 and c2 = 2. So we have
u(t) = 2e
In particular if we evaluate it at t =
u(
3⇡
) = 2e
2
3( 3⇡
2 )
3⇡
2 ,
sin
3t
sin t.
we then get
3⇡
= 2e
2
9⇡
2
( 1) =
2e
9⇡
2
.
e) What is the quasi-frequency of this system?
Solution µ = 1 by looking at the angel of trig functions of the solution above.
Example 120 (Fall 2000 Midterm Exam II).
A mass of 2 kg stretches a spring 1 m. The mass is in a medium that exerts a viscous
resistance of 8 newtons when the velocity of the mass is 2 m/sec. The mass ispstretched from
its equilibrium position 1 m and set in motion with a downward velocity of (3 3 1) m/sec.
Let g = 10 m/sec2 .
a) Set up and solve the initial value problem for u(t), the displacement of the mass from
its equilibrium position.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
147
Solution From the information above, we have m = 2 kg, g = 10 m/sec2 , L = 1 m,
Fd = 8 N and u0 = 2 m/sec. Then we also have
mg
2(10)
=
= 20
L
1
Fd
8
= 0 = = 4.
u
2
k=
p
Moreover the initial displacement is 1, and the initial velocity is 3 3
p
u(0) = 0 and u0 (0) = 3 3 1.
1. So
Both of them are positive since the direction of the action is downward. Therefore the
required initial value problem for u(t) is
p
2u00 + 4u0 + 20u = 0, u(0) = 1, u0 (0) = 3 3 1.
To solve the above equation, we consider its characteristic equation.
p
2 ± 22 4(1)(10)
2 ± 6i
2
r + 2r + 10 = 0 ) r =
=
=
2
2
1 ± 3i.
So the general solution is
t
u(t) = c1 e
cos 3t + c2 e
t
sin 3t.
Note that
u0 (t) = c1 ( 3e
t
sin 3t
e
t
cos 3t) + c2 (3e
3t
cos 3t
e
3t
sin 3t).
Then we impose two initial conditions here,
u(0) = 0
p
u0 (0) = 3 3
Thus c1 = 1 and c2 =
p
)
1
)
p
3 3
1 = c1
1=
c1 + 3c2 .
3. Consequently its solution is
p
u(t) = e t cos 3t + 3e t sin 3t.
b) What is the limit of the solution as t ! 1?
p
Solution lim u(t) = lim (e t cos 3t + 3e t sin 3t) = 0.
t!1
t!1
c) How many times does the mass pass through its equilibrium position? Justify your
answer.
Solution The system is underdamped, therefore, the mass will oscillate (i.e., pass
through the equilibrium position) infinitely many times.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
148
Example 121 (Summer 2002 Midterm Exam II).
5
A mass of 1 kg stretches a spring m. The system has a damping constant of 4 kg/s. The
8
mass is pulled down 1 m from its equilibrium position and released. You may take g = 10
m/s2 .
a) Set up an initial value problem modeling this system.
5
Solution We have m = 1 kg, L = m, = 4 kg/s. Moreover,
8
mg
(1)(10)
=
= 16.
5
L
8
k=
So the initial value problem modeling the above system is
u00 + 4u0 + 16u = 0,
u0 (0) = 0.
u(0) = 1,
b) Solve this initial value problem.
Solution Write down its characteristic equation,
p
4 ± 42 4(1)(16)
r2 +4r+16 = 0 ) r =
=
2
p
4 ± 48
=
2
p
4±4 3
=
2
p
2±2 3i.
So the general solution of this system is
2t
u(t) = c1 e
p
cos 2 3t + c2 e
2t
p
sin 2 3t.
Note that
p
u0 (t) = c1 ( 2 3e
2t
p
sin 2 3t 2e
2t
p
p
cos 2 3t)+c2 (2 3e
2t
p
cos 2 3t 2e
2t
p
sin 2 3t).
Apply initial conditions,
)
u(0) = 1
0
)
u (0) = 0
1 = c1
0=
p
2c1 + 2 3c2 .
1
Thus c1 = 1 and c2 = p . Therefore the solution is
3
u(t) = e
2t
p
1
cos 2 3t + p e
3
2t
p
sin 2 3t
c) At what quasi-frequency does the system oscillate?
p
Solution µ = 2 3 rad/sec by looking at the angel of trig functions of the solution
above.
d) In the absence of damping (i.e.
= 0), what is the systems natural frequency?
Solution The natural frequency is
r
w0 =
k
=
m
r
16
= 4 rad/sec.
1
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
149
Example 122 (Spring 2004 Midterm Exam II).
Consider a spring-mass system where the spring has a Hooke’s constant of 2 N/m, and where
the damping constant is 2 N·s/m
a) Give a specific condition on a mass m, in kg, which guarantees that the system will be
underdamped.
Solution Note that k = 2 and
2
4mk < 0
= 2. The system is underdamped when
22
)
4m(2) < 0
)
4 < 8m
)
m>
1
(kg).
2
b) For m = 1 kg and an initial displacement of 10 cm , set up and solve an initial value
problem for the equation of motion if the mass is simply released at t = 0.
1
Solution We have m = 1 kg, u(0) = 10 centimeters =
meter and u0 (0) = 0. So the
10
IVP for this system is
u00 + 2u0 + 2u = 0,
u(0) =
1
,
10
u0 (0) = 0.
In order to solve it, let’s first consider its characteristic equation.
p
2 ± 22 4(1)(2)
2 ± 2i
2
r + 2r + 2 = 0 ) r =
=
=
2
2
1 ± i.
Hence the general solution is
u(t) = c1 e
t
cos t + c2 e
t
t
cos t) + c2 (e
sin t.
Its first derivative is
u(t) = c1 ( e
t
sin t
e
t
cos t
e
t
sin t).
Then we impose initial conditions,
1
10
u0 (0) = 0
u(0) =
Thus c1 =
)
)
1
= c1
10
0 = c1 + c2 .
1
1
and c2 =
. Then its solution is
10
10
u(t) =
1
e
10
t
cos t +
1
e
10
t
sin t.
c) What is the quasi-frequency of the motion you determined in b)?
Solution µ = 1 rad/sec.
d) If the same 1 kg mass were attached to a weaker spring with Hooke’s constant
would the system be overdamped, underdamped, or critically damped?
Solution It’s overdamped since
2
4mk = 22
1
4(1)( ) = 4
3
4
> 0.
3
1
3
N/m,
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.7.3
150
Undamped Forced Vibrations
Now we assume that the system has no damping ( = 0) but we allow external force to a↵ect the
system (F (t) 6= 0). The motion equation now then becomes
mu00 + ku = F (t).
We are interested in the case where F (t) is a periodic function. Let us assume that F (t) = F0 cos wt
(or F (t) = F0 sin wt). So now we have
mu00 + ku = F0 cos wt.
This is a nonhomogeneous linear equation with the complementary solution
uc (t) = c1 cos w0 t + c2 sin w0 t,
r
k
. (It is the solution to mu00 + ku = 0, undamped free vibration.) The form of
m
the particular solution depends on the value of w by the undetermined coefficients method.
where w0 =
1. Beat (when w 6= w0 )
In this case, the form of the particular solution corresponding to the forcing function is
Y (t) = A cos wt + B sin wt.
After finding A and B, we will get
Y (t) =
F0
m(w02
w2 )
cos wt.
So the general solution of the displacement function is
u(t) = c1 cos w0 t + c2 sin w0 t +
F0
cos wt.
m(w02 w2 )
If the initial conditions are u(0) = 0 and u0 (0) = 0, we get
c1 =
F0
,
m(w02 w2 )
c2 = 0.
Therefore the particular solution to the system is
u(t) =
F0
(cos wt
m(w02 w2 )
cos w0 t).
Then we use this trig identity
sin A sin B =
cos(A
B)
cos(A + B)
2
,
we will change the form of u to
u(t) =
2F0
(w0 w)t
(w0 + w)t
sin
sin
.
m(w02 w2 )
2
2
This type of motion, possessing a periodic variation of amplitude, exhibits what is called a
beat.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
151
Figure 3.5: Beat; sol of u00 + u = 0.5 cos 0.8t, u(0) = 0, u0 (0) = 0; u = 2.77778(sin 0.1t)(sin 0.9t).
2. Resonance (when w = w0 )
In this case, since w = w0 , the form of the particular solution is
Y = (A cos w0 t + B sin w0 t)t = At cos w0 t + Bt sin w0 t.
Then we find that A = 0 and B =
F0
2mw0 .
Hence the general solution is
u(t) = c1 cos w0 t + c2 sin w0 t +
F0
t sin w0 t.
2mw0
The first two terms can be combine to R cos(w0 t
) representing the steady oscillation.
While the last term will become unbounded as t ! 1 regardless of the values of c1 and c2 .
So
F0
u(t) = R cos(w0 t
)+
t sin w0 t.
2mw0
Figure 3.6: Resonance; solution of u00 + u = 0.5 cos t, u(0) = 0, u0 (0) = 0; u = 0.25t sin t.
This phenomenon is called resonance.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
152
Example 123 (Summer 2013 Midterm Exam I).
True or false: a mass-spring system is described by the equation
2u00 + 2u = 5 cos t,
then the mass-spring system is undergoing resonance?
Solution Since F (t) = 5 cos t = F0 cos wt, we have w = 1. On the other hand,
r
r
k
2
w0 =
=
= 1.
m
2
Since w = w0 , the mass-spring system is undergoing resonance.
Example 124 (Spring 2000 Midterm Exam I).
A 2 kg mass is placed on a spring with k = 8. At t = 0, the system is suddenly set in motion
from its equilibrium position by an external force given by 2 cos(wt), where w is a positive
constant. For which value of w will the system have resonance?
Solution The system has resonance when w = w0 . Therefore
r
r
k
8
w=
=
= 2.
m
2
Example 125 (Spring 2005 Midterm Exam II).
Assume that acceleration due to gravity g is equal to 10 meter/sec2 . An object with mass 2
kg stretches a spring 2.5 meters to the equilibrium position. Assume that there is no damping
device attached and also assume that at time t = 0 the object is released 1 meter below its
equilibrium position with an upward velocity of 4 meter/sec.
a) Write down a di↵erential equation with initial conditions for y(t) for the displacement
of the object below its equilibrium position.
5
Solution We have m = 2, L = 2.5 = , = 0 (no damping e↵ect) and g = 10. Then
2
k=
mg
2(10)
= 5 = 8.
L
2
Also the initial displacement is 1 (positively downward) and the initial velocity is 4
(negatively upward). So two initial conditions are u(0) = 1, u0 (0) = 4. In conclusion,
the IVP governing the motion of the mass is
2u00 + 8u = 0,
u(0) = 1
u0 (0) =
4.
b) Find a formula for u(t).
Solution Consider the characteristic equation of 2u00 + 8u = 0,
2r2 + 8 = 0
)
r2 + 4 = 0
)
r = ±2i.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
So u(t) = c1 cos 2t + c2 sin 2t. Also u0 (t) =
initial conditions here,
)
u(0) = 1
0
u (0) =
)
4
153
2c2 sin 2t + 2c2 cos 2t. Now let’s apply two
1 = c1
4 = 2c2
)
c2 =
2.
Hence
y = cos 2t
2 sin 2t.
c) Find the maximum value of u(t)
Solution Recall that we can rewrite the above solution in the form
u(t) = R cos(w0 t
).
Since the value of cosine function is from 1 to 1, the maximum value of u is R. Also
q
p
p
R = c21 + c22 = 12 + ( 2)2 = 5.
So the maximum value of u is
p
5.
d) If a periodic external force equal to 3 cos wt Newtons is applied, then for what positive
value of w does resonance occur?
Solution The resonance phenomenon happens when
r
r
k
8
w = w0 =
=
= 2.
m
2
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
154
Exercises 3.7
1. A mass weighing 0.25 kg stretches a spring 1.25 m. The mass-spring system has a damping
constant of 12 kg/s. At t = 0 the mass is displaced an additional 50 cm downward from its
equilibrium position and set in motion with an upward velocity of 2 m/s. (You may use
g = 10 m/s2 as the gravitational constant. )
(a) Write an initial value problem that describes the motion of the mass.
(b) Determine the system’s quasi frequency and quasi period.
2. A mass-spring system is described by the equation
5u00 + u0 + ku = F (t).
(a) Suppose the mass originally stretched the spring 2 m to reach its equilibrium position.
What is the spring constant k? (Assume g = 10 m/s2 to be the gravitational constant)
(b) Suppose k = 45. For what value(s) of
(c) Suppose
would this system be critically damped?
= 0 and k = 400. What is the natural frequency of this system?
A second mass-spring system is described by the initial value problem
2u00 + 12u0 + 20u = 0,
(a) Find the position of the mass when t =
u0 (0) = 2.
u(0) = 0,
3⇡
2 .
(b) What is the quasi-frequency of this system?
3. A 2 kg mass is placed on a spring with k = 8. At t = 0, the system is suddenly set in motion
from its equilibrium position by an external force given by 2 cos(!t), where ! is a positive
constant. For which value of ! will the system have resonance?
4. Consider a mass-spring system described by the equation
3u00 + 12u0 + ku = 0,
k > 0.
Answer the following questions. Be sure to justify your answer. Full credit will not be given
without supporting work.
(a) For what value(s) of k would the system be overdamped?
(b) When k = 12, determine whether the system is overdamped, underdamped or critically
damped.
(c) The system would oscillate when: k = 9 or k = 15. (Circle the correct value and justify
your answer.)
(d) Find the quasi-period of the system whose k-value you found in part iii).
(e) True or false: When k = 3 the mass will never cross the system’s equilibrium position
more than once.
5. A mass of 2 kg stretches a spring 0.4 m. The system has no damping. At t = 0, the mass
is pulled down 1 m from its equilibrium position and set in motion with with an initial
downward velocity of 4 m/s. You may use g = 10 m/s2 as the gravitational constant.
(a) Find the Hooke’s constant, k, of the spring.
(b) Set up, but do not solve, an initial value problem to find the system’s displacement
function u(t).
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
155
6. When an object with mass 5 kg is attached to a spring, the object stretches the spring by 2
m. A damper with damping coefficient of 4 N/m is attached to the system. Assume there
is no external force acting on the system and that acceleration due to gravity 10 m/s2 . If
the object is released 1 m above its equilibrium position and is given an initial downward
velocity of 3 m/s, which initial value problem describes the displacement of the mass from
its equilibrium position? Take the downward direction to be positive for all displacements
and forces.
(a) 5y 00 + 4y 0 + 25y = 0,
00
0
00
0
(b) 5y + 4y + 25y = 0,
y 0 (0) =
y(0) = 1,
y(0) =
1,
(c) 5y + 4y + 10y = 0,
y(0) = 1,
(d) 5y 00 + 4y 0 + 10y = 0,
y(0) =
3
0
y (0) = 3
0
y (0) =
1,
3
y 0 (0) = 3
7. A spring-mass system is modeled by the initial value problem
2y 00 + y 0 + 8y = F (t),
0,
y(0) = 3,
y 0 (0) =
4.
(a) If
= 0 and F (t) = 0, what is the amplitude of displacement?
(b) If
= 0 and F (t) = 3 cos(!t), for which value(s) of ! will the system undergo resonance?
(c) If F (t) = 0, for which value(s) of
will the system not oscillate?
8. A spring-mass system, subject to an external force of 10 cos(2t) Newtons, is equipped with
a spring with a Hooke’s constant 12 Newtons per meter. For what mass will the resonance
occur?
(a) 2 kg
(b) 3 kg
(c) 6 kg
(d) 10 kg
9. Consider a mass-spring system described by the equation
6u00 + u0 + ku = 0,
0,
k > 0.
Answer the following questions. Be sure to justify your answer. Full credit will not be given
without supporting work.
(a) Suppose the spring was stretched 4 meters by the mass to its equilibrium position. Find
the value of k. You may use g = 0 as the gravitational constant.
(b) Suppose k = 20. For what value(s) of
would this system be critically damped?
(c) Suppose = 9 and k = 6. Will any nonzero solution of the equation cross the equilibrium position exactly once?
(d) Suppose
= 10 and k = 10. Find the quasi frequency of the system.
(e) Suppose a force of F (t) = 100 cos(↵t) is applied to the system and given k = 150. What
are the values of and ↵ if the system exhibits resonance?
10. A mass-spring system is described by the equation
5u00 + u0 + 80u = 0.
(a) When
= 0, what is the system’s natural period?
(b) When
= 40, find the displacement u(t) that satisfies u(0) = 6 and u0 (0) =
(c) When
= 40, is the system underdamped, overdamped, or critically damped?
4.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
156
(d) True or False: Regardless of the initial conditions, no solution of the system described
in (b) can cross the equilibrium position more than once.
11. A mass-spring system is described by the initial value problem
4u00 + u0 + 20u = 0,
(a) Suppose
u(0) = 0,
u0 (0) = 4.
= 8. Find the real-valued particular solution of this initial value problem.
(b) What is the quasi-period of this mass-spring system described in a)?
(c) True or false: Some, but not at all, nonzero solutions of this mass-spring system, with
= 8 (regardless of initial conditions), will cross the equilibrium position more than
once.
(d) Find all the value(s) of
that would make the system to be overdamped.
12. Suppose a mass-spring system described by the equation
u00 + ku = 2 cos 4t
sin 4t
is undergoing resonance. What is the value of the spring constant k?
(a) 0
(b) 2
(c) 4
(d) 16
Answers
1. (a) 14 u00 + 12 u0 + 2u = 0,
2. (a) k = 25, (b)
u(0) = 12 ,
u0 (0) =
p
= 30, (c) !0 = 4 5, (d) u =
2, (b) µ =
2e
9⇡
2
p
7, T =
2⇡
p
7
, (e) µ = 1
3. 2
4. (a) 0 < k < 12, (b) critically damped, (c) k = 15, (d) T = 2⇡, (e) True
5. (a) k = 50, (b) 2u00 + 50u = 0,
u(0) = 1,
u0 (0) = 4
6. (b)
7. (a)
p
13, (b) 2, (c)
8
8. (b)
p
9. (a) 15,( b) 4 30, (c) No, (d)
10. (a)
⇡
2,
4t
+ 20te
35
6 ,
4t
(e) 5
, (c) critically damped, (d) T
p
11. (a) u(t) = 2e t sin 2t, (b) ⇡, (c) F, (d) > 8 5
12. (d)
(b) u(t) = 6e
p
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
3.8
157
Nonlinear Equation*
In this section, we study special type of second order equations solvable by reduction to a system
of two first order equations.
3.8.1
Equations of the Type F (t, y 0 , y 00 ) = 0*
In this case, the dependent variable y is missing in di↵erential equation. To solve it, we let
u = y0 .
Therefore u0 = y 00 . By using these two equations, we substitute them back into original equation.
So
F (t, y 0 , y 00 ) = 0 ) F (t, u, u0 ) = 0,
which is a first order equation.
Example 126. Solve the following initial value problem
y 00 = 2ty 0 ,
y 0 (0) = 1.
y(0) = 0,
Solution Let u = y 0 and so u0 = y 00 , we change the equation to
y 00 = 2ty 0
This is a separable equation and so
Z
Z
du
du
= 2tu )
= 2t dt
dt
u
)
u0 = 2tu.
)
ln u = t2 + c0
)
2
u(t) = cet .
Also, let’s change the second initial condition,
y 0 (0) = 1
)
u(0) = 1.
Now we invoke this condition,
1 = ce0
2
)
c=1
Since u = y 0 , we get
2
u(t) = et .
)
t2
y 0 (t) = e 2 .
Integrating both sides, we obtain
y(t) =
Z
t
2
et dt.
0
We can’t simplify further and so this is the solution. Notice that it also satisfies y(0) = 0.
Example 127. Solve the following initial value problem
y 00 = 2e
2t
(y 0 )2 ,
y(0) =
1,
y 0 (0) = 1.
Solution Let u = y 0 and thus u0 = y 00 , we obtain new equation
y 00 = 2e
2t
(y 0 )2
)
u0 = 2e
2t 2
u .
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
This is a first order separable equation and so
Z
Z
du
du
= 2e 2t u2 )
=
2e 2t dt )
dt
u2
1
=
u
2t
e
158
+c
)
u(t) =
1
e
2t
c
.
One of the initial condition now changes to,
y 0 (0) = 1
)
u(0) = 1.
)
c=0
Apply this new condition,
1=
1
e
2(0)
c
)
1=
1
1
c
Replace u = y 0 ,
)
u(t) =
1
e
2t
= e2t .
y 0 (t) = e2t .
Integrating both sides, we get
y(t) =
Z
e2t dt =
Now we apply another initial condition y(0) =
1=
e0
+c
2
e2t
+ c.
2
1,
)
3
.
2
c=
Therefore the particular solution is
y(t) =
3.8.2
e2t
3
2
.
Equations of the Type F (y, y 0 , y 00 ) = 0*
In this case, we consider the second order di↵erential equation in which the independent variable
t is missing. In order to solve it, we let u = y 0 . By chain rule, we obtain
y 00 =
dy 0
du
du dy
du
=
=
=
u.
dt
dt
dy dt
dy
Substituting back into the original equation,
F (y, y 0 , y 00 ) = 0
)
F (y, u, u
du
).
dy
Now we view u as a function of y. Once we solve u, we integrate and obtain the solution.
Example 128. Solve the following IVP
yy 00 = y 0 (y 0 + 2),
y(0) = 1,
y 0 (0) =
1.
Solution Substitute y 0 by u and y 00 by u du
dy , the equation becomes
yy 00 = y 0 (y 0 + 2)
)
✓
◆
du
y u
= u(u + 2).
dy
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
We solve this by the method of separable equation,
Z
Z
1
1
du =
dy ) ln(u + 2) = ln y + c0
u+2
y
)
159
u = cy
Now let’s change one of the initial condition,
y 0 (t = 0) =
)
1
u(y(t = 0)) =
1
)
u(y = 1) =
1.
Apply this new condition,
1 = c(1)
2
)
Replace u = y 0 ,
)
c=1
y0 = y
u(t) = y
2.
2.
Again, this is a separable equation.
Z
Z
dy
1
=y 2 )
dy =
dt
dt
y 2
)
ln |y
2| = t + c.
Apply another initial condition y(0) = 1,
ln |1
2| = 0 + c
)
c = ln 1 = 0.
Therefore the particular solution is
ln |y
2| = t
or
y = et + 2.
Example 129. Solve the following IVP
y 00 = 2yy 0 ,
y 0 (0) = 1.
y(0) = 0,
Solution Substitute y 0 by u and y 00 by u du
dy , the equation becomes
y 00 = 2yy 0
)
By the method of separable equation,
Z
Z
du = 2y dy
u
du
= 2yu.
dy
u = y 2 + c.
)
Now let’s change one of the initial condition,
y 0 (t = 0) = 1
)
u(y(t = 0)) = 1
)
)
u = y 2 + 1.
u(y = 0) = 1.
Invoke this condition,
1 = 02 + c
Plug in u = y 0 ,
c=1
)
y 0 = y 2 + 1.
2.
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
This is again a separable equation. Let’s solve it:
Z
Z
dy
1
= y2 + 1 )
dy
=
dt
dt
y2 + 1
)
160
arctan y = t + c.
Another initial condition y(0) = 1 is used here,
)
arctan 0 = 0 + c
c = 0.
Therefore the particular solution is
arctan y = t,
or simply,
y = tan t.
3.8.3
Equations of the Type F (t, y, y 0 , y 00 ) = 0 with F homogeneous*
A function F is homogeneous of degree k with respect to y, y 0 , y 00 , if
F (t, y, y 0 , y 00 ) =
F (t, y, y 0 , y 00 ),
k
for all real number . If F is a homogeneous function of degree k, we write y 0 (t) = y(t)u(t). Then
y 00 = (yu)0 = y 0 u + yu0 = (yu)u + yu0 = yu2 + yu0 .
Hence
F (t, y, y 0 , y 00 ) = 0
F (t, y, yu, yu2 + yu0 ).
)
By the homogeneity of F ,
F (t, y, yu, yu2 + yu0 ) = uk F (t, u, u2 + u0 ),
results
F (t, u, u2 + u0 ) = 0.
After we solve for u, we integrate and then the solution will be obtained.
Example 130. Solve the following IVP
yy 00
Solution We have
(y 0 )2
2ty 2 = 0,
y(0) = 1,
F (t, y, y 0 , y 00 ) = yy 00
(y 0 )2
y 0 (0) = 0.
2ty 2 .
It is a homogeneous function of degree 2 since
F (t, y, y 0 , y 00 ) = ( y)( y)00
(( y)0 )2
=
2
(yy 00
(y 0 )2
=
2
F (t, y, y 0 , y 00 ).
2t( y)2
2ty 2 )
Substitute y 0 by yu and y 00 by yu2 + yu0 , the equation becomes
yy 00
(y 0 )2
2ty 2 = 0
)
y(yu2 + yu0 )
)
(yu)2
)
y 2 u2 + y 2 u0
y 2 u0
2ty 2 = 0
)
y 2 (u0
2t) = 0
y 2 u2
2ty 2 = 0
2ty 2 = 0
CHAPTER 3. SECOND ORDER DIFFERENTIAL EQUATIONS
Hence u0
2t = 0. Therefore
u(t) =
Two initial conditions give,
Z
u(0) =
161
2t dt = t2 + c.
y 0 (0)
0
= = 0.
y(0)
1
Invoke this condition, c = 0. Therefore u(t) = t2 and so
y 0 = yu = yt2 .
This can be solve by the method of separable equation.
Z
Z
dy
1
t3
2
= yt
)
dy = t2 dt ) ln y =
+ c0
dt
y
3
Invoking the initial condition y(0) = 1,
y = ce
Therefore the particular solution is
03
3
)
c = 1.
t3
y(t) = e 3 .
)
t3
y = ce 3 .
Chapter 4
Higher Order Linear Equation
4.1
General Theory of nth Order linear Equations
The mathematical theory of nth order di↵erential equation is completely analogous to those for
the 2nd order equation.
Theorem 7 (The Existence and Uniqueness Theorem for nth order linear equation).
If the
functions p1 , p2 , . . . , pn , and g are continuous on the open interval I, then there exists exactly
one solution y = (t) of the di↵erential equation
y (n) + p1 (t)y (n
1)
+ · · · + pn
1 (t)y
0
+ pn (t)y = g(t)
that also satisfies
y 0 (t0 ) = y00 ,
y(t0 ) = y0 ,
...
,
y (n
1)
(n 1)
(t0 ) = y0
,
where t0 is any point in I. This solution exists throughout the interval I.
Example 131. Determine intervals in which solutions are sure to exist.
4)y (6) + t2 y 000 + 9y = 0.
(t2
Solution Writing the equation in standard form,
y (6) +
(t
t2
y 000 +
2)(t + 2)
(t
9
y = 0.
2)(t + 2)
Then pi and g have discontinuities at t = 2, 2.
Hence the intervals of validity are ( 1, 2), ( 2, 2) and (2, 1).
Also we have the generalization of the principle of superposition.
Theorem 8. If y1 (t), . . . , yn (t) are solutions to
y (n) + p1 (t)y (n
1)
+ · · · + pn
1 (t)y
0
+ pn (t)y = 0
on an interval I, then so is their linear combination y = c1 y1 + . . . + cn yn .
162
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
163
It is then natural to ask whether every possible solution of
y (n) + p1 (t)y (n
1)
+ · · · + pn
1 (t)y
0
+ pn (t)y = 0
can be expressed as a linear combination of y1 , . . . , yn ,
c 1 y 1 + c 2 y 2 + . . . + c n yn .
Just as for second order linear equation, we have the following theorem involving the Wronskian.
Theorem 9. If the functions p1 , p2 , . . . , pn , and g are continuous on the open interval I, if
the functions y1 , . . . , yn are solutions of
y (n) + p1 (t)y (n
1)
+ · · · + pn
1 (t)y
0
+ pn (t)y = 0,
and if W (y1 , y2 , . . . , yn )(t) 6= 0 for at least one point in I, then every solution of the above
homogeneous equation can be expressed as a linear combination of the solutions y1 , y2 , . . . , yn .
In this theorem, the Wronskian of y1 , y2 , . . . , yn is defined by
y1
y10
..
.
W (y1 , y2 , . . . , yn )(t) :=
(n 1)
y1
···
···
y2
y20
..
.
(n 1)
y2
···
yn
yn0
..
.
(n 1)
yn
and a set of solutions y1 , y2 , . . . , yn of
y (n) + p1 (t)y (n
1)
+ · · · + pn
1 (t)y
0
+ pn (t)y = 0,
whose Wronskian is not a zero function is referred to as a fundamental set of solutions.
Linearly Independent vs Linearly Dependent:
• The functions f1 , f2 , . . . , fn are said to be linearly dependent on an interval I if there
exists a set of constants k1 , k2 , . . . , kn , not all zero, such that
k1 f1 (t) + k2 f2 (t) + · · · + kn fn (t) = 0
for all t in I.
• The functions f1 , f2 , . . . , fn are said to be linearly independent on I if they are not linearly
dependent there.
• In special case n = 2,
f and g are linearly dependent on I if we can find a constant c such that f = cg on I.
f and g are linearly independent on I if we assume c1 f + c2 g = 0 then it implies that
c1 = c2 = 0.
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
164
Wronskian vs Linear (In)dependence:
There is a close relationship between the Wronkian and linear (in)dependence we’ve just talked
about.
Theorem 10. If f and g are di↵erentiable functions on an open interval I such that
W (f, g)(t0 ) 6= 0 for some point t0 in I, then f and g are linearly independent on I. Equivalently, if f and g are linearly dependent on I, then W (f, g)(t) = 0 for all t in I.
It is worth noting that
• The converse of theorem above does not hold in general.
• That is to say, it is not true in general that two di↵erentiable functions f and g with a
vanishing Wronskian on an open interval must be linearly dependent on the same interval.
• But the converse may hold if additional conditions are imposed on the pair of functions
involved. The next Theorem is one of them.
Theorem 11. If f and g are di↵erentiable functions on an open interval I such that f (t) 6= 0
and W (f, g)(t) = 0 for all t in I, then f and g are linearly dependent which means f (t) = cg(t)
for a constant c.
In a nutshell, we have
To determine whether f, g are linearly independent or linearly dependent
1. Compute their Wronskian.
2. If their Wronskian is not a zero function then they are linearly independent.
3. Otherwise, find a constant c such that f = cg, then they are linearly dependent. (since
a vanishing Wronskian does not necessarily imply linear dependence)
Example 132. Determine whether each pair of functions is linearly independent or linearly
dependent.
a) e2t
1,
2
2e2t
Solution They are linearly dependent since we can write
2
b) 2 cos 4t,
2e2t = ( 2)(e2t
1).
3 sin 4t
Solution Consider their Wronskian,
W (2 cos 4t, 3 sin 4t) =
2 cos 4t
8 sin 4t
3 sin 4t
= 24 cos2 4t + 24 sin2 4t = 24 6= 0.
12 cos 4t
Therefore they are linearly independent.
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
c) 2e5t ,
e5t
165
5
Solution Consider their Wronskian,
W (2e5t , e5t
5) =
2e5t
10e5t
e5t 5
= 10e10t
5e5t
(10e10t
50e5t ) = 50e5t 6= 0.
So they are linearly independent.
d) 1,
e
2t
Solution Consider their Wronskian,
W (1, e
2t
1
0
)=
e 2t
=
2e 2t
2e
2t
6= 0.
So linear independence follows.
e) e2t ,
e2t+3
Solution They are linearly dependent since
e2t+3 = (e3 )(e2t ).
f ) sin 3t,
0
Solution They are linearly dependent since
0 = (0)(sin 3t).
g) e t ,
e4t
Solution Consider their Wronskian,
W (e t , e4t ) =
e t
e t
e4t
= 4e3t + e4t = 5e3t 6= 0.
4e4t
Hence they are linearly independent.
h) sin 2t,
cos t sin t
Solution They are linearly dependent since
sin 2t = (2)(cos t sin t),
as the trig identity.
i) sin( ⇡2
t),
2 cos t
Solution They are linearly dependent since
sin(
⇡
2
t) = sin
⇡
cos t
2
cos
⇡
1
sin t = cos t = ( )(2 cos t).
2
2
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
166
Exercises 4.1
1. Which pair of functions below is linearly independent?
(a) et ,
et
(b) sin 3t,
2t
(c) e ,
0
e
t
(d) e ,
2t+1
t
2+e
2. Indicate whether the two functions are linearly independent on the interval ( 1, 1):
(a) e2t ,
e2t + 2
(b) sin(2t),
(c) t + 1,
sin(2t + 2⇡)
2t + 2
3. All of the following pairs of functions are linearly independent on ( 1, 1) EXCEPT
a) e t ,
b) et ,
e4t
et
5
e
3t+3
3t
c) e ,
d) cos(2t),
sin(2t + 4⇡)
4. Which of the following pairs of the functions is not linearly independent on ( 1, 1)?
1
1
3t
(a) e 3 t ,
e
(b) e t ,
2e
(c) 3 cos(⇡t),
(d) 5,
e
(t 4)
2 sin(⇡t)
5t
5. Let y1 (t) and y2 (t) be two solutions of a second order homogeneous, linear di↵erential equation. Suppose the Wronskian W (y1 , y2 ) = e t . Which of the following is FALSE?
(a) y1 (t) and y2 (t) are linearly independent functions.
(b) 2y1 (t)
3y2 (t) is also a solution of the di↵erential equation.
(c) y1 (t) and y2 (t) do not constitute a fundamental set of solutions.
(d) All solutions of the di↵erential equation can be expressed as c1 y1 + c2 y2 , where c1 and
c2 are constants.
(e) W (2y1 (t), 3y2 (t)) = 6e t .
6. Which of the following pairs of functions is linearly dependent?
(a) sin 5t,
cos 5t
2t
(b) e sin t,
(c) e2t ,
e4t
(d) et ,
et+4
e2t cos t
7. Are the functions y1 (t) = t3 and y2 (t) =
1
t
linear independent? Justify your answer.
8. Show y1 (t) = t2 and y2 (t) = t3 , t > 0 are linear independent by calculating the Wronskian.
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
Answers
1. (d)
2. (a) yes, (b) no, (c) no
3. (c)
4. (b)
5. (c)
6. (d)
7. yes
8. -
167
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
4.2
168
Homogeneous Equations with Constant Coefficients
Consider the nth order linear homogeneous equation with constant coefficients,
a0 y (n) + a1 y (n
1)
where a0 6= 0, a1 , . . . , an are constant.
Its characteristic equation is
a0 r n + a1 r n
1
+ . . . + an y = 0
+ . . . + an = 0.
It is an nth degree polynomial equation1 which gives us n roots r1 , r2 , . . . , rn .
To get the general solution, apply the followings:
• If r is a distinct real root, then ert is a solution.
• If r =
± µ are complex conjugate roots, then
e cos µt,
e sin µt are solutions
• If r is a real root appearing k times then
ert ,
• If r =
tert ,
...
,
tk
1 rt
e
are solutions
± µ are complex conjugate roots appearing k times then,
tk
e cos µt,
e sin µt
te cos µt,
..
.
te sin µt
..
.
1
tk
e cos µt,
1
e sin µt
are solutions.
• The general solution is the linear combination of all above solutions.
1 The first phase in the history of algebra was the search for solutions of polynomial equations. The degree of
difficulty of an equation corresponds rather neatly to the degree of the corresponding polynomial.
Linear equations are easily solved, and 2000 years ago the Chinese were even able to solve n equations in n
unknowns by the method we now call Gaussian elimination.
Quadratic equations are harder to solve, because they generally require the square root operation. But the
solution-essentially the same as that taught in high schools today-was discovered independently in many cultures
more than 1000 years ago.
The first really hard case is the cubic equation, whose solution requires both square roots and cube roots. The
method for solving the cubic equation was apparently discovered by Scipione dal Ferro (1465 1526) about 1500,
although it was first published in 1545 by Girolamo Cardano (1501 1576) in his Ars Magna. This book also contains
a method for solving quartic equations that Cardano attributes to his pupil Ludovico Ferrari (1522 1565). Formulas
for the cubic and quartic equations are analogous to the formula for quadratic equations but more complicated.
The question of whether analogous formulas exist for the roots of higher degree equations remained open for more
than two centuries, until 1826, when Niels Abel showed that no general solution formulas can exist for polynomial
equations of degree five or higher. A more general theory was developed by Evariste Galois (1811 1832) in 1831,
but unfortunately it did not become widely known for several decades.
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
Example 133. Find the general solution of y (4)
169
y = 0.
Solution Its characteristic equation is
r4
1=0
)
(r2
1)(r2 + 1) = 0
)
(r
1)(r + 1)(r2 + 1) = 0.
So r = 1, 1, ±i are roots. Thus the general solution is
y = c1 et + c2 e
t
+ c3 cos t + c4 sin t.
Example 134 (Summer 2010 Midterm Exam I).
What is the general solution of
y (4) + 16y 00 = 0?
Solution The characteristic equation of y (4) + 16y 00 = 0 is
r4 + 16r2 = 0
r2 (r2 + 16) = 0
)
)
r = 0, 0, ±2i.
Therefore the general solution is
y(t) = c1 e0t + c2 te0t + c3 cos 2t + c4 sin 2t = c1 + c2 t + c3 cos 2t + c4 sin 2t.
Example 135 (Spring 2008 Midterm Exam I).
Find the general solution of
y (5)
2y (4) + y 000 = 0.
Solution The corresponding characteristic equation is
r5
2r4 + r3 = 0
)
r3 (r2
2r + 1) = 0
)
r3 (r
1)2 = 0.
So r = 0, 0, 0, 1, 1 are all roots. Hence the general solution is
y = c1 e0t + c2 te0t + c3 t2 e0t + c4 et + c5 tet = c1 + c2 t + c3 t2 + c4 et + c5 tet .
Example 136. Let (r2 + 6r + 13)3 = 0 be the characteristic equation of certain di↵erential
equation. What is the general solution to that equation?
Solution From
2
r + 6r + 13 = 0
)
So for (r2 + 6r + 13)3 = 0, we have r =
general solution is
y = c1 e
3t
cos 2t+c2 e
3t
sin 2t+t(c3 e
3t
r=
6±
p
36
2
52
=
3 ± 2i.
3 ± 2i, 3 ± 2i, 3 ± 2i as all roots. Therefore the
cos 2t+c4 e
3t
sin 2t)+t2 (c5 e
3t
cos 2t+c6 e
3t
sin 2t).
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
170
Example 137 (Fall 2008 Midterm Exam II).
What is the general real-valued solution of the following ODE:
y 000 + y 0 = 0?
(a) y(t) = c1 et + c2 e
t
+ c3
(b) y(t) = c1 cos t + c2 sin t + c3 et
(c) y(t) = c1 + c2 cos t + c3 sin t
(d) y(t) = c1 t2 + c2 t + c3
Solution The characteristic equation is
r3 + r = 0
r(r2 + 1) = 0
)
)
r = 0, ±i.
So the general solution is
y(t) = c1 e0t + c2 cos t + c3 sin t = c1 + c2 cos t + c3 sin t.
Therefore (c) is the right answer.
Example 138 (Spring 2009 Midterm Exam II).
Consider the third order linear equation
y 000 + 4y 00 = 0.
What is its general solution?
(a) y(t) = c1 + c2 cos 2t + c3 sin 2t
(b) y(t) = c1 + c2 t + c3 e
4t
(c) y(t) = c1 + c2 e2t + c3 e
4t
(d) y(t) = c1 + c2 t + c3 t4
Solution Note that the corresponding characteristic equation of y 000 + 4y 00 = 0 is
r3 + 4r2 = 0
)
r2 (r + 4) = 0
)
r = 0, 0, 4.
So the general solution of this third order linear equation is
y(t) = c1 e0t + c2 te0t + c3 e
4t
= c1 + c2 t + c3 e
Hence (b) is the correct answer.
Example 139 (Summer 2011 Midterm Exam II).
Find the general solution of the linear equation
y (5) + 2y (4) + 5y 000 = 0.
4t
.
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
171
Solution The characteristic equation is
5
4
3
r +2r +5r = 0
3
)
2
)
r (r +2r +5) = 0
2±
r = 0, 0, 0,
p
22
2
4(1)(5)
=
1±2i.
Consequently, the solution is
y = c 1 + c 2 t + c 3 t2 + c 4 e
t
cos 2t + c5 e
t
sin 2t.
Example 140 (Fall 2011 Midterm Exam II).
Consider the fourth order linear equation
y (4)
8y 00 + 16y = 0.
What is its general solution?
(a) y(t) = c1 cos 2t + c2 sin 2t + c3 t cos 2t + c4 t sin 2t
(b) y(t) = c1 + c2 t + c3 cos 4t + c4 sin 4t
(c) y(t) = c1 e2t + c2 e
2t
+ c3 te2t + c4 te
2t
(d) y(t) = c1 + c2 t + c3 e4t + c4 te4t
Solution Notice that its characteristic equation is
r4
8r2 + 16 = 0
)
(r2
4)2 = 0
)
(r
2)2 (r + 2)2 = 0
)
r = 2, 2, 2, 2.
So the general solution is
y(t) = c1 e2t + c2 e
2t
+ c3 te2t + c4 te
2t
which gives (c) the right choice.
Example 141 (Summer 2012 Midterm Exam II).
Find the general solution of the linear equation
y (5) + 16y 000 = 0.
Solution The characteristic equation of y (5) + 16y 000 = 0 is
r5 + 16r3 = 0
)
r3 (r2 + 16) = 0
)
r = 0, 0, 0, ±4i.
Then the solution is
y = c1 + c2 t + c3 t2 + c4 cos 4t + c5 sin 4t.
Example 142 (Fall 2012 Midterm Exam II).
Consider the fourth order linear equation
y (4)
What is its general solution?
2y (3) + y 00 = 0.
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
172
(a) y(t) = c1 + c2 t + c3 et + c4 tet
(b) y(t) = c1 et + c2 e
t
+ c3 cos t + c4 sin t
(c) y(t) = c1 + c2 t + c3 cos t + c4 sin t
(d) y(t) = c1 + c2 t + c3 e
t
+ c4 te
t
Solution Its characteristic equation is
r4
2r3 + r2 = 0
)
r2 (r2
2r + 1) = 0
Consequently, the general solution to y (4)
)
r2 (r
2y (3) + y 00 = 0 is
y(t) = c1 + c2 t + c3 et + c4 tet .
Then (a) is the correct answer.
1)2 = 0
)
r = 0, 0, 1, 1.
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
173
Exercises 4.2
1. Find the general solution of the linear equation
y (5) + 16y 000 = 0.
2. Find the general solution of the equation below by examining its characteristic equation.
y 000 + 4y 0 = 0
3. Consider the fourth order linear equation
y (4)
2y (3) + y 00 = 0.
What is its general solution?
(a) y(t) = c1 + c2 t + c3 et + c4 tet
(b) y(t) = c1 et + c2 e
t
+ c3 cos t + c4 sin t
(c) y(t) = c1 + c2 t + c3 cos t + c4 sin t
(d) y(t) = c1 + c2 t + c3 e
t
t
+ c4 te
4. Consider the fourth order linear equation
2y (4) + 50y 00 = 0.
What is its general solution?
(a) y(t) = c1 et + c2 e
t
+ c3 cos 5t + c4 sin 5t
(b) y(t) = c1 + c2 t + c3 cos 5t + c4 sin 5t
p
p
p
p
(c) y(t) = c1 cos 5t + c2 sin 5t + c3 t cos 5t + c4 t sin 5t
(d) y(t) = c1 + c2 et + c3 cos 5t + c4 sin 5t
5. Find the general solution of the linear equation
y (5) + 2y (4) + 5y 000 = 0.
6. Consider the fourth order linear equation
y (4)
8y 00 + 16y = 0.
What is its general solution?
(a) y(t) = c1 cos 2t + c2 sin 2t + c3 t cos 2t + c4 t sin 2t
(b) y(t) = c1 + c2 t + c3 cos 4t + c4 sin 4t
(c) y(t) = c1 e2t + c2 e
2t
+ c3 te2t + c4 te
(d) y(t) = c1 + c2 t + c3 e4t + c4 te4t
2t
CHAPTER 4. HIGHER ORDER LINEAR EQUATION
7. Consider the third order linear equation
y 000 + 4y 00 = 0.
What is its general solution?
(a) y(t) = c1 + c2 cos 2t + c3 sin 2t
4t
(b) y(t) = c1 + c2 t + c3 e
(c) y(t) = c1 + c2 e2t + c3 e
2t
(d) y(t) = c1 + c2 t + c3 t4
8. Which of the following functions is the general solution of
y 000 + y 00 + y 0 + y = 0.
What is its general solution?
(a) y(t) = c1 e
t
t
+ c2 te
+ c 3 t2 e
t
(b) y(t) = c1 et + c2 tet + c3 t2 et
(c) y(t) = c1 cos t + c2 sin t + c3 e
(d) y(t) = c1 et + c2 e
t
+ c3 te
t
t
9. Consider the fourth order linear equation
y 0000 + 6y 00 + 9y = 0.
What is its general solution?
(a) y(t) = c1 + c2 t + c3 e3t + c4 te3t
(b) y(t) = c1 e
p
3t
+ c2 e
p
3t
+ c3 te
p
3t
+ c4 te
p
3t
(c) y(t) = c1 + c2 t + c3 cos 3t + c4 sin 3t
p
p
p
p
(d) y(t) = c1 cos 3t + c2 sin 3t + c3 t cos 3t + c4 t sin 3t
Answers
1. y = c1 + c2 t + c3 t2 + c4 cos 4t + c5 sin 4t
2. y = c1 + c2 cos 2t + c3 sin 2t
3. (a)
4. (b)
5. y = c1 + c2 t + c3 t2 + c4 e
6. (c)
7. (b)
8. (c)
9. (d)
t
cos 2t + c5 e
t
sin 2t
174
Chapter 5
Laplace Transform
Laplace transform can be used to solve IVPs that we can’t use any previous methods we’ve learned
so far.
5.1
Definition of Laplace Transform
The Laplace transform1 of f , which we denote it by L{f (t)} or F (s), is defined by the equation
Z 1
L{f (t)} =
e st f (t) dt,
0
whenever this improper integral converges.
Example 143. Compute L{1}.
Solution By the definition of Laplace transform,
L{1} =
Z
1
e
st
1 dt = lim
A!1
0
So L{1} =
Z
A
st
e
e
dt = lim
s
A!1
0
st
t=A
t=0
= lim
A!1
e
sA
s
+
1
1
= , if s > 0.
s
s
1
, if s > 0.
s
Example 144. Compute L{eat }.
Solution By the definition of Laplace transform,
L{eat } =
Z
1
e
st at
1
s
a
A!1
0
e(a
= lim
A!1 a
So L{eat } =
e dt = lim
s)A
s
1
a
s
Z
A
e(a
0
=
1
s
a
s)t
e(a s)t
A!1 a
s
dt = lim
t=A
t=0
, if s > a.
, if s > a.
1 The Laplace transform is named for the eminent French mathematician Pierre-Simon de Laplace. However,
the techniques described in this chapter were not developed until a century or more later. We owe them mainly to
Oliver Heaviside (1850 1925), an innovative but unconventional English electrical engineer, who made significant
contributions to the development and application of electromagnetic theory.
175
CHAPTER 5. LAPLACE TRANSFORM
176
Example 145. Compute L{sinat}.
R
Solution First, let’s find e st sin at dt by integration-by-parts. We have
Z
Z
e
st
sin at dt =
e
st
cos at dt =
e
st
sin at
+
s
e st cos at
s
Z
a
e
s
Z
a
e
s
st
cos at dt
st
sin at dt
Substitute the second equation into the first one,
Z
Z
e st sin at a e st cos at a
e st sin at dt =
+ (
e st sin at dt)
s
s
s
s
Z
e st sin at
a st
a2
e
cos
at
e st sin at dt
=
s
s2
s2
Z
a2
e st sin at
a st
(1 + 2 ) e st sin at dt =
e
cos at
s
s
s2
Z
s2 + a2
e st sin at
a st
st
(
)
e
sin
at
dt
=
e
cos at
2
s
s
s2
Z
s2
e st sin at
a st
e st sin at dt = 2
(
e
cos at) + c
s + a2
s
s2
Hence
L{sin at} =
Z
1
e
st
sin at dt = lim
0
A!1
Z
A
e
st
e st sin at
a
s2
(
e
2
2
A!1 s + a
s
s2
s2
s2
a
= 2
(0
+
0)
(
)
s + a2
s 2 + a2 s 2
a
= 2
, if s > 0.
s + a2
= lim
Therefore L{sin at} =
sin at dt
0
st
cos at)
t=A
t=0
a
if s > 0.
s 2 + a2
• Notice that we had to put a restriction on s in order to actually compute the Laplace
transform. All Laplace transform will have restrictions on s. From now on we omit this but
we really shouldn’t forget that it’s there for our convenience.
• Properties of Laplace transform
1. L{cf (t)} = cL{f (t)}.
2. L{f (t) ± g(t)} = L{f (t)} ± L{g(t)}.
3. But L{f (t)g(t)} is not necessary the same as L{f (t)}L{g(t)}.
4. L{0} = 0.
• As above examples, computing Laplace transform of even simple functions is often messy. So
usually we just use a table of Laplace transform when actually computing it without using
the definition directly.
CHAPTER 5. LAPLACE TRANSFORM
1
f (t) = L
{F (s)}
1. 1
2. eat
3. tn ;
n = positive integer
4. tk ,
k>
1
5. sin at
6. cos at
eat
7. sinh at :=
e
2
eat + e
8. cosh at :=
2
at
at
9. eat sin bt
10. eat cos bt
11. tn eat , n = positive integer
12. uc (t)
13. uc (t)f (t
177
c)
F (s) = L{f (t)}
1
, s>0
s
1
, s>a
s a
n!
, s>0
sn+1
(k + 1)
, s>0
sk+1
a
, s>0
s 2 + a2
s
, s>0
s 2 + a2
a
, s > |a|
2
s
a2
s
, s > |a|
2
s
a2
b
, s>a
(s a)2 + b2
s a
, s>a
(s a)2 + b2
n!
, s>a
(s a)n+1
e cs
, s>0
s
e
cs
F (s)
14. ect f (t)
F (s
15. f (ct)
1
s
c F ( c ),
16.
Rt
0
17. (t
f (t
⌧ )g(⌧ ) dt
c)
c>0
F (s)G(s)
e
18. f (n) (t)
c)
cs
sn F (s)
sn
n = 1, f 0 (t)
sF (s)
f (0)
n = 2, f 00 (t)
s2 F (s)
19. ( t)n f (t)
n = 1, tf (t)
n = 2, t2 f (t)
1
f (0)
sf (0)
···
f 0 (0)
F (n) (s)
F 0 (s)
F 00 (s)
Table 5.1: Elementary LaplaceTransforms
f (n
1)
(0)
CHAPTER 5. LAPLACE TRANSFORM
178
Let’s see how we apply properties of Laplace transform and formulas in Laplace transform table.
Example 146 (Fall 2009 Midterm Exam II).
Evaluate
Z
1
e
st
t dt.
0
Solution By the definition of Laplace transform,
Z 1
1
e st t dt = L{t} = 2 .
s
0
Example 147 (Fall 2010 Midterm Exam II).
Evaluate the integral
Z
1
e
(s+1)t
sin 2t dt.
0
Solution Using the definition of Laplace transform,
Z 1
Z 1
(s+1)t
e
sin 2t dt =
e st e t sin 2t dt = L{e t sin 2t} =
0
0
(s
2
2
= 2
.
( 1))2 + 22
s + 2s + 5
Example 148 (Spring 2009, 2011 Midterm Exam II).
True or false:
• L{f (t)
6g(t)} = L{f (t)}
6{g(t)}
• L{4f (t)g(t)} = 4L{f (t)}{g(t)}
• L{(t + 5)2 } = L{t + 5}L{t + 5}
Solution True for the first one by properties of Laplace transform and false for the last two
since the Laplace transform of the product does not necessarily equal the product of Laplace
transforms.
Example 149. Compute
1. L{5 + 3e
2t
+ 4 sin 6t}
Solution
L{5 + 3e
2t
+ 4 sin 6t} = L{5} + L{3e
2t
} + L{4 sin 6t}
2t
= 5L{1} + 3L{e } + 4L{sin 6t}
1
1
6
= 5( ) + 3
+ 4( 2
)
s
s ( 2)
s + 62
5
3
24
= +
+
.
s s + 2 s2 + 36
CHAPTER 5. LAPLACE TRANSFORM
2. L{6e
5t
+ e3t + 5t3
179
9}
Solution
L{6e
5t
+ e3t + 5t3
3. L{ 9 sin 4t + 2 cos 10t
} + L{e3t } + 5L{t3 } 9L{1}
1
1
3!
1
= 6(
)+
+ 5 3+1 9( )
s ( 5)
s 3
s
s
6
1
30 9
=
+
+ 4
.
s+5 s 3
s
s
9} = 6L{e
5t
e3t cos 6t}
Solution
L{ 9 sin 4t + 2 cos 10t
e3t cos 6t} =
=
=
4. L{5 sinh 2t
9L{sin 4t} + 2L{cos 10t} L{e3t cos 6t}
4
s
s 3
9( 2
)+2 2
2
2
s +4
s + 10
(s 3)2 + 62
36
2s
s 3
+
.
s2 + 16 s2 + 100 s2 6s + 45
2 sin 2t}
Solution
L{5 sinh 2t
2 sin 2t} = 5L{sinh 2t} 2L{sin 2t}
2
2
= 5( 2
) 2 2
2
s
2
s + 22
10
4
= 2
.
s
4 s2 + 4
5. L{t cosh 3t}
Solution
L{t cosh 3t} =
=
=
=
(L{cosh 3t})0
s
s
( 2
)0 = ( 2
)0
2
s
3
s
9
(s2 9)(1) (s)(2s)
(s2 9)2
2
s
9
s2 + 9
=
.
(s2 9)2
(s2 9)2
6. L{t2 sin 2t}
Solution
2
)00
+4
4s
12s2 16
0
=( 2
)
=
.
(s + 4)2
(s2 + 4)3
L{t2 sin 2t} = (L{sin 2t})00 = (
s2
CHAPTER 5. LAPLACE TRANSFORM
180
Exercises 5.1
1. Which of the following is the Laplace transform of t sin(2t)?
2
(s2 + 4)2
4s
(b) 2
(s + 4)2
s
(c) 2
(s + 4)
3
(d) 2 2
s (s + 4)2
(a)
2. For each part below, determine whether the statement is true or false. You must justify your
answers.
(a) L{f (t)g(t)} = L{f (t)}L{g(t)}
(b) L{cos(2t) sin(2t)} =
6 L{cos(2t)}L{sin(2t)}
(c) Suppose L{f (t)} =
s2
(s 2)2
, then L{e2t f (t)} =
.
+1
(s 2)4 + 1
s4
3. Evaluate the following definite integral
Z
1
e
st 2
t dt
2
2
s3
2e 2s
b)
s3
2
4
4
c) 3 + 2 +
s
s
s
2
4
4
d) e 2s ( 3 + 2 + )
s
s
s
a)
4. Evaluate the following definite integral
Z 1
e(1
s)t
cos 3t dt.
0
1
2s + 10
s 1
b) e s 2
s
2s + 10
s+1
s
c) e
s2 + 2s + 10
s+1
d) 2
s + 2s + 10
a)
s
s2
5. Given that L{f (t)} =
6s
. Use properties of the Laplace transform to determine L{tf (t)}.
s3 + 64
CHAPTER 5. LAPLACE TRANSFORM
181
6. Suppose
L{f (t)} =
s3
s2
.
+ 27
What is L{7e5t f (t)}?
(a)
(b)
(c)
(d)
7(s 5)2
(s 5)3 + 27
7s2
5)(s3 + 27)
(s
7s2
5)(s3 + 27)
s(s
7(s + 5)2
(s + 5)3 + 27
7. (a) What is the definition of the Laplace transform L{e3t }?
(b) Use the answer to Part (a) to calculate L{e3t }. (Be sure to explain why this exists only
when s > 3).
8. Find the Laplace transform of the following functions.
3t
(a) e
(b) te
cos 5t
5t
9. Given that L{f (t)} = F (s), what is L{e5t tf (t)}?
a)
1
s2 (s
5)
F (s)
1 0
F (s)
s 5
1
c) 2 F (s + 5)
s
d) F 0 (s 5)
b)
10. Which of the following functions has
a) t2
R1
0
e
st 2
t dt as its Laplace transform?
b) e t t2
c) t
d) (t
11.
R1
0
e
2)t2
(s 2)t
dt is the Laplace transform of which of the following functions?
a) t2
b) e2t
c) u(t
2)
d) (t
2)
12. (a) Evaluate the following definite integral
2
(b) Suppose L{f (t)} =
L{e
⇡t
f (t)}.
s4
R1
0
e(4
s)t
cos(3t) dt.
2s
. Use properties of the Laplace transform to determine
+ 100
CHAPTER 5. LAPLACE TRANSFORM
182
13. Evaluate the following definite integral
Z 1
e
st
(t3 + sin 2t) dt
0
3
s4
6
b) 4
s
6
c) 4
s
3
d) 2
s
a)
2
s2 + 4
2
+ 2
s +4
1
+ 2
s + 22
1
+ 2
s + 22
+
14. For each part below, determine whether the statement is true or false. State a brief reason
that justifies each answer.
(a) L{e
2t+3
(sin(t) + cos(t))} = e3 L{e
s
(b) Suppose L{f (t)} = e
t
2t
sin(t)} + e3 L{e
s+1
s then L{e f (t)} = e
s e
1
2s
(c) Suppose L{f (t)} =
, then L{tf (t)} =
.
1 + s2
(1 + s2 )2
Answers
1. (b)
2. (a) F, (b) T, (c) T
3. (d)
4. (a)
5.
12s3 384
(s3 + 64)2
6. (a)
7. (a)
8. (a)
R1
0
e
1
st 3t
e dt, (b)
s
3
s+3
1
, (b)
s2 + 6s + 34
(s 5)2
9. (d)
10. (a)
11. (b)
12. (a)
s
s2
4
2(s + ⇡)2
, (b)
8s + 25
(s + ⇡)4 + 100
13. (b)
14. (a) T, (b) T, (c) T
2t
cos(t)}.
s+1
.
CHAPTER 5. LAPLACE TRANSFORM
5.2
183
The Gamma Function*
The gamma function of k, written as (k), is defined by the improper integral
Z 1
(k) =
xk 1 e x dx, k > 0.
0
This function satisfies
1.
(1) = 1.
2.
(k) =
3.
(n + 1) = n!,
4.
(k) =
Proof.
(k + 1)
,
k
k > 0.
where n is a positive integer.
(k + n)
k(k + 1)(k + 2) · · · (k + n
1)
k 6= 0, 1, 2, . . . , (n
,
1).
(1) = 1 since
(1) =
Z
1
e
x
dx = lim
A!1
0
Z
A
x
e
dx = lim
e
A!1
0
x A
0
= lim
From the definition of gamma function, integration by parts with u = e
(k) = lim
A!1
= lim
A!1
= lim
A!1
Z
x
A
+ 1 = 1.
, dv = xk
1
dx gives
A
xk
0
✓
✓
e
x k
x
k
e
1
e
◆A
0
A
Ak
k
(k + 1)
.
k
=
e
A!1
x
dx
Z
1 1 k x
x e dx
k 0
◆
1
0 +
(k + 1)
k
+
Therefore (k + 1) = k (k). Now let’s use this identity along with (1) = 1 to obtain,
(2) = 1 (1) = 1 = 1!
(3) = 2 (2) = 2 · 1 = 2!
(4) = 3 (3) = 3 · 2 · 1 = 3!
(5) = 4 (4) = 4 · 3 · 2 · 1 = 4!..
In general, (n+1) = n! where n is a positive integer. It can be proved by mathematical induction
(verify!). Write again the second identity, if k 6= 0,
(k) =
(k + 1)
.
k
(5.1)
Substitute k by k + 1, we get
(k + 1) =
(k + 2)
,
k+1
k 6=
1.
(5.2)
Substitute (5.2) into (5.1),
(k) =
(k + 1)
(k + 2)
=
,
k
k(k + 1)
k 6= 0, 1.
(5.3)
CHAPTER 5. LAPLACE TRANSFORM
184
Substitute k by k + 1 in (5.2),
(k + 3)
,
k+2
(k + 2) =
k 6=
2.
(5.4)
Substitute (5.4) into (5.3),
(k) =
(k + 2)
(k + 3)
=
,
k(k + 1)
k(k + 1)(k + 2)
k 6= 0, 1, 2.
In general, we have
(k) =
(k + n)
k(k + 1)(k + 2) · · · (k + n
1)
k 6= 0, 1, 2, . . . , (n
,
1).
It is worth noting that identity 2 and 4 help us to extend the definition of gamma function, (k) to
negative values of k if k 6= 0, 1, 2, . . . (it was defined only for k > 0). For example, by identity
1
2, with k =
,
2
✓
◆
✓ ◆
1
1
1
2 +1
=
=
2
.
1
2
2
2
With k =
3
2,
✓
3
2
◆
=
3
2
+1
3
2
✓
2
3
=
1
2
◆
2
3
=
✓
✓ ◆◆
✓ ◆
1
4
1
2
=
.
2
3
2
Here we have another identity that we will use shortly.
✓ ◆
p
1
5.
= ⇡,
2
Proof. By the definition of gamma function,
✓ ◆ Z 1
1
1
=
x 2 1e
2
0
Making the substitution x = u2 gives
Z 1
Z
1
x 2 e x dx =
0
It is possible to show that
Therefore,
Z
1
u
1
x
e
dx =
u2
Z
1
x
(2u du) = 2
0
1
e
u2
du =
0
1
2
x
e
dx.
0
Z
1
e
u2
0
p
⇡
.
2
✓ ◆
p
1
= ⇡.
2
By using the identity above, we finally obtain
✓
◆
p
1
= 2 ⇡, and
2
✓
3
2
◆
=
4p
⇡.
3
du.
CHAPTER 5. LAPLACE TRANSFORM
185
Another advantage of the gamma function is that it gives meaning to n! where n is any number
(not only integers) excepting a negative integer by using identity 3 (since at these points gamma
function isn’t defined as we mentioned above). For examples,
0! = (0 + 1) = (1) = 1,
◆
✓
◆
✓ ◆
p
1
1
1
!=
= ⇡,
+1 =
2
2
2
✓
◆
✓
◆
✓
◆
p
3
3
1
!=
+1 =
= 2 ⇡,
2
2
2
✓ ◆
✓
◆
✓ ◆
✓ ◆
1
1
3
1
1
!=
+1 =
=
=
2
2
2
2
2
✓ ◆
✓
◆
✓ ◆
✓ ◆
3
3
5
3
3
!=
+1 =
=
=
2
2
2
2
2
✓
In fact, for any positive integer n,
✓
◆ ✓
◆
1
1
6.
+n =
+n !=
2
2
✓
◆ ✓
◆
1
1
7.
n =
n !=
2
2
we have
(2n)! p
⇡.
4n n!
( 4)n n! p
⇡.
(2n)!
We also obtain Laplace transforms of tk , where k >
8. L{tk } =
9. L{tn } =
(k + 1)
,
sk+1
n!
,
sn+1
1p
⇡,
2
✓
◆
3 1p
3p
⇡ =
⇡.
2 2
4
1 is any number, in term of gamma function.
s > 0.
n is a positive integer.
Proof. From the definition of Laplace transform, we have
Z 1
k
L{t } =
tk e st dt.
0
By changing of variables with st = x, we have
Z 1
Z 1 ⇣ ⌘k
x
tk e st dt =
e
s
0
0
But by the definition of gamma function,
Z 1
xk e
x
x
dx
1
= k+1
s
s
dx = (k + 1).
0
Therefore we obtain an identity 8,
L tk =
(k + 1)
.
sk+1
If k = n is an integer, we apply an identity 3,
L {tn } =
to receive an identity 9.
(n + 1)
n!
= n+1
sn+1
s
Z
1
0
e
x k
x dx.
CHAPTER 5. LAPLACE TRANSFORM
186
Example 150. Show that
1
2
L{t
r
}=
⇡
,
s
p
1
L{t 2 } =
and
⇡
3
2s 2
.
Solution Apply an identity 8,
1
2
L{t
1
2
}=
1
2
L{t } =
n
10. L t
1
2
e
at
o
11. L tn eat =
=
r
(s
Proof.
⇡
s
a
L t
1
2
e
at
o
n>
Z
=
Making the substitution x = (s
Z
1
0
t
1
2
e
(s a)t
s
1
2
s
s
p
3
2
=
1
2 +1
p
=
=
3
2
=
1
2
⇡
2
s
⇡
3
2
=
r
p
⇡
⇡
s
.
3
2s 2
.
n!
,
a)n+1
n
s
+1
1
2
=
1
2 +1
s
1
2
+1
dt =
1 an integer.
1
0
a)t,
Z
1
0
⇣
✓
1
2
t
e
x
s
a
at
⌘
◆
st
e
1
2
s
L t
1
2
e
at
o
a
1
1
2
t
e
(s a)t
dt.
0
=p
1
s a
Z
1
x
1
2
e
x
dx.
0
✓ ◆
p
1
= ⇡.
2
0
n
dx
x
e
But we computed earlier in the identity 5 that
Z 1
1
x 2 e x dx =
Hence
dt =
Z
=
r
⇡
s
a
.
Next we’ll show an identity 11 by using similar idea of what we just did.
Z 1
Z 1
L tn eat =
tn eat e st dt =
tn e (s a)t dt.
0
Making the substitution x = (s a)t,
◆n
Z 1
Z 1✓
x
tn e (s a)t dt =
e
s a
0
0
0
dx
x
s
a
=
1
(s
By the definition of gamma function and an identity 3,
Z 1
xn e x dx = (n + 1) = n!.
0
Therefore
L tn eat =
(s
n!
.
a)n+1
a)n
Z
1
0
xn e
x
dx.
CHAPTER 5. LAPLACE TRANSFORM
5.3
187
Inverse Laplace Transform
We learn from the first section that Laplace transform sends a function of t to the function of s.
The inverse Laplace transform is simply the inverse function of Laplace transform which does
the opposite. It changes a function of s to the function of t.
L{f (t)} = F (s)
,
L
1
{F (s)} = f (t).
Similar to the Laplace transform, the inverse one has the following properties:
1. L
1
{cF (s)} = cL
1
2. L
1
{F (s) ± G(s)} = L
3. L
1
{0} = 0.
{F (s)}.
1
{F (s)} ± L
1
{G(s)}.
The key to compute the inverse Laplace transform is to
• first look at the denominator and try to identify what you’ve got based on that.
• Correct the numerator (if necessary) to get it in the correct form and then use the table to
find the inverse Laplace transform.
Example 151. Find the inverse Laplace transform of F (s).
1. F (s) =
2
s
3
s
8
4
s+1
+
Solution
L
1
{
2
s
3
s
8
+
4
} = 2L
s+1
1
= 2(1)
=2
2. F (s) =
2
3s
5
1
{ }
s
3L
3e8t + 4e
3e8t + 4e
1
{
1
s
8
} + 4L
1
{
s
t
t
11
s5
Solution
L
3. F (s) =
6s
s2 + 9
1
{
2
3s
4
s2 + 25
5
11
1
1
} = 2L 1 {
} + 11L 1 { 4+1 }
s5
s
3(s 53 )
2
1
11
4!
= L 1{
} + L 1 { 4+1 }
3
4!
s
s 53
2 5
11
= e 3 t + t4 .
3
24
1
}
( 1)
CHAPTER 5. LAPLACE TRANSFORM
188
Solution
L
4. F (s) =
3s2
1
6s
s2 + 9
{
4
} = 6L
s2 + 25
s
1
} 4L 1 { 2
}
s 2 + 32
s + 52
s
4 1
5
= 6L 1 { 2
}
L { 2
}
2
s +3
5
s + 52
4
= 6 cos 3t
sin 5t.
5
1
{
4
3
+ 2
+ 12 s
49
Solution
L
5. F (s) =
1
{
3s2
4
3
+ 2
}=
+ 12 s
49
1
1
} + 3L 1 { 2
}
2
+2 )
s
72
4
2
3
7
=
L 1{ 2
} + L 1{ 2
}
3(2)
s + 22
7
s
72
2
3
=
sin 2t + sinh 7t.
3
7
4L
1
{
3(s2
2s 5
s2 + 7
Solution
L
1
{
2s 5
}=L
s2 + 7
2s
5
}
2
+7 s +7
s
1
1
= 2L 1 {
p 2 } 5L {
p 2}
2
2
s + 7
s + 7
p
s
5
7
1
1
= 2L {
p 2} p L {
p 2}
7
s2 + 7
s2 + 7
p
p
5
= 2 cos 7t p sin 7t.
7
1
{
s2
Example 152. Find the inverse Laplace transform of F (s).
1. F (s) =
2s 1
s2 + 4s + 5
Solution Note that s2 + 4s + 5 = (s2 + 4s + 22 )
L
1
{
s2
2s 1
}=L
+ 4s + 5
2(s + 2 2) 1
2(s + 2) 5
} = L 1{
}
2
2
(s + 2) + 1
(s + 2)2 + 12
s+2
1
= 2L 1 {
} 5L 1 {
}
(s + 2)2 + 12
(s + 2)2 + 12
= 2e
2. F (s) =
s2
1 3s
+ 8s + 21
22 + 5 = (s + 2)2 + 12 .
1
{
2t
cos t
5e
2t
sin t.
CHAPTER 5. LAPLACE TRANSFORM
189
Solution We have s2 + 8s + 21 = (s2 + 8s + 42 )
L
3. F (s) =
1
{
s2
1 3s
}=L
+ 8s + 21
1
{
1
3(s + 4
p
4)
2
42 + 21 = (s + 4)2 +
p
2
5 .
3(s + 4) + 12
p 2 }
(s + 4)2 + 5
3(s + 4)
p 2}
(s + 4)2 + 5
}=L
1
{
1
(s + 4)2 + 5
13
= L 1{
p 2
(s + 4)2 + 5
p
13
5
s+4
1
1
=p L {
p 2 } 3L {
p 2}
2
5
(s + 4) + 5
(s + 4)2 + 5
p
p
13
= p e 4t sin 5t 3e 4t cos 5t.
5
2s 3
s2 + 6s + 13
Solution We have s2 + 6s + 13 = (s2 + 6s + 32 )
L
1
{
2s 3
}=L
s2 + 6s + 13
32 + 13 = (s + 3)2 + 22 .
2(s + 3 3) 3
2(s + 3) 9
} = L 1{
}
(s + 3)2 + 22
(s + 3)2 + 22
s+3
9 1
2
= 2L 1 {
}
L {
}
(s + 3)2 + 22
2
(s + 3)2 + 22
9 3t
= 2e 3t cos 2t
e
sin 2t.
2
1
{
Now it’s time to remind ourselves how to use the partial fraction decomposition.
Factor in denominator
Terms in fraction decomposition
s
A
s
sk
s+a
(s + a)k
s2 + as + b
(s2 + as + b)k
A1
A2
Ak
+ 2 + ··· + k
s
s
s
A
s+a
A1
A2
Ak
+
+ ··· +
s + a (s + a)2
(s + a)k
As + B
2
s + as + b
A1 s + B 1
A2 s + B 2
Ak s + B k
+ 2
+ ··· + 2
2
2
s + as + b (s + as + b)
(s + as + b)k
CHAPTER 5. LAPLACE TRANSFORM
190
Example 153. Find the inverse Laplace transform of F (s).
1. F (s) =
s3
2s2 1
3s2 10s
Solution First we would like to use partial fraction decomposition to rewrite F as
follows
2s2 1
A
B
C
= +
+
.
3
2
s
3s
10s
s
s 5 s+2
Then we have,
s3
)
s3
2s2 1
A
= +
3s2 10s
s
A(s
=
s(s
2s2 1
A(s
=
3s2 10s
B
s
C
+
5 s+2
5)(s + 2)
Bs(s + 2)
Cs(s 5)
+
+
5)(s + 2)
s(s 5)(s + 2) s(s 5)(s + 2)
5)(s + 2) + Bs(s + 2) + Cs(s 5)
.
s(s 5)(s + 2)
And so
2s2
1 = A(s
5)(s + 2) + Bs(s + 2) + Cs(s
5)
(5.5)
In order to find the coefficients A, B and C, we can do in di↵erent ways.
Method 1: Impose sample points into the Equation (5.5):
s=0
)
2(0)2
1 = A(0
s=5
)
2(5)2
1 = B(5)(5 + 2)
)
2( 2)2
s=
2
)
5)(0 + 2)
)
1 = C( 2)( 2
)
5)
1
10
7
B=
5
1
C=
2
A=
Method 2: Comparison of the coefficients, by (5.5), we have
2s2
2
2s + 0s
1 = A(s
5)(s + 2) + Bs(s + 2) + Cs(s
= A(s
2
= (As
2
2
3s
10) + B(s + 2s) + C(s
3As
5)
2
5s)
2
10A) + (Bs + 2Bs) + (Cs2
2
1 = (A + B + C)s + ( 3A + 2B
5Cs)
5C)s + ( 10A)
Hence
A + B + C = 2,
which also give us A =
L
1
3A + 2B
5C = 0,
10A =
1
1
7
1
, B = and C = . So
10
5
2
{F (s)} = L
1
=L
1
{2s2
{
1
10
s
1}
+
7
5
s
5
+
1
{ 10 } + L 1 {
s
s
1
7 5t 1
=
(1) + e + e
10
4
2
=L
1
1
2
s+2
7
5
5
2t
}
}+L
=
1
{
1
2
s+2
}
1
7
1
+ e5t + e
10 4
2
2t
.
CHAPTER 5. LAPLACE TRANSFORM
2. F (s) =
191
s2 + 1
s(s + 1)2
Solution Get a new form for F by using partial fraction decomposition,
s2 + 1
A
B
C
= +
+
s(s + 1)2
s
s + 1 (s + 1)2
A(s + 1)2
Bs(s + 1)
Cs
=
+
+
s(s + 1)2
s(s + 1)2
s(s + 1)2
A(s + 1)2 + Bs(s + 1) + Cs
=
s(s + 1)2
And so
s2 + 1 = A(s + 1)2 + Bs(s + 1) + Cs
(5.6)
Method 1: Impose sample points into the Equation (5.5):
s=0
s=
s=1
1
)
)
)
(0)2 + 1 = A(0 + 1)2
2
( 1) + 1 = C( 1)
2
2
)
)
A=1
C=
2
1 + 1 = A(1 + 1) + B(1)(1 + 1) + C(1)
)
B = 0.
Method 2: Now let’s find the coefficients A, B and C, by comparison of the coefficients
of (5.6),
s2 + 1 = A(s + 1)2 + Bs(s + 1) + Cs
= A(s2 + 2s + 1) + B(s2 + s) + Cs
s2 + 0s + 1 = (A + B)s2 + (2A + B + C)s + A
Hence
A + B = 1,
2A + B + C = 0,
which give us A = 1, B = 0 and C =
A=1
2.
Therefore, we have
L
1
s2 + 1
}
s(s + 1)2
1
0
2
= L 1{ +
+
}
s s + 1 (s + 1)2
1
1
= L 1 { } 2L 1 {
}
s
(s + 1)2
= 1 2te t .
{F (s)} = L
1
{
CHAPTER 5. LAPLACE TRANSFORM
3. F (s) =
192
5s2 20s + 34
(s 6)(s2 + 11)
Solution Using partial fraction decomposition to rewrite F as follows
5s2 20s + 34
A
Bs + C
=
+ 2
(s 6)(s2 + 11)
s 6
s + 11
2
A(s + 11) + (Bs + C)(s 6)
=
(s 6)(s2 + 11)
(As2 + 11A) + (Bs2 6Bs + Cs
=
(s 6)(s2 + 11)
(A + B)s2 + ( 6B + C)s + (11A
=
(s 6)(s2 + 11)
6C)
6C)
.
Then we have
5s2
20s + 34 = (A + B)s2 + ( 6B + C)s + (11A
6C).
(5.7)
By the comparison of the coefficients of (5.7),
A + B = 5,
6B + C =
which gives us A = 2, B = 3 and C =
20,
11A
6C = 34
2.
Therefore, we have
L
4. F (s) =
1
5s2 20s + 34
}
(s 6)(s2 + 11)
2
3s 2
+
}
= L 1{
s 6 s2 + 11
2
3s
2
= L 1{
+ 2
}
2
s 6 s + 11 s + 11
1
s
= 2L 1 {
} + 3L 1 {
p 2}
s 6
s2 + 11
p
p
2
= 2e6t + 3 cos 11t p sin 11t.
11
{F (s)} = L
1
{
4s4 + s3 4s2 14s
s3 (s2 + 4s + 5)
2
p L
11
1
{
p
11
p 2}
s2 + 11
5
Solution Rewrite F by using partial fraction decomposition,
4s4 + s3 4s2 14s 5
A
B
C
Ds + E
= + 2+ 3+ 2
s3 (s2 + 4s + 5)
s
s
s
s + 4s + 5
As2 (s2 + 4s + 5) + Bs(s2 + 4s + 5) + C(s2 + 4s + 5) + (Ds + E)s3
=
s3 (s2 + 4s + 5)
(A + D)s4 + (4A + B + E)s3 + (5A + 4B + C)s2 + (5B + 4C)s + 5C
=
.
s3 (s2 + 4s + 5)
Then we have
4s4 +s3 4s2 14s 5 = (A+D)s4 +(4A+B +E)s3 +(5A+4B +C)s2 +(5B +4C)s+5C.
(5.8)
CHAPTER 5. LAPLACE TRANSFORM
193
Therefore by the comparison of the coefficients of (5.8),
A + D = 4,
4A + B + E = 1,
which give us A = 1, B =
2, C =
5A + 4B + C =
4,
1, D = 3 and E =
5B + 4C =
14,
5C =
1.
Therefore, we have
L
1
{F (s)} = L
=L
=L
=1
=1
=1
=1
4s4 + s3 4s2 14s 5
}
s3 (s2 + 4s + 5)
2
1
3s 1
1 1
{
+ 2
}
s s2
s3
s + 4s + 5
1
1
3s 1
1 1
{ } 2L 1 { 2 } L 1 { 3 } + L 1 {
}
s
s
s
(s + 2)2 + 12
1 1 2!
3(s + 2 2) 1
2t
L { 3 } + L 1{
}
2
s
(s + 2)2 + 12
1 2
3(s + 2) 7
2t
t + L 1{
}
2
(s + 2)2 + 12
t2
s+2
1
2t
+ 3L 1 {
} 7L 1 {
}
2
(s + 2)2 + 12
(s + 2)2 + 12
t2
2t
+ 3e 2t cos t 7e 2t sin t.
2
1
{
5.
CHAPTER 5. LAPLACE TRANSFORM
194
Exercises 5.3
1. Find the inverse Laplace transform of:
s2
(a)
2et cos(2t)
(b)
2et cos(2t) + 3et sin(2t)
(c)
2et cos(2t) + et sin(2t)
(d)
2et cos(2t)
1 t
2e
2s + 1
2s + 5
sin(2t)
2. What is the inverse Laplace transform of
2s + 7
s2 + 8s + 65
(a) 2e
8t
cos t
(b) 2e
4t
cos 7t
(c) 2e
7t
cos 4t
(d) 3 cos 7t
11e
8t
1
4t
7e
1
7t
4e
sin t
sin 7t
sin 4t
4 sin 7t
3. The inverse Laplace transform of
(a)
2 + e2t + e
(b)
2t + e2t + e
8
s3
4s
is
2t
2t
(c) 2t + cos 2t
(d)
2+
1
2
sin 2t
4. Find the inverse Laplace transform of each function:
s+4
s2 4s + 13
2s2 10s 18
(b)
s3 s2 6s
2s + 2
(c)
(s 3)2
(a)
5. True or false. L
1
{F (s) + 7} = L
1
{F (s)} + 7.
6. Find the inverse Laplace transform of the function given below
F (s) =
s2 6s + 10
.
(s 1)(s 2)(s 3)
7. Find the function y(t) whose Laplace transform is the expression
F (s) =
7s2 8s + 12
.
(s 2)(s2 + 4)
CHAPTER 5. LAPLACE TRANSFORM
195
8. Which of the following is the Laplace transform of t sin(2t)?
2
+ 4)2
4s
(b) 2
(s + 4)2
s
(c) 2
s +4
3
(d) 2 2
s (s + 4)
(a)
(s2
9. Find the inverse Laplace transform of
3s + 4
+ 4s + 20
s2 2s 1
(b)
s(s + 1)2
(a)
s2
3s2 2s + 8
s3 + 4s
p
4s + 10
(d) 7 + 2
s
4s + 20
2
2s
1
(e) 3
s
3s2 10s
s2 + 11s 31
(f)
(s + 5)(s2 + 36)
(c)
Answers
1. (d)
2. (b)
3. (a)
4. (a) e2t cos 3t + 2e2t sin 3t, (b) 3 + e
2t
2e3t , (c) 2e3t + 8te3t
5. F
6.
5 t
e
2
1
2e2t + e3t
2
7. 3e2t + 4 cos 2t
8. (b)
9. (a) 3e 2t cos 4t 12 e 2t sin 4t, (b) 1+2e
9 2t
1
7 5t
1
2t
, f) e
2 e sin 4t, (e) 10 + 5 e + 2 e
t
5t
2te t , (c) 2+cos 2t sin 2t, (d) (t)+4e2t cos 4t+
+ 2 cos 6t + 16 sin 6t
CHAPTER 5. LAPLACE TRANSFORM
5.4
196
Solving Initial Value Problems with Laplace Transform
We can derive the Laplace transform formula for nth derivative of y as follows:
L{y (n) (t)} = sn L{y(t)}
sn
1
y(0)
y (n
...
1)
(0).
Most of the time we use this formula in special cases (when n = 1, 2 and 3).
L{y 0 (t)} = sL{y(t)}
00
y(0),
2
L{y (t)} = s L{y(t)}
y 0 (0)
sy(0)
L{y 000 (t)} = s3 L{y(t)}
sy 0 (0)
s2 y(0)
y 00 (0).
Proof. First identity is followed by the integration by parts,
L{y 0 (t)} = lim
A!1
= lim
A!1
= lim
A!1
= lim
A!1
=0
Z
A
st 0
e
y (t) dt
0
e
⇣
e
e
st
y(t)
sA
sA
Z
t=A
t=0
y(A)
y(A)
e
A
st
( s)e
0
s(0)
⌘
y(0) + s lim
y(0)
A!1
s lim
A!1
y(0) + sL{y(t)}
= sL{y(t)}
Now we apply L{y 0 (t)} = sL{y(t)}
y(t) dt
Z
A
e
Z
A
e
st
y(t) dt
0
st
y(t) dt
0
y(0).
y(0) twice to obtain the second formula,
L{y 00 (t)} = sL{y 0 (t)}
y 0 (0)
= s [sL{y(t)}
= s2 L{y(t)}
y(0)]
y 0 (0)
sy(0)
y 0 (0).
Similarly, we can derive last formula by putting two previous formulas together,
L{y 000 (t)} = s2 L{y 0 (t)}
sy 0 (0)
y(0)]
sy 0 (0)
y 00 (0)
= s3 L{y(t)}
s2 y(0)
sy 0 (0)
y 00 (0).
= s2 [sL{y(t)}
y 00 (0)
Notice two functions evaluations, y(0) and y 0 (0), that appear in these formula are often what we’ve
been using for initial conditions in our initial value problem. So this mean if we want to apply
these formulas, we need initial conditions at t = 0 only.
• When we solve the initial value problem in the past, we need to first find a general solution,
di↵erentiate it, plug in the initial conditions and then solve for the constants to get the
particular solution.
• But by using Laplace transform, the initial conditions are applied during the first step and
at the end we get the particular solution instead of a general solution.
CHAPTER 5. LAPLACE TRANSFORM
197
The process of solving initial value problem with Laplace transform,
1. Take Laplace transform to both sides of the equation.
2. Substitute two initial conditions, y(0) and y 0 (0).
3. Find L{y(t)} = Y (s).
4. Take inverse Laplace transform both sides of the equation to get y(t) = L
5.4.1
Equations with Constant Coefficients
Example 154. Use the Laplace transform to solve the IVP
y 00 + y 0
6y = 1,
y(0) = 0,
y 0 (0) = 2.
Solution First, we take the Laplace transform to both sides of the equation,
s2 L{y}
sy(0)
s2 L{y}
L{y 00 + y 0 6y} = L{1}
L{y 00 } + L{y 0 } 6L{y} = L{1}
1
y 0 (0) + (sL{y} y(0)) 6L{y} =
s
1
s(0) 2 + (sL{y} 0) 6L{y} =
s
1
2
s L{y} + sL{y} 6L{y} 2 =
s
1
2
(s + s 6)L{y} = + 2
s
2s + 1
(s + 3)(s 2)L{y} =
s
2s + 1
L{y} =
s(s + 3)(s 2)
Finally, we take an inverse Laplace transform to get the particular solution,
y=L
1
{
=L
1
{
)y=
1
6
2s + 1
}
s(s + 3)(s 2)
1
6
s
1
e
3
+
1
3
+
1
2
s+3 s
1
3t
+ e2t .
2
2
}
Example 155 (Spring 2004 Midterm Exam II).
Solve the IVP
y 00 + 2y 0
3y = 3et ,
y(0) = 0,
y 0 (0) =
5
.
4
1
{Y (s)}.
CHAPTER 5. LAPLACE TRANSFORM
198
Solution Taking the Laplace transform to both sides of the equation,
L{y 00 + 2y 0
3y} = L{3et }
L{y 00 } + 2L{y 0 }
3L{y} = 3L{et }
3
s2 L{y} sy(0) y 0 (0) + 2 (sL{y} y(0)) 3L{y} =
s 1
✓
◆
5
3
2
s L{y} s(0) (
) + 2 (sL{y} 0) 3L{y} =
4
s 1
5
3
s2 L{y} + 2sL{y} 3L{y} + =
4
s 1
3
5
(s2 + 2s 3)L{y} =
s 1 4
5
1)
3
4 (s
(s + 3)(s 1)L{y} =
s 1
s 1
5
17
4s + 4
L{y} =
2
(s 1) (s + 3)
By partial fraction decomposition,
(s
5
17
4s + 4
2
1) (s +
3)
A
B
C
+
+
s 1 (s 1)2
s+3
A(s 1)(s + 3)
B(s + 3)
C(s 1)2
=
+
+
2
2
(s 1) (s + 3)
(s 1) (s + 3) (s 1)2 (s + 3)
A(s2 + 2s 3) + B(s + 3) + C(s2 2s + 1)
=
(s 1)2 (s + 3)
(A + C)s2 + (2A + B 2C)s + ( 3A + 3B + C)
=
(s 1)2 (s + 3)
=
By the comparison of the coefficients of the numerator, we have
A + C = 0,
2A + B
1
2, B
Then we can find that A =
Hence
=
L{y} =
3
4
5
,
4
2C =
3A + 3B + C =
17
.
4
and C = 12 .
1
2
s
1
+
3
4
(s
1)2
+
1
2
s+3
.
Taking an inverse Laplace transform to get the particular solution,
y=L
=
)y=
1
{
1
2
1
1
{
3
4
(s
Example 156 (Summer 2003 Final Exam ).
1)2
}+L
1
{
1
2
}
s+3
1 1
1
3
1
1
1
L {
} + L 1{
} + L 1{
}
2
2
s 1
4
(s 1)
2
s+3
1 t 3 t 1 3t
e + te + e .
2
4
2
s
}+L
CHAPTER 5. LAPLACE TRANSFORM
199
Consider the initial value problem
y 0 + y = tet ,
y(0) = 1.
a) Solve it using the method of integrating factor.
Solution The equation is already in standard form with p(t) = 1. So the integrating
factor is
R
R
e p(t) dt = e 1 dt = et .
Multiply both side to the equation by et ,
et y 0 + et y = te2t
Now we change y 0 =
dy
dt
)
(et y)0 = te2t .
and use the technique of separable equation to solve it,
d(et y)
= te2t
dt
Z
)
d(et y) =
Z
te2t dt.
Integration by parts gives
Z
te2t dt =
1 2t
te
2
1 2t
e + c.
4
So we have
et y =
1 2t
te
2
1 2t
e +c
4
)
y(t) =
1 t
te
2
1 t
e + ce t .
4
Then we apply the given initial condition, y(0) = 1,
1=0
1
+c
4
)
c=
5
.
4
Substitute the obtained value of c back into the general solution to get the particular
solution,
1
1 t 5 t
y(t) = tet
e + e .
2
4
4
b) Solve it using the Laplace transform.
Solution Firstly, we take the Laplace transform to both sides of the given DE,
L{y 0 + y} = L{tet }
L{y 0 } + L{y} = L{tet }
1
(sL{y} y(0)) + L{y} =
(s 1)2
1
(s + 1)L{y} =
(s 1)2
1
1
) L{y} =
+
.
(s 1)2 (s + 1) s + 1
So by taking inverse Laplace transform both sides, we then get
y(t) = L
1
{
(s
1
}+L
1)2 (s + 1)
1
{
1
}.
s+1
CHAPTER 5. LAPLACE TRANSFORM
200
By partial fraction decomposition, we write
(s
1
A
B
C
=
+
+
.
1)2 (s + 1)
s 1 (s 1)2
s+1
Now we rewrite all the RHS and then compare it with the LHS,
(s
So A + C = 0, B
1
A
B
C
=
+
+
1)2 (s + 1)
s 1 (s 1)2
s+1
A(s 1)(s + 1) + B(s + 1) + C(s 1)2
=
(s 1)2 (s + 1)
2
As
A + Bs + B + Cs2 2Cs + C
=
(s 1)2 (s + 1)
2
(A + C)s + (B 2C)s + ( A + B + C)
=
.
(s 1)2 (s + 1)
2C = 0 and
A=
A + B + C = 1 which give
1
,
4
B=
Therefore
(s
1
=
1)2 (s + 1)
s
1
2
1
4
1
and
+
C=
1
2
(s
1)2
1
.
4
+
1
4
s+1
Hence the particular solution is
⇢
⇢
1
1
1
1
y(t) = L
+L
(s 1)2 (s + 1)
s+1
⇢ 1
1
1
1
4
2
4
=L
+
+
+L
s 1 (s 1)2
s+1
1 t 1 t 1 t
=(
e + te + e ) + e t
4
2
4
1 t 1 t 5 t
) y(t) =
e + te + e .
4
2
4
1
.
⇢
1
s+1
CHAPTER 5. LAPLACE TRANSFORM
201
Exercises 5.4.1
1. Find L{(e⇡t sin 3t)000 }. Hint:
(e⇡t sin 3t)0 = ⇡e⇡t sin 3t + 3e⇡t cos 3t
(e⇡t sin 3t)00 = (⇡ 2 + 9)e⇡t sin 3t + 6⇡e⇡t cos 3t
2. Suppose that the Laplace transform of y is Y . If y(0) = 2 and y 0 (0) = 3, then find the
Laplace transform of y 00 .
3. Suppose y(t) is the solution of the second order linear initial value problem
y 00 + 4y = t,
y(0) = 1,
y 0 (0) = 0.
What is the Laplace transform of y(t)?
(a) Y (s) =
1
s2 (s2 + 4)
(b) Y (s) =
1 s2
s2 (s2 + 4)
(c) Y (s) =
s2 + 1
s2 (s2 + 4)
(d) Y (s) =
s3 + 1
+ 4)
s2 (s2
4. Suppose y(t) is the solution of the first order linear initial value problem
y 0 + 2y = t3 e
4t
,
What is the Laplace transform of y(t)?
6
3
(s + 2)(s + 4)4
s+4
6
3
(b) Y (s) =
+
(s + 2)(s + 4)4
s+4
3
3
(c) Y (s) =
+
(s + 2)(s 4)4
s+2
3
3
(d) Y (s) =
(s + 2)(s 4)4
s+2
(a) Y (s) =
Answers
1.
(s
2. s2 Y
3. (d)
4. (a)
3s3
⇡)2 + 9
2s + 3
3s
6⇡
y(0) =
3.
CHAPTER 5. LAPLACE TRANSFORM
5.4.2
202
Equations with Variable Coefficients*
In some special cases, we can use the Laplace transform to solve IVP for di↵erential equation with
nonconstant coefficients. Let’s consider some identities that we’re gonna use shortly. We have
L{ty 0 (t)} =
d
L{y 0 (t)} =
ds
d
(sY (s)
ds
d
(sY ) =
ds
y(0)) =
sY 0
Y,
where L{y(t)} = Y (s). Similarly,
L{ty 00 (t)} =
=
d
d 2
L{y 00 (t)} =
s Y (s) sy(0) y 0 (0)
ds
ds
d 2
s Y (s) sy(0) =
s2 Y 0 + 2sY y(0) =
ds
s2 Y 0
2sY + y(0).
Example 157. Solve the initial value problem
y 00
2ty 0 + 4y =
2,
y(0) = 0,
y 0 (0) = 0.
Solution As usual, we take Laplace transform to both sides of the equation,
L{y 00 }
s2 Y
2L{ty 0 } + 4L{y} = L{ 2}
2
y 0 (0)
2( sY 0 Y ) + 4Y =
s
2
2
0
s Y + 2sY + 2Y + 4Y =
s
2
0
2
2sY + (s + 6)Y =
.
s
sy(0)
Then we divide both sides by 2s,
s2 + 6
Y =
2s
Y0+
1
.
s2
We’ll use the method of an integrating factor. Consider
µ=e
R
s2 +6
2s
ds
R
2
s
3
s
= e ( 2 + s ) ds = e 4
+3 ln s
= s3 e
s2
4
.
Multiply both sides of the equation by µ and grouping two terms on the LHS,
Z
2
2
s2
s2
3 s4
0
3 s4
4
(s e Y ) = se
) s e Y =
se 4 ds
⇣ 2⌘
Z
d s4
s2
s2
) s3 e 4 Y =
se 4
s
2
)
)
3
s e
s2
4
s2
4
Y = 2e + c
2
c
Y =
+
.
s2
3
s
s3 e 4
By the technical issue of Laplace transform, we can’t transform Y unless c = 0. Therefore
Y (s) =
2
s3
)
L{y(t)} =
is the particular solution we’re looking for.
2
s3
)
y(t) =
L
1
{
2
}=
s3
t2
CHAPTER 5. LAPLACE TRANSFORM
203
Example 158. Solve the initial value problem
ty 00 + 2ty 0
2y =
2,
y 0 (0) = 2.
y(0) = 1,
Solution Take Laplace transform to both sides of the equation,
L{ty 00 } + 2L{ty 0 }
s2 Y 0
2L{y} = L{ 2}
2
2sY + y(0) + 2( sY 0 Y ) 2Y =
s
2
2 0
0
s Y
2sY + 1 2sY
2Y 2Y =
s
2
2
0
(s + 2s)Y
(2s + 4)Y =
1
s
2
s+2
(s2 + 2s)Y 0 + (2s + 4)Y = + 1 =
s
s
s+2
0
s(s + 2)Y + 2(s + 2)Y =
.
s
Dividing by s(s + 2), it becomes
2
1
Y 0 + Y = 2.
s
s
This is a first order di↵erential equation which is solvable by the method of integrating factor.
We have
R 2
µ = e s ds = e2 ln s = s2 .
Multiply both sides of the equation by s2 ,
2
0
s Y + 2sY = 1
)
2
0
(s Y ) = 1
Therefore
Y (s) =
)
2
s Y =
Z
1 ds
)
s2 Y = s + c.
s+c
1
c
= + 2.
s2
s s
Recall that Y (s) = L{y(t)},
L{y(t)} =
1
c
+ 2
s s
)
y(t) = L
1
1
c
{ + 2 } = 1 + ct.
s s
Note that y 0 (t) = c. After we use an initial condition y 0 (0) = 2, c = 2. Therefore
y(t) = 1 + 2t
is the particular solution to the given IVP.
CHAPTER 5. LAPLACE TRANSFORM
5.5
204
Step Functions
The unit step function or Heaviside function, denoted by uc (t) or u(t
(
0 if t < c
uc (t) = u(t c) =
, where c 0.
1 if t c.
c), is denoted by
Figure 5.1: Graph of y = uc (t).
Example 159. By the definition,
u⇡ (t) =
(
0
1
if t < ⇡
.
if t ⇡.
And so u⇡ (1) = 0, u⇡ (3.14) = 0 but u⇡ (⇡) = 1, u⇡ (2) = 1 and u⇡ (100) = 1.
Example 160 (Spring 2011 Midterm Exam II).
Suppose f (t) = 2 + u3 (t)(t
1) + u5 (t)t2 . What is f (4)?
Solution
f (4) = 2 + u3 (4) · (4
1) + u5 (4) · 42 = 2 + 1 · 3 + 0 · 16 = 5.
Example 161 (Fall 2008 Midterm Exam II).
If f (t) = u(t
2)
2u(t
3) + 3u(t
4)
4u(t
5). What is f (⇡)?
Solution
f (⇡) = u2 (⇡)
2u3 (⇡) + 3u4 (⇡)
4u5 (⇡) = 1
2·1+3·0
4·0=
1.
CHAPTER 5. LAPLACE TRANSFORM
205
Example 162. Write the following function in terms of unit step function.
f (t) =
Solution
Then f (t) = 3 + ( 1
3)u2 (t) + (2
8
>
<3,
>
:
1,
2,
t<2
2t<7.
t 7.
( 1))u7 (t) = 3
4u2 (t) + 3u7 (t).
Example 163. Rewrite the following piecewise continuous function f (t) in terms of unit step
functions.
Solution Then f (t) = t2 + (1
8
2
>
<t ,
f (t) = 1 t,
>
: 2t
2e ,
t
t<1
1t<3.
t 3.
t2 )u1 (t) + (2e2t
(1
Example 164. Sketch the graph of f (t) = u1 (t) + 2u3 (t)
t))u3 (t).
6u4 (t)
Solution It is easier to graph if we can rewrite the function as a piecewise continuous one.
In order to do that, we consider its value on each subintervals.
On
t<1
On 1  t < 3
On 3  t < 4
On
t
4
)
Pick t = 0
)
Pick t = 3
)
Pick t = 2
)
Pick t = 5
)
f (0) = u1 (0) + 2u3 (0)
6u4 (0) = 0
)
f (2) = u1 (2) + 2u3 (2)
6u4 (2) = 1
)
f (3) = u1 (3) + 2u3 (3)
6u4 (2) = 1 + 2 = 3
)
f (5) = u1 (5) + 2u3 (5)
6u4 (5) = 1 + 2
6=
3
CHAPTER 5. LAPLACE TRANSFORM
And so we have
206
8
0,
>
>
>
<1,
f (t) =
>
3,
>
>
:
t<1
1t<3
.
3t<4
t 4.
3,
Now let’s consider the Laplace transform formulas involving unit step functions. By the definition,
Z 1
Z c
Z 1
L{uc (t)} =
e st uc (t) dt =
e st (0) dt +
e st (1) dt
=
Z
0
0
1
st
e
A!1
st
e
= lim
s
A!1
=
e
t=A
t=c
cs
c
A
e st dt

e sA
= lim
A!1
s
dt = lim
c
Z
c
(
e
sc
)
s
, if s > 0.
s
And also by changing the variable technique, let ⇠ = t c(or t = ⇠ + c), we have
Z 1
Z t=1
L{uc (t)f (t c)} =
e st uc (t)f (t c) dt =
e st f (t c) dt
=
Z
=e
0
t=c
⇠=1
e
s(⇠+c)
f (⇠) d⇠ = e
⇠=0
cs
L{f (t)} = e
cs
F (s).
cs
Z
1
e
s⇠
f (⇠) d⇠
0
So
L
1
{e
cs
F (s)} = uc (t)f (t
In conclusion, we have
c) = uc (t)f (t)t7!t
c
= uc (t)L
1
{F (s)}t7!t
c.
Laplace transform formulas involving unit step functions:
⇢ cs
e cs
e
• L {uc (t)} =
, L 1
= uc (t)
s
s
• L {uc (t)f (t
c)} = e
cs
L{f (t)}
,
L
1
{e
cs
F (s)} = uc (t)L
1
{F (s)}t7!t
c
.
CHAPTER 5. LAPLACE TRANSFORM
207
Example 165 (Spring 2011 Midterm Exam II).
Find the Laplace transform of u2 (t)(t2 + 1).
2) = t2 + 1. Substitute t by t + 2, we have
Solution Here c = 2, so we let f (t
2) = (t + 2)2 + 1 = (t2 + 4t + 4) + 1 = t2 + 4t + 5.
f (t) = f ((t + 2)
Therefore, by using L {u2 (t)f (t
2)} = e
L{u2 (t)(t2 + 1)} = e
2s
2s
L{f (t)},
L{t2 + 4s + 5} = e
2s
(
2
4
5
+ 2 + ).
s3
s
s
Example 166 (Fall 2010 Midterm Exam II).
2)2 .
Find the Laplace transform of u4 (t)(t
Solution Now c = 4, let f (t
4) = (t
f (t) = f ((t + 4)
So, by using L {u4 (t)f (t
L{u4 (t)(t
4)} = e
2)2 . Then we substitute t by t + 4,
4) = (t + 4
4s
2)2 } = e
2)2 = (t + 2)2 = t2 + 4t + 4.
L{f (t)},
4s
L{t2 + 4t + 4} = e
2s
(
2
4
4
+ 2 + ).
3
s
s
s
Example 167 (Summer 2011 Midterm Exam II).
Rewrite the following piecewise continuous function f (t) in terms of the unit step function.
Then find its Laplace transform.
(
2t2 + t, 0  t < 3
f (t) =
e4t ,
3t
Solution We can rewrite f as 2t2 + t + (e4t
L{f (t)} = L 2t2 + t + (e4t
2
2t2
2t2
t)u3 (t). And so
t)u3 (t)
4t
= 2L{t } + L{t} + L (e
2t2 t)u3 (t)
n
o
2!
1
= 2 · 3 + + e 3s L (e4(t+3) 2(t + 3)2 (t + 3)
s
s
4
1
= 3 + + e 3s L e4t · e12 2(t2 + 6t + 9) t 3
s
s
4
1
= 3 + + e 3s L e4t · e12 2t2 12t 18 t 3
s
s
4
1
= 3 + + e 3s L e4t · e12 2t2 13t 21
s
s
4
1
= 3 + + e 3s e12 L{e4t } 2L{t2 } 13L{t} 21L{1}
s
s
✓
◆
4
1
1
2
1
1
3s
12
= 3 + +e
e ·
2 · 3 13 · 2 21 ·
s
s
s 4
s
s
s
✓ 12
◆
4
1
e
4
13 21
) L{f (t)} = 3 + + e 3s
.
s
s
s 4 s3
s2
s
CHAPTER 5. LAPLACE TRANSFORM
208
Example 168.
4s
Find the inverse Laplace transform of e
Solution We apply L
L
1
⇢
e
4s
1
2s 8
s2 + 2s + 26
e
4s
2s 8
.
s2 + 2s + 26
F (s) = u4 (t)L
= u4 (t)L
= u4 (t)L
1
1
⇢
1
{F (s)}t7!t
2s 8
s2 + 2s + 26
⇢
t7!t 4
2(s + 1) 10
= u4 (t)L 1
(s + 1)2 + 52 t7!t
✓
⇢
s+1
= u4 (t) 2L 1
(s + 1)2 + 52
2s 8
) L 1 {e 4s 2
s + 2s + 26
= u4 (t) 2e
⇣
} = u4 (t) 2e
t
cos 5t
(t 4)
2e
t
cos 5(t
to get
t7!t 4
2(s + 1 1) 8
(s + 1)2 + 52
⇢
4
sin 5t
4)
4
10
L
5
1
⇢
5
(s + 1)2 + 52
t7!t 4
2e
(t 4)
sin 5(t
◆
t7!t 4
⌘
4) .
Example 169 (Fall 2010 Midterm Exam II).
Find the inverse Laplace transform of e
Solution By L
L
1
⇢
e
9s
1
e
9s
F (s) = u9 (t)L
s+3
2
s + 6s + 25
= u9 (t)L
1
= u9 (t)L
1
= u9 (t)L
= u9 (t)
= u9 (t)
)L
1
{e
4s
s+3
} = u9 (t)
s2 + 6s + 25
✓
✓
✓
⇢
⇢
⇢
9s
1
s+3
.
s2 + 6s + 25
{F (s)}t7!t
s2
s+3
+ 6s + 25
9,
t7!t 9
(s + 3 3) + 3
(s + 3)2 + 42
t7!t 9
(s + 3) + 6
1
(s + 3)2 + 42 t7!t 9
⇢
⇢
◆
s+3
6 1
4
L 1
+
L
(s + 3)2 + 42
4
(s + 3)2 + 42
t7!t
◆
3 3t
3t
e
cos 4t + e
sin 4t
2
t7!t 9
◆
3
3(t 9)
3(t 9)
e
cos 4(t 9) + e
sin 4(t 9)
2
9
CHAPTER 5. LAPLACE TRANSFORM
209
Example 170 (Summer 2010 Midterm Exam II).
Find the inverse Laplace transform of
e
3s
.
+ s2
Solution Using partial fraction decomposition,
s3
1
A
B
C
= + 2+
+ 1)
s
s
s+1
As(s + 1) + B(s + 1) + Cs2
=
s2 (s + 1)
2
2
0s + 0s + 1
(A + C)s + (A + B)s + B
=
2
s (s + 1)
s2 (s + 1)
s2 (s
Hence
A + C = 0,
which give us A =
Therefore
Using L
1
e
3s
A + B = 0,
B = 1,
1, B = 1 and C = 1.
1
1
1
1
=
+ 2+
.
s3 + s2
s
s
s+1
1
F (s) = u3 (t)L
L
1
⇢
e
{F (s)}t7!t
3s
s3 + s2
= u3 (t)L
= u3 (t)L
= u3 (t)
= u3 (t)
)L
1
{
e
3s
s3 + s
⇣
⇣
}
=
u
(t)
t
3
2
3,
1
1
⇢
⇢
1
s3 + s2
t7!t 3
1
1
1
+ 2+
s
s
s+1
1+t+e
1 + (t
t
t7!t 3
3) + e
⌘
4 + e t+3) .
(t 3)
t7!t 3
⌘
Next we’ll use Laplace transform to solve initial value problems containing unit step functions.
This is the benefit of Laplace transform since we can not solve these kind of initial value problems
with the method of undetermined coefficients.
CHAPTER 5. LAPLACE TRANSFORM
210
Example 171 (Spring 2008 Midterm Exam II).
Solve the following IVP
y 00
3y 0
4y = u2 (t)
u6 (t),
y(0) = 0,
y 0 (0) = 0.
Solution We use the same strategy as before by firstly taking Laplace transform to both sides
of the equation.
00
L{y }
s2 L{y}
sy(0)
s2 L{y}
y 0 (0)
s(0)
L{y 00
0
3L{y }
3 (sL{y}
0
3y 0
y(0))
3 (sL{y}
s2 L{y}
0)
3sL{y}
(s2
(s
3s
4y} = L{u2 (t)
u6 (t)}
4L{y} = L{u2 (t)}
4L{y} =
4L{y} =
4L{y} =
4)L{y} =
4)(s + 1)L{y} =
e
2s
e
s
e
2s
s
e
s
e
2s
2s
e
2s
6s
s
e
s
e
6s
s
s
e
L{u6 (t)}
6s
6s
s
e
6s
s
s
e 2s
L{y} =
s(s 4)(s + 1)
e 6s
s(s 4)(s + 1)
By taking an inverse Laplace transform, we have
⇢
⇢
e 2s
e 6s
y=L 1
L 1
s(s 4)(s + 1)
s(s 4)(s + 1)
⇢
⇢
1
1
= u2 (t)L 1
u6 (t)L 1
s(s 4)(s + 1) t7!t 2
s(s 4)(s + 1) t7!t 6
⇢ 1
⇢ 1
1
1
1
1
1
1
4
20
5
4
= u2 (t)L
u6 (t)L
+
+
+ 20 + 5
s
s 4 s + 1 t7!t 2
s
s 4 s + 1 t7!t
✓
◆
✓
◆
1
1 4t 1 t
1
1 4t 1 t
= u2 (t)
+ e + e
u6 (t)
+ e + e
4 20
5
4 20
5
t7!t 2
t7!t 6
✓
◆
✓
◆
1
1 4(t 2) 1 (t 2)
1
1 4(t 6) 1 (t 6)
) y = u2 (t)
+ e
+ e
u6 (t)
+ e
+ e
.
4 20
5
4 20
5
6
CHAPTER 5. LAPLACE TRANSFORM
211
Example 172.
Solve the IVP
y 00 + 3y 0 + 2y = f (t),
y(0) = 0,
Solution We can rewrite f as f (t) = 1
sides of the equation,
00
y 0 (0) = 0,
f (t) =
sy(0)
L{y 00 + 3y 0 + 2y} = L{1
2
s L{y}
1
s
1
0 + 3 (sL{y} 0) + 2L{y} =
s
1
(s2 + 3s + 2)L{y} =
s
1
(s + 2)(s + 1)L{y} =
s
y 0 (0) + 3 (sL{y}
s(0)
1,
0,
0 < t < 10
t 10.
u10 (t). Then take the Laplace transform to both
u10 (t)}
L{y } + 3L{y 0 } + 2L{y} = L{1}
s2 L{y}
(
y(0)) + 2L{y} =
e
L{u10 (t)}
10s
s
e
10s
s
e
10s
s
e
10s
s
1
L{y} =
s(s + 1)(s + 2)
e 10s
.
s(s + 1)(s + 2)
Taking an inverse Laplace transform to get the particular solution
⇢
⇢
1
e 10s
y=L 1
L 1
s(s + 1)(s + 2)
s(s + 1)(s + 2)
⇢
⇢
1
1
=L 1
u10 (t)L 1
s(s + 1)(s + 2)
s(s + 1)(s + 2) t7!t 10
⇢1
⇢1
1
1
1
1
=L 1 2 +
+ 2
u10 (t)L 1 2 +
+ 2
s
s+1 s+2
s
s + 1 s + 2 t7!t
✓
◆
✓
◆
1
1
1
1
=
e t + e 2t
u10 (t)
e t + e 2t
2
2
2
2
t7!t 10
✓
◆
✓
◆
1
1
1
1
t
2t
(t 10)
2(t 10)
=
e + e
u10 (t)
e
+ e
.
2
2
2
2
10
CHAPTER 5. LAPLACE TRANSFORM
212
Exercises 5.5
1. Rewrite the following function f (t) in terms of step functions, and find its Laplace transform.
(
3t2 ,
0  t < 2⇡
f (t) =
2
3t + sin t, t 2⇡.
2. (a) Sketch the function
f (t) = t
u1 (t)(t
1) + u3 (t)(2
(b) Write the given function in terms of unit step
8
2
>
<t ,
f (t) = 6 t,
>
:
0,
t)
functions and find its Laplace transform.
0t<2
2t<6
6t
3. Find the solution to the following initial value problem
y 00 + 2y 0
8y = e3t u3 (t),
y(0) = 0,
y 0 (0) = 1.
4. Find the Laplace transform of u ⇡2 (t) cos(2t). The following identity may be useful:
cos(↵ ± ) = cos ↵ cos
s⇡
s
e 2
s2 + 4
s⇡
2
(b)
e 2
2
s +4
s⇡
1
(c) 2
e 2
s +4
s
(d)
e
s2 + 4s + 4
⌥ sin ↵ sin
(a)
5. Suppose f (t) = 2
s⇡
2
u2 (t) + t(u3 (t)
u6 (t)), what is f (5)?
(a) 2
(b) 4
(c) 6
(d) 7
6. Rewrite the following function in terms of unit step functions and find it’s Laplace transform.
(
t+1
0t<3
f (t) = 2
t
t +e , 3t
7. Let g(t) = tu1 (t)
t2 u3 (t) + tu5 (t). What is g(4)?
(a) 4
(b)
12
(c)
16
(d)
e s
s2
2e 3s
e 5s
+ 2
3
s
s
CHAPTER 5. LAPLACE TRANSFORM
213
8. Which of the following functions corresponds to this graph?
(a) tu1 (t)
1
(b) t + u1 (t)
(c) (1
t)u2 (t) + t
(d) t + (1
t)u1 (t)
9. True or false:
1
+1
u2 (t) + 3u4 (t), then f (3) = 1.
(a) L{u⇡ (t) sin t} = e
(b) If f (t) = 2
⇡s
s2
10. Find the inverse Laplace transform of each function given below:
(a)
e
5s
s2 (s
2)
1 + 4s
(b) e 3s 2
s + 10s + 41
5s + 4
(c) F (s) = e 101s 2
s
2s + 10
11. Find the Laplace transform of f (t) = u3 (t)t2 e t .
2
(s + 1)3
2
(b) e 3s
s(s + 1)3
2
3
(c) e 3s+3 (
+
)
(s + 1)3
s+1
2
6
9
(d) e 3s 3 (
+
+
)
(s + 1)3
(s + 1)2
s+1
(a) e
3s
12. Rewrite the following piecewise continuous function f (t) in terms of the unit step function.
Then find its Laplace transform.
(
t2 + 3t, 0  t < 7
f (t) = 2
t + 3t + te t + 5, t 7
CHAPTER 5. LAPLACE TRANSFORM
214
13. Let y(t) be the solution of the initial value problem
y 00 + 2y 0 + 5y = 6u⇡ (t),
y 0 (0) = 0.
y(0) = 2,
Find its Laplace transform Y (s) = L{y(t)}.
(a) Y (s) =
s(s2
6e ⇡s
+ 2s + 5)
2s + 2
s2 + 2s + 5
(b) Y (s) =
6e ⇡s + 2s
s(s2 + 2s + 5)
(c) Y (s) =
6e ⇡s
s(s2 + 2s + 5)
(d) Y (s) =
2s + 4
6e ⇡s
+ 2
2
s(s + 2s + 5) s + 2s + 5
14. Find L{u⇡ (t) sin( 2t )}.
Hint: sin(↵ ± ) = sin(↵) cos( ) ± cos(↵) sin( ).
15. Find the Laplace transform L{u1 (t)tet
(a) e
1
s
s
}.
1)2
1
(b) e s
s(s 1)2
1
(c) e s 1
s(s 1)2
s
(d) e s 1
(s 1)2
(s
Answers
1. f (t) = 3t2 + u2⇡ (t) sin t, F (s) =
2. (a) -, (b) f (t) = t2 + (6
3.
1
e
6
4t
1
1
+ e2t + ( e3(t
6
7
t
6
+e
s3
1
s2 + 1
t2 )u2 (t) + (t
1 2(t
e
6
3)
2⇡s
3)
+
6)u6 (t),
1
e
42
4(t 3)
2
+e
s3
2s
(
2
s3
5
)+e
s2
6s
(
1
)
s2
)e9 u3 (t)
4. (a)
5. (c)
6. f (t) = (t + 1) + u3 (t)(t2
t
1 + et ), F (s) =
1
1
+ +e
s2
s
3s
(
2
5
5
e3
+ 2+ +
)
3
s
s
s s 1
7. (b)
8. (d)
9. (a) F, (b) T
1 1
1
(t 5) + e2(t 5) ), (b) u3 (t)(4e
4 2
4
(c) u101 (t)(5et 101 cos(3t 303) + 3et 101 sin(3t
10. (a) u5 (t)(
11. (d)
5(t 3)
cos 4(t
303))
3)
19
e
4
5(t 3)
sin 4(t
3)),
CHAPTER 5. LAPLACE TRANSFORM
2
3
12. 3 + 2 + e
s
s
13. (d)
14. e
⇡s
15. (a)
4s
4s2 + 1
7s
✓
e 7
7e 7
5
+
+
2
(s + 1)
s+1 s
215
◆
CHAPTER 5. LAPLACE TRANSFORM
5.6
216
Impulse Functions
The idealized impulse function is called Dirac Delta2 function at t = c if it has two following
properties:
1. (t c) = 0,
if t 6= c,
R1
2.
(t c) dt = 1.
1
Note here that
• In elementary calculus, there is no such function to satisfy above two properties.
• The function
is an example of what are known as generalized function or distribution.
So if this is an ideal function, how do mathematicians work with it? The answer is that mathematicians use Dirac Delta function in terms of the limit of tangible function d⌧ ,
(t
c) = lim+ d⌧ (t
⌧ !0
c).
(5.9)
Here is how d⌧ is defined,
d⌧ (t
c) =
(
1
2⌧
0
if
⌧ < t c < ⌧,
if t c  ⌧ or t c
⌧,
where ⌧ is a positive constant. For simplicity, we consider its graph when c = 0.
Figure 5.2: Graph of y = d⌧ (t).
2 Paul Adrien Maurice Dirac (1902
1984), English mathematical physicist, received his Ph.D. from Cambridge
in 1926 and was professor of mathematics there until 1969. He was awarded the Nobel Prize in 1933 (with Erwin
Schrdinger) for fundamental work in quantum mechanics. His most celebrated result was the relativistic equation
for the electron, published in 1928. From this equation he predicted the existence of an “anti-electron,” or positron,
which was first observed in 1932. Following his retirement from Cambridge, Dirac moved to the United States and
held a research professorship at Florida State University.
In the early 1930’s he developed a very controversial method for dealing with impulsive functions. Most mathematicians usually ridiculed this method. “How can you make believe that (t t0 ) is an ordinary function if it
is obviously not,” they asked. However, they never laughed too loud since Dirac and his followers always obtained
the right answer. In the late 1940’s, the French mathematician Laurent Schwartz succeeded in placing the delta
function a firm mathematical foundation. He accomplished this by enlarging the class of all functions so as to
include the delta function. This new object is called distribution or generalized function.
CHAPTER 5. LAPLACE TRANSFORM
217
Figure 5.3: Graphs of y = d⌧ (t) as ⌧ ! 0+ .
By using Equation (5.9), we have the following results,
Laplace transform formulas involving Dirac Delta functions:
• L { (t)} = 1
cs
• L { (t
c)} = e
• L { (t
c)f (t)} = f (c)e
cs
,
L
1
{1} = (t).
,
L
1
{e
,
L
1
{f (c)e
cs
} = (t
cs
c).
} = (t
c)f (t).
Example 173 (Fall 2008 Midterm Exam II).
Solve the IVP
y 0 + 3y = tu(t
2) + (t
3),
y(0) = 1.
Solution Again firstly taking Laplace transform to both sides of the equation.
L{y 0 + 3y} = L{tu(t 2) + (t 3)}
L{y 0 } + 3L{y} = L{tu2 (t)} + L{ (t 3)}
L {t + 2} + e 3s
✓
◆
1
2
2s
(sL{y} 1) + 3L{y} = e
+
+ e 3s
s2
s
2s + 1
(s + 3)L{y} 1 = e 2s ·
+ e 3s
s2
2s + 1
e 3s
1
L{y} = e 2s 2
+
+
.
s (s + 3) s + 3 s + 3
(sL{y}
y(0)) + 3L{y} = e
2s
CHAPTER 5. LAPLACE TRANSFORM
218
By taking an inverse Laplace transform, we have
⇢
⇢
⇢
1
1
2s 2s + 1
1
3s 1
1
y=L
e
+L
e
+L
s2 (s + 3)
s+3
s+3
⇢
⇢
⇢
2s + 1
1
1
1
1
1
= u2 (t)L
+ u3 (t)L
+L
s2 (s + 3) t7!t 2
s + 3 t7!t 3
s+3
⇢5
1
5
9
= u2 (t)L 1 9 + 32 +
+ u3 (t) e 3t t7!t 3 + e 3t
s
s
s + 3 t7!t 2
✓
◆
5 1
5 3t
= u2 (t)
+ t
e
+ u3 (t)e 3(t 3) + e 3t
9 3
9
t7!t 2
✓
◆
5 t 2 5 3(t 2)
) y = u2 (t)
+
e
+ u3 (t)e 3(t 3) + e 3t .
9
3
9
Example 174 (Fall 2007 Midterm Exam II).
Use Laplace transforms to solve the following initial value problem
y 0 + 2y = et
3
u(t
3) + (t
3),
y(0) = 3.
Solution Firstly, we take the Laplace transform to both sides of the equation,
L{y 0 + 2y} = L{et
0
(sL{y}
L{y } + 2L{y} = L{e
y(0)) + 2L{y} = e
e
s
3
u(t
3) + (t
t 3
u(t
3)} + L{ (t
3s
t
L{e } + e
3)}
3)}
3s
3s
+ e 3s
1
e 3s
e 3s
3
) L{y} =
+
+
.
(s 1)(s + 2) s + 2 s + 2
(s + 2)L{y}
3=
Then we take the inverse Laplace transform to both sides to get the particular solution,
⇢
e 3s
e 3s
3
y(t) = L 1
+
+
(s 1)(s + 2) s + 2 s + 2
⇢
⇢
⇢
1
1
3
= u3 (t)L 1
+ u3 (t)L 1
+L 1
(s 1)(s + 2) t7!t 3
s + 2 t7!t 3
s+2
⇢ 1
1
1
3
3
= u3 (t)L
+
+ u3 (t) e 2t t7!t 3 + 3e 2t
s 1 s + 2 t7!t 3
✓
◆
1 t 1 2t
= u3 (t)
e
e
+ u3 (t)e 2(t 3) + 3e 2t
3
3
t7!t 3
✓
◆
1 t 3 1 2(t 3)
) y(t) = u3 (t)
e
e
+ u3 (t)e 2(t 3) + 3e 2t .
3
3
CHAPTER 5. LAPLACE TRANSFORM
219
We can also simplify further as follows:
1
1
u3 (t)et 3
u3 (t)e 2(t 3) + u3 (t)e
3
3
1
2
= u3 (t)et 3 + u3 (t)e 2(t 3) + 3e 2t
3
3
1
t 3
) y(t) = u3 (t)(e
+ 2e 2t+6 ) + 3e 2t .
3
y(t) =
2(t 3)
+ 3e
2t
Example 175 (Spring 2005 Midterm Exam II).
Solve the following IVP:
y 0 + 3y = (t
1) + u(t
2),
y(0) =
4
Solution In order to solve the above IVP, we firstly take the Laplace transform to both sides
of the equation,
L{y 0 + 3y} = L{ (t
1) + u(t
0
L{y } + 3L{y} = L{ (t
(sL{y}
y(0)) + 3L{y} = e
(s + 3)L{y} + 4 = e
) L{y} =
s
s
+
+
e
2)}
1)} + L{u(t
2)}
2s
s
e
2s
s
s
e
e 2s
+
s + 3 s(s + 3)
4
.
s+3
Now let’s take the inverse Laplace transform to both sides, we then get
⇢
e s
e 2s
4
1
y(t) = L
+
s + 3 s(s + 3) s + 3
⇢
⇢
1
1
1
1
= u1 (t)L
+ u2 (t)L
L
s + 3 t7!t 1
s(s + 3) t7!t 2
⇢1
1
3t
1
3
3
= u1 (t)(e )t7!t 1 + u2 (t)L
+
4e 3t
s
s + 3 t7!t 2
1 1 3t
= u1 (t)e 3(t 1) + u2 (t)(
e )t7!t 2 4e 3t
3 3
1 1 3(t 2)
) y(t) = u1 (t)e 3(t 1) + u2 (t)(
e
) 4e 3t .
3 3
Example 176.
Solve the IVP
y 00 + y = (t
2⇡) cos t,
y(0) = 0,
y 0 (0) = 1..
1
⇢
4
s+3
CHAPTER 5. LAPLACE TRANSFORM
220
Solution Apply the Laplace transform to both sides of the equation.
L{y 00 + y} = L{ (t
00
2
s L{y}
2
1 + L{y} = e
2⇡s
0
y (0) + L{y} = e
sy(0)
s L{y}
L{y } + L{y} = e
2⇡s
s(0)
(s2 + 1)L{y}
1=e
L{y} =
2⇡) cos t}
cos 2⇡
2⇡s
2⇡s
e 2⇡s
1
+ 2
2
s +1 s +1
Hence, by taking the inverse Laplace transform to both sides of the equation,
⇢
⇢
1
1
y = L 1 e 2⇡s · 2
+L 1
s +1
s2 + 1
⇢
⇢
1
1
1
= u2⇡ (t)L 1
+
L
2
2
2
s + 1 t7!t 2⇡
s + 12
= u2⇡ (t) (sin t)t7!t
= u2⇡ (t) sin(t
2⇡
+ sin t
2⇡) + sin t
) y = u2⇡ (t) sin(t) + sin t
Example 177 (Fall 2009 Midterm Exam II).
Solve using Laplace transform
y 00 + 2y 0 + 2y = (t
5),
y(0) = 1,
y 0 (0) = 2,
where (t) is the Dirac delta function.
Solution The first step to solve this IVP is to take the Laplace transform to both sides of the
equation,
00
2
(s L{y}
sy(0)
(s2 L{y}
L{y 00 + 2y 0 + 2y} = L{ (t
0
L{y } + 2L{y } + 2L{y} = e
0
y (0)) + 2(sL{y}
s
y(0)) + 2L{y} = e
2) + 2(sL{y}
1) + 2L{y} = e
(s2 + 2s + 2)L{y}
s
4=e
) L{y} =
s2
5)}
5s
5s
5s
5s
e 5s
s+4
+
.
+ 2s + 2 s2 + 2s + 2
CHAPTER 5. LAPLACE TRANSFORM
221
Then we take inverse Laplace transform to both sides,
⇢
⇢
s+4
e 5s
1
1
y(t) = L
+L
s2 + 2s + 2
s2 + 2s + 2
⇢
⇢
1
s+4
1
1
= u5 (t)L
+L
s2 + 2s + 2 t7!t 5
(s + 1)2 + 12
⇢
⇢
1
s+1
3
1
1
= u5 (t)L
+L
+
2
2
2
2
(s + 1) + 1 t7!t 5
(s + 1) + 1
(s + 1)2 + 12
= u5 (t)(e
) y(t) = u5 (t)e
t
sin t)t7!t
(t 5)
sin(t
5
+e
t
5) + e
cos t + 3e
t
t
cos t + 3e
sin t
t
sin t.
Example 178 (Spring 2004 Midterm Exam II).
Use the Laplace transform to solve the initial value problem
y 00
2y 0
15y = e2t + 2 (t
1),
y(0) = 0,
y 0 (0) = 0.
Solution The first step is to take the Laplace transform to both sides of the equation,
00
L{y }
(s2 L{y}
sy(0)
y 0 (0))
L{y 00
2y 0
0
2L{y }
15y} = L{e2t + 2 (t
1)}
2t
15L{y} = L{e } + 2L{ (t 1)}
1
2(sL{y} y(0)) 15L{y} =
+ 2e s
s 2
1
(s2 2s 15)L{y} =
+ 2e s
s 2
1
) L{y} =
+
(s 2)(s 5)(s + 3) (s
Then we take inverse Laplace transform to both sides,
⇢
1
1
y(t) = L
+L
(s 2)(s 5)(s + 3)
1
⇢
(s
2e s
5)(s + 3)
We apply partial fraction decomposition here,
⇢
⇢
1
2e s
y(t) = L 1
+L 1
(s 2)(s 5)(s + 3)
(s 5)(s + 3)
⇢
⇢
1
2
=L 1
+ u1 (t)L 1
(s 2)(s 5)(s + 3)
(s 5)(s + 3)
⇢ 1
⇢ 1
1
1
1
1
1
15
24
40
4
4
=L
+
+
+ u1 (t)L
+
s 2 s 5 s+3
s 5 s+3
.
t7!t 1
t7!t 1
2e s
.
5)(s + 3)
CHAPTER 5. LAPLACE TRANSFORM
222
Hence the particular solution is
⇢ 1
1
1
1
15
y(t) = L
+ 24 + 40
s 2 s 5 s+3
=
1 2t
1
1
e + e5t + e
15
24
40
3t
) y(t) =
1 2t
1
1
e + e5t + e
15
24
40
3t
+ u1 (t)L
+ u1 (t)
+ u1 (t)
✓
✓
1
1 5t
e
4
⇢
1
4
s
1
e
4
1 5(t
e
4
1)
1
4
+
5
◆
s+3
t7!t 1
3t
1
e
4
t7!t 1
3(t 1)
◆
.
Example 179 (Summer 2007 Midterm Exam II).
Consider the initial value problem
y 000
2y 00 + y 0 + 4y = e
2t
+ (t
⇡),
y 0 (0) = 1,
y(0) = 0,
y 00 (0) = 0.
Find L{y(t)} = Y (s), the Laplace transform of its solution.
Solution Taking the Laplace transform to both sides of the equation,
000
L{y }
L{y 000
00
2y 00 + y 0 + 4y} = L{e
0
2L{y } + L{y } + 4L{y} = L{e
2t
2t
+ (t
⇡)}
} + L{ (t
⇡)}
Before we proceed further, let’s consider:
L{y 000 } = s3 L{y}
00
2L{y } =
0
s2 y(0)
2
2(s L{y}
L{y } = sL{y}
sy(0)
sy 0 (0)
0
y 00 (0) = s3 Y (s)
y (0)) =
y(0) = sY (s)
2
2(s Y (s)
s2 (0)
s(0)
0 = s3 Y (s)
s(1)
2
1) =
2s Y (s) + 2
(0) = sY (s)
4L{y} = 4Y (s).
Combine altogether, we have
(s3 Y (s)
1
+e
s+2
1
2s2 + s + 4)Y (s) s + 2 =
+e
s+2
1
(s3 2s2 + s + 4)Y (s) =
+e
s+2
s) + ( 2s2 Y (s) + 2) + sY (s) + 4Y (s) =
(s3
⇡s
⇡s
⇡s
+s
2
Therefore
Y (s) =
(s +
2)(s3
1
+
2s2 + s + 4) s3
e ⇡s
+
2s2 + s + 4 s3
s 2
.
2s2 + s + 4
s
CHAPTER 5. LAPLACE TRANSFORM
223
Exercises 5.6
1. Solve the initial value problem,
y 00
4y = (t
2⇡),
y 0 (0) = 1.
y(0) = 0,
2. Solve the initial value problem
y 00 + 5y 0 + 4y = 2 (t
3),
y(0) = 1,
y 0 (0) = 0.
4),
y(0) = 1,
y 0 (0) = 0
3. Solve the initial value problem:
y 00
4y 0 + 4y = 2 (t
4. Use the Laplace transform to solve the initial value problem
y 00
5. Evaluate L{ (t
2y 0
15y = e2t + 2 (t
1),
y 0 (0) = 0.
y(0) = 0,
1)t10 tan2 (t⇡/4)}
6. (a) Use the Laplace transform to solve the following initial value problem
y 00 + 9y = t
u2 (t) + (t
1),
y(0) = 0,
y 0 (0) = 1.
(b) Evaluate y(⇡/3).
7. True or false. Justify your answers.
L{ (t
2⇡)et cos(t)} = e
2⇡(s+1)
8. (a) Use the Laplace transform to solve the following initial value problem
y 00 + 3y 0 + 2y = u6 (t)
2 (t
1),
y 0 (0) = 2.
y(0) = 0,
(b) Evaluate y(ln 2).
9. Find L{ (t
t
⇡) sin( )(3t2
2
1)}.
10. Solve the following using Laplace transforms.
y 00 + 2y 0 + 2y = (t
3),
y 0 (0) = 0.
y(0) = 0,
11. Solve the initial value problem:
y 00 + 2y 0 + 5y = 3 (t
2⇡),
y 0 (0) = 0.
y(0) = 1,
12. Solve the following IVP:
y 0 + 3y = (t
1) + u(t
2),
y(0) =
4.
13. Use the Laplace transform to solve the following
y 0 + 3y = tu(t
2) + (t
2),
y(0) = 1.
14. Use the Laplace transform to solve the initial value problem
y 00 + 4y 0 + 8y = (t
⇡),
y(0) = 0,
y 0 (0) =
1.
CHAPTER 5. LAPLACE TRANSFORM
224
15. Consider the initial value problem
y 000
2y 00 + y 0 + 4y = e
2t
+ (t
⇡),
y(0) = 0,
y 0 (0) = 1,
y 00 (0) = 0.
Find L{f (t)} = Y (s), the Laplace transform of its solution. You do not need to simplify
your answer. Do not solve for its inverse transform, y(t)!
16. Use the Laplace transform to solve the initial value problem
y 00 + 6y 0 + 9y = (t)
17. Evaluate the following definite integral
Z 1
e
2u5 (t),
st
(t
0
y(0) = 0,
y 0 (0) =
1.
⇡
) cos t dt.
2
(a) 0
1
s2 + 1
⇡
s
(c) e 2 s 2
s +1
⇡ ⇡
(d) e 2 s
2
⇡
2s
(b) e
18. Use the Laplace transform to solve the following initial value problem.
y 00 + 4y 0 + 3y = (t)
19. Evaluate the following definite integral
Z 1
e
u6 (t),
st
(t
0
y 0 (0) = 0.
⇡
) sin 3t dt.
6
⇡
6s
(a) e
⇡
6s
(b) e
⇡
(c) e 6 s
(d)
y(0) = 2,
s2
3
+9
s
s2 + 9
⇡
e6s
20. True or false.
(a) L{ (t
1)t10 tan2 ( t⇡
4 )} = e
s
(b) Suppose f (2) = 0, then L{ (t
.
2)ef (t) } = e
2s
21. (a) Use the Laplace transform to solve the following initial value problem
y 00 + 9y = t
(b) Evaluate y( ⇡3 )
u2 (t) + (t
1),
y(0) = 0,
y 0 (0) = 1.
CHAPTER 5. LAPLACE TRANSFORM
225
Answers
1. y(t) =
1 2t
e
4
2. y(t) =
1
e
3
1
e
4
4t
3. y(t) = e2t
4. y(t) =
5. e
2t
1
+ u2⇡ (t)[ e2(t
4
4
+ e
3
t
2
+ u3 (t)( e
3
2te2t + u4 (t)2(t
1 2t
1
e + e
15
40
3t
+
1
e
4
2⇡)
2
e
3
(t 3)
4)e2(t
2(t 2⇡)
4(t 3)
]
)
4)
1 5t
1
e + u1 (t)( e5(t
24
4
1
e
4
1)
3(t 1)
)
s
6. (a)
t
9
sin 3t
27
u2 (t)(
1
9
cos 3(t
9
2)
) + u1 (t)
sin 3(t
3
1)
+
sin 3t
⇡
sin 3
, (b)
+
3
27
3
7. F
t
8. (a) y(t) = 2e
9. (3⇡ 2
2e
2t
2(t 1)
+ 2u1 (t)(e
e
(t 1)
) + u6 (t)(
10. y(t) = u3 (t)e
(t 3)
sin(t
14. y(t) =
(t 6)
1
+ e
2
2(t 6)
), (b)
1
2
1
cos 2t + e
2
3)
3
sin 2t + u2⇡ (t)e t+2⇡ sin 2t
2
⌘
1⇣
12. y(t) = 4e 3t + u(t 1)e 3(t 1) +
1 e 3(t 2) u(t
3
✓
◆
2 3(t 2) 7 1
13. y(t) = u(t 2)
e
+
(t 2)
9
9 3
t
e
⇡s
1)e
11. y(t) = e
1
2
1
e
2
2t
t
1
sin 2t + u⇡ (t)e
2
2(t ⇡)
1
+
2s2 + s + 4) s3
15. Y (s) =
(s + 2)(s3
✓
2 2
16. y(t) = u5 (t)
+ e
9 9
3t+15
2
+ (t
3
sin(2t
2)
2⇡)
e ⇡s
+
2s2 + s + 4 s3
◆
3t+15
5)e
s 2
2s2 + s + 4
17. (a)
7
18. y(t) = e
2
t
3
e
2
3t
+ u6 (t)
✓
1
e
2
t+6
1
3
1
e
6
3t+18
◆
19. (a)
20. (a) T, (b) T
t
21. (a) y(t) =
9
sin(⇡ 3)
3
sin 3t
27
u2 (t)
✓
1
9
cos 3(t
9
2)
◆
+ u1 (t)
sin 3(t
3
1)
+
sin 3t
⇡
, (b)
+
3
27
CHAPTER 5. LAPLACE TRANSFORM
5.7
226
The Convolution Integral*
In this section, we form a new operator ⇤ which resembles multiplication, and satisfy this wonderful
property
L{(f ⇤ g)(t)} = L{f (t)}L{g(t)}.
This operation appears quite often in application, and is called as the “convolution” of f with g.
The convolution (f ⇤ g)(t) of f with g is defined by
(f ⇤ g)(t) =
Z
t
f (t
⌧ )g(⌧ ) d⌧.
0
The convolution operator has many of the properties of ordinary multiplication. Here are some of
them
1. f ⇤ g = g ⇤ f
(commutative law)
2. f ⇤ (g1 + g2 ) = f ⇤ g1 + f ⇤ g2
(distributive law)
3. (f ⇤ g) ⇤ h = f ⇤ (g ⇤ h)
(associative law)
4. f ⇤ 0 = 0 = 0 ⇤ f
5. (cf ) ⇤ g = c(f ⇤ g), for any constant c
Proof. To show the commutative law, we make the substitution s = t ⌧ in the definition,
Z t
Z 0
(f ⇤ g)(t) =
f (t ⌧ )g(⌧ ) d⌧ =
f (s)g(t s) ( ds)
0
=
Z
t
0
f (s)g(t
s) ds =
t
Z
t
g(t
0
s)f (s) ds = (g ⇤ f )(t).
On the other hand, the convolution operator has di↵erent properties from the multiplication
operator. For example,
f ⇤ 1 6= f, and f ⇤ f 6= f 2 .
These can be seen in the following example,
Example 180. Compute the convolution f ⇤ 1 and f ⇤ f , where f (t) = cos t.
Solution By the definition
Z t
(cos t ⇤ 1)(t) =
cos(t
0
⌧ ) · 1 d⌧ = ( sin(t
⌧ =t
⌧ ))⌧ =0 = 0
( sin t) = sin t.
Apply the trig identities
cos(A + B) + cos(A B)
2
cos( A) = cos(A) and sin( A) =
cos A cos B =
sin(A),
CHAPTER 5. LAPLACE TRANSFORM
227
we obtain
(cos t ⇤ cos t)(t) =
=
=
=
=
=
Z
t
cos(t
⌧ ) cos t d⌧
0
Z
1 t
(cos(t 2⌧ ) + cos( ⌧ )) d⌧
2 0
Z
1 t
(cos(t 2⌧ ) + cos(⌧ )) d⌧
2 0
✓
◆⌧ =t
1
1
sin(t 2⌧ ) + ⌧ cos t
2
2
✓
✓ ⌧ =0
◆◆
1
1
1
sin( t) + t cos t
sin t + 0
2
2
2
1
(sin(t) + t cos t)
2
Now we consider the most important identity/theorem in this section.
Theorem 12.
L{(f ⇤ g)(t)} = L{f (t)}L{g(t)}.
Proof. By definition,
L{(f ⇤ g)(t)} =
Z
This integral equals the double integral
ZZ
e
1
e
st
0
st
f (t
✓Z
t
f (t
⌧ )g(⌧ ) d⌧
0
◆
dt
⌧ )g(⌧ ) d⌧ dt,
R
where R is the region described below.
Figure 5.4: Region of integration .
We reverse the order of integration by integrating first with respect to t, instead of u. It gives
✓Z 1
◆
Z 1
L{(f ⇤ g)(t)} =
g(⌧ )
e st f (t ⌧ ) dt d⌧
0
⌧
Making the change of variables by setting ⇠ = t ⌧ ,
Z 1
Z 1
e st f (t ⌧ ) dt =
e s(⇠+⌧ ) f (⇠) d⇠ = e
⌧
0
s⌧
Z
1
0
e
s⇠
f (⇠) d⇠.
CHAPTER 5. LAPLACE TRANSFORM
228
Therefore
◆
e s⇠ f (⇠) d⇠ d⌧
0
0
✓Z
◆ ✓Z 1
◆
1
=
e s⌧ g(⌧ ) d⌧
e s⇠ f (⇠) d⇠
L{(f ⇤ g)(t)} =
Z
1
✓
g(⌧ ) e
s⌧
Z
1
0
0
= L{f (t)}L{g(t)}.
Let’s see how we apply this wonderful formula below.
Example 181. Find the inverse Laplace transform of the function
F (s) =
a
.
s2 (s2 + a2 )
Solution Notice that
a
1
a
= 2· 2
= L{t}L{sin at}.
s2 (s2 + a2 )
s (s + a2 )
By the theorem above,
L{t}L{sin at} = L{t ⇤ sin at}.
Therefore
a
= L{t ⇤ sin at}
s2 (s2 + a2 )
)
L
1
⇢
a
s2 (s2 + a2 )
= t ⇤ sin at.
Integration by parts give
(t ⇤ sin at)(t) =
Z
t
(t
⌧ ) sin(a⌧ ) d⌧
0
◆⌧ =t
sin a⌧
=
cos a⌧
a
a2
✓
◆ ⌧ =0
sin at
t
= 0
(
0)
a2
a
sin at
t
at sin at
=
+ =
.
a2
a
a2
Hence
L
1
⇢
✓
t
⌧
a
s2 (s2 + a2 )
=
at
sin at
.
a2
Chapter 6
Systems of First Order Linear
Equations
6.1
Introduction
A system of n-linear first order di↵erential equations in n unknowns has the following general
form:
x01 = a11 x1 + a12 x2 + · · · + a1n xn + g1
x02 = a21 x1 + a22 x2 + · · · + a2n xn + g2
..
.
x0n = an1 x1 + an2 x2 + · · · + ann xn + gn
where aij ’s and gi ’s are arbitrary functions of t.
• If every gi is a zero function, then the system is called to be homogeneous .
• Otherwise, it is a nonhomogeneous system if one of the gi ’s is a nonzero function.
Actually a system of n-linear first order di↵erential equations is equivalent to, mathematically the
same thing, an nth order di↵erential equation.The followings are how we can transform them back
and forth.
We start with the method to transform an nth order di↵erential equation to a system of n-first
order linear equations. Given
an y (n) (t) + an
1y
(n 1)
(t) + . . . + a1 y 0 (t) + a0 y(t) = g(t)
be an nth order linear di↵erential equation.
229
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
230
How to transform nth order equation to a system of n first order linear equations:
1. Introduce n new variables x1 , x2 , . . . , xn defined by
x1 = y,
x2 = y 0 ,
xn = y (n
...,
1)
.
2. Then it follows that
x01 = x2 ,
are n
x02 = x3 ,
x0n
...,
1
= xn
1 first order linear equations.
3. Rewrite the given equation in the following form,
an 1 (n
y
an
y (n) =
1)
a1 0
y
an
...
a0
g(t)
y+
.
an
an
Then change the equations to new variables introduced in step 1 to receive
an 1
xn
an
x0n =
a1
x2
an
...
a0
g(t)
x1 +
.
an
an
4. Equations from step 2 and 3 form a system of n first order linear equations.
Example 182 (Summer 2010 Midterm Exam II).
Transform the equation below into a system of 1st order equations.
2y (4) + t2 y 000
2y 0 + 6ty = 3 sin t.
Solution Introduce 4 new variables since the given equation is the 4th order one,
x1 = y,
Then it follows that
x2 = y 0 ,
x01 = x2 ,
x3 = y 00 ,
x02 = x3 ,
x4 = y 000 .
x03 = x4 .
Now rewrite the given equation and change its variables to new ones introduced earlier
y (4) =
Then
t2 000
y + y0
2
3ty +
8 0
>
> x1
>
< x0
2
>x03
>
>
: 0
x4
3
sin t
2
)
= x2
= x3
= x4
= 3tx1 + x2
x04 =
t2
2 x4
+
t2
x4 + x2
2
3
2
3tx1 +
3
sin t.
2
sin t
form the system of 1st order linear equations.
Example 183 (Spring 2010 Midterm Exam II).
Find the system of 1st order linear equations which is equivalent to the 3rd order linear
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
equation
y 000 + y 00
231
2y 0 + 3y = 0.
Solution The equation above is the third order one so we need to introduce 3 new variables,
x2 = y 0 ,
x1 = y,
Then it follows that
x01 = x2 ,
x3 = y 00 .
x02 = x3 .
Rewrite the given equation and change its variables to new ones introduced earlier
y 000 =
y 00 + 2y 0
)
3y
x03 =
Hence the system of 1st order equations is
8
0
>
< x1 = x2
x02 = x3
>
: 0
x3 = 3x1 + 2x2
x3 + 2x2
3x1 .
x3 .
The reverse is also true. Given a system of n 1st order linear equations, it can be rewritten into
a single nth order linear equation. We take a look here for the special case when n = 2.
Let
x01 = ax1 + bx2
x02 = cx1 + dx2 .
be a given system of 1st order linear equations.
How to transform a system of two 1st order linear equations to a 2nd order
equation:
1. From the first equation, solve for x2 :
x2 =
x01
b
ax1
.
b
2. Then substitute x2 we get from step 1 to the second equation.
Example 184 (Fall 2007 Midterm Exam II).
Find 2nd order equation which is equivalent to the given linear system
x01 = x2
x02 = 3x1
2x2 .
Solution From the first equation, we have x2 = x01 . Substitute it back into the second one,
x02 = 3x1
2x2
)
(x01 )0 = 3x1
2(x1 )0
Hence the equivalent second order equation is
x001 + 2x01
3x1 = 0.
)
x001 = 3x1
2x01 .
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
232
Example 185. Transform the given system into a single equation of second order
5
x1 +
4
3
x02 = x1 +
4
x01 =
3
x2 ,
4
5
x2 ,
4
x1 (0) =
2
x2 (0) = 1.
Solution From the first equation, we have
x01
5
3
= x1 + x2
4
4
4
x2 =
3
)
✓
x01
5
x1
4
◆
)
x2 =
4 0
x
3 1
5
x1 .
3
Substitute it back into the second equation,
x02 =
3
5
x1 + x2
4
4
)
)
)
)
4
5
3
5 4
5
( x01
x1 )0 = x1 + ( x01
x1 )
3
3
4
4 3
3
4 00 5 0
3
5
25
x1
x1 = x1 + x01
x1
3
3
4
3
12
4 00 10 0
4
x1
x1 + x1 = 0
3
3
3
4x001 10x01 + 4x1 = 0
)
Therefore we have
2x001
2x001
5x01 + 2x1 = 0.
5x01 + 2x1 = 0
as an equivalent single equation of second order. Now you can see that the equation we just
derived is in terms of x1 . So an initial condition x1 (0) = 2 works but we need to change the
second one x2 (0) = 1 to variable x1 . To do this, we use the first equation from the problem,
x01 =
5
3
x1 + x2
4
4
)
x01 (0) =
5
3
x1 (0) + x2 (0)
4
4
)
x01 (0) =
5
3
( 2) + (1) =
4
4
In the end, we have the following IVP
2x001
5x01 + 2x1 = 0,
x1 (0) =
2,
x01 (0) =
7
.
4
7
.
4
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
233
Exercises 6.1
1. Which system of first order linear equations below is equivalent to the third order linear
equation
y 000 + y 00 3y 0 y = 0?
8
0
>
< x1 = x2
(a) x02 = x3
>
: 0
x3 = x1 3x2 x3
8
0
>
< x1 = x2
(b) x02 = x3
>
: 0
x3 = x1 + 3x2 + x3
8
0
>
< x1 = x2
(c) x02 = x3
>
: 0
x3 = x1 3x2 + x3
8
0
>
< x1 = x2
(d) x02 = x3
>
: 0
x3 = x1 + 3x2 x3
2. Rewrite the following third order linear equation into an equivalent system of first order
linear equations.
y 000 + 3y 00 2y 0 + 4y = sin 2t
3. Determine whether the statement is true or false. The system of first order linear equations
8
0
>
< x1 = x2
x02 = x3
>
3
: 0
x3 = 38 x1 t3 x2 + 13 sin(8t)
is equivalent to the third order linear equation 3y 000 + t3 y 0
8y = sin(8t).
4. Which system of first order linear equations below is equivalent to the second order linear
equation
y 00 4y 0 + 7y = 10t3 ?
(
x01 = x2
(a)
x02 = 7x1 + 4x2 + 10t3
(
x01 = x2
(b)
x02 = 7x1 4x2 + 10t3
(
x01 = x2
(c)
x02 = 4x1 + 7x2 + 10t3
(
x01 = x2
(d)
x02 = 4x1 7x2 + 10t3
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
234
5. Which second order di↵erential equation below is equivalent to the system
✓
◆
✓
◆
0 1
0
x0 =
x+
?
1
2
sin t
(a)
(b)
(c)
(d)
y 00 + 2y 0 y = sin t
y 00 + sin ty 0 + y = 2
y 00 y 0 + 2y = sin t
y 00 + y 0 2y = sin t
6. Which system of first order linear equations below is equivalent to the second order linear
equation
y 00 5y 0 + 6y = t2 t?
(
x01 = x2
(a)
x02 = 5x1 + 6x2 + t2 t
(
x01 = x2
(b)
x02 = 5x1 6x2 t2 + t
(
x01 = x2
(c)
x02 = 6x1 + 5x2 + t2 t
(
x01 = x2
(d)
x02 = 6x1 5x2 t2 + t
7. Rewrite the following equation into an equivalent system of first order linear equations.
2y 00 + 6y 0
4y = e
t
cos 9t
8. What second order di↵erential equation is equivalent to the system
✓
◆
✓ ◆
0
1
0
x0 =
x+ 2 ?
5t sin t
t
(a)
(b)
(c)
(d)
y 00 + 5ty 0 sin ty = t2
y 00 5ty 0 + sin ty = t2
y 00 t2 y 0 + 5ty = sin t
y 00 sin ty 0 + 5ty = t2
Answers
1. (d)
8
0
>
< x1
2. x02
>
: 0
x3
= x2
= x3
= 4x1 + 2x2
3x3 + sin 2t
3. True
4. (a)
5. (a)
6. (c)
(
x01
7.
x02
8. (d)
= x2
= 2x1
3x2 + 12 e
t
cos 9t
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
6.2
235
Review of Matrices
An m ⇥ n (called the size or dimension of the matrix) matrix is a rectangular array of numbers
with m rows and n columns and the entry in the ith row and jth column is denoted by aij .
0
1
a11 a12 · · · a1n
B a21 a22 · · · a2n C
B
C
A=B .
= (aij )m⇥n
..
.. C
..
@ ..
.
.
. A
am1 am2 · · · amn m⇥n
Example 186.
✓
1
2
1
3
◆
4
, 2
5 2⇥3
0
3
4
0
1
1
5 1⇥5 , @ 2 A
3 3⇥1
Le’t consider special-name matrices here:
1. Square matrix is any matrix which has
0
a11
B a21
B
A=B .
@ ..
am1
the same number of rows as columns.
1
a12 · · · a1n
a22 · · · a2n C
C
..
.. C
..
.
.
. A
am2
···
ann
n⇥n
The diagonal that start in the upper left and end in the lower right is called the main
diagonal. The sum of the entires in the main diagonal is called trace and write it as tr. So
here
n
X
tr A =
aii = a11 + a22 + . . . + ann .
i=1
2. Zero matrix, denoted 0m⇥n or 0, is a matrix which all entries are zeros. It satisfies two
properties:
A+0=A=0+A
A0 = 0 = 0A.
3. Identity matrix is a square matrix, denoted by
and other entries are zero.
0
1 0 ···
B0 1 · · ·
B
B .. .. . .
@. .
.
0
It satisfies AI = A = IA.
0
In or I, whose main diagonals are all 1’s
1
0
0C
C
.. C
.A
··· 1
4. Column(row) matrix or Column(row) vector is a matrix consisting of a single column
(row).
0 1
x1
B x2 C
B C
x = B . C is a column vector and y = y1 y2 · · · yn is a row vector.
@ .. A
xn
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
236
Next let’s consider the arithmetic of matrices.
1. Addition, subtraction and scalar multiplication:
✓
◆ ✓
◆ ✓
◆
a b
e f
a±e b±f
±
=
c d
g h
c±g d±h
✓
◆ ✓
◆
a b
↵a ↵b
↵
=
c d
↵c ↵d
2. Matrix multiplication of A and B: It is defined only when A has the same number of
columns as B has rows. Otherwise, we can’t compute AB.
Am⇥p Bp⇥n = ABm⇥n
where its ijth entry (AB)ij can be found by the dot product of row vector i of matrix A
and the transpose of column vector j of matrix B. Here are some examples:
✓
◆
✓
◆
✓
◆
a b
e f
ae + bg af + bh
=
c d 2⇥2 g h 2⇥2
ce + dg cf + dh 2⇥2
✓
◆
✓ ◆
✓
◆
a b
e
ae + bf
=
c d 2⇥2 f 2⇥1
ce + df 2⇥1
✓
◆
a b
e f 1⇥2 is not defined.
But
c d 2⇥2
Unlike multiplication of real numbers , matrix multiplication does not commute. In other
words, it is NOT true that AB equals to BA.
3. Determinant, denoted by det, is a function that takes a square matrix into a number in a
certain way. In particular for 2 ⇥ 2 matrix, we define determinant as follows
✓
◆
a b
a b
det
=
:= ad bc
c d
c d
• If the determinant of a matrix is zero, we call that matrix singular.
• We call the matrix nonsingular if its determinant is nonzero.
4. Inverse matrix of a square matrix: We call B the inverse of A, denoted by A
1
, if
AB = BA = In .
If A
1
exists, then we say A is invertible.
✓
◆
a b
• For 2 ⇥ 2 matrix if A =
then
c d
✓
1
d
A 1=
c
ad bc
b
a
◆
=
1
det A
✓
d
c
◆
b
.
a
• It is the fact that a square matrix is invertible if and only if its determinant is nonzero.
Example 187. Let A =
det A = 2(1)
✓
2
3
◆
✓
1
2
and B =
1
6
3( 1) = 5 6= 0,
◆
1
. Then
3
det B = 2(3)
( 6)( 1) = 0.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
So B is not invertible while A is invertible with
✓
◆ ✓
1 1 1
A 1=
=
3 2
5
1
5
3
5
1
5
2
5
◆
237
.
5. Eigenvalues and eigenvectors: Given a square matrix A, suppose there are constant r
and a nonzero column vector x such that
Ax = rx,
then r is called eigenvalue of A and x is called eigenvector of A corresponding to r. How
do we find them? First, we rewrite the equation,
Ax = rx
)
Ax
rx = 0
) (A
rI)x = 0.
Then apply the following theorem,
Theorem 13. Let A be n ⇥ n matrix then the system Ax = 0 has
• a unique solution x = 0 if and only if A is invertible (det A 6= 0).
• infinitely many solutions if and only if A is not invertible (det A = 0).
Since we’re looking for a nontrivial solution x, we receive
det(A
rI) = 0.
Now take a look in the particular case of 2 ⇥ 2 matrix A,
✓
◆
✓
◆ ✓
a b
1 0
a r
A rI =
r
=
c d
0 1
0
And so det(A
d
r
◆
.
rI) = 0 means
(a
r)(d
r)
0(0) = 0
)
)
We call r2
equation
0
r2
r
2
(a + d)r + (ad
bc) = 0
tr(A)r + det A = 0.
tr(A)r + det A the characteristic polynomial corresponding to A and the
r2
tr(A)r + det A = 0
the characteristic equation of matrix A.
✓
◆
a b
In conclusion, for A =
, we have
c d
How to find eigenvalues and eigenvectors for A:
1. Write the characteristic equation
r2
tr(A)r + det A = 0.
Its roots give eigenvalues of A.
2. Plug in each eigenvalue back to (A
eigenvectors.
rI)x = 0 and solve for x to find the corresponding
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
238
Example 188. Find the eigenvalues and eigenvectors of
✓
◆
3
1
A=
.
4
2
Solution First, find the roots of its characteristic equation:
r2
(3
2)r + 3( 2)
4( 1) = r2
Hence the eigenvalues are 2 and
r
2=0
)
)
)
)
)
r = 2, 1.
x1
x2
=
✓
)
✓
◆✓ ◆ ✓ ◆
x1
0
=
4
2 2
x2
0
✓
◆✓ ◆ ✓ ◆
1
1
x1
0
=
4
4
x2
0
(
x1 x2 = 0,
4x1 4x2 = 0.
3
2
1
x1 = x2 .
◆
✓ ◆
x2
1
= x2
. For simplicity, we choose x2 = 1. Then
x2 ✓ ◆
1
1
eigenvector corresponding to r = 2 is
.
1
So the eigenvectors are
◆
2)(r + 1) = 0
1.
For r = 2: Consider (A 2I)x = 0,
✓✓
◆
✓
◆◆ ✓ ◆ ✓ ◆
3
1
1 0
x1
0
2
=
4
2
0 1
x2
0
✓
(r
For r = 1: Consider (A ( 1)I)x = 0,
✓✓
◆
✓
◆◆ ✓ ◆ ✓ ◆
3
1
1 0
x1
0
( 1)
=
4
2
0 1
x2
0
)
)
)
x1
x2
◆
=
✓
✓
3+1
4
4
4
(
4x1
4x1
◆✓ ◆ ✓ ◆
1
x1
0
=
2+1
x2
0
◆✓ ◆ ✓ ◆
1
x1
0
=
1
x2
0
x2 = 0,
x2 = 0.
x2 = 4x1 .
◆
✓ ◆
x1
1
= x1
. For simplicity, we choose x1 = 1. Then
4x1 ✓ ◆
4
1
eigenvector corresponding to r = 1 is
.
4
So the eigenvectors are
✓
)
✓
Note that for each value of an eigenvalue, its corresponding eigenvector is not unique since nonzero
scalar multiplication of an eigenvector is also an eigenvector.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
6.3
239
Solutions to Systems of First Order Linear Equations
Recall a system of n-linear first order di↵erential equations is in the form
x01 = a11 x1 + a12 x2 + · · · + a1n xn + g1
x02 = a21 x1 + a22 x2 + · · · + a2n xn + g2
..
.
x0n = an1 x1 + an2 x2 + · · · + ann xn + gn
which can be rewritten into a matrix-vector equation,
0 01 0
10 1 0 1
x1
a11 a12 · · · a1n
x1
g1
B x02 C B a21 a22 · · · a2n C B x2 C B g2 C
B C B
CB C B C
B .. C = B ..
..
.. C B .. C + B .. C ,
..
@ . A @ .
.
.
. A@ . A @ . A
x0n
am1 am2 · · · ann
xn
gn
or just simply
x0 = Ax + g.
In this class, we consider only for homogeneous system
x0 = Ax.
• Equilibrium solutions or critical points are solutions x for which x0 = 0 (or Ax = 0).
Since we’re going to assume that det A 6= 0, by the theorem above, we will have only one
equilibrium solution, namely x = 0. So in this chapter, the only critical point is the origin.
• A sketch of a particular solution in the phase plane x1 x2 plane is called the trajectory of
the solution.
• A plot that shows a representative sample of trajectories for a given system is called a phase
portrait.
• There are three types of stability of the system based on the behavior of trajectories:
1. It is asymptotically stable if all trajectories converge to zero.
2. It is unstable if all or all but a few trajectories move away from origin.
3. It is (neutrally) stable if all trajectories stay in a fixed orbit around the origin.
• Solutions to the homogeneous system x0 = Ax: For n = 1 and write A = [r], we know that
the solution to x0 = rx is x = kert . So let’s use this as a guideline for general n. If x = kert
is a solution of a system, then x0 = rkert and so
x0 = Ax
)
rkert = Akert
)
Ak = rk,
where r is an eigenvalue of A and k is an eigenvector of A corresponding to r.
• For this class, we consider when A is 2 ⇥ 2 matrix. Just like the solution of a second order
homogeneous linear equation, there are three possibilities of solutions depending on a type
of eigenvalues of A.
1. Two distinct real eigenvalues.
2. Complex conjugate eigenvalues.
3. Repeated eigenvalues.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
6.3.1
240
Real and Distinct Eigenvalues
If the coefficient matrix A has two real distinct eigenvalues r1 and r2 with corresponding eigenvectors k1 , k2 respectively. Then the system x0 = Ax has a general solution
x = c 1 k 1 e r1 t + c 2 k 2 e r2 t .
Case 1.1: r1 , r2 have opposite signs.
Example 189.
a) Solve the following IVP
✓
◆
1 2
0
x =
x,
3 2
✓
x(0) =
◆
0
.
4
Solution Firstly, we find the eigenvalues of the coefficient matrix A =
characteristic equation is
r2
(1 + 2)r + 1(2)
3(2) = r2
Hence the eigenvalues are
For r1 =
3r
)
4=0
(r + 1)(r
4) = 0
✓
)
1
3
r=
1 and 4.
1: Consider (A ( 1)I)k1 = 0,
✓
◆✓ ◆ ✓ ◆
1 ( 1)
2
m1
0
=
3
2 ( 1)
m2
0
)
)
)
So the eigenvector corresponding to r1 = 1 is
✓ ◆ ✓
◆
m1
m2
Choose m =1
=
=======2=)
m2
m2
For r2 = 4: Consider (A 4I)k2 = 0,
✓
◆✓ ◆ ✓ ◆
1 4
2
n1
0
=
3
2 4
n2
0
)
)
)
✓
(
3
3
✓
◆✓ ◆ ✓ ◆
2 2
m1
0
=
3 3
m2
0
(
2m1 + 2m2 = 0,
3m1 + 3m2 = 0.
m1 =
k1 =
2
2
✓
◆
1
.
1
◆✓ ◆ ✓ ◆
n1
0
=
n2
0
n1 =
2
n2 .
3
✓ ◆
2
k2 =
.
3
The general solution is then
t
✓
3n1 + 2n2 = 0,
3n1 2n2 = 0.
So the eigenvector corresponding to r2 = 4 is
✓ ◆ ✓2 ◆
n1
n
Choose n =3
= 3 2
=======2=)
n2
n2
x(t) = c1 e
m2 .
◆
✓ ◆
1
2
+ c2 e4t
.
1
3
◆
2
. Its
2
1, 4.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
241
Apply the given initial condition to determine the values c1 and c2 ,
(
(
✓ ◆
✓ ◆
✓ ◆
c1 + 2c2 = 0,
c1 =
0
1
2
= c1
+ c2
)
)
4
1
3
c1 + 3c2 = 4
c2 =
Therefore the particular solution is
✓ ◆
8 t
1
x(t) =
e
1
5
b) Sketch the phase portrait of x0 =
✓
1
3
✓ ◆ ✓
4 4t 2
e
=
3
5
8
t
5e
8
t
5e
8 4t
5e
12 4t
5 e
◆
8
5,
4
5
.
.
◆
2
x. Then find its type and stability.
2
Type: The critical point (0, 0) is called a saddle point.
Stability: Since some solutions move away from the origin (0, 0) as t increases, it is
unstable.
Case 1.2: r1 , r2 have same signs. (they are either both positive or both negative)
Example 190.
a) Solve the following IVP
✓
◆
5 1
x0 =
x,
4
2
x(0) =
✓ ◆
1
.
2
Solution Firstly, we find the eigenvalues of the coefficient matrix A =
characteristic equation is
r2 ( 5 2)r+( 5)( 2) 4(1) = r2 +7r+6 = 0
)
(r+1)(r+6) = 0
✓
5
4
)
◆
1
. Its
2
r=
1, 6.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
So the eigenvalues are
For r1 =
✓
1 and
242
6.
1: Consider (A
5
( 1)
4
( 1)I)k1 = 0,
◆✓ ◆ ✓ ◆
1
m1
0
=
2 ( 1)
m2
0
✓
)
(
)
)
◆✓ ◆ ✓ ◆
m1
0
=
m2
0
4m1 + m2 = 0,
4m1 m2 = 0.
k1 =
✓ ◆
1
.
4
✓
◆✓ ◆ ✓ ◆
1 1
n1
0
=
4 4
n2
0
(
n1 + n2 = 0,
4n1 + 4n2 = 0.
)
)
)
So the eigenvector corresponding to r2 = 6 is
✓ ◆ ✓
◆
n1
n2
Choose n =1
=
=======2=)
n2
n2
1
1
m2 = 4m1 .
So the eigenvector corresponding to r1 = 1 is
✓ ◆ ✓
◆
m1
m1
Choose m =1
=
=======1=)
m2
4m1
For r2 = 6: Consider (A ( 6)I)k2 = 0,
✓
◆✓ ◆ ✓ ◆
5 ( 6)
1
n1
0
=
4
2 ( 6)
n2
0
4
4
n1 =
n2 .
✓
◆
1
.
1
k2 =
Then the general solution is
x(t) = c1 e
t
✓ ◆
1
+ c2 e
4
6t
✓
◆
1
.
1
Apply the given initial condition to get the values c1 and c2 ,
(
✓ ◆
✓ ◆
✓ ◆
c1 c2 = 1,
1
1
1
= c1
+ c2
)
)
2
4
1
4c1 + c2 = 2
Therefore the particular solution is
✓ ◆
3
2
1
x(t) = e t
e
4
5
5
b) Sketch the phase portrait of x0 =
✓
5
4
6t
✓
◆ ✓3
1
5e
= 12
1
5 e
+ 25 e
2
t
5e
t
(
c1 = 35 ,
c2 = 25 .
6t
6t
◆
.
.
◆
1
x. Then find its type and stability.
2
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
243
Type: The critical point (0, 0) is called a node.
Stability: Since all trajectories move towards (0, 0) as t increases, it is asymptotically
stable since both r1 , r2 are negative. (But this node can be unstable if both r1 , r2
are positive numbers.)
6.3.2
Complex Conjugate Eigenvalues
If the coefficient matrix A has complex conjugate eigenvalues r1,2 =
first case, the general solution is
x = c1 k1 e(
+µi)t
+ c 2 k2 e(
µi)t
± µi, then similar to the
.
But what we want is to rewrite it in the form of real valued solution. In order to do this, we again
use the help from Euler’s formula. Let k1 = a + ib be the eigenvector corresponding to r1 = + µi.
Then it follows that k2 = a ib is the eigenvector corresponding to r2 =
µi. Then we can
derive new form of general solution as follows:
x = c1 k1 e(
+µi)t
+ c2 k2 e(
µi)t
= c1 (a + ib)e t eµit + c2 (a
ib)e t e
µit
= c1 (a + ib)e t (cos µt + i sin µt) + c2 (a
t
= c1 e (a cos µt
+ c2 e (a cos µt
ib)e t (cos µt
i sin µt)
b sin µt + i(a sin µt + b cos µt))
b sin µt
t
= (c1 + c2 )e (a cos µt
i(a sin µt + b cos µt))
b sin µt) + (c1
c2 )ie (a sin µt + b cos µt)
After renaming the constant, the new form of the general solution is
x(t) = c1 e t (a cos µt
where the eigenvalues of A is
b sin µt) + c2 e (a sin µt + b cos µt) ,
± µi and the eigenvector corresponding to
Case 2.1: Real part is zero. (r1,2 = ±µi)
+ µi is a + ib.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
Example 191.
a) Solve the following IVP
✓
◆
3
9
x0 =
x,
4
3
x(0) =
✓
244
◆
2
.
4
Solution The characteristic equation is
p
p
4( 9) = r2 + 27 = 0 ) r = ± 27i = ±3 3i.
p
p
So the eigenvalues are r1,2 = 0 ± 3 3i. Note that = 0 and µ = 3 3.
p
p
For r1 = 3 3i: Consider (A (3 3i)I)k1 = 0,
(
p
p
✓
◆✓ ◆ ✓ ◆
(3 3 3i)m1 9m2 = 0,
3 3 3i
9p
m1
0
p
=
)
m2
0
4
3 3 3i
4m1 + ( 3 3 3i)m2 = 0.
p
1
) m2 = (1
3i)m1 .
3
p
So the eigenvector corresponding to r1 = 3 3i is
✓ ◆ ✓
◆
✓
◆
m
3p
m1
Choose m =3
p1
= 1
=======1=) k1 =
.
m2
3i)m1
1
3i
3 (1
r2
(3
3)r + (3)( 3)
Write k1 in terms of a + ib,
✓
◆ ✓
◆ ✓ ◆
✓
◆
3p
3 +p0i
0
3
p
k1 =
=
=
+i
= a + ib.
1
1
3i
1
3i
3
So the general solution is
✓✓ ◆
✓
◆
◆
✓✓ ◆
✓
◆
◆
p
p
p
p
0
0
3
3
p
p
sin 3 3t + c2 e0t
cos 3 3t
x(t) = c1 e0t
cos 3 3t
sin 3 3t +
1
1
3
3
p
p
✓
◆
✓
◆
3 3t p
3 3t p
p3 cos p
p 3 sinp
= c1
+ c2
.
cos 3 3t + 3 sin 3 3t
sin 3 3t
3 cos 3 3t.
Apply initial condition,
✓
2
4
◆
✓ ◆
✓
◆
0
3
p
= c1
+ c2
1
3
)
(
3c1 = 2,
p
c1
3c2 =
4.
)
(
c1 = 23 ,
p .
c2 = 314
3
Therefore the particular solution is
p
p
✓
◆
✓
◆
2
14
3 3t p
3 3t p
p3 cos p
p 3 sin p
x(t) =
+ p
3 cos 3 3t
3 cos 3 3t + 3 sin 3 3t
3 3 sin 3 3t
!
!
p
p
14
p sin 3 3t
2 cos 3 p3t
3
p
p
p
p
= 2
+ 14
14
2 3
p sin 3 3t
3 cos 3 3t
3 cos 3 3t + 3 sin 3 3t
3 3
!
p
p
14
2 cos 3 3t + p
sin 3 3t
3
p
p
p
=
2 3
14
p
( 23 14
3 ) cos 3 3t + ( 3 + 3 3 ) sin 3 3t
p
p !
14
2 cos 3 3t + p
sin
3
3t
3
p
p
) x(t) =
.
20
4 cos 3 3t + 3p3 sin 3 3t
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
0
b) Sketch the phase portrait of x =
✓
3
4
245
◆
9
x. Then find its type and stability.
3
Type: The critical point (0, 0) is called a center.
Stability: It is (neutrally) stable since the trajectories of the solutions are circles or
ellipses centered at the origin. To check the orientation, pick a sample point and
see what we get.
✓ ◆
✓
◆✓ ◆ ✓ ◆
1
3
9
1
3
=
.
x=
) x0 =
0
4
3
0
4
At the point (0, 1), the trajectory points in upward direction. So the trajectories
travel in counter clockwise direction.
Case 2.2: Real part is nonzero. (r1,2 =
Example 192.
± µi, where
a) Solve the following IVP
✓
◆
3
13
0
x =
x,
5
1
6= 0)
x(0) =
✓
◆
3
.
10
Solution Consider the characteristic equation,
r2
(3 + 1)r + (3)(1)
5( 13) = r2
So the eigenvalues are r1,2 = 2 ± 8i. Here
4r + 68 = 0
= 2 and µ = 8.
)
r = 2 ± 8i.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
For r1 = 2 + 8i: Consider (A
✓
3
(2 + 8i)
5
1
13
(2 + 8i)
246
(2 + 8i)I)k1 = 0,
◆✓ ◆ ✓ ◆
m1
0
=
m2
0
)
)
So the eigenvector corresponding to r1 = 2 + 8i is
✓ ◆ ✓ 1+8i ◆
m1
Choose m =5
5 m2
=
=======2=)
m2
m2
(
(1 8i)m1
5m1 + ( 1
13m2 = 0,
8i)m2 = 0.
m1 =
1 + 8i
m2 .
5
k1 =
✓
◆
1 + 8i
.
5
Write k1 in terms of a + ib,
✓
◆ ✓
◆ ✓ ◆
✓ ◆
1 + 8i
1 + 8i
1
8
k1 =
=
=
+i
= a + ib.
5
5 + 0i
5
0
So the general solution is
✓✓ ◆
✓ ◆
◆
✓✓ ◆
✓ ◆
◆
1
8
1
8
2t
2t
x(t) = c1 e
cos 8t
sin 8t + c2 e
sin 8t +
cos 8t
5
0
5
0
✓
◆
✓
◆
cos 8t 8 sin 8t
sin 8t + 8 cos 8t
= c1 e2t
+ c2 e2t
.
5 cos 8t
5 sin 8t
Apply the given initial condition,
✓
◆
✓ ◆
✓ ◆
3
1
8
= c1
+ c2
10
5
0
)
(
c1 + 8c2 = 3
5c1 = 10
)
(
c1 =
c2 =
2,
5
8
Hence the particular solution is
✓
◆
✓
◆
5
cos 8t 8 sin 8t
sin 8t + 8 cos 8t
x(t) = 2e2t
+ e2t
5 cos 8t
5 sin 8t
8
✓ 2t
◆
133 2t
3e cos 8t + 8 e sin 8t
=
.
2t
10e2t cos 8t + 25
8 e sin 8t
b) Sketch the phase portrait of x0 =
✓
3
5
◆
13
x. Then find its type and stability.
1
.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
247
Type: The critical point (0, 0) is called a spiral point.
Stability: It is unstable since real part = 2 of the eigenvalues is positive one. (But this
spiral point can be asymptotically stable if real part of the eigenvalues is negative
one). The trajectories spiral into (if < 0) or out of the origin (if > 0) which
in the case in this example. To check the orientation, pick a sample point and see
what we get.
✓ ◆
✓
◆✓ ◆ ✓ ◆
1
3
13
1
3
x=
) x0 =
=
.
0
5
1
0
5
At the point (0, 1), the trajectory points in upward direction. So the trajectories
travel in counter clockwise direction.
6.3.3
Repeated Real Eigenvalues
Suppose the coefficient matrix A has repeated real eigenvalues r1 = r2 = r.
✓
◆
r 0
Case 3.1: A is in the form of
where r 6= 0. If this happens, we can choose arbitrary two
0 r
linearly independent eigenvectors k1 and k2 (Choose in such a way that k1 6= ck2 ). So the general
solution to x0 = Ax is
x(t) = c1 k1 ert + c2 k2 ert .
Example 193.
0
a) Solve the following system x =
Solution Consider the characteristic equation,
r2
(3 + 3)r + 3(3)
0(0) = r2
So the eigenvalues are r1 = r2 = 3.
✓
3
0
6r + 9 = (r
◆
0
x.
3
3)2 = 0
)
r = 3, 3.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
For r1 = r2 = 3: Consider (A 3I)k1 = 0,
✓
◆✓ ◆ ✓ ◆
3 3
0
m1
0
=
0
3 3
m2
0
)
248
0 = 0,
which gives us nothing. So we can choose arbitrary two linearly independent eigenvectors, for example,
✓ ◆
✓ ◆
1
0
k1 =
, k2 =
.
0
1
Therefore the general solution is
✓ ◆
✓ ◆
✓ 3t ◆
1 3t
0 3
c e
x(t) = c1
e + c2
e t = 1 3t .
0
1
c2 e
b) Sketch the phase portrait of x0 =
✓
3
0
◆
0
x. Then find its type and stability.
3
The phase portrait has a star-burst shape.
Type: The critical point (0, 0) is called a proper node or star point.
Stability: It is unstable since repeated eigenvalues, r = 3, are positive. (But this proper
node can be asymptotically stable if repeated eigenvalues are negative ones). The
trajectories move towards the origin (if r < 0) or move away from the origin (if
r > 0) which is the case in this example.
✓
◆
r 0
Case 3.2: Matrix A gives repeated eigenvalues but is not in the form of
where r 6= 0.
0 r
In this case, we will find only one eigenvector k corresponding to repeated eigenvalues r. So for
sure, one solution is kert . We need to find the second solution. As we studied before, it is natural
to guess that the second one is ktert . To check whether it work or not, plug it back into x0 = Ax,
x0 = Ax
) (ktert )0 = Aktert
)
rktert + kert = Aktert .
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
249
Comparing the coefficients,
(
rk = Ak )
This is the same as saying k is the eigenvector corresponding to r.
kert ) k = 0 )
This is a contradiction. Since an eigenvector k can’t be zero.
So our first guess is incorrect. Let’s choose something new but not far from the first guess. How’s
about ktert + ⌘ert . Checking whether it work or not, substitute it back into x0 = Ax,
x0 = Ax
)
(ktert + ⌘ert )0 = A(ktert + ⌘ert )
)
rktert + (k + r⌘)ert = Aktert + A⌘ert
rktert + kert + r⌘ert = Aktert + A⌘ert
)
Comparing the coefficients,
(
rk = Ak ) This is again the same as saying k is the eigenvector corresponding to r.
k + r⌘ = A⌘ ) (A rI)⌘ = k )
This is new and it gives us how to find the vector ⌘.
So the general solution is
x(t) = c1 kert + c2 ktert + ⌘ert ,
where ⌘ can be found by (A
Example 194.
rI)⌘ = k.
a) Solve the following IVP
✓
◆
7 1
x0 =
x,
4 3
x(0) =
✓
◆
2
5
Solution Its characteristic equation is,
r2
(7 + 3)r + 7(3)
( 4)(1) = r2
10r + 25 = (r
5)2 = 0
)
r = 5, 5.
So the eigenvalues are r1 = r2 = 5.
For r1 = r2 = 5: Consider (A
5I)k = 0,
(
✓
◆✓ ◆ ✓ ◆
2m1 + m2 = 0
7 5
1
m1
0
=
)
4 3 5
m2
0
4m1 2m2 = 0
Hence the eigenvector corresponding to r = 5 is
✓ ◆ ✓
◆
m1
m1
Choose m =1
=
=======1=)
m2
2m1
Solve for ⌘: Consider (A
✓
7
5
1
4 3 5
Hence ⌘ is
n1
n2
◆
✓
m2 =
2m1 .
◆
1
.
2
5I)⌘ = k,
◆✓ ◆ ✓ ◆
n1
1
=
n2
2
✓
k=
)
=
✓
)
n1
1 2n1
◆
(
2n1 + n2 = 1
4n1 2n2 =
Choose n1 =0
========)
2
)
✓ ◆
0
⌘=
.
1
n2 = 1
2n1 .
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
250
Therefore the general solution is
✓ ◆
✓✓ ◆
✓ ◆ ◆
1
1
0 5t
5t
5t
x(t) = c1
e + c2
te +
e
.
2
2
1
Apply the given initial condition,
✓
◆
✓ ◆
✓ ◆
2
1
0
= c1
+ c2
5
2
1
)
(
c1 = 2,
2c1 + c2 =
5
)
(
c1 = 2,
c2 = 1.
.
So the particular solution is
✓ ◆
✓✓ ◆
✓ ◆ ◆
1
1
0 5t
x(t) = 2
e5t
te5t +
e
2
2
1
✓ 5t ◆ ✓
◆ ✓
◆
2e
te5t
2e5t te5t
=
=
.
4e5t
2te5t + e5t
5e5t + 2te5t
b) Sketch the phase portrait of x0 =
✓
7
4
◆
1
x. Then find its type and stability.
3
Type: The critical point (0, 0) is called a improper node.
Stability: It is unstable since repeated eigenvalues, r = 5, are positive. (But this
improper node can be asymptotically stable if repeated eigenvalues are negative
ones). The trajectories move towards the origin (if r < 0) or move away from
the origin (if r > 0) which is the case in this example. They keep paralleling to
eigenvector. To check the orientation, pick a sample point and see what we get.
✓ ◆
✓
◆✓ ◆ ✓ ◆
1
7 1
1
7
) x0 =
=
.
x=
0
4 3
0
4
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
(a) Case 1.1: Saddle Point
(b) Case 1.2: Node
(c) Case 2.1: Center
(d) Case 2.2: Spiral Point
(e) Case 3.1: Proper Node
(f) Case 3.2: Improper Node
Figure 6.1: Phase Portraits
251
Node
r1 < r2 < 0
+ iµ
where ⌘ can be found from (A
x(t) = c1 kert + c2 (ktert + ⌘ert )
rI)⌘ = k.
where k1 and k2 are linearly independent vectors.
x(t) = c1 k1 ert + c2 k2 ert
where a + ib is the eigenvector to
b sin µt) + c2 e t (a sin µt + b cos µt),
b sin µt) + c2 (a sin µt + b cos µt)
x(t) = c1 e t (a cos µt
x(t) = c1 (a cos µt
where ki is the eigenvector corresponding to ri
x(t) = c1 k1 er1 t + c2 k2 er2 t ,
General Solutions
Table 6.1: Systems of First Order Linear Equations
Asymptotically Stable
if r1 = r2 = r < 0,
Improper node
Unstable
0C
C
A
r
if r1 = r2 = r > 0,
Br
3.2) A 6= B
@
0
Asymptotically Stable
if r1 = r2 = r < 0,
0
1
Asymptotically Stable
Unstable
Unstable
Proper node
Spiral Point
Stable
Asymptotically Stable
Unstable
Unstable
Stability
if r1 = r2 = r > 0,
0C
C
A
r
< 0,
0
1
if
Br
3.1) A = B
@
0
> 0,
± µi
if
2.2) r1,2 =
Center
Node
1.2) r1 > r2 > 0
2.1) r1,2 = µi
Saddle Point
Type
1.1) r2 < 0 < r1
Eigenvalues
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
252
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
253
Example 195 (Spring 2007 Final Exam).
Consider systems of the form x0 = Ax. For each pair of eigenvalues for the matrix A listed
below, state the type and stability of the critical point at (0, 0)?
Eigenvalues
Case
Type
Stability
r=
3, 2
1.2
Node
Asymptotically Stable
r=
3, 2
1.1
Saddle Point
Unstable
2.2
Spiral Point
Unstable
r = 2i, 2i
2.1
Center
Stable
r = 3, 5
1.2
Node
Unstable
r =1+
p
p
5i, 1
5i
Example 196 (Summer 2013 Final Exam).
Consider systems of linear di↵erentiable equations of the form x0 = Ax, where A is 2 ⇥ 2
matrix. For each pair of eigenvalues of the matrix A listed below, state the type and stability
of the critical point at (0, 0)?
Eigenvalues
Case
Type
Stability
1.2
Node
Unstable
2.2
Spiral Point
Asymptotically Stable
r1 = 6, r2 = 6 with
two linearly independent eigenvectors
3.1
Proper Node
Unstable
r1 =
1.1
Saddle Point
Unstable
2.1
Center
Stable
r1 = 1, r2 = 5
r1 =
2+
p
7i, r2 =
4, r2 = 8
r1 = 3i, r2 =
3i
2
p
7i
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
254
Example 197 (Fall 2005 Midterm Exam II).
Given the general solution of the homogeneous linear system x0 = Ax below. State the name
associated with the critical point at (0, 0) and state whether it is stable, asymptotically stable
or unstable?
General Solution
0 1
0
x(t) = c1 e
1
B1C
B 1C
B C + c2 e t B C
@ A
@ A
1
1
0 1
0 1
3t
B1C
B 1C
C + c2 e t B C
x(t) = c1 e3t B
@ A
@ A
1
1
0 1
0 1
B 1C
B1C
3t B
C
C
x(t) = c1 e3t B
@ A + c2 e @ A
1
1
Case
Type
Stability
1.2
Node
Asymptotically Stable
1.1
Saddle Point
Unstable
3.1
Proper Node
Unstable
Example 198 (Spring 2009 Midterm Exam II).
Determine the type and stability of the critical point at the origin for each of the 2 ⇥ 2 linear
systems x0 = Ax whose general solutions are given below.
General Solution
0
1
0
1
B 2 cos t C
B2 sin tC
C + c2 B
C
x(t) = c1 B
@
A
@
A
sin t
cos t
0 0 1
0
11
x(t) = e
t
x(t) = c1 e
x(t) = c1 e
B B1C
B t + 1 CC
Bc1 B C + c2 B
CC
@ @ A
@
AA
2
2t + 1
0 1
0 1
t
B1C
B2C
B C + c2 et B C
@ A
@ A
2
1
0 1
0 1
B1C
B C + c2 e
@ A
0
0
1
t
t
B0C
B C
@ A
1
0
1
B 2 cos t C
B2 sin tC
C + c 2 et B
C
x(t) = c1 et B
@
A
@
A
sin t
cos t
Case
Type
Stability
2.1
Center
Stable
3.2
Improper Node
Asymptotically Stable
1.1
Saddle Point
Unstable
3.1
Proper Node
Asymptotically Stable
2.2
Spiral Point
Unstable
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
255
Example 199 (Fall 2011 Midterm Exam II).
For each part below, consider a certain system of two first order linear di↵erential equations
in two unknowns, x0 = Ax, where A is a 2 ⇥ 2 matrix of real numbers. Base solely on the
information given in each part, determine the type and stability of the system’s critical point
at (0, 0).
✓
◆
1 2
1. The coefficient matrix is A =
.
2 1
Solution Its characteristic equation is
0 = r2
2r
3 = (r
3)(r + 1)
)
r = 3, 1.
So this is case 1.1.
Type: Saddle point. Stablity: Unstable.
✓p
◆
7 p0
2. The coefficient matrix is A =
.
0
7
p
Solution This is case 3.1 in which eigenvalues are repeated real roots. (r1 = r2 = 7)
p
Type: Proper node or star point. Stablity: Unstable since both eigenvalues are 7 < 0.
3. Eigenvalues of A are
5 + 2i and
5
2i.
Solution This is case 2.2.
Type: Sprial point. Stablity: Asymptotically stable since the real part of eigenvalues is
5 < 0.
4. One of the eigenvalues of A is
4i.
Solution This implies that 4i is another eigenvalue of A. This is case 2.1.
Type: Center. Stablity: Stable.
5. The general solution is x(t) = c1 e
t
19
✓ ◆
2
+ c2 e
1
t
19
✓
◆
2t
.
5+t
Solution This implies that eigenvalues are repeated. (r1 = r2 =
3.2.
1
19 ).
This is case
Type: Improper node. Stablity: Asymptotically stable because the repeated eigenvalues
are negative.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
256
Exercises 6.3
1. Match the sketches of phase portraits for 2 ⇥ 2 homogeneous linear systems x0 = Ax with
the names of their critical points at the origin
(a) saddle
(b) node
(c) proper node
(d) center
(e) spiral
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
257
2. Match the following formulas for general solutions of 2 ⇥ 2 homogeneous linear systems
x0 = Ax with the sketches of the phase portraits given in Problem 1:
✓ ◆
✓ ◆
1
1
(a) c1 e2t
+ c2 e t
1
1
✓
◆
✓
◆
cos t
sin t
(b) c1 et
+ c2 et
2 sin t
2 cos t
✓ ◆
✓ ◆
1
0
(c) c1 e t
+ c2 e t
0
1
✓ ◆
✓ ◆
1
0
(d) c1 et
+ c2 e t
0
1
✓
◆
✓
◆
cos t
sin t
(e) c1
+ c2
2 sin t
2 cos t
✓ ◆
0
3. Match the three adjectives for the critical point
of a 2 ⇥ 2 homogeneous linear systems
0
x0 = Ax with the five general solutions given in the table below by placing one of the letters
A, U, or S in each of the five blanks. Use A for asymptotically stable, U for unstable and S
for stable.
✓ ◆
✓ ◆
1
1
(a) c1 e2t
+ c2 e t
1
1
✓
◆
✓
◆
cos t
sin t
t
t
(b) c1 e
+ c2 e
2 sin t
2 cos t
✓ ◆
✓ ◆
1
0
(c) c1 e t
+ c2 e t
0
1
✓ ◆
✓ ◆
1
0
(d) c1 et
+ c2 e t
0
1
✓
◆
✓
◆
cos t
sin t
(e) c1
+ c2
2 sin t
2 cos t
4. Suppose the phase portrait of the system
0
x =
✓
2
1
◆
1
x
d
has an improper node at (0, 0). Find the value (or range of values) of d.
5. Find the general solution of the system
x0 =
✓
5
0
◆
0
x.
5
6. Consider the system of linear equations
✓
◆
✓ ◆
3
2
1
0
x =
x, x(0) =
.
4
1
4
Solve the initial value problem and classify the type and stability of the critical point at
(0, 0).
7. Consider the 2 ⇥ 2 linear homogeneous system with constant coefficients x0 = Ax.
(a) If the eigenvalues of A are ±i, classify the type and stability of the critical point (0, 0).
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
258
✓ ◆
2
(b) If in addition an eigenvector corresponding to the eigenvalue i is
, then write down
i
the real-valued general solution x(t).
8. Consider the system
0
x =
✓
3
1
◆
4
x.
2
(a) Find the general solution x(t)
✓ ◆
6
(b) If x(0) =
and lim x(t) = 0, what is ?
t!1
9. (a) Solve the initial value problem
x0 =
✓
◆
✓ ◆
1
4
x, x(0) =
3
1
2
2
(b) What is lim |x(t)|?
t!1
10. Consider a certain linear system x0 = Ax, where A is a matrix of real numbers with complex
conjugate eigenvalues. Suppose some of its solutions do not reach a limit either as t ! +1,
or as t ! 1. Then the critical point (0, 0) must be a(n)
(a) unstable spiral point.
(b) unstable saddle point.
(c) asymptotically stable improper node.
(d) (neutrally) stable center.
11. Suppose the linear system x0 =
✓
↵2
2
0
2↵
0
Determine all possible vlaue(s) of ↵.
◆
1
x has an unstable proper node at (0, 0).
12. For each part below, consider a certain system of two first order linear linear di↵erential
equations in two unknowns, x0 = Ax, where A is a 2 ⇥ 2 matrix of real numbers. Based
solely on the information given in each part, determine the type and stability of the system’s
critical point at (0, 0).
(a) Eigenvalues of A are
1 and
6.
(b) Eigenvalues of A are 3 + 7i and 3
7i.
(c) Eigenvalues of A are 9i and 9i.
p
11 and ⇡ 2 .
(d) Eigenvalues of A are
◆
✓ ◆
1
5t 1
(e) The general solution is x(t) = c1 e
+ c2 e
1
0
✓ ◆
✓
◆
1
p t
(f) The general solution is x(t) = c1 et
+ c2 et
1
3+t
5t
✓
13. Consider a certain linear system x0 = Ax, where A is a matrix of real numbers with distinct
nonzero real eigenvalues. Suppose all of its solutions have a finite limit as t ! +1. Then
the critical point (0, 0) must be a(n)
(a) (neutrally) stable center.
(b) asymptotically stable spiral point.
(c) unstable saddle point.
(d) asymptotically stable node.
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
259
14. Consider the initial value problem
x0 =
✓
2
1
◆
✓ ◆
1
2
x, x(0) =
0
6
(a) Solve the initial value problem.
(b) Classify the type and stability of the critical point at (0, 0).
15. For each part below, consider a certain system of two first order linear linear di↵erential
equations in two unknowns, x0 = Ax, where A is a 2 ⇥ 2 matrix of real numbers. Based
solely on the information given in each part, determine the type and stability of the system’s
critical point at (0, 0).
✓
◆
5
2
(a) The coefficient matrix is
x.
0 5
(b) Eigenvalues of A are
e and
e2 .
(c) One of the eigenvalues of A is r = 1 49i.
✓
◆
✓
◆
2 cos t
2 sin t
(d) The general solution is x(t) = c1
+ c2
.
cos t 2 sin t
sin t + 2 cos t
✓ ◆
✓ ◆
p
p
1
2
(e) The general solution is x(t) = c1 e 7t
+ c2 e 7t
.
2
1
16. Consider a certain system of two first order linear di↵erential equations in two unknowns,
x0 = Ax, where A is a matrix of real numbers. Suppose one of the eigenvalues
✓
◆ of the
4
coefficient matrix A is r = 5 + 7i, which has a corresponding eigenvector
. Write
5 + 6i
down the system’s real valued general solution.
17. (a) Find the general solution of the system of linear equations
✓
◆
2
12
0
x =
x.
3
11
✓ ◆
5
(b) Find the solution satisfying x(0) =
.
0
18. For each part below, consider a certain system of two first order linear di↵erential equations
in two unknowns, x0 = Ax, where A is a 2 ⇥ 2 matrix of real numbers. Based solely on
the information given in each part, determine the type and stability of the system’s critical
point at (0, 0).
✓
◆
3
2
(a) The coefficient matrix is
.
2
1
(b) One of the eigenvalues of A is r =
2 + 11i.
(c) The characteristic equation of A can be rewritten as r2 + 4 = 0.
✓ ◆
✓ ◆
4
2
(d) The general solution is x(t) = c1 e2t
+ c2 e2t
.
1
3
19. For what value(s) of
(a) 0
(b) 2
will the linear system below have a (neutrally) stable center at (0, 0)?
✓
◆
2
5
x0 =
2 x
5
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
(c)
260
2, 2
(d) 0, 2
20. Consider a certain system of two first order linear di↵erential equations in two unknowns,
x0 = Ax, where A is a matrix of real numbers. Suppose one of the eigenvalues
of the
✓
◆
2
5i
0
coefficient matrix A is r = 3 + i, which has a corresponding eigenvector x =
.
4
What is the system’s real-valued general solution?
✓
◆
✓
◆
2 cos t 5 sin t
5 cos t 2 sin t
(a) x(t) = c1 e3t
+ c2 e3t
4 cos t
4 sin t
✓
◆
✓
◆
2 cos t + 5 sin t
5 cos t 2 sin t
3t
3t
(b) x(t) = c1 e
+ c2 e
4 cos t
4 sin t
✓
◆
✓
◆
2 cos t + 5 sin t
5 cos t 2 sin t
3t
3t
(c) x(t) = c1 e
+ c2 e
4 sin t
4 cos t
✓
◆
✓
◆
2 cos t 5 sin t
2 sin t
3t
3t 5 cos t
(d) x(t) = c1 e
+ c2 e
4 sin t
4 cos t
✓
◆
4
1
0
21. True or false. Consider the linear system x =
x. The critical point (0, 0) is an
5 0
asymptotically stable node.
✓
◆
0 4
22. True or false. Consider the linear system x0 =
x. The critical point (0, 0) is an
4 0
unstable saddle point.
23. For what range of values of
will the linear system below have a saddle point at (0, 0)?
✓
◆
1
x0 =
x
1
(a) ( 1, 1)
(b) ( 1, 1) [ (1, 1)
(c) ( 1, 1)
(d) (1, 1)
24. (a) Find the general solution of the system of linear equations
✓
◆
0
1
x0 =
x.
1 2
(b) Find the solution satisfying x(0) =
✓
◆
1
.
2
(c) Classify the type and stability of the critical point at (0, 0).
25. For each part below, consider a certain system of two first order linear di↵erential equations
in two unknowns, x0 = Ax, where A is a 2 ⇥ 2 matrix of real numbers. Based solely on the
information given in each part, answer each question.
(a) Suppose one of the eigenvalues
✓ of the
◆ coefficient matrix A is r = 2 + i, which has
3
a corresponding eigenvector
. Write down the system’s real-valued general
1 6i
solution.
(b) Classify the type and stability of the critical point at (0, 0) for the system described in
(a).
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
(c) Suppose one of the eigenvalues of the coefficient matrix A is r =
and stability of the critical point at (0, 0) for the system.
261
9i, classify the type
(d) Suppose the coefficient matrix A ✓
only
eigenvalue, r = 8, which has
◆ has one
✓ distinct
◆
0
2
corresponding eigenvectors both
and
. Write down the system’s general
4
5
solution.
(e) Classify the type and stability of the critical point (0, 0) for the system described in
(d).
Answers
1. (a) 5, (b) 1, (c) 4, (d) 2, (e) 3
2. (a) none, (b) 3, (c) 4, (d) 5, (e) 2
3. (a) U, (b) U, (c) A, (d) U, (e) S
4. 0, 4
✓ ◆
✓ ◆
1 5t
0 5t
5. c1
e + c2
e
0
1
✓
◆
✓
◆
cos 2t
sin 2t
t
t
6. e
+ 5e
, sprial point, unstable
cos 2t + sin 2t
sin 2t cos 2t
✓
◆
✓
◆
2 cos t
2 sin t
7. (a) center, (neutrally) stable, (b) x(t) = c1
+ c2
sin t
cos t
✓ ◆
✓ ◆
1
4 2t
8. (a) x(t) = c1
e t + c2
e , (b) = 6
1
1
✓ ◆
✓ ◆
3
1
t
9. (a) x(t) =
e +
e 4t , (b) 0
3
2
10. (d)
11. ↵ =
1 only
12. (a) node, asymptotically stable, (b) spiral point, unstable, (c) center, (neutrally) stable, (d)
node, unstable, (e) proper node, asymptotically stable, (f) improper node, unstable
13. (d)
14. (a) x(t) =
✓
◆
4t 2
et , (b) improper node, unstable
4t + 6
15. (a) improper node, unstable, (b) node, asymptotically stable, (c) spiral point, unstable, (d)
center, stable, (e) proper node, asymptotically stable
✓
◆
✓
◆
4 cos 7t
4 sin 7t
16. x(t) = c1 e 5t
+ c2 e 5t
5 cos 7t 6 sin 7t
6 cos 7t + 5 sin 7t
✓ ◆
✓ ◆
✓
◆
3
4
9e 2t 4e 7t
2t
7t
17. (a) x(t) = c1
e
+ c2
e , (b) x(t) =
1
3
3e 2t + 3e 7t
18. (a) improper node, unstable, (b) spiral point, asymptotically stable, (c) center, (neutrally)
stable, (d) proper node, unstable
19. (d)
CHAPTER 6. SYSTEMS OF FIRST ORDER LINEAR EQUATIONS
262
20. (b)
21. False
22. False
23. (b)
◆
✓
◆
✓
◆
1 t
t+1 t
tet et
e +c2
e , (b) x(t) =
, (c) improper node, unstable
1
t
tet + 2et
✓
◆
✓
◆
3 cos t
3 sin t
25. (a) x(t) = c1 e2t
+ c2 e2t
, (b) spiral, unstable, (c) center,
cos t + 6 sin t
6 cos t + sin t
✓ ◆
✓ ◆
0
2
(neutrally) stable, (d) x(t) = c1
e 8t + c2
e 8t , (e) proper node, asymptotically
4
5
stable
24. (a) x(t) = c1
✓
Chapter 7
Nonlinear Di↵erential Equations
and Stability
7.1
Autonomous Systems
Consider a nonlinear system of two simultaneous di↵erential equations of the form
dx
= F (x, y),
dt
dy
= G(x, y).
dt
Observe that the functions F and G do not depend on the independent variable t but only on the
dependent variables x and y.
• A system in this form is said to be autonomous.
• The critical points are points (x, y) such that x0 = y 0 = 0. (or F (x, y) = G(x, y) = 0)
• Unlike a linear system which has only one critical point, namely (0, 0), a nonlinear system
could have none, one, two, three or any number of critical points.
Example 200 (Spring 2011 Midterm Exam II).
Find all critical points of the nonlinear system:
x0 = (2x + y)(2x
0
y = (x
y)
2)(y + 2)
Solution Critical points can be found by setting x0 = 0 and y 0 = 0. So
x0 = 0
0
y =0
)
)
(2x + y)(2x
(x
y) = 0
)
)
2)(y + 2) = 0
2x + y = 0 or 2x
x
Hence we have four pairs of equations to solve:
(
2x + y = 0
) x = 2, y = 2x = 2(2) = 4
x 2=0
(
2x + y = 0
y
2
) y = 2, x =
=
=1
2
2
y+2=0
263
y = 0.
2 = 0 or y + 2 = 0.
)
(2, 4) is a critical point.
)
(1, 2) is a critical point.
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
(
(
2x y = 0
x 2=0
)
x = 2,
2x y = 0
y+2=0
)
y=
2,
y = 2x = 2(2) = 4
x=
y
2
=
=
2
2
1
)
(2, 4) is a critical point.
)
( 1, 2) is a critical point.
264
From all 4 cases, (2, 4), (1, 2), (2, 4) and ( 1, 2) are all critical points of the given nonlinear system.
Example 201 (Spring 2010 Midterm Exam II).
Find all critical points of the nonlinear system:
x0 = x2
xy
0
y = xy + 2y 2
6y.
Solution Note that the nonlinear system can be written as
x0 = x(x y)
y 0 = y(x + 2y
6).
To find all critical points, we set x0 = 0 and y 0 = 0. Then
x0 = 0
0
y =0
Hence we have four
(
x=0
y=0
(
x=0
x + 2y 6 = 0
(
x y=0
y=0
(
x y=0
x + 2y 6 = 0
)
)
x(x
y) = 0
y(x + 2y
6) = 0
)
)
x = 0 or x
y = 0.
y = 0 or x + 2y
6 = 0.
pairs of equations to solve:
)
(0, 0) is a critical point.
)
x = 0, y =
)
)
x+6
=
2
0+6
=3
2
)
(0, 3) is a critical point.
y = 0, x = y = 0
)
(0, 0) is a critical point.
x = 2, y = 2
)
(2, 2) is a critical point.
From all 4 cases, (0, 0), (0, 3) and (2, 2) are all critical points of the above system.
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
Exercises 7.1
1. Which of the points below is not a critical point of the nonlinear system of equation
x0 = x2
y2
y 0 = xy + x + y + 1?
a)
b)
c)
d)
(1, 1)
(1, 1)
( 1, 1)
( 1, 1)
2. True or false. The point (0, 1) is a critical point of the nonlinear system
x0 =
x + 2xy
0
y =y
x2
y2 .
3. Which point below is NOT a critical point of the system
a)
b)
c)
d)
x0 = y
y2
y0 = x
x3 ?
(1, 0)
( 1, 1)
(0, 1)
(1, 1)
4. Find the critical points of the following nonlinear system.
x0 = (x + 1)(y
2)
y 0 = y(x + y).
5. Find all critical points of the following nonlinear system.
dx
= (1 y)(2x y)
dt
dy
= (2 + x)(x 2y).
dt
6. Find all critical points of the following autonomous nonlinear system.
x0 = x2
xy
0
y = 4xy + y 2
Answers
1. (a)
2. True
3. (d)
4. ( 1, 1), ( 1, 0) and ( 2, 2)
5. ( 2, 1), (2, 1), ( 2, 4) and (0, 0)
2 2
6. (0, 2), (0, 0) and ( , )
5 5
2y.
265
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
7.2
266
Locally Linear Systems
Consider an autonomous nonlinear system,
x0 = F (x, y),
y 0 = G(x, y).
Since there might be multiple critical points presenting on the phase portrait, the type and stability
of each critical point need to be determined locally (in a small neighborhood around each point).
We’ll approximate the behavior of the trajectories using the linearizations of F and G about each
critical point. This will convert the globally nonlinear system into a locally linear system around
each critical point. First, using Taylor series1 about the critical point (x, y) = (↵, ),
x0 = F (x, y) ⇡ F (↵, ) + Fx (↵, )(x
y 0 = G(x, y) ⇡ G(↵, ) + Gx (↵, )(x
↵) + Fy (↵, )(y
),
↵) + Gy (↵, )(y
).
Since (↵, ) is a critical point, F (↵, ) = G(↵, ) = 0. Then
x0 = F (x, y) ⇡ Fx (↵, )(x
0
y = G(x, y) ⇡ Gx (↵, )(x
Using translation by substitution x
↵ by x and y
↵) + Fy (↵, )(y
)
↵) + Gy (↵, )(y
).
by y, we have
x0 = F (x, y) ⇡ Fx (↵, )x + Fy (↵, )y
y 0 = G(x, y) ⇡ Gx (↵, )x + Gy (↵, )y.
Write it in matrix form,
✓ 0◆ ✓
◆
✓ ◆
x
Fx F y
x
=
.
y0
Gx Gy (↵, ) y
✓
◆
Fx F y
The coefficient matrix
is called the Jacobian matrix2 and denoted by J.
Gx Gy
To consider the type and stability of each critical point of nonlinear system, we first compute
Jacobian matrix, evaluate it at each critical point and then use its eigenvalues to determine their
types and stabilities.
1 Brook Taylor (1685
1731) was a Cambridge graduate, an enthusiastic admirer of Newton, and secretary of
the Royal Society. He was the first to use integration by parts, was one of the founders of the calculus of finite
di↵erences and was the first to recognize the existence of singular solutions of di↵erential equations.
Today his name is recalled almost exclusively in connection with the Taylor series
f (x + a) = f (a) + f 0 (a)x + f 00 (a)
x2
xn
+ · · · + f (n) (a)
+ ···
2!
n!
which appeared in Methodus Incrementorum Directa et Inversa in 1715. The series becomes Maclaurin series on
substituting zero for a. The general Taylor series had been known many years prior to James Gregory and also, in
essence, to Jean Bernoulli, but Taylor was unaware of this.
2 Carl Gustav Jacob Jacobi (1804
1851), a German analyst who was professor and lecturer at the Universities
of Konigsberg and Berlin, made important contributions to the theory of elliptic functions.
He worked on determinants and studied the functional determinant now called the Jacobian. Jacobi was not the
first to study the functional determinant which now bears his name, it appears first in a 1815 paper of Cauchy.
However Jacobi wrote a long memoir De determinantibus functionalibus in 1841 devoted to this determinant. He
proved, among many other things, that if a set of n functions in n variables are functionally related then the
Jacobian is identically zero, while if the functions are independent the Jacobian cannot be identically zero.
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
267
Example 202 (Summer 2002 Final Exam).
Consider the system:
x0 = x + y
y 0 = x2 + y 2
8.
a) Find all critical points of the system.
Solution Set x0 = 0 and y 0 = 0. Then
x0 = 0
)
0
From the first equation, we have x =
8=0
)
)
Use x =
x2 + y 2
)
y =0
x2 + y 2
x+y =0
8 = 0.
y. Apply this condition into the second equation,
( y)2 + y 2
(y
8=0
2)(y + 2) = 0
)
)
2y 2
8=0
y = 2, 2.
y again to find the corresponding values x,
y=2
y=
2
)
)
x=
y=
x=
y=2
2
)
)
( 2, 2) is a critical point.
(2, 2) is a critical point.
In conclusion, ( 2, 2) and (2, 2) are all critical points of the above system.
b) Find the linearized matrix of the system.
Solution We have F (x, y) = x + y, G(x, y) = x2 + y 2 8. Then the Jacobian matrix is
✓
◆ ✓
◆
F x Fy
1
1
J=
=
.
Gx Gy
2x 2y
At the critical point ( 2, 2), the linearized matrix of the system is
✓
◆ ✓
◆
1
1
1 1
A1 = J|( 2,2) =
=
.
2( 2) 2(2)
4 4
At the critical point (2, 2), the linearized matrix of the system is
✓
◆ ✓
◆
1
1
1 1
A2 = J|(2, 2) =
=
.
2(2) 2( 2)
4
4
c) Classify the type and stability of each critical point.
Solution
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
268
At ( 2, 2), the characteristic equation of A1 is
0=r
2
(1 + 4)r + 1(4)
( 4)1 = r
2
5r + 8
)
p
5 ± 7i
r=
.
2
Since we get complex eigenvalues with positive real part (case 2.2), ( 2, 2) is an unstable
spiral point. While at (2, 2), the characteristic equation of A2 is
p
3 ± 41
2
2
0=r
(1 4)r + 1( 4) 4(1) = r + 3r 8 ) r =
.
2
Since we get real distinct eigenvalues with opposite signs (case 1.1), (2, 2) is an unstable saddle point.
Example 203 (Summer 2010 Midterm Exam II).
Consider the nonlinear system:
x0 = x(1
x + y)
y 0 = y(x + y).
a) Verify that (1, 0) is critical point.
Solution Substitute x = 1 and y = 0,
x0 = x(1
x + y) = 1(1
1 + 0) = 0
0
y = y(x + y) = 0(1 + 0) = 0.
Therefore (1, 0) is a critical point.
b) Linearize the system about the point (1, 0). Clearly identify the matrix for the linearized
system.
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
269
Solution Note F (x, y) = x(1 x + y) = x x2 + xy and G(x, y) = y(x + y) = yx + y 2 .
Then we compute Jacobian matrix,
✓
◆ ✓
◆
Fx F y
1 2x + y
x
J=
=
.
G x Gy
y
x + 2y
Now we evaluate the Jacobian matrix at the critical point (1, 0) to get the matrix for
the linearized system,
✓
◆ ✓
◆
1 2(1) + 0
1
1 1
A = J|(1,0) =
=
.
0
1 + 2(0)
0 1
(This means that near the point (1, 0), the nonlinear system acts like the linear system
✓
◆
1 1
x0 =
x.)
0 1
c) Classify the type and stability of the critical point (1, 0) by examining the linearized
system obtained in part b)
Solution Find the eigenvalues from the characteristic equation of A,
0 = r2
( 1 + 1)r + ( 1)1
0(0) = r2
1 = (r
1)(r + 1)
)
r = 1, 1.
Since we get two distinct real eigenvalues with opposite signs (case 1.1), (1, 0) is an
unstable saddle point.
In this example, we can check further that there are two more critical points (three in total):
• (1, 0) is an unstable saddle point.
• (0, 0). At this critical point, the coefficient matrix has zero determinant.
• ( 12 , 12 ) is an asymptotically stable spiral point. (case 2.2, we have complex eigenvalues
with negative real part)
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
270
Example 204 (Fall 2007 Midterm Exam II).
Linearize the system
x0 = 1
0
y =x
y
2
y2
around its critical point (1, 1) and classify its type and stability.
Solution Here F (x, y) = 1
y and G(x, y) = x2 y 2 . So the Jacobian matrix is,
✓
◆ ✓
◆
Fx Fy
0
1
J=
=
.
Gx Gy
2x
2y
Then the linearized system at (1, 1) has coefficient matrix,
✓
◆ ✓
0
1
0
A = J|(1,1) =
=
2(1)
2(1)
2
◆
1
.
2
Now let’s find the eigenvalues from the characteristic equation of A,
0=r
2
(0
2)r + 0( 2)
2
2( 1) = r + 2r + 2
)
r=
2±
p
2
4
8
=
1 ± i.
Since we get complex eigenvalues with negative real part (case 2.2), (1, 1) is an asymptotically
stable spiral point.
(You can check further that another critical point is ( 1, 1). And it is an unstable saddle
point since its linearized matrix gives distinct real eigenvalues with opposite signs. (case 1.1))
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
271
Example 205 (Summer 2008 Midterm Exam II).
Linearize the following nonlinear system about its critical point (4, 2) and classify its type and
stability.
x0 = x2
4y 2
y 0 = xy
2x.
Solution We have F (x, y) = x2
4y 2 and G(x, y) = xy 2x. Then the Jacobian matrix is,
✓
◆ ✓
◆
F x Fy
2x
8y
J=
=
.
Gx G y
y 2
x
Then the linearized system at (4, 2) has a coefficient matrix,
✓
◆ ✓
2(4)
8(2)
8
A = J|(4,2) =
=
2 2
4
0
◆
16
.
4
So the characteristic equation of A is
0 = r2
(8 + 4)r + 8(4)
0( 16) = r2
12r + 32 = (r
8)(r
4)
)
r = 4, 8.
Since we have a pair of positive real distinct eigenvalues (case 1.2), (4, 2) is an unstable node.
In this example, we can check further that there are two more critical points (three in total):
• (4, 2) is an unstable node.
• (0, 0). At this critical point, the coefficient matrix has vanishing determinant.
• ( 4, 2) is an asymptotically stable node. (case 1.2: negative real distinct eigenvalues)
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
272
Exercises 7.2
1. (a) Find the critical points of the following nonlinear system
(
x0 = x(x + y)
y 0 = y(2 x + y)
(b) Linearize the following nonlinear system about its critical point (0, 2) and classify its
type and stability.
(
x0 = xy 6x
?
y 0 = xy 2x + y 2
2. Given that the point (0, 1) is a critical point of the nonlinear system of equations
x0 = x2 y
xy 2
y 0 = xy
x
3y + 3
The critical point (0, 1) is a(n)
(a) unstable saddle point.
(b) unstable spiral point.
(c) asymptotically stable node.
(d) asymptotically stable proper node (star point).
3. Consider the nonlinear system:
x0 = x
0
y = 4y
y
x2 y.
(a) One of the critical points of the system is (2, 2). Verify that (2, 2) is indeed a critical
point. That is, show that (2, 2) satisfies the condition(s) of being a critical point.
(b) Besides (2, 2), there are 2 other critical points. Find those other 2 critical points of the
system.
(c) Linearize the system about the point (2, 2). Classify the type and stability of the critical
point at (2, 2) by examining the linearized system. Be sure to clearly state the linearized
system’s matrix and its eigenvalues.
4. Consider the nonlinear system:
x0 = x2
0
y = xy
xy
3x + 2.
(a) One of the critical points of the system is (1, 1). There is another critical point. Find
it.
(b) Linearize the system about the point (1, 1). Classify the type and stability of the critical
point at (1, 1) by examining the linearized system. Be sure to clearly state the linearized
system’s matrix and its eigenvalues.
5. Consider the nonlinear system:
x0 =(x + 1)(y
0
y =(x + y)(2x
2) = xy
2
2x + y
y) = 2x + xy
y
2
2
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
273
(a) The system has 4 critical points. One of the critical points of the system is ( 1, 1).
Find the other 3 critical points of the system.
(b) Linearize the system about the critical point ( 1, 1). Identify the coefficient matrix of
the linearized system.
(c) What are eigenvalues of the coefficient matrix? Classify the type and stability of the
critical point at ( 1, 1) by examining the linearized system found in (b).
6. Consider the autonomous nonlinear system:
x0 =4(x
0
y =y(2x
1)
2
y
4
y)
(a) The system has 3 critical points. One of the critical points is (1, 0). Verify that (1, 0)
is indeed a critical point. Then find the other 2 critical points.
(b) Linearize the system about the point (1, 0). Classify the type and stability of this
critical point by examining the linearized system. Be sure to clearly state the linearized
system’s matrix and its eigenvalues.
7. Given that ( 2, 4) is a critical point of the nonlinear system
x0 = xy + 2y
y 0 = xy
4x.
Linearize this system about ( 2, 4) to determine the critical point as a(n)
(a) unstable saddle point.
(b) asymptotically stable spiral point.
(c) unstable node.
(d) asymptotically stable proper node.
8. Given that (3, 2) is a critical point of the nonlinear system
x0 = 3y
xy
y 0 = xy + 2x.
Linearize this system about (3, 2) to determine the critical point as a(n)
(a) unstable saddle point.
(b) asymptotically stable spiral point.
(c) unstable node.
(d) asymptotically stable improper node.
9. Given that (1, 1) is a critical point of the system
x0 = x2
0
y =2
y2
2xy
Which of the following statement is TRUE regarding (1, 1)?
(a) It is an asymptotically stable spiral point.
(b) It is an unstable node.
(c) It is an asymptotically stable proper node.
(d) It is an unstable saddle point.
CHAPTER 7. NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
274
Answers
1. (a) (0, 0), (0, 2), (1, 1), (b) saddle point, unstable
2. (c)
3. (a) Substitute (2, 2) into the equations directly to see that x0 = y 0 = 0, (b) (0, 0) and
( 2, 2), (c) saddle point, unstable
4. (a) (2, 2), (b) saddle point, unstable
✓
1
5. (a) ( 1, 2), ( 2, 2) and (1, 2), (b)
3
◆
0
, (c)
3
6. (a) (2, 4), (0, 4), (b) saddle point, unstable
7. (a)
8. (c)
9. (d)
1, 3, asymptotically stable node
Chapter 8
Partial Di↵erential Equation
Recall that a partial di↵erential equation is any di↵erential equation that contains two or more
independent variables. Therefore the derivatives appearing in the equation are partial derivative.
In this class, we’ll examine the second order partial di↵erential equation with two independent
variables such as
↵2 uxx =ut
(Heat equation)
2
a uxx =utt
(Wave equation)
The goal of this chapter is to solve the above two partial di↵erential equations by using the method
of separation of variables.
8.1
8.1.1
Boundary Value Problem
BVP vs IVP
Recall that the initial value problem for second order di↵erential equation is in the form of
y 00 + p(t)y 0 + q(t)y = g(t),
y(t0 ) = y0 ,
y 0 (t0 ) = y00 .
Notice that initial conditions of IVP evaluate at the same point. (in this case at t0 )
If the conditions are specified at two di↵erent points, such conditions are called boundary conditions. A di↵erential equation with boundary conditions form a (two-point) boundary value
problem, or simply, BVP. A typical example is the di↵erential equation
y 00 + p(t)y 0 + q(t)y = g(t)
(8.1)
with boundary conditions
y(↵) = y0 ,
y( ) = y1
0
(8.2)
( or y(↵) = y0 ,
y ( ) = y1
(8.3)
or y 0 (↵) = y0 ,
y 0 ( ) = y1 )
(8.4)
where ↵ 6= . If the function g in Equation (8.1) is zero and if the boundary values y0 and y1 in
Equation (8.2)-(8.4) are also zero, then BVP is called homogeneous. Otherwise, the problem is
nonhomogeneous.
275
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
276
Example 206. The followings are all BVPs since the boundary conditions evaluate at different points. (at 0 and ⇡)
y 00 + 2y = 0,
y(0) = 1,
y(⇡) = 0
y 00 + y = 2,
y(0) = 0,
y 0 (⇡) = 0
y 0 (0) = 0,
y 0 (⇡) = 0
y 00
2y = 0,
Although IVP and BVP may superficially appear to be quite similar, their situations di↵er in
some very important way. Under mild conditions on the coefficients of the di↵erential equation,
IVP are certain to have a unique solution by the Existence and Uniqueness Theorem below. (for
second order ODEs)
Theorem 14 (Existence and Uniqueness Theorem). Consider the initial value problem
y 00 + p(t)y 0 + q(t)y = g(t),
y(t0 ) = y0 ,
y 0 (t0 ) = y00 ,
where p, q and g are continuos on an open interval I that contains the point t0 . Then there is
exactly one solution y = (t) of this problem, and the solution exists throughout the interval
I.
On the other hand, BVP under similar conditions may have a unique solution, no solution or even
infinitely many solutions. Let’s take a look in the next example.
Example 207. Consider the solution to y 00 + y = 0
y(0) = 0
y(⇡) = 0.
2
Solution The characteristic equation is 0 = r + 1 and so r = ±i. So the general solution is
y = c1 cos t + c2 sin t.
Invoke condition y(0) = 0, we have c1 = 0, and y(⇡) = 0 also gives us c1 = 0. Hence solution
is y(t) = c2 sin t where c2 remains arbitrary. So this BVP has infinitely many solutions.
Example 208 (Fall 2013 Final Exam). Which initial or boundary value problem below is
guaranteed to have a unique solution according to the Existence and Uniqueness Theorems?
p
a) t2 y 00 + ty 0 + e2t y = 2, y(2) = 3, y 0 (2) = 1e .
b) y 00 + sin(t)y 0 + 10ty = ln t,
c) y 0 + tan(t)y = sec t,
d) y 00
t3 y 0 + e
3t
y = 0,
y( 5) = 1,
y 0 ( 5) = 4.
y( 3⇡
2 ) = 9.
y 0 (1) =
1,
y 0 (⇡) = 1.
Solution Notice that di↵erential equations in multiple choices a) to c) are IVP while the one
in d) is BVP. So d) is not a correct answer since there is no theorem to guarantee about
existence and uniqueness of solution for BVP.
b) also is not the correct answer since t0 = 5 is not defined for the function ln while in c),
1
t0 = 3⇡
2 is a point of discontinuity of sec t = cos t .
So a) is the correct answer by the Existence and Uniqueness Theorem since t0 = 2 is not the
discontinuous point. (The only discontinuity for a) is 0.)
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
277
Exercises 8.1.1
1. Which of the following has a unique solution on the whole interval (0, ⇡)?
(a) y 00 + y = 0,
y(0) = 0,
00
(b) y + 4y = 0,
y(⇡) = 0.
0
y (0) = 0,
y 0 (⇡) = 0.
(c) (t + 1)y 00 + ty = 0,
y(1) = 1,
y 0 (1) = 0.
1)y 0 + 2y = 0,
y(0) = 0,
y 0 (0) = 1.
(d) (t
2. Of the four initial or boundary value problems below, only one is guaranteed to have a unique
solution according to the Existence and Uniqueness Theorems. Which one is it?
(a) ty 00
t2 y 0 + et y = tan(t),
00
(b) ty + 2y
0
3y = 0,
00
(c) y + 9y = 0,
y(0) = 2,
y(0) = 0,
0
y(1) = 0,
(d) y + sec(t)y = sin(2t),
y 0 (1) = ⇡.
0
y (0) = 0.
y(3⇡) = 0.
y( ⇡2 ) = 0.
3. Consider the two initial/ boundary value problems below. Which is certain to have a unique
solution for every value of ↵?
(I)
(II)
y 00 + 9y = 0,
00
y + 9y = 0,
y(↵) = ↵2 ,
y 0 (↵) =
y(0) = 0,
0
↵.
2
y (↵ ) = 0.
(a) I only.
(b) II only.
(c) Both I and II.
(d) Neither.
4. Consider the second order equation
y 00 + y = 0,
Where
is a real number. Which statement below about the equation is false?
(a) For any value of , its general solution is in the form of y = C1 y1 + C2 y2 , where y1 and
y2 are two fundamental solutions of the equation.
(b) If
= 0, then its general solution is y(t) = C1 + C2 t.
(c) For any value of , there is a unique solution satisfying boundary conditions y(0) = 0
and y 0 (2⇡) = 0.
(d) For any value of , there is a unique solution satisfying initial conditions y(⇡) = 0 and
y 0 (⇡) = 32.
5. Consider the following two di↵erential equations:
(I)
y 00 + ay 0 + by = 0,
y(0) = 0,
y 0 (0) = 2.
(II)
y 00 + ay 0 + by = 0,
y(0) = 0,
y(⇡) = 2.
where a, b are real numbers. Which of the following statements are true?
(a) Both I and II always have a unique solution on some interval.
(b) Only I always has a solution on some interval.
(c) Only II always has a solution on some interval.
(d) None of the above
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
Answers
1. (c)
2. (a)
3. (a)
4. (c)
5. (b)
278
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.1.2
279
Eigenvalues Problems
Next let’s consider a certain BVP
y 00 + y = 0
together with two boundary conditions, where
is a constant real number.
• We call it the eigenvalue problem.
• The values of
for which nontrivial solutions y occur are called eigenvalues.
• The nontrivial solutions (y 6= 0) themselves are called eigenfunctions.
For eigenvalue problems, the following trig identities are helpful:
sin(x) = 0
)
x = n⇡ for all integer n
cos(x) = 0
)
x=
(2n
1)⇡
2
for all integer n
Example 209. Find the eigenvalues and eigenfunctions of the given BVP
y 00 + y = 0,
y(0) = 0,
y(L) = 0
where L is a positive real number. (L > 0)
Solution We will consider into three cases:
> 0,
= 0 and
< 0.
= µ2 , where µ > 0. So the
Case 1: > 0. To avoid the radical signs, it’s convenient to let
equation becomes
y 00 + µ2 y = 0.
So its characteristic equation is 0 = r2 + µ2 and so r = ±µi. So the general solution is
y(x) = c1 cos µx + c2 sin µx.
(8.5)
Invoking the boundary conditions,
y(0) = 0
y(L) = 0
)
)
0 = c1 cos(0) + c2 sin(0) = c1
0 = c1 cos(µL) + c2 sin(µL) = c2 sin(µL)
)
)
c1 = 0.
(8.6)
c2 = 0 or sin(µL) = 0.
(8.7)
In Equation (8.7), if c2 = 0 together with c1 = 0 from Equation (8.6), then the general solution
is a trivial solution y = 0 by Equation (8.5). Since we’re seeking nontrivial solutions (by the
definition of eigenfunctions), so we must require that c2 6= 0. So by Equation (8.7), we get
sin(µL) = 0
)
µL = n⇡
)
µn =
n⇡
,
L
for n = 1, 2, . . .
The last step we use the fact that L > 0. So in this first case, the eigenvalues are
n
= (µn )2 = (
n⇡ 2
) ,
L
and the corresponding eigenfunctions are
yn (x) = c2 sin µn x = c2 sin
n⇡
x,
L
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
280
where n = 1, 2, . . . by Equation (8.5).
Case 2:
= 0. Then the equation is y 00 = 0 and so y(x) = c1 x + c2 . Now let’s use the
boundary conditions:
y(0) = 0
y(L) = 0
)
)
0 = c1 (0) + c2 = c2
)
c2 = 0.
)
0 = c1 L + c2 = c1 L
c1 = 0 since L 6= 0.
So there is only the trivial solution y0 = 0 in this case. Hence
and there is no eigenfunction.
Case 3:
< 0. Similar to case 1, let
0
= 0 is not an eigenvalue
µ2 , where µ > 0. So the equation becomes
=
y 00
µ2 y = 0.
So its characteristic equation is 0 = r2 µ2 = (r µ)(r + µ) and so r = ±µ. Thus the general
solution is
y(x) = c1 eµx + c2 e µx .
(8.8)
By using the given boundary conditions, we have
y(0) = 0
y(L) = 0
)
0 = c1 + c2
)
c1 = 0 and hence c2 = 0.
)
0 = c1 e
Here we use the fact that eµL
eµL = e
µL
e
µL
+ c2 e
µL
)
)
c2 =
µL
= c1 e
c1 .
µL
c1 e
µL
= c1 (eµL
e
µL
)
6= 0 since if it is, we get
µL =
µL
)
=(
n⇡ 2
) ,
L
which is a contradiction.
For all 3 cases, the eigenvalues are
n
2µL = 0
)
L = 0,
and the corresponding eigenfunctions are
yn (x) = sin
n⇡
x,
L
where n = 1, 2, . . .
Example 210 (Fall 2005, Final Exam).
Determine all positive eigenvalues
two point boundary value problem:
and the corresponding eigenfunctions for the following
y 00 + y = 0,
Solution As usual, write
y 0 (0) = 0,
y(3) = 0.
= µ2 , where µ > 0. So the equation becomes
y 00 + µ2 y = 0.
So its characteristic equation is 0 = r2 + µ2 and so r = ±µi. So the general solution is
y(x) = c1 cos µx + c2 sin µx.
(8.9)
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
Note here that
y 0 (x) =
281
c1 µ sin µx + c2 µ cos µx.
Invoking the boundary conditions,
y 0 (0) = 0
y(3) = 0
)
)
)
0 = c2 µ
c2 = 0 since µ 6= 0.
)
0 = c1 cos(3µ)
c1 = 0 or cos(3µ) = 0.
Again if both c1 and c2 are zero, then the general solution is a trivial solution y = 0. Since
we’re seeking nontrivial solutions, we need that c1 6= 0. So we get
cos(3µ) = 0
)
3µ =
(2n
1)⇡
)
2
µn =
(2n
1)⇡
6
,
for n = 1, 2, . . . .
Therefore the eigenvalues are
n
= (µn )2 = (
(2n
1)⇡
6
)2 ,
and the corresponding eigenfunctions are
yn (x) = c1 cos µn x = c1 cos
(2n
1)⇡
6
x,
where n = 1, 2, . . . by Equation (8.9).
Example 211 (Summer 2014 Final Exam ).
Consider the 2-point boundary value problem:
⇡2
X = 0,
4
X 00 +
X 0 (0) = 0,
X 0 (2) = 0.
Find ALL eigenvalues and the corresponding eigenfunctions. Check whether there are any
eigenvalues for > 0, for = 0, for < 0. Show your work!
Solution Case 1:
> 0. Write
= µ2 . Then the equation is
X 00 +
µ2 ⇡ 2
X = 0.
4
Its characteristic equation is
r2 +
µ2 ⇡ 2
=0
4
So the solution is
X(x) = c1 cos
Note that
)
r=±
µ⇡
i.
2
µ⇡x
µ⇡x
+ c2 sin
.
2
2
µ⇡
µ⇡x
µ⇡
µ⇡x
sin
+ c2
cos
.
2
2
2
2
Then we impose the boundary conditions,
X 0 (x) =
X 0 (0) = 0
)
X 0 (2) = 0
)
c1
µ⇡
= 0 ) c2 = 0 since µ 6= 0.
2
µ⇡
c1
sin µ⇡ = 0 ) c1 sin µ⇡ = 0
2
c2
)
c1 = 0 or sin µ⇡ = 0.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
282
If c1 = 0, then X is a trivial solution. Since we’re seeking eigenfunctions which by definition
is a nonzero function, c1 6= 0. And so
sin µ⇡ = 0
So eigenvalues are
n
)
µn = n.
= µ2n = n2 and its corresponding eigenfunctions are
Xn (x) = cos
Case 2:
)
µ⇡ = n⇡
µ⇡x
n⇡x
= cos
.
2
2
= 0. Then the equation becomes X 00 = 0 which gives
X(x) = c1 x + c2
as a solution. Then X 0 (x) = c1 . Now we apply the boundary conditions. Both equations give
c1 = 0. So X(x) = c2 where c2 is arbitrary. So = 0 is an eigenvalue and its corresponding
eigenfunction is X0 (x) = 1.
Case 3: < 0. Write = µ2 . Now the equation is
µ2 ⇡ 2
X = 0.
4
X 00
Its corresponding characteristic equation is
r2
µ2 ⇡ 2
=0
4
)
µ⇡
µ⇡
)(r +
)=0
2
2
(r
So the solution is
X(x) = c1 e
µ⇡x
2
+ c2 e
µ⇡x
2
)
r=±
µ⇡
.
2
.
Note that its first derivative is
X 0 (x) = c1
µ⇡ µ⇡x
e 2
2
c2
µ⇡
e
2
µ⇡x
2
.
Now we apply the boundary conditions,
X 0 (0) = 0
)
X 0 (2) = 0
)
µ⇡
µ⇡
c2
=0
2
2
µ⇡
µ⇡
c1 eµ⇡ c2 e
2
2
)
c1
(c1
µ⇡
)
µ⇡
= 0 ) c1 = c2
2
µ⇡ µ⇡
c2
e
e µ⇡ = 0 )
2
c2 )
c2 = 0.
Then c1 = c2 = 0. So there are neither eigenvalues nor eigenfunctions.
Example 212 (Fall 2001 Final Exam ).
Consider the 2-point boundary value problem:
X 00 + ⇡ 2 X = 0,
X(0) = 0,
X 0 (1) = 0.
Is = 14 and eigenvalue? If not, justify your answer. If so, find a corresponding eigenfunction.
2
Solution With = 14 , the equation is X 00 + ⇡4 X = 0. Its characters equation is
r2 +
⇡2
=0
4
)
⇡
r = ± i.
2
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
Hence the general solution is
X(x) = c1 cos
⇡
⇡
x + c2 sin x.
2
2
Note here that its first derivative is
X 0 (x) =
⇡
⇡
⇡
⇡
c1 sin x + c2 cos x.
2
2
2
2
Now we impose the boundary conditions,
X(0) = 0
0
X (1) = 0
So
=
1
4
)
)
c1 cos 0 + c2 sin 0 = 0 ) c1 = 0.
⇡
⇡
c2 cos = 0 ) 0 = 0 ) c2 is arbitrary.
2
2
is an eigenvalue with the corresponding eigenfunction sin ⇡2 x.
283
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
284
Exercises 8.1.2
1. Consider the two-point boundary value problem
X 00 + X = 0,
X(0) = 0,
X 0 (3) = 0
(a) Find all positive eigenvalues and corresponding eigenfunctions of the boundary value
problem.
(b) Is = 0 an eigenvalue of this problem? If yes, find its corresponding eigenfunction. If
no, briefly explain why it is not an eigenvalue.
2. Find all eigenvalues and corresponding eigenfunctions of the boundary value problem
X 00 + X = 0,
X 0 (0) = 0,
X 0 (⇡) = 0.
Make sure to consider, and show your work for, all three possibilities:
> 0.
< 0,
= 0, and
3. Consider the 2-point boundary value problem:
X 00 + ⇡ 2 X = 0,
Is
=
X 0 (1) = 0.
X(0) = 0,
1
an eigenvalue? If not, justify your answer. If so, find a corresponding eigenfunction.
4
4. (a) For what positive values of
does the boundary value problem
X 00 + X = 0,
X 0 (0) = 0,
X(⇡) = 0
have a nontrivial solution?
(b) What are the corresponding eigenfunctions?
5. Find the eigenvalues and eigenfunctions of the boundary value problem
X 00 + X = 0,
Show your work in all three cases:
< 0,
X(0) = 0,
= 0, and
⇡
X( ) = 0.
2
> 0.
Answers
1. (a) The positive eigenvalues are n = (2n 361)
functions are Xn (x) = sin( 2n6 1 ⇡x) (b) No.
2
⇡2
, n = 1, 2, 3, . . . ; The corresponding eigen-
2. Eigenvalues are
= 0 and
= n2 , n = 1, 2, 3, . . .; Eigenfunctions are X0 (x) = 1, and
Xn (x) = cos(nx), n = 1, 2, 3, . . .
3. Yes, it is. Eigenfunction is sin
4. (a) Eigenvalues are
n
=
⇡
x.
2
(2n 1)2
,
4
(b) eigenfunctions are Xn (x) = cos( (2n 2 1)x ), for all n 2 N.
5. Eigenvalues are n = 4n2 , n = 1, 2, 3, . . . and the corresponding eigenfunctions are Xn (x) =
sin(2nx), n = 1, 2, 3, . . .
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.2
285
Fourier Series
A function f is said to be periodic with period T > 0 if
f (x + T ) = f (x)
for every value of x.
• Notice here that a period of T is not unique, because if T is a period of f , then also 2T and
so indeed is any integral multiple of f .
• We call the smallest value of T such that f (x + T ) = f (x), the fundamental peiod of f .
• The fundamental period does not necessarily exist but if it does, it is unique.
• Moreover the sum of any finite terms or even the sum of a convergent infinite series of
functions of period T is also periodic with period T .
Example 213.
• sin x, cos x are periodic functions with fundamental period 2⇡.
• Constant function is a periodic function with an arbitrary period but no fundamental
period.
A function f is said to be piecewise continuous on an interval a  x  b if the interval can be
partitioned by a finite number of point a = x0 < x1 < . . . < xn = b so that f is continuous on
each open subinterval xi 1 < x < xi for all i.
8.2.1
The Fourier Convergence Theorem
Next let’s consider the most important theorem in this chapter involving Fourier series1 .
1 Fourier series are named for Jean Baptiste Joseph Fourier (1768
1830). Fourier introduced this series for
the purpose of solving the heat equation. His work on the theory of heat began around 1804 and by 1807 he had
completed his important memoir On the Propagation of Heat in Solid Bodies. The memoir was read to Institut de
France on 21 December 1807. This was so promising that the Academy encouraged Fourier to continue by setting
a contribution to the mathematical theory of heat as its problem for the Grand Prize in 1812. Fourier won the
prize, but not without some criticism which he resented deeply but which was well taken.
Committee consisting of Lagrange, Laplace, Monge and Lacroix was set up to report on the work. The objection,
made by Lagrange and Laplace in 1808, was to Fourier’s expansions of functions as Fourier series. While admitting
the novelty and importance of Fourier’s work they pointed out that the mathematical treatment was not rigorous;
the later clarification of Fourier’s ideas was to some extent the reason that the nineteenth century came to be called
the age of rigor. This paper did not appear in print until 1822, reworked into book entitled Thorie analytique de
la chaleur (The Analytic Theory of Heat). It was one of the great scientific works of all time.
It was not until many years later that Fourier’s ideas were finally put on a rigorous and mathematically reliable
footing. Certainly Johann Peter Gustav Lejeune Dirichlet (1805 1859) and Georg Friedrich Bernhard Riemann
(1826 1866) played key roles in working out some of the basic ideas of Fourier series. The Fourier series has many
such applications in electrical engineering, vibration analysis, acoustics, optics, signal processing, image processing,
quantum mechanics, econometrics, thin-walled shell theory, etc.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
Theorem 15 (The Fourier Convergence Theorem).
286
Suppose that f and f 0 are piecewise
continuous on the interval L  x < L. Further suppose that f is defined outside the interval
L  x < L so that it is periodic with period 2L, i.e., f (x+2L) = f (x). Then f has a Fourier
series, denoted by F ,
F (x) =
1
X
a0
m⇡x
m⇡x
+
(am cos
+ bm sin
),
2
L
L
m=1
whose coefficients are given by Euler-Fourier Formulas,
a0 =
am =
bm =
1
L
1
L
1
L
Z
Z
Z
L
f (x) dx,
L
L
f (x) cos
m⇡x
dx,
L
m = 1, 2, . . . ,
f (x) sin
m⇡x
dx,
L
m = 1, 2, . . . .
L
L
L
The Fourier series converges to f (x) at all points where f is continuous and to
at all points where f is discontinuous. In other words,
8
if f is continuous at x0
<f (x0 ),
F (x0 ) = f (x+ ) + f (x )
.
0
0
:
, if f is discontinuous at x0
2
Recall that
f (x0 ) = lim f (x),
x!x0
f (x+ )+f (x )
2
f (x+
0 ) = lim f (x)
x!x0 +
is the left-hand (right-hand) limit of f at x0 . Let’s consider the following example to recall
ourselves about it.
8
>
< x 2
Example 214. Let f (x) =
1
>
: 2
x +1
if x < 0
if x = 0
if x > 0.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
287
Then we have
f (0) =
1,
f (0 ) =
2
and
f (0+ ) = 1.
Example 215 (Spring 2004, Final Exam). Let
8
>
< 2 x,
f (x) = x,
>
:
2 x,
if
2<x< 1
if
1x1
if 1 < x < 2.
and f (x + 4) = f (x).
Set up, but do not integrate, the integral(s) to find the Fourier coefficients of this function.
Solution: With L = 2, for m = 0, 1, 2, . . ., we have
Z
1 2
m⇡x
am =
f (x) cos
dx
2 2
2
✓Z 1
Z 1
Z 2
1
m⇡x
m⇡x
=
( 2 x) cos
dx +
x cos
dx +
(2
2
2
2
2
1
1
and, for m = 1, 2, . . . ,
Z
1 2
m⇡x
f (x) sin
dx
bm =
2 2
2
✓Z 1
Z
1
m⇡x
=
( 2 x) sin
dx +
2
2
2
1
x sin
1
m⇡x
dx +
2
Z
◆
m⇡x
x) cos
dx ,
2
2
(2
x) sin
1
◆
m⇡x
dx .
2
Example 216 (Fall 2003, Final Exam).
Consider the Fourier series of function given below:
(
x2 + 1, if
2 < x < 12
f (x) =
1
sin 2⇡x, if 2  x < 2
and f (x + 4) = f (x).
To what value does the Fourier series converges to at x = 12 .
Solution Notice that f is discontinuous at x =
Fourier series of f converges to
1
2.
By the Fourier convergence theorem,
+
sin 2⇡( 12 ) + ( 12 )2 + 1
f ( 12 ) + f ( 12 )
sin ⇡ +
=
=
2
2
2
at x = 12 .
The following trig identities are very helpful in this topic:
sin(m⇡) = 0
for all integer m
sin( m⇡) = 0
cos(m⇡) = ( 1)
for all integer m
m
cos( m⇡) = ( 1)m
for all integer m
for all integer m.
5
4
=
5
,
8
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
288
Example 217 (Spring 2005, Final Exam).
Consider the function
8
>
<0
f (x) = 3
>
:
0
if
2x<0
if 0  x < 1
if 1  x  2.
a) Sketch the graph of f for three periods.
b) Find the Fourier series of f (x) on [ 2, 2].
Solution Note that L = 2. Then
Z
Z
1 2
1 1
3
a0 =
f (x) dx =
3 dx = (x)
2 2
2 0
2
x=1
=
x=0
3
.
2
And for any positive integer m, we have
Z
Z
Z
1 2
m⇡x
1 1
m⇡x
3 1
m⇡x
am =
f (x) cos
dx =
3 cos
dx =
cos
dx
2 2
2
2 0
2
2 0
2
✓
◆ x=1
x=1
3
2
m⇡x
3 ⇣
m⇡x ⌘
3
m⇡
=
sin
=
sin
=
sin
.
2 m⇡
2
m⇡
2
m⇡
2
x=0
x=0
Also,
Z
Z
m⇡x
1 1
m⇡x
3 1
m⇡x
dx =
3 sin
dx =
sin
dx
2
2 0
2
2 0
2
2
✓
◆ x=1
x=1
3
2
m⇡x
3 ⇣
m⇡x ⌘
=
cos
=
cos
2
m⇡
2
m⇡
2
x=0
x=0
⇣
⌘
⇣
⌘
3
m⇡
3
m⇡
=
cos
1 =
1 cos
.
m⇡
2
m⇡
2
bm =
1
2
Z
2
f (x) sin
So the Fourier series is
◆
1 ✓
3 X
3
m⇡
m⇡x
3 ⇣
+
sin
cos
+
1
4 m=1
m⇡
2
2
m⇡
cos
m⇡ ⌘
m⇡x
sin
.
2
2
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
289
c) Find value that the Fourier series in a) converge at x = 9.
Solution Since f is a periodic function of period 4, so is its Fourier series. So the the
value that the Fourier series converge at x = 9 is the same one at x = 1. Since at x = 1,
f is discontinuous. By the Fourier convergence theorem, its Fourier series converge to
f (1+ ) + f (1 )
0+3
3
=
= .
2
2
2
d) Sketch the graph of the function of which the series converges for three periods.
Example 218 (Fall 2007, Final Exam).
Let f be the periodic function with period 2⇡ such that
(
2
if
⇡x<0
f (x) =
2 if 0  x < ⇡.
a) Sketch the graph of the given function of three periods.
b) Find the Fourier series of the function f .
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
290
Solution We have L = ⇡. Then
✓Z 0
◆
Z
Z ⇡
1 ⇡
1
a0 =
f (x) dx =
2 dx +
2 dx
⇡
⇡
⇡
⇡
0
⌘
1⇣ 0
1
⇡
=
2x ⇡ + ( 2x) 0 = [(0 ( 2⇡)) + (( 2⇡)
⇡
⇡
and for any positive inter m, we have
✓Z 0
Z
Z ⇡
1 ⇡
m⇡x
1
am =
f (x) cos
dx =
2 cos mx dx +
⇡
⇡
⇡
⇡
⇡
0
✓
◆
1 2
2
0
⇡
=
sin mx ⇡ + (
sin mx) 0 = 0,
⇡ m
m
0)] = 0
2 cos mx dx
◆
and also,
✓Z 0
◆
Z
Z ⇡
1 ⇡
m⇡x
1
f (x) sin
dx =
2 sin mx dx +
2 sin mx dx
⇡
⇡
⇡
⇡
⇡
0
✓
◆
1
2
2
0
⇡
=
cos mx ⇡ +
cos mx 0
⇡
m
m
✓
◆
✓
◆
1
2
2
2
2
1
4
4
=
(
cos( m⇡)) +
cos m⇡
=
+
cos(m⇡)
⇡
m
m
m
m
⇡
m m
8
✓
◆ <0
if m is even
1
4
4
m
=
+ ( 1)
=
8
:
⇡
m m
if m is odd.
m⇡
bm =
Therefore the Fourier series of f is
✓
◆
1
X
1
4
4
m
f (x) =
+ ( 1)
sin mx
⇡
m m
m=1
=
1
X
m=1
m odd
1
X
8
sin mx =
m⇡
m=1
8
(2m
1)⇡
sin(2m
1)x.
c) To what values does the Fourier series in a) converges at x = ⇡?
Solution At x = ⇡, f is discontinuous. By the Fourier convergence theorem, its Fourier
series converge to
f (⇡ + ) + f (⇡ )
2 + ( 2)
=
= 0,
2
2
at x = ⇡.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
291
d) Sketch the graph of the function of which the series converges for three periods.
Just because a Fourier series could have infinitely many terms does not mean that it will always
have that many terms. If a periodic function f can be expressed by finitely many terms normally
found in a Fourier series, then f must be the Fourier series of itself.
Example 219.
1. The Fourier series (period 2⇡) of f (x) = 5 + cos 4x
sin 5x is f itself.
2. Let f be the periodic function 6 cos x sin x with period 2⇡. Even though the product
cos x sin x is not a term in a Fourier series, we can use the double angle formula to get
6 cos x sin x = 3 sin 2x.
So the Fourier series is f itself.
Example 220 (Fall 2005 Final Exam).
Which of the following already has the form of a Fourier series on the interval [ 2, 2]?
f (x) = 4 sin
⇡
x,
3
g(x) =
1
+ 2 cos 3⇡x.
4
Solution Here L = 2. Note that the angle of trig functions in the Fourier series is
m⇡x
m⇡x
=
.
L
2
For function f , the angle is
⇡x
. If it is already has the form of a Fourier series, then
3
⇡x
m⇡x
=
3
2
)
m=
2
,
3
which is not an integer. So f is not in the form of Fourier series yet. On the other hand, for
g, we have
m⇡x
3⇡x =
) m = 6.
2
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
292
So g is already in the form of Fourier series.
8.2.2
Even and Odd Functions and their Fourier Series
Even Function vs Odd Function
• An even function is any function f such that
f (x) = f ( x),
for all x in its domain. Common examples of even functions are
cos x,
|x|,
x 2 , x4 , . . . , x
2
4
,x
, . . . (or any xm where m is even integer)
• An odd function is any function f such that
f (x) =
f ( x),
for all x in its domain. Common examples of odd functions are
sin x,
x, x3 , . . . , x
1
3
,x
, . . . (or any xm where m is odd integer)
• Unlike integer which is either even or odd, most functions are neither even nor odd. For
instance,
f (x) = ex .
• Also there is only one function which is both even and odd, namely zero function. This is
the case since, if f is both even and odd, then
f (x) = f ( x) =
)
f (x)
2f (x) = 0
)
f (x) = 0.
The first equality holds since f is an even and the second one is true since f is an odd
function.
Properties of Even and Odd Functions
• Elementary properties of even and odd functions are displayed in the following table:
+/
⇥/÷
Even & Even
Even
Even
Odd & Odd
Odd
Even
Even & Odd
Neither
Odd
Functions
• The integral properties of even and odd functions:
– If f is an even function, then
Z
– If f is an odd function, then
L
f (x) dx = 2
L
Z
Z
L
f (x) dx.
0
L
f (x) dx = 0.
L
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
293
• By using this property, we are able to evaluate the integration faster, for examples,
Z 2
x3 dx = 0
Z
2
8
x6 dx = 2
8
Z
8
x6 dx.
0
since x3 is an odd function while x6 is an even function.
Fourier Series of Even and Odd Functions
• Now we will apply what we learn about even and odd functions to help us find Fourier series
faster. Suppose now that f is an even periodic function with period 2L. Then, by using the
above table, we have
m⇡x
is even and
L
m⇡x
f (x) sin
is odd.
L
f (x) cos
So
am
1
=
L
bm
1
=
L
Z
Z
L
m⇡x
2
f (x) cos
=
L
L
L
L
m⇡x
f (x) sin
= 0,
L
L
Z
L
f (x) cos
0
m⇡x
dx,
L
for all m. Hence the Fourier series of f is
1
X
a0
m⇡x
+
am cos
.
2
L
m=1
Since cosine is the only trig function that appears in the Fourier series, we call it the Fourier
cosine series.
• Now let f be an odd periodic function with period 2L. Then, by using the above table, we
have
m⇡x
is odd and
L
m⇡x
f (x) sin
is even.
L
f (x) cos
Hence
am
1
=
L
bm =
1
L
Z
Z
L
f (x) cos
m⇡x
= 0,
L
f (x) sin
m⇡x
2
=
L
L
L
L
L
Z
L
f (x) sin
0
for all m. Hence the Fourier series of f is
1
X
m=1
bm sin
m⇡x
.
L
We call this Fourier series the Fourier sine series.
m⇡x
dx,
L
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
294
We summarize what we have done here:
The Fourier Cosine Series: If f is an even periodic function with period 2L, then
1
X
a0
m⇡x
+
am cos
,
2
L
m=1
is its Fourier cosine series where
Z
2 L
a0 =
f (x) dx,
L 0
am =
2
L
Z
L
f (x) cos
0
m⇡x
dx,
L
m = 1, 2, . . .
The Fourier Sine Series: If f is an odd periodic function with period 2L, then
1
X
bm sin
m=1
m⇡x
,
L
is its Fourier sine series where
bm =
2
L
Z
L
f (x) sin
0
m⇡x
dx,
L
m = 1, 2, . . .
Example 221 (Fall 2003 Final Exam).
Find the Fourier series for the following function
f (x) = ⇡ 2
x2 ,
⇡  x  ⇡,
f (x + 2⇡) = f (x).
Solution Here L = ⇡. Note here that f is an even function because
f ( x) = ⇡ 2
( x)2 = ⇡ 2
x2 = f (x),
for all x. By the property of an even function, we have bm = 0 for all m. Also
Z
x=⇡
2 ⇡ 2
2
x3
2
⇡3
4
a0 =
(⇡
x2 ) dx = (⇡ 2 x
)
= (⇡ 3
) = ⇡2 ,
⇡ 0
⇡
3 x=0
⇡
3
3
and for m = 1, 2, . . . , by using integration-by-parts,
Z
2 ⇡ 2
am =
(⇡
x2 ) cos mx dx
⇡ 0
✓
◆ x=⇡
2
sin mx
cos mx
sin mx
=
(⇡ 2 x2 )
2x
+
2
⇡
m
m2
m3
x=0
2
cos m⇡
4
4
4
m
= ( 2⇡
)=
cos m⇡ =
( 1) = ( 1)m+1 2 .
⇡
m2
m2
m2
m
So the Fourier cosine series of f is
1
2 2 X
4
⇡ +
( 1)m+1 2 cos mx.
3
m
m=1
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
295
Example 222 (Spring 2007 Final Exam). Consider the following periodic functions
a(x) = x,
2
a(x + 4) = a(x)
⇡  x < ⇡,
c(x + 2⇡) = c(x)
1  x < 1,
b(x) = x ,
c(x) = 1 + cos x,
b(x + 2) = b(x)
⇡  x < ⇡,
d(x) = 1 + sin x,
e(x) = x
2  x < 2,
3
d(x + 2⇡) = d(x)
2  x < 2,
x ,
e(x + 4) = e(x).
a) List all functions that can be expressed as a Fourier sine series.
Solution a is clearly an odd function. So is e since
e( x) = ( x)
( x)3 =
x
( x3 ) =
(x
x3 ) =
e(x),
for all x. Therefore both a and e can be expressed as a Fourier sine series.
b) List all functions that can be expressed as a Fourier cosine series.
Solution b is obviously an even function. So is c since
c( x) = 1 + cos( x) = 1 + cos x = c(x),
for all x. Then both b and c can be expressed as Fourier cosine series.
Note here that d is neither even nor odd function, since
d( x) = 1 + sin( x) = 1
is neither d(x) = 1 + sin x nor
d(x) =
1
sin x.
sin x
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.2.3
296
The Cosine and Sine Series Extensions
If f and f 0 are piecewise continuous functions defined on the interval 0  t  L then f can be
expressed into an even(odd) periodic function, f˜, of period 2L, such that f (x) = f˜(x) on [0, L]
and whose Fourier series is therefore a cosine(sine) Fourier series, respectively.
Even (cosine series) extension of f (x)
Let f be defined on [0, L]. Then its even extension of period 2L is
(
f (x)
if 0  x  L
˜
f (x) =
, f˜(x + 2L) = f˜(x),
f ( x) if
Lx<0
where the Fourier cosine series is
1
X
a0
m⇡x
+
am cos
,
2
L
m=1
and
a0 =
2
L
Z
L
f (x) dx,
am =
0
2
L
Z
L
f (x) cos
0
m⇡x
dx,
L
m = 1, 2, . . .
Odd (sine series) extension of f (x) If f is defined on (0, L). Then its odd extension of
period 2L is
8
>
<f (x)
˜
f (x) = 0
>
:
f ( x)
where the Fourier sine series is
if 0 < x < L
,
if x = 0, L
if
L<x<0
1
X
bm sin
m=1
and
bm =
2
L
Z
L
f (x) sin
0
f˜(x + 2L) = f˜(x),
m⇡x
,
L
m⇡x
dx,
L
m = 1, 2, . . .
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
297
Example 223 (Fall 2001 Final Exam). Consider the following function, define for 0  x  3:
(
1 ,0  x  2
f (x) =
.
2
,2 < x  3
a) Graph the even extension of f (x) on the interval
b) Graph the odd extension of f (x) on the interval
3  x  3.
3  x  3.
Example 224 (Summer 2003 Final Exam). Let
f (x) =
(
1
x
1
,0  x < 2
.
,2  x  3
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
298
a) Sketch the even, period 6 extension of f (x) on the interval [ 9, 9].
b) Sketch the odd, period 6 extension of f (x) on the interval [ 9, 9].
c) Set up, but do not evaluate, the integral(s) that will give the Fourier sine coefficients of
the odd extension.
Solution We know that for Fourier sine series, a0 and all am are zero. For m =
1, 2, . . .,
Z
2 3
m⇡x
bm =
f (x) sin
dx
3 0
3
✓Z 2
◆
Z 3
2
m⇡x
m⇡x
=
(1 x) sin
dx +
( 1) sin
dx .
3
3
3
0
2
d) To what value does the sine series above converge to at x =
3, x = 0 and x = 2?
Solution f is discontinuous at x = 3 and x = 0 and continuous at x = 2. So by
Fourier Convergence Theorem, we have
f ( 3+ ) + f ( 3 )
1 + ( 1)
=
=0
2
2
f (0+ ) + f (0 )
1 + ( 1)
F (0) =
=
=0
2
2
F (2) = f (2) = 1.
F ( 3) =
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
299
Example 225 (Spring 2003 Final Exam). Consider the following function, defined on the
interval (0, 2).
(
x ,0  x < 1
f (x) =
.
2 ,1  x  2
a) Graph the even, period 4, extension of f (x) on [ 4, 4].
b) Graph the odd, period 4, extension of f (x) on [ 4, 4].
c) Which of the above two has a cosine series, and which has a sine series?
Solution The extension in Part a) has a cosine series because it is an even function
while the one in Part b) gives a sine series since it is an odd one.
d) What does the Fourier series representing part b) (the odd extension) converge to at
x = 2, x = 12 , and x = 3?
Solution f is discontinuous at x = 2 and x = 3 and continuous at x =
Fourier Convergence Theorem, we have
f ( 2+ ) + f ( 2 )
2+2
=
=0
2
2
1
1
1
F( ) = f( ) =
2
2
2
f (3+ ) + f (3 )
1 + ( 2)
3
F (3) =
=
=
.
2
2
2
F ( 2) =
1
2.
So by
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
300
Exercises 8.2
1. True or false:
(a) The constant term of the Fourier series representing f (x) = x2 , 2 < x < 2, f (x + 4) =
a0
4
f (x), is
= .
2
3
(b) The Fourier series (of period 2⇡) representing f (x) = 3 7 sin2 (x) is a Fourier sine
series.
(c) The Fourier series of
f (x) = 3x2
4 cos 2x,
⇡ < x < ⇡,
f (x + 2⇡) = f (x)
is a cosine series.
(d) Every Fourier sine series converges to 0 at x = 0.
(e) Every Fourier sine series has 0 as its constant term.
(f) The constant term of the Fourier series representing
f (x) = 2x3 ,
1 < x < 1,
f (x + 2) = f (x),
is zero.
2. Consider the Fourier series (of period 2⇡) representing
5 cos(4x) + 2 sin2 (6x).
f (x) = 13
Which statement below is true? (Hint: Is f (x) an even or an odd function?)
(a) The Fourier series is a cosine series.
(b) The Fourier series is a sine series.
(c) The Fourier series is neither a cosine series nor a sine series.
(d) The function does not have a Fourier series because it is not periodic.
3. Consider the Fourier series (of period 20) representing
f (x) = x5 ,
10 < x < 10,
f (x + 20) = f (x).
Which statement below is true?
(a) The Fourier series is a cosine series.
(b) The Fourier series is a sine series.
(c) The Fourier series is neither a cosine series nor a sine series.
(d) The function does not have a Fourier series because it is not periodic.
4. Consider the Fourier series (of period 6⇡) representing
f (x) = 5x3 + 7,
3⇡ < x < 3⇡,
f (x + 6⇡) = f (x).
Which statement below is true?
(a) The Fourier series is a cosine series.
(b) The Fourier series is a sine series.
(c) The Fourier series is neither a cosine series nor a sine series.
(d) The function does not have a Fourier series because it is not periodic.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
301
5. Find the Fourier cosine coefficient corresponding to n = 2, a2 , of the Fourier series of period
T = 2⇡ representing the function f (x) = 6 cos 2x.
(a) a2 = 0
(b) a2 = 6
3
(c) a2 =
⇡
(d) a2 =
1
⇡
6. Consider the 2⇡-periodic function
f (x) = |x|, when
⇡ < x < ⇡, and f (x + 2⇡) = f (x).
Find the fourth sine coefficient b4 of the Fourier series.
(a)
(b)
1
2.
1
4.
(c) 0.
(d)
1
2.
7. Find the Fourier sine coefficient corresponding to n = 6, b6 , of the Fourier series of period
T = 4⇡ representing the function
(
1,
2⇡ < x < 0
f (x) =
, f (x + 4⇡) = f (x).
3, 0 < x < 2⇡
(a) b6 = 0.
4
(b) b6 =
.
5⇡
3
(c) b6 =
.
2⇡
3
(d) b6 =
.
4⇡
8. Let
f (x) =
(
2,
1,
0<x<1
1x2
(a) Sketch the odd, period 4 extension of f (x) on the interval
(b) Sketch the even, period 4 extension of f (x) on the interval
(c) Find the Fourier series of the even extension.
(d) What does the series above converge to when x =
9. Let f (x) = 2 + sin x,
6  x  6.
6  x  6.
2, x = 0 and x = 1?
0 < x < ⇡.
(a) Consider the odd periodic extension, of period T = 2⇡, of f (x). Sketch 3 periods, on
the interval 3⇡ < x < 3⇡, of this odd periodic extension.
(b) Which of the integrals below can be used to find the Fourier sine coefficients of the odd
periodic extension in a)?
Z ⇡
1
i.
(2 + sin x) sin nx dx
2⇡
⇡
Z
2 ⇡
ii.
(2 + sin x) sin nx dx
⇡ 0
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
302
Z ⇡
1
nx
iii.
(2 + sin x) sin
dx
2⇡ 0
2
Z ⇡
2
nx
iv.
(2 + sin x) sin
dx
⇡
2
⇡
(c) To what value does the Fourier series of this odd periodic extension converge at x = ⇡?
3⇡
At x =
?
2
(d) Consider the even periodic extension, of period T = 2⇡ of f (x). Sketch 3 periods, on
the interval 3⇡ < x < 3⇡, of this even periodic extension.
(e) To what value does the Fourier series of this even periodic extension converge at x = ⇡?
10. Let f (x) = 2
x,
0 < x < 2.
(a) Consider the odd periodic extension, of period T = 4, of f (x). Sketch 3 periods, on the
interval 6 < x < 6, of this odd periodic extension.
(b) True of False: The Fourier sine coefficients of the odd periodic extension in (a) can be
found by
✓Z 0
◆
Z 2
1
n⇡x
n⇡x
bn =
( 2 x) sin
dx +
(2 x) sin
dx
2
2
2
2
0
(c) To what value does the Fourier series of this odd periodic extension converge at x =
At x = 4?
1?
(d) Consider the even periodic extension, of period T = 4 of f (x). Sketch 3 periods, on the
interval 6 < x < 6, of this even periodic extension.
(e) Find the constant term of the Fourier series of the periodic function described in (d).
(f) To what value does the Fourier series of this even periodic extension converge at x = 4?
Answers
1. (a) T, (b) F, (c) T, (d) T, (e) T, (f) T
2. (a)
3. (b)
4. (c)
5. (b)
6. (c)
7. (a)
8. (c) F (x) =
1
X
3
2
m⇡
m⇡x
+
sin
cos
, (d) At x =
2 m=1 m⇡
2
2
converges to 2. At x = 1, it converges to
9. (b) ii), (c) At x =
10. (b) T, (c)
2, it converges to 1. At x = 0, it
3
2
⇡, it converges to 0. At x = 32 ⇡, it converges to
1 and 0, e() 1, (f) 2
3, (e) 2.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.3
303
Heat Equations & Separation of Variables
Consider a straight bar of uniform cross section and homogeneous material. Let the x-axis be
chosen to lie along the axis of the bar, and let x = 0 and x = L denote the ends of the bar.
Suppose further that the sides of the bar are perfectly insulated so that no heat passes through
them. We also assume that the cross sectional dimension are so small that the temperature can
be considered constant on any given crosse section. Then the temperature distribution U is a
function only of the axial coordinate x and the time t.
Figure 8.1: A heat-conducting solid bar.
The equation describing this function u(x, t) is
↵2 uxx = ut ,
where ↵2 is a constant known as the thermal di↵usivity. This equation is called heat conduction equation or heat equation. The constant ↵2 depends only on material from which bar is
made. Typical values of ↵2 are given in the table below.
Material
↵2 (cm2 /s)
Silver
1.71
Copper
1.14
Aluminum
0.86
Cast iron
0.12
Granite
0.011
Brick
0.0038
Water
0.00144
Table 8.1: Values of the Thermal Di↵usivity for Some Common Materials
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.3.1
304
Type I: Homogeneous Boundary Conditions & Separation of Variables
In addition, we assume that the initial temperature distribution in the bar is given
u(x, 0) = f (x),
0 < x < L.
Assume that the ends of the bar are held at zero temperature at all time, so
u(0, t) = 0,
u(L, t) = 0,
t > 0.
So the following is the problem we would like to solve
Heat Conduction Problem Type I: Homogeneous Boundary Conditions
(Heat equation)
↵2 uxx = ut ,
(Initial condition)
u(x, 0) = f (x),
(Boundary conditions)
u(0, t) = 0,
u(L, t) = 0,
0  x  L, t > 0
0 < x < L,
t > 0.
We call this type of problems which contain both initial condition(s) and boundary conditions an
initial-boundary value problem or IBVP. Alternatively, we can consider this problem in the
xt-plane as shown below. The solution u(x, t) is sought in the semi-infinite strip 0 < x < L, t > 0,
subject to the requirement that u(x, t) must assume a prescribed value at each point on the
boundary of this strip.
Figure 8.2: Boundary value problem for the heat conduction equation.
The method we’ll use to solve this heat equation is called the separation of variables. Assume
that u(x, t) can be written as a product of two functions, one depending only on x and the other
depending only on t. So we write
u(x, t) = X(x)T (t).
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
305
Example 226 (Spring 2010 Final Exam). True or false.
a) It is possible to separate the PDE,
x3 utt
t3 uxx = 0,
into two ODEs.
Solution Let u(x, t) = X(x)T (t). Then utt = XT 00 and uxx = X 00 T . Then we have
x3 utt
t3 uxx = 0
)
x3 XT 00
t3 X 00 T = 0
)
T 00
X 00
=
t3 T
x3 X
x3 XT 00 = t3 X 00 T
)
So the statement is true since the given PDE is separable.
b) It is possible to separate the PDE,
uxx
2utx + utt = 0,
into two ODEs.
Solution Let u(x, t) = X(x)T (t). Then uxx = X 00 T, utx = X 0 T 0 and utt = XT 00 . So
we have
uxx 2utx + utt = 0 ) X 00 T 2X 0 T 0 + XT 00 = 0.
Since there is no common term in each pair of the expression, we can’t separate this
PDE. So the argument is false.
Example 227 (Spring 2009 Final Exam). True or false.
It is possible to separate the PDE,
uxx
2utt = 3ut ,
into two ODEs.
Solution Let u(x, t) = X(x)T (t). Then uxx = X 00 T , utt = XT 00 and ut = XT 0 . Then we
have
uxx
2utt = 3ut
)
X 00 T
2XT 00 = 3XT 0
)
X 00 T = (3T 0 + 2T 00 )X
00
)
X T = 3XT 0 + 2XT 00
)
X 00
3T 0 + 2T 00
=
.
X
T
So the statement is true since the given PDE is separable.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
306
Example 228 (Spring 2004 Final Exam).
Separate the PDE,
2utx + 5x3 ut = 0
uxx
into a system of two equations of one independent variable each
Solution Again using the method of separation of variables by writing u(x, t) = X(x)T (t).
Then uxx = X 00 T, utx = X 0 T 0 and ut = XT 0 . Then we have
uxx
2utx + 5x3 ut = 0
)
X 00 T
2X 0 T 0 + 5x3 XT 0 = 0
)
X 00 T = 2X 0 T 0
)
X 00 T = (2X 0
)
2X 0
)
X 00 + 2 X 0
)
T 0 + T = 0,
5x3 XT 0
5x3 X)T 0
X 00
T0
=
=
3
5x X
T
So we get
2X 0
X 00
=
5x3 X
T0
=
T
5x3 x = 0
as two equations of one independent variable.
Example 229 (Summer 2014 Final Exam ).
Consider the partial di↵erential equation with its boundary conditions,
uxt
uxx
u(0, t) = 0,
ut = 0,
ux (5, t) = 0.
a) Use the technique of separation of variables to change the above PDE into two ordinary
di↵erential equations. Namely, let u(x, t) = X(x)T (t) and find the ordinary di↵erential
equations X(x) and T (t) satisfy.
Solution Let u(x, t) = X(x)T (t). Then uxt = X 0 T 0 , uxx = X 00 T and ut = XT 0 . So the
new equation is
uxt
uxx
ut = 0
)
X 0T 0
X 00 T
X 0T 0
XT 0 = X 00 T
)
(X 0
)
X0
)
X 00 + X 0
)
T 00 + T = 0
)
XT 0 = 0
X)T 0 = X 00 T
X 00
X
=
T0
=
T
Then we have
X 00
X0
=
X
T0
=
T
X=0
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
307
b) What new boundary conditions must the equation of X(x) satisfy after using the technique of separation of variables?
Solution Again we use the technique of separation of variables, u(x, t) = X(x)T (t),
u(0, t) = 0
ux (5, t) = 0
)
)
X(0)T (t) = 0
0
X (5)T (t) = 0
)
X(0) = 0 or T (t) = 0
X 0 (5) = 0 or T (t) = 0.
)
So the new boundary conditions are X(0) = 0 and X 0 (5) = 0.
c) True or false. If new boundary conditions are u(0, t) = 1 and ux (5, t) = 1, then it
can be rewritten as X(0) = 1 and X 0 (5) = 1 after using the technique of separation of
variables.
Solution This is a false statement.
u(0, t) = 1
ux (5, t) = 1
)
)
X(0)T (t) = 1. It is not true that X(0) = 1 or T (t) = 1
X 0 (5)T (t) = 1. It is not true that X 0 (5) = 1 or T (t) = 1.
Example 230 (Fall 2009 Final Exam ).
Let u(x, y) be the solution of the Laplace’s equation
uxx + uyy = 0,
u(x, 0) = 0,
0 < x < 7,
uy (x, 3) = 0,
0 < y < 3,
u(0, y) = 0,
u(7, y) = sin(⇡y).
Suppose it has the following form u(x, y) = X(x)Y (y). Then X(x) or Y (y) satisfies one of
the following pairs of boundary conditions. Find the pair.
(a) X(0) = 0, X(7) = 0.
(b) X(0) = 0, X(3) = 0.
(c) Y (0) = 0, Y 0 (7) = 0.
(d) Y (0) = 0, Y 0 (3) = 0.
Solution Suppose u(x, y) = X(x)Y (y), then
u(x, 0) = 0
uy (x, 3) = 0
u(0, y) = 0
)
)
)
u(7, y) = sin(⇡y)
X(x)Y (0) = 0
X(x)Y 0 (3) = 0
X(0)Y (y) = 0
)
)
)
)
X(x) = 0 or Y (0) = 0
X(x) = 0 or Y 0 (3) = 0
X(0) = 0 or Y (y) = 0
X(7)Y (y) = sin(⇡y)
)
No conclusion.
We avoid X(x) = 0 and Y (y) = 0 since they give us trivial solution u(x, y). In the end, we
have Y (0) = 0, Y 0 (3) = 0 and X(0) = 0. So (d) is the correct answer.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
308
Now let’s comeback to our Heat problem. Write u(x, t) = X(x)T (t) by the separation of variables
method. From this we have uxx = X 00 T, ut = XT 0 . Substituting these into the heat equation,
↵2 uxx = ut
↵2 X 00 T = XT 0
)
)
X 00
T0
= 2 =
X
↵ T
.
We put negative sign here because it makes our life easier later. Next step is to consider them
separately,
X 00
=
X
T0
=
↵2 T
)
X 00 + X = 0.
)
T 0 + ↵2 T = 0.
Now let’s take a look at the boundary conditions. By using u(x, t) = X(x)T (t), we have
u(0, t) = 0
u(L, t) = 0
)
)
X(0)T (t) = 0
X(L)T (t) = 0
)
X(0) = 0 or T (t) = 0
)
X(L) = 0 or T (t) = 0
If T (t) = 0, both conditions would be satisfied but u(x, t) = X(x)T (t) = 0 would be a trivial
function which is totally uninteresting solution. Hence we have to let the new boundary conditions
to be
X(0) = 0, X(L) = 0.
For now, after using heat equation and boundary conditions, we transform the problem to
X 00 + X = 0,
0
T +↵
2
X(0) = 0,
X(L) = 0
(8.10)
T = 0.
(8.11)
Recall that we solved BVP (8.10) before in Example 209, the eigenvalues are
n
=(
n⇡ 2
) ,
L
and the corresponding eigenfunctions are
Xn (x) = sin
n⇡x
,
L
where n = 1, 2, . . ..
n⇡
Now substitute n = ( )2 into Equation (8.11),
L
T 0 + ↵2 T = 0
)
T0 +
↵ 2 n2 ⇡ 2
T =0
L2
)
Tn (t) = exp(
↵ 2 n2 ⇡ 2 t
),
L2
n = 1, 2, . . .
Thus, for each n = 1, 2, . . .,
un (x, t) = Xn (x)Tn (t) = exp(
↵ 2 n2 ⇡ 2 t
n⇡x
) sin
.
L2
L
Then functions {un } are sometimes called fundamental solutions of the heat conduction problem. Then the linear combination of un for the general solution of the heat equation,
u(x, t) =
1
X
cn un (x, t) =
n=1
where cn are as yet undetermined.
1
X
n=1
cn exp(
↵ 2 n2 ⇡ 2 t
n⇡x
) sin
,
2
L
L
(8.12)
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
309
To find the value of cn , apply initial condition u(x, 0) = f (x) into Equation (8.12),
f (x) =
1
X
cn sin
n=1
n⇡x
.
L
So we need to choose cn so that the series of sine functions converges to f (x). Notice that the
series on the right hand side is just the Fourier sine series for f , so its coefficients are given by
c n = bn =
2
L
Z
L
f (x) sin
0
n⇡x
dx.
L
In conclusion, we get the following result
Heat Conduction Problem Type I: Homogeneous Boundary Conditions
(Heat equation)
↵2 uxx = ut ,
(Initial condition)
u(x, 0) = f (x),
(Boundary conditions)
u(0, t) = 0,
The solution is u(x, t) =
1
X
0  x  L,
0 < x < L,
u(L, t) = 0,
cn exp(
n=1
t > 0.
↵ 2 n2 ⇡ 2 t
n⇡x
) sin
,
L2
L
Z
2 L
n⇡x
where cn =
f (x) sin
dx.
L 0
L
( The steady state is lim u(x, t) = 0.)
t!1
Example 231. Solve
8uxx = ut ,
u(0, t) = 0,
0  x  5,
t>0
u(5, t) = 0,
u(x, 0) = x.
Solution We have ↵2 = 8, L = 5 and f (x) = x. So the general solution is
u(x, t) =
=
1
X
n=1
1
X
n=1
cn exp(
n⇡x
8n2 ⇡ 2 t
) sin
52
5
cn exp(
8n2 ⇡ 2 t
n⇡x
) sin
25
5
To find the particular solution, let’s find cn by using integration-by-parts.
Z
2 5
n⇡x
cn =
x sin
dx.
5 0
5
✓
◆ x=5
2
5
n⇡x
25
n⇡x
=
x cos
+ 2 2 sin
5
n⇡
5
n ⇡
5
x=0
2
25
10
= (
cos n⇡) =
cos n⇡
5 n⇡
n⇡
10
10
=
( 1)n = ( 1)n+1 .
n⇡
n⇡
t>0
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
310
So the particular solution is
u(x, t) =
1
X
( 1)n+1
n=1
10
8n2 ⇡ 2 t
n⇡x
exp(
) sin
.
n⇡
25
5
Example 232 (Spring 2008, Final Exam).
Suppose the temperature distribution function u(x, t) of a rod that has both ends constantly
kept at 0 degree is given by the heat conduction problem
0  x  6,
4uxx = ut ,
u(0, t) = 0,
t>0
u(6, t) = 0,
⇡x
u(x, 0) = 2 sin
+ 4 sin ⇡x
3
10 sin
3⇡x
.
2
a) Find the particular solution of the above initial-boundary value problem.
Solution Here ↵2 = 4 and L = 6. So the general solution is
u(x, t) =
=
1
X
n=1
1
X
cn exp(
4n2 ⇡ 2 t
n⇡x
) sin
2
6
6
cn exp(
n2 ⇡ 2 t
n⇡x
) sin
9
6
n=1
So when t = 0 we have
u(x, 0) =
1
X
cn sin
n=1
n⇡x
.
6
(8.13)
On the other hand, it is given by the initial condition that
u(x, 0) = 2 sin
⇡x
+ 4 sin ⇡x
3
10 sin
3⇡x
2
(8.14)
By Equations (8.13) and (8.14),
1
X
cn sin
n=1
n⇡x
⇡x
= 2 sin
+ 4 sin ⇡x
6
3
10 sin
3⇡x
.
2
To find cn , comparing the angles of the expressions,
⇡x
n⇡x
=
3
6
n⇡x
⇡x =
6
3⇡x
n⇡x
=
2
6
)
n=2
)
c2 = 2
)
n=6
)
c6 = 4
)
n=9
)
c9 =
10.
And cn = 0 where n 6= 2, 6, 9. Substitute all cn back into the general solution to get a
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
particular solution,
u(x, t) =
1
X
cn exp(
n=1
n2 ⇡ 2 t
n⇡x
) sin
9
6
22 ⇡ 2 t
2⇡x
62 ⇡ 2 t
6⇡x
92 ⇡ 2 t
9⇡x
) sin
+ c6 exp(
) sin
+ c9 exp(
) sin
9
6
9
6
9
6
4⇡ 2 t
⇡x
3⇡x
2
2
= 2 exp(
) sin
+ 4 exp( 4⇡ t) sin ⇡x 10 exp( 9⇡ t) sin
.
9
3
2
= c2 exp(
b) What is lim u(x, t)?
n!1
Solution The limit is 0.
311
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
312
Exercises 8.3.1
1. (a) Use separation of variables to rewrite the partial di↵erential equation below into a pair
of ordinary di↵erential equations.
utt + 5u = 4uxx .
(b) Suppose the above partial di↵erential equation has boundary condition ux (0, t) =
0, u(20, t) = 0. Use separations of variables to determine the corresponding boundary conditions that the ordinary di↵erential equations found in (a) must satisfy.
(c) (Yes or no) Could the partial di↵erential equation, uxx
two ordinary di↵erential equations?
2uxt = 5utt , be separated into
2. Consider the following partial di↵erential equation
uyy + 2uy = y 2 uxx
(a) Separate this equation into two ordinary di↵erential equations.
(b) Translate the following boundary conditions on the above partial di↵erential equation
to conditions on the ordinary di↵erential equations found above.
u(0, y) = 0, u(L, y) = 0
3. True or false:
(a) It is possible to separate the partial di↵erential equation, 4uxx = utt + t2 ut , into two
ordinary di↵erential equations.
(b) Using the formula u(x, t) = X(x)T (t), the boundary conditions u(0, t) = 0 and ux (9, t) =
0 can be rewritten as X(0) = 0 and X 0 (9) = 0.
(c) Using the formula u(x, t) = X(x)T (t), the boundary conditions ux (0, t) = 0 and
ux (5, t) = 0 can be rewritten as T 0 (0) = 0 and T 0 (5) = 0.
(d) Using the formula u(x, t) = X(x)T (t), the boundary conditions ux (0, t) = 0 and
u(1, t) = 1 can be rewritten as X 0 (0) = 0 and X(1) = 1.
4. Consider the two linear partial di↵erential equations.
(I)
uxx = uxt
(II)
uxx = 2utt
2ut
x4 u
Use the substitution u(x, t) = X(x)T (t) and attempt to separate each equation into two
ordinary di↵erential equations. Which statement below is true?
(a) Neither equation is separable.
(b) Only (I) is separable.
(c) Only (II) is separable.
(d) Both equations are separable.
5. Consider the two linear partial di↵erential equations.
(I)
(II)
uxx + 4uxt + 4utt = 0
uxx
4utx
5u = 0
Use the substitution u(x, t) = X(x)T (t) and attempt to separate each equation into two
ordinary di↵erential equations. Which statement below is true?
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
313
(a) Neither equation is separable.
(b) Only (I) is separable.
(c) Only (II) is separable.
(d) Both equations are separable.
6. Using the substitution u(x, t) = X(x)T (t), where u(x, t) is not the trivial solution, consider
the two statements below.
(I) The equation uxx = utt + 4u can be separated into two ordinary di↵erential equations.
(II) The boundary conditions ux (0, t) = 0 and u(4, t) = 0 can be rewritten into X 0 (0) = 0
and X(4) = 0.
Which statement below is true?
(a) Neither is true.
(b) Only (I) is true.
(c) Only (II) is true.
(d) Both are true.
7. Using the substitution u(x, t) = X(x)T (t), where u(x, t) is not the trivial solution. Consider
the two statements below.
(I) The equation uxt = utt + 4ut can be separated into two ordinary di↵erential equations.
(II) The boundary conditions ux (0, t) = 0 and u(6, t) = 0 can be rewritten into X 0 (0) = 0
and X(6) = 0.
What can you say regarding the truthfulness of these statements?
(a) Only (I) is true.
(b) Only (II) is true.
(c) Both are true.
(d) Neither is true.
Answers
1. (a) 4X 00 + X = 0, and T 00 + ( + 5)T = 0, (b) X 0 (0) = 0, X(20) = 0, (c) No
2. (a) X 00 + X = 0, and Y 00 + 2Y 0 + y 2 Y = 0, (b) X(0) = X(L) = 0
3. (a) T, (b) T, (c) F, (d) F
4. (d)
5. (c)
6. (d)
7. (c)
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.3.2
314
Type II: Nonhomogeneous Boundary Conditions
Suppose now that one end of the bar is held at a constant temperature T1 and the other is
maintained at a constant temperature T2 where T1 and T2 are not necessarily zero. So now our
IBVP becomes
Heat Conduction Problem Type II: Nonhomogeneous Boundary Conditions
(Heat equation)
↵2 uxx = ut ,
(Initial condition)
u(x, 0) = f (x),
(Boundary conditions)
u(0, t) = T1 ,
0  x  L,
u(L, t) = T2 ,
t>0
(8.15)
0 < x < L,
(8.16)
t > 0.
(8.17)
This problem is only slightly more difficult to solve since we can reduce it to a problem of Type I
by introducing the concept of steady-state solution, V (x).
After a long period of time (t ! 1), we anticipate that a steady temperature distribution V (x)
will be reached, which is independent of time t. Thus, by definition, we have
V (x) = lim u(x, t).
t!1
So V must satisfy Equations (8.15) and (8.17). Note that
Vxx = V 00 ,
Vt = 0.
Hence
(8.15) becomes lim ↵2 uxx = lim ut
)
↵2 Vxx = Vt
(8.17) becomes lim u(0, t) = lim T1
)
V (0) = T1 ,
)
V (L) = T2 .
t!1
t!1
t!1
t!1
lim u(L, t) = lim T2 ,
t!1
t!1
V 00 = 0,
)
Now we have IVP of homogeneous equation,
V 00 = 0,
V (0) = T1 ,
V (L) = T2 .
Let’s solve it.
V 00 = 0
)
V (x) = c1 x + c2
V (0) = T1
)
c 2 = T1
V (L) = T2
)
T2 = c 1 L + T1
)
c1 =
T2
T1
L
.
Hence we finally derive the formula for steady state function V ,
V (x) =
T2
T1
L
x + T1 .
Now let’s return to the original problem, express u(x, t) as
u(x, t) = V (x) + W (x, t),
where W (x, t) is called transient part. Substitute this into Equations (8.15)-(8.17), we have
↵2 uxx = ut
u(x, 0) = f (x)
u(0, t) = T1
u(L, t) = T2
)
↵2 (V + W )xx = (V + W )t
)
V (0) + W (0, t) = T1
)
V (x) + W (x, 0) = f (x)
)
V (L) + W (L, t) = T2
)
↵2 (V 00 + Wxx ) = Vt + Wt
)
W (x, 0) = f (x)
)
T2 + W (L, t) = T2
)
V (x)
T1 + W (0, t) = T1
)
↵2 Wxx = Wt
)
W (0, t) = 0
)
W (L, t) = 0.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
315
We use these facts, V 00 = 0, Vt = 0, for the first equation. Now we can see that all equations of W
is exactly the Type I heat problem. So we have the general solution for W ,
W (x, t) =
1
X
n⇡x
↵ 2 n2 ⇡ 2 t
) sin
,
2
L
L
cn exp(
n=1
where
2
cn =
L
Z
L
(f (x)
V (x)) sin
0
n⇡x
dx.
L
Finally, we then get the general solution of U , as follow
u(x, t) = V (x) + W (x, t)
✓
◆ X
1
T 2 T1
↵ 2 n2 ⇡ 2 t
n⇡x
=
x + T1 +
cn exp(
) sin
.
2
L
L
L
n=1
In conclusion, we then have
Heat Conduction Problem Type II: Nonhomogeneous Boundary Conditions
(Heat equation)
↵2 uxx = ut ,
(Initial condition)
u(x, 0) = f (x),
(Boundary conditions)
u(0, t) = T1 ,
The solution is u(x, t) = V (x) +
where
cn =
Z
2
L
T2
0  x  L, t > 0
0 < x < L,
u(L, t) = T2 ,
1
X
cn exp(
n=1
L
(f (x)
n⇡x
↵ 2 n2 ⇡ 2 t
) sin
,
L2
L
V (x)) sin
0
t > 0.
n⇡x
dx,
L
T1
x + T1 .
L
( The steady state is lim u(x, t) = V (x).)
and
V (x) =
t!1
Example 233 (Summer 2014 Final Exam ).
Suppose a thin rod 10 cm long is insulated along its sides and made of a silver (its thermal
di↵usivity is 1.71 cm2 /sec.) Also suppose that the left end is held at constant temperature
of 30 C and the right end is held at constant temperature of 80 C. Assume that the initial
temperature distribution is 6x, where x is a position inside the rod from the left end.
a) What is the temperature at x = 2 at the beginning of the experiment (t = 0)?
Solution We know that u(x, 0) = 6x. At x = 2, u(2, 0) = 6(2) = 12.
b) Find the temperature u(x, t) of the rod at any time t and at any point x inside the rod.
Solution Here we have L = 10, ↵2 = 1.71, T1 = 30 and T2 = 80. Firstly, the steady
solution is
V (x) =
=
T2
T1
L
80
30
10
x + T1
x + 30 = 5x + 30.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
316
Then the general solution is
u(x, t) = V (x) +
1
X
cn exp(
n=1
1
X
= 5x + 30 +
↵ 2 n2 ⇡ 2 t
n⇡x
) sin
2
L
L
1.71n2 ⇡ 2 t
n⇡x
) sin
.
100
10
cn exp(
n=1
Moreover,
cn =
=
=
=
=
=
=
2
L
Z
L
(f (x)
V (x)) sin
0
Z
n⇡x
dx
L
10
2
n⇡x
(6x (5x + 30)) sin
dx
10 0
10
Z
1 10
n⇡x
(x 30) sin
dx
5 0
10
✓
◆x=10
1
10
n⇡x
100
n⇡x
(x 30) cos
+ 2 2 sin
5
n⇡
10
n ⇡
10 x=0
✓
◆
1
10
10
( 20) cos n⇡ (
( 30))
5
n⇡
n⇡
✓
◆
1 200
300
cos n⇡
5 n⇡
n⇡
40
60
( 1)n
.
n⇡
n⇡
So the particular solution is
u(x, t) = 5x + 30 +
1 ✓
X
40
n=1
n⇡
60
n⇡
( 1)n
◆
exp(
1.71n2 ⇡ 2 t
n⇡x
) sin
.
100
10
c) Find the steady state solution at x = 2.
Solution V (2) = 5(2) + 30 = 40.
Example 234 (Spring 2009, Final Exam).
Suppose the temperature distribution function u(x, t) of a rod is given by initial-boundary
value problem
3uxx = ut ,
u(0, t) = 30,
0  x  5,
t>0
u(5, t) = 50,
u(x, 0) = 5x + 30.
a) Find its steady-state solution.
Solution We have T1 = 30, T2 = 50 and L = 5. Then
V (x) =
T2
T1
L
x + T1 =
50
30
5
x + 30 = 4x + 30.
b) Find the general solution of the above initial-boudary value problem.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
317
Solution We have ↵2 =3. So the general solution is
u(x, t) = V (x) +
1
X
↵ 2 n2 ⇡ 2 t
n⇡x
) sin
2
L
L
cn exp(
n=1
1
X
= 4x + 30 +
3n2 ⇡ 2 t
n⇡x
) sin
.
25
5
cn exp(
n=1
c) Find the formula for cn .
Solution
2
cn =
L
2
=
5
2
=
5
Z
Z
Z
L
(f (x)
V (x)) sin
0
5
((5x + 30)
n⇡x
dx
L
(4x + 30)) sin
0
5
x sin
0
✓
n⇡x
dx
5
n⇡x
dx
5
2
5
n⇡x
25
n⇡x
x cos
+ 2 2 sin
5
n⇡
5
n ⇡
5
2
25
= (
cos n⇡)
5 n⇡
10
= ( 1)n+1 .
n⇡
=
◆
x=5
x=0
d) What is lim u(1, t)?
n!1
Solution lim u(1, t) = V (1) = 4(1) + 30 = 34.
n!1
Example 235 (Fall 2010, Final Exam).
Suppose the temperature distribution function u(x, t) of a rod that has both ends constantly
kept at di↵erent temperatures is given by initial-boundary value problem
4uxx = ut ,
u(0, t) = 60,
u(x, 0) = 60
0  x  1,
t>0
u(1, t) = 40,
20x
30 sin 2⇡x + 50 sin 7⇡x.
a) Find the steady-state solution of the above initial-boundary value problem.
Solution We have T1 = 60, T2 = 40 and L = 1. Then
V (x) =
T2
T1
L
x + T1 =
40
60
1
x + 60 =
20x + 60
b) Based on the given boundary conditions, state the general form of the solution. Then
find the particular solution of the initial-boudary value problem.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
318
Solution Here we have ↵2 = 4. So the general solution is
u(x, t) = V (x) +
1
X
cn exp(
n=1
=
20x + 60 +
1
X
↵ 2 n2 ⇡ 2 t
n⇡x
) sin
2
L
L
cn exp( 4n2 ⇡ 2 t) sin n⇡x.
n=1
Notice that by substituting t = 0 into above equation, we get
u(x, 0) =
20x + 60 +
1
X
cn sin n⇡x.
(8.18)
n=1
But it is given that
u(x, 0) = 60
20x
30 sin 2⇡x + 50 sin 7⇡x.
(8.19)
So by comparing Equations (8.18) with (8.19), we find cn as follows:
2⇡x = n⇡x
7⇡x = n⇡x
)
)
n=2
n=7
)
)
c2 =
30,
c7 = 50.
And cn = 0 where n 6= 2, 7. Substitute all cn back into the general solution to get a
particular solution,
u(x, t) =
20x + 60 +
1
X
cn exp( 4n2 ⇡ 2 t) sin n⇡x
n=1
=
20x + 60 + c2 exp( 4(2)2 ⇡ 2 t) sin 2⇡x + c7 exp( 4(7)2 ⇡ 2 t) sin 7⇡x
=
20x + 60
30 exp( 16⇡ 2 t) sin 2⇡x + 50 exp( 196⇡ 2 t) sin 7⇡x.
c) What is lim u(0.9, t)?
n!1
Solution lim u(0.9, t) = V (0.9) =
n!1
20(0.9) + 60 =
18 + 60 = 42.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.3.3
319
Type III: Bar with Both Ends Insulated
Now let’s consider the situation where the two ends of the bar are also sealed with perfect insulation
so that no heat could escape to the outside environment, or vice versa. The new boundary
conditions are
ux (0, t) = 0, ux (L, t) = 0,
reflecting the fact that there will be no heat transferring at x = 0, x = L. Now we would like to
solve
Heat Conduction Problem Type III: Bar with Both Ends Insulated
↵2 uxx = ut ,
(Heat equation)
(Initial condition)
u(x, 0) = f (x),
(Boundary conditions)
ux (0, t) = 0,
0  x  L,
t > 0,
0 < x < L,
ux (L, t) = 0,
t > 0.
By using the method of separation of variables to solve. Two ordinary di↵erential equations we
get are the same as before which are
X 00 + X = 0,
T 0 + ↵2 T = 0.
The only di↵erent conditions we have this time come from new boundary conditions,
X 0 (0)T (t) = 0
)
ux (0, t) = 0
0
)
ux (L, t) = 0
X (L)T (t) = 0
)
)
X 0 (0) = 0 or T (t) = 0
X 0 (L) = 0 or T (t) = 0
Again we discard the case when T (t) = 0 since it’s gonna gives us an interesting trivial solution
u(x, t) = X(x)T (t) = 0. Hence we have to let the new boundary conditions to be
X 0 (0) = 0,
X 0 (L) = 0.
So now we have transformed the problem to
X 00 + X = 0,
0
T +↵
2
X 0 (0) = 0,
X 0 (L) = 0,
(8.20)
T = 0.
(8.21)
The first equation (8.20) is the Eigenvalue problem. The result is summarized below:
Case 1: > 0, the eigenvalues are
n⇡ 2
) ,
n =(
L
and the corresponding eigenfunctions are
Xn (x) = cos
n⇡x
,
L
n = 1, 2, . . . .
Case 2: 0 = 0, it is indeed an eigenvalue, with corresponding eigenfunction X0 (x) = 1.
Case 3: < 0, there are neither eigenvalues nor eigenfunctions in this case.
For 0 = 0, substitute its value in Equation (8.21),
T0 = 0
For
n
=(
)
T0 = constant. (We will use 1 for simplicity.)
n⇡ 2
) , we get
L
T 0 + ↵2 T = 0
)
T0 +
↵ 2 n2 ⇡ 2
T =0
L2
)
Tn (t) = exp(
↵ 2 n2 ⇡ 2 t
),
L2
n = 1, 2, . . .
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
320
Therefore we have the fundamental set of solutions as follows:
u0 (x, t) = X0 (x)T0 (t) = 1
un (x, t) = Xn (x)Tn (t) = exp(
↵ 2 n2 ⇡ 2 t
n⇡x
) cos
.
L2
L
Then the general solution which is their linear combination is
u(x, t) =
1
X
cn un (x, t) = c0 +
n=0
1
X
cn exp(
n=1
↵ 2 n2 ⇡ 2 t
n⇡x
) cos
,
L2
L
where cn are as yet undetermined.
In order to find the value of cn , apply initial condition u(x, 0) = f (x) into the general solution,
f (x) = c0 +
1
X
cn cos
n=1
n⇡x
,
L
which is a Fourier cosine series. So we then have
!
Z
Z
a0
1 2 L
1 L
c0 =
=
f (x)dx =
f (x) dx,
2
2 L 0
L 0
Z
2 L
n⇡x
c n = bn =
f (x) cos
dx, n = 1, 2, . . . .
L 0
L
In conclusion, we get the following result
Heat Conduction Problem Type III: Bar with Both Ends Insulated
(Heat equation)
↵2 uxx = ut ,
(Initial condition)
u(x, 0) = f (x),
(Boundary conditions)
ux (0, t) = 0,
The solution is u(x, t) = c0 +
where
1
c0 =
L
Z
0  x  L,
0 < x < L,
ux (L, t) = 0,
1
X
cn exp(
n=1
L
↵ 2 n2 ⇡ 2 t
n⇡x
) cos
,
L2
L
f (x) dx,
0
Z
2 L
n⇡x
f (x) cos
dx,
L 0
L
( The steady state is lim u(x, t) = c0 .)
and
cn =
t!1
t > 0.
n = 1, 2, . . . .
t > 0,
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
321
Example 236 (Summer 2014 Final Exam ).
Suppose a thin homogeneous rod 5 cm long is insulated along its sides and made of an aluminum (its thermal di↵usivity is 0.86 cm2 /sec.) Also assume that both ends of this rod are
insulated.
a) If the initial temperature distribution of the rod is x2 + 5 then find the eventual temperature.
Solution We have L = 5 and ↵2 = 0.86 and u(x, 0) = x2 + 5. This is type III heat
equation. As we know the eventual temperature of type III heat equation is
lim u(x, t) = c0 .
t!1
It can be found by the following formula
1
L
c0 =
Z
L
f (x) dx
0
Z 5
✓
◆x=5
1
1 x3
2
=
(x + 5) dx =
+ 5x
5 0
5 3
x=0
1 125
200
40
= (
+ 25) =
=
.
5 3
15
3
b) If the initial temperature distribution of the rod is 337 + 338 cos(⇡x), then what is the
temperature of the rod at x-units from the left end at time t > 0?
Solution Here we have u(x, 0) = 337 + 338 cos(⇡x). The general solution is
u(x, t) = c0 +
= c0 +
1
X
n=1
1
X
cn exp(
↵ 2 n2 ⇡ 2 t
n⇡x
) cos
L2
L
cn exp(
0.86n2 ⇡ 2 t
n⇡x
) cos
.
25
5
n=1
When t = 0, we have
u(x, 0) = c0 +
1
X
cn cos
n=1
n⇡x
.
5
Compare this with the given initial temperature distribution u(x, 0) = 337 + 338 cos(⇡x),
c0 +
1
X
n=1
cn cos
n⇡x
= 337 + 338 cos(⇡x).
5
And so,
c0 = 337
n⇡x
= ⇡x ) n = 5
5
cn = 0 if n 6= 0, 5.
)
c5 = 338
So the particular solution is
u(x, t) = c0 + c5 exp(
0.86(5)2 ⇡ 2 t
5⇡x
) cos
= 337 + 338 exp( 0.86⇡ 2 t) cos ⇡x.
25
5
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
322
Example 237 (Spring 2010, Final Exam).
Suppose the temperature distribution function u(x, t) of a rod that both ends perfectly insulated
is given by the initial-boundary value problem
3uxx = ut ,
0  x  2⇡,
t>0
ux (0, t) = 0,
ux (2⇡, t) = 0,
u(x, 0) = 20
10 cos x + 30 cos 4x.
a) Find the particular solution of the above initial-boundary value problem.
Solution For this problem we have ↵2 = 3, L = 2⇡. So the general solution is
u(x, t) = c0 +
= c0 +
1
X
n=1
1
X
cn exp(
3n2 ⇡ 2 t
n⇡x
) cos
(2⇡)2
2⇡
cn exp(
3n2 t
nx
) cos
4
2
n=1
Notice that by substituting t = 0 into above equation, we get
u(x, 0) = c0 +
1
X
cn cos
n=1
nx
.
2
(8.22)
But it is provided that
u(x, 0) = 20
10 cos x + 30 cos 4x.
(8.23)
So by comparing Equations (8.22) with (8.23), we find cn as follows:
c0 = 20,
nx
x=
2
nx
4x =
2
)
n=2
)
c2 =
)
n=8
)
c8 = 30.
10,
And cn = 0 where n 6= 0, 2, 8. Substitute all cn back into the general solution to get a
particular solution,
3(2)2 t
2x
3(8)2 t
8x
) cos
+ c8 exp(
) cos
4
2
4
2
10 exp( 3t) cos x + 30 exp( 48t) cos 4x.
u(x, t) = c0 + c2 exp(
= 20
b) What is lim u(1, t)?
t!1
Solution
lim u(x, t) = lim (20
n!1
n!1
10 exp( 3t) cos x + 30 exp( 48t) cos 4x) = 20,
or simply lim u(x, t) = c0 = 20. In particular, lim u(1, t) = 20.
t!1
t!1
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
323
Example 238 (Spring 2003 Final Exam ).
Consider a thing rod of length 10 cm with thermal di↵usivity ↵2 = 2 cm2 /s and perfectly
insulated ends. Using the variable x as the distance from the left end of the rod the initial
temperature of the rod is
⇡x
⇡x
f (x) = 3 + cos
5 cos
.
5
2
a) Construct the partial di↵erential equation for this situation, also give boundary conditions.
Solution This is type III heat equation with ↵2 = 2 since both ends are perfectly insulated. Also L = 10 reflecting the length of the thin rod. So we have
(Heat equation)
2uxx = ut
(Boundary conditions) ux (0, t) = 0,
ux (10, t) = 0,
⇡x
⇡x
u(x, 0) = 3 + cos
5 cos
.
5
2
(Initial condition)
b) Solve this partial di↵erential equation.
Solution The general solution is
u(x, t) = c0 +
= c0 +
= c0 +
1
X
n=1
1
X
n=1
1
X
cn exp(
↵ 2 n2 ⇡ 2 t
n⇡x
) cos
2
L
L
cn exp(
2n2 ⇡ 2 t
n⇡x
) cos
100
10
cn exp(
n2 ⇡ 2 t
n⇡x
) cos
50
10
n=1
If t = 0, we have
u(x, 0) = c0 +
1
X
cn cos
n=1
n⇡x
.
10
Compare this with the given initial condition, we have
c0 +
1
X
cn cos
n=1
n⇡x
⇡x
= 3 + cos
10
5
5 cos
⇡x
.
2
Therefore
c0 = 3
n⇡x
⇡x
=
10
5
n⇡x
⇡x
=
10
2
)
n=2
)
c2 = 1
)
n=5
)
c5 =
5.
And cn = 0 where n 6= 0, 2, 5. Then the particular solution is
22 ⇡ 2 t
2⇡x
52 ⇡ 2 t
5⇡x
) cos
+ c5 exp(
) cos
50
10
50
10
2⇡ 2 t
⇡x
⇡2 t
⇡x
= 3 + exp(
) cos
5 exp(
) cos
.
25
5
2
2
u(x, t) = c0 + c2 exp(
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
324
c) Determine the steady-state temperature.
Solution lim u(x, t) = c0 = 3.
t!1
8.3.4
More Problems
If boundary conditions of IBVP are not ones in Type I, II or III, we use the same idea we did in
the proof of Type II to find the steady state solution. In other words, we use
V (x) = lim u(x, t).
t!1
It’s always the case that, after we use this definition, we will get
V (x) = c1 x + c2 .
Example 239 (Spring 2007 Final Exam ).
Find the steady-state solution, V (x), of the heat conduction equation
↵2 uxx = ut ,
given the nonhomogeneous boundary conditions
ux (0, t) + 4u(0, t) = 0,
ux (5, t) = 10.
Solution Notice that the given boundary conditions are not neither Type I, II nor III. So we
need to approach this problem directly from the definition of the steady state solution which is
V (x) = lim u(x, t).
t!1
By using the above relation, we can change the heat equation as follows:
↵2 uxx = ut
)
lim ↵2 uxx = lim ut
t!1
t!1
)
↵2 Vxx = Vt .
But V is a function of x only (without t), so Vxx = V 00 and Vt = 0. Then we get
↵2 V 00 = 0
V 00 = 0
)
)
V (x) = c1 x + c2 .
Note that V 0 (x) = c1 . Now let’s use the same technique to change the given boundary conditions,
lim (ux (0, t) + 4u(0, t)) = lim 0
t!1
t!1
lim ux (5, t) = lim 10
t!1
t!1
)
V 0 (0) + 4V (0) = 0
)
V 0 (5) = 10
And so
c1 + 4c2 = 0
)
c2 =
So the steady-state solution is
V (x) = 10x
1
c1 =
4
5
.
2
)
10
=
4
)
c1 + 4c2 = 0
c1 = 10.
5
.
2
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.3.5
325
Summary of Heat Conduction Problems
Heat Conduction Problem Type I: Homogeneous Boundary Conditions
(Heat equation)
↵2 uxx = ut ,
(Initial condition)
u(x, 0) = f (x),
(Boundary conditions)
u(0, t) = 0,
The solution is u(x, t) =
0  x  L,
0 < x < L,
u(L, t) = 0,
1
X
t>0
cn exp(
n=1
t > 0.
↵ 2 n2 ⇡ 2 t
n⇡x
) sin
,
L2
L
Z
2 L
n⇡x
f (x) sin
dx.
L 0
L
( The steady state is lim u(x, t) = 0.)
where
cn =
t!1
Heat Conduction Problem Type II: Nonhomogeneous Boundary Conditions
(Heat equation)
↵2 uxx = ut ,
(Initial condition)
u(x, 0) = f (x),
(Boundary conditions)
u(0, t) = T1 ,
The solution is u(x, t) = V (x) +
0  x  L, t > 0
0 < x < L,
u(L, t) = T2 ,
1
X
cn exp(
n=1
t > 0.
↵ 2 n2 ⇡ 2 t
n⇡x
) sin
,
2
L
L
Z
2 L
n⇡x
where cn =
(f (x) V (x)) sin
dx,
L 0
L
T2 T1
and V (x) =
x + T1 .
L
( The steady state is lim u(x, t) = V (x).)
t!1
Heat Conduction Problem Type III: Bar with Both Ends Insulated
(Heat equation)
↵2 uxx = ut ,
(Initial condition)
u(x, 0) = f (x),
(Boundary conditions)
ux (0, t) = 0,
The solution is u(x, t) = c0 +
where
c0 =
1
L
Z
0  x  L,
0 < x < L,
ux (L, t) = 0,
1
X
cn exp(
n=1
L
↵ 2 n2 ⇡ 2 t
n⇡x
) cos
,
L2
L
f (x) dx,
0
Z
2 L
n⇡x
f (x) cos
dx,
L 0
L
( The steady state is lim u(x, t) = c0 .)
and
cn =
t!1
t > 0.
n = 1, 2, . . . .
t > 0,
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
326
Exercises 8.3
1. Solve the heat conduction problem described by:
ut = 9uxx ,
0 < x < 5,
t>0
u(0, t) = u(5, t) = 0, for t > 0
2⇡x
7⇡x
u(x, 0) = sin
3 sin ⇡x + 13 sin
5
5
2. Solve the nonhomogeneous heat equation
10uxx = ut ,
0 < x < 6,
t>0
u(0, t) = 20,
u(6, t) = 80,
t>0
u(x, 0) = 20 + 10x + 5 sin ⇡x
10 sin
3⇡x
2
3. Find the steady-state solution of the heat conduction problem
uxx = ut ,
0 < x < 10,
u(0, t) = 100,
t>0
u(10, t) = 60,
u(x, 0) = f (x).
(a) v(x) =
40x + 60
(b) v(x) =
4x + 100
(c) v(x) = 4x + 60
(d) v(x) = 40x + 100
4. Suppose the temperature distribution function u(x, t) of a rod that has both ends constantly
kept at 0 degree is given by the heat conduction problem
4uxx = ut ,
0 < x < 6,
t>0
u(0, t) = 0,
u(6, t) = 0,
⇡x
u(x, 0) = 2 sin
+ 4 sin ⇡x
3
10 sin
3⇡x
.
2
(a) Find the particular solution of the above initial-boundary value problem.
(b) What is lim u(x, t)?
t!1
5. Suppose the temperature distribution function u(x, t) of a rod that has both ends perfectly
insulated is given by the initial-boundary value problem
9uxx = ut ,
0 < x < 4,
ux (0, t) = 0,
ux (4, t) = 0,
u(x, 0) = 2
cos ⇡x
t>0
7 cos 5⇡x.
(a) Find the particular solution of the above initial-boundary value problem.
(b) What is lim u(3, t)?
t!1
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
327
6. Consider the following two heat conduction initial-boundary value problem
uxx = ut ,
0 < x < 10,
u(0, t) = 0,
t>0
u(10, t) = 0,
I.
u(x, 0) = 200.
uxx = ut ,
0 < x < 10,
ux (0, t) = 0,
t>0
ux (10, t) = 0,
II.
u(x, 0) = 200.
(a) Which problem (I or II) models the temperature distribution of a rod with both ends
perfectly insulated?
(b) Suppose u1 (x, t) is the solution of problem I and u2 (x, t) is the solution of problem II.
Compare the temperatures at the middle of the rod (that is, at x = 5) as t ! 1.
i. lim u1 (5, t) > lim u2 (5, t)
t!1
t!1
ii. lim u1 (5, t) < lim u2 (5, t)
t!1
t!1
iii. lim u1 (5, t) = lim u2 (5, t)
t!1
t!1
iv. There is not enough available information to compare them.
7. Find the steady-state solution, v(x), of the heat conduction problem
6uxx = ut ,
u(0, t) = 5,
0 < x < 4,
t>0
ux (4, t) = 2,
u(x, 0) = f (x).
(a) v(x) = 34 x
(b) v(x) =
5
2x
5
(c) v(x) = 2x + 5
(d) v(x) =
3
4x
+5
8. . Find the steady-state solution, v(x), of the heat conduction problem with nonhomogeneous
boundary conditions:
↵2 uxx = ut ,
0 < x < 6,
u(0, t) = 100,
u(6, t)
t>0
2ux (6, t) = 20,
u(x, 0) = f (x).
(a) v(x) =
20x + 100
(b) v(x) =
40
3 x + 100
40
3 x + 100
(c) v(x) =
(d) v(x) = 20x + 100
Answers
1. u(x, t) = e
36 2
25 ⇡ t
sin 2⇡x
5
2. u(x, t) = 20 + 10x + 5e
3. (b)
3e
10⇡ 2 t
9⇡ 2 t
sin ⇡x + 13e
sin ⇡x
10e
45 2
2 ⇡ t
441 2
25 ⇡ t
sin 7⇡x
5
sin 3⇡x
2
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
4. (a) u(x, t) = 2e
5. (a) u(x, t) = 2
6. (a) II, (b) (ii)
7. (c)
8. (a)
4 2
9⇡ t
e
sin ⇡x
3 + 4e
2
9⇡ t
cos ⇡x
4⇡ 2 t
7e
sin ⇡x
2
225⇡ t
10e
328
9⇡ 2 t
sin 3⇡x
2 , (b) lim u(x, t) = 0
t!1
cos 5⇡x, (b) lim u(3, t) = 2
t!1
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.4
329
Wave Equation
Suppose that an elastic string of length L is tightly stretched between two supports at the same
horizontal level so that the x-axis lies along the string.
Figure 8.3: A vibrating string.
Let u(x, t) be the vertical displacement experienced by the string at the point x at time t. If
damping e↵ects, such as air resistance, are neglected and if the amplitude of the motion is not too
large, then u(x, t) satisfies
a2 uxx = utt ,
where a, horizontal propagation speed of the wave motion, is a positive constant. The equation
is known as wave equation. Since the ends are assumed to remain fixed and therefore the
boundary conditions are
u(0, t) = 0, u(L, t) = 0, t > 0.
In addition, the equation also come with two initial conditions,
u(x, 0) = f (x)
ut (x, 0) = g(x)
(initial displacement)
(initial velocity).
Alternatively, it can be considered as a boundary value problem in the semi-infinite strip 0 < x <
L, t > 0 of the xt-plane. One condition is imposed at each point on the semi-infinite sides, and
two are imposed at each point on the finite base.
Figure 8.4: Boundary value problem for the wave equation.
So the IBVP we would like to solve is
Wave Equation Problem
(Wave equation)
a2 uxx = utt ,
(Initial conditions)
u(x, 0) = f (x),
(Boundary conditions)
u(0, t) = 0,
ut (x, 0) = g(x)
u(L, t) = 0,
0  x  L,
0 < x < L,
t > 0.
t>0
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.4.1
330
Solutions to Wave Equation Problem
Let’s use the method of separation of variables by writing
u(x, t) = X(x)T (t).
Thus uxx = X 00 T, ut = XT 00 . Substituting these into the wave equation,
a2 uxx = utt
where
a2 X 00 T = XT 00
)
X 00
T 00
= 2 =
X
a T
)
,
is constant. Now consider them separately,
X 00
=
X
T 00
=
a2 T
)
X 00 + X = 0.
)
T 00 + a2 T = 0.
For boundary conditions, again we use u(x, t) = X(x)T (t),
)
u(0, t) = 0
X(0)T (t) = 0
)
u(L, t) = 0
X(L)T (t) = 0
)
X(0) = 0 or T (t) = 0
)
X(L) = 0 or T (t) = 0
After we discard the case of T (t) = 0, as we did before, we obtain
X(0) = 0,
X(L) = 0.
Thus our problem changes to
X 00 + X = 0,
00
T +a
2
X(0) = 0,
X(L) = 0
(8.24)
T = 0.
(8.25)
Recall once again that the result of BVP (8.24) is that the eigenvalues are
n
=(
n⇡ 2
) ,
L
and the corresponding eigenfunctions are
Xn (x) = sin
where n = 1, 2, . . .
Now let’s substitute
n
=(
T 00 + a2 T = 0
n⇡x
,
L
n⇡ 2
) into Equation (8.25),
L
)
)
a 2 n2 ⇡ 2
T =0
L2
an⇡t
an⇡t
Tn (t) = An cos
+ Bn sin
,
L
L
T 00 +
n = 1, 2, . . .
an⇡
Notice that the above equation is the second order homogeneous equation with r = ±
i, as
L
the roots of its characteristic equation.
So, for each n = 1, 2, . . .,
✓
◆
an⇡t
an⇡t
n⇡x
un (x, t) = Xn (x)Tn (t) = An cos
+ Bn sin
sin
.
L
L
L
Then the linear combination of un form the general solution of the heat equation,
◆
1 ✓
X
an⇡t
an⇡t
n⇡x
u(x, t) =
An cos
+ Bn sin
sin
,
L
L
L
n=1
(8.26)
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
331
where An and Bn are not yet determined.
To find the value of An , apply initial condition u(x, 0) = f (x) into Equation (8.26),
f (x) =
1
X
An sin
n=1
n⇡x
.
L
Hence f needs to be a Fourier sine series and so we expand it into its odd periodic extension to
receive,
Z
n⇡x
2 1
An = b n =
f (x) sin
dx.
L 0
L
Note here that
1 ✓
X
ut (x, t) =
n=1
an⇡
an⇡t
an⇡
an⇡t
An
sin
+ Bn
cos
L
L
L
L
◆
sin
n⇡x
.
L
(8.27)
Plug in another initial condition ut (x, 0) = g(x) into (8.27),
g(x) =
1
X
Bn
n=1
an⇡
n⇡x
sin
.
L
L
So g needs to be also a Fourier sine series. By expanding it into odd periodic extension,
1
X
bn sin
n=1
1
X
an⇡
n⇡x
n⇡x
= g(x) =
Bn
sin
.
L
L
L
n=1
It immediate follows that
L
L
Bn =
bn =
an⇡
an⇡
2
L
Z
L
0
n⇡x
g(x) sin
dx
L
!
2
=
an⇡
Z
L
g(x) sin
0
n⇡x
dx.
L
We conclude what we get here:
Wave Equation Problem
(Wave equation)
(Initial conditions)
a2 uxx = utt ,
u(x, 0) = f (x),
(Boundary conditions)
ut (x, 0) = g(x)
0  x  L, t > 0,
0 < x < L,
u(0, t) = 0, u(L, t) = 0,
t > 0.
✓
◆
1
X
an⇡t
an⇡t
n⇡x
The solution is u(x, t) =
An cos
+ Bn sin
sin
,
L
L
L
n=1
Z
2 L
n⇡x
where An =
f (x) sin
dx,
L 0
L
Z L
2
n⇡x
and Bn =
g(x) sin
dx.
an⇡ 0
L
( The steady state is 0 and lim u(x, t) doesn’t exist. )
t!1
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.4.2
332
Special Cases of Wave Equation Problems
We take a look here in two particular cases which make our solutions in much simpler form.
Special case 1: If f (x) 6= 0, g(x) = 0, then all Bn = 0. Hence
Wave Equation Problem: Special Case 1 (f (x) 6= 0, g(x) = 0)
a2 uxx = utt ,
(Wave equation)
(Initial conditions)
u(x, 0) = f (x),
(Boundary conditions)
u(0, t) = 0,
The solution is u(x, t) =
1
X
ut (x, 0) = 0
u(L, t) = 0,
An cos
n=1
0  x  L, t > 0,
0 < x < L,
t > 0.
an⇡t
n⇡x
sin
,
L
L
Z
2 L
n⇡x
f (x) sin
dx.
L 0
L
( The steady state is 0 and lim u(x, t) doesn’t exist. )
where
An =
t!1
Special case 2: If f (x) = 0, g(x) 6= 0, then all An = 0. Hence
Wave Equation Problem: Special Case 2 (f (x) = 0, g(x) 6= 0)
a2 uxx = utt ,
(Wave equation)
(Initial conditions)
u(x, 0) = 0,
ut (x, 0) = g(x)
(Boundary conditions)
u(0, t) = 0,
u(L, t) = 0,
The solution is u(x, t) =
1
X
Bn sin
n=1
0  x  L,
0 < x < L,
t > 0.
an⇡t
n⇡x
sin
,
L
L
Z L
2
n⇡x
g(x) sin
dx.
an⇡ 0
L
( The steady state is 0 and lim u(x, t) doesn’t exist. )
where
Bn =
t!1
t>0
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
333
Example 240 (Summer 2010, Final Exam).
Solve the one-dimensional wave problem
utt = 4uxx ,
u(0, t) = 0,
0  x  5,
t>0
u(5, t) = 0,
u(x, 0) = 2 sin ⇡x
3 sin 2⇡x,
ut (x, 0) = 0.
Solution We have a2 = 4(a = 2), L = 5, f (x) 6= 0 and g(x) = 0. This is a wave equation
special case 1. So the general solution is
u(x, t) =
1
X
An cos
n=1
1
an⇡t
n⇡x X
2n⇡t
n⇡x
sin
=
An cos
sin
.
L
L
5
5
n=1
Notice that by substituting t = 0 into the above equation, we have
u(x, 0) =
1
X
n⇡x
.
5
(8.28)
3 sin 2⇡x.
(8.29)
An sin
n=1
But it is provided that
u(x, 0) = 2 sin ⇡x
So by comparing Equations (8.28) with (8.29), we find cn as follows:
n⇡x
= ⇡x
5
n⇡x
= 2⇡x
5
)
n=5
)
A5 = 2,
)
n = 10
)
A10 =
3.
And An = 0 where n 6= 5, 10. Substitute all An back into the general solution to get a
particular solution,
2(5)⇡t
5⇡x
2(10)⇡t
10⇡x
sin
+ A10 cos
sin
5
5
5
5
= 2 cos 2⇡t sin ⇡x 3 cos 4⇡t sin 2⇡x.
u(x, t) = A5 cos
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
334
Example 241 (Fall 2009 Final Exam ).
The displacement u(x, t) of a vibrating string of length ⇡ cm satisfies the following wave
equation and boundary conditions:
8
>
0 < x < ⇡, t > 0,
<utt = uxx
u(0, t) = 0, u(⇡, t) = 0,
t > 0,
>
:
u(x, 0) = 0, ut (x, 0) = 4 sin(2x), 0 < x < ⇡.
Which of the following three statements is not true?
(a) The ends of the string are fixed.
(b) The string is set in motion with no initial velocity.
(c) u(x, t) = 2 sin(2t) sin(2x).
(d) All three statements are true.
Solution
• The boundary conditions u(0, t) = 0, u(⇡, t) = 0 tell us that both ends of the string are
fixed and motionless at all time.
• ut (x, 0) = 4 sin(2x) tells us the initial velocity at any position x. It is non-zero. Since
(b) is the false statement, it is the correct answer.
• This is the special case 2 wave equation with a = 1 and L = ⇡. Then the general
solution is
u(x, t) =
=
=
1
X
n=1
1
X
n=1
1
X
Bn sin
an⇡t
n⇡x
sin
L
L
Bn sin
1n⇡t
n⇡x
sin
⇡
⇡
Bn sin nt sin nx.
n=1
Its first derivative is
ut (x, t) =
At t = 0, ut (x, 0) =
4 sin(2x). So
P1
1
X
nBn cos nt sin nx.
n=1
n=1
nBn sin nx. But the given initial velocity is ut (x, 0) =
1
X
nBn sin nx = 4 sin(2x).
n=1
Compare the angle of sine function,
nx = 2x
)
n=2
)
2B2 = 4
)
B2 = 2.
Other Bn = 0 if n 6= 2. So the particular solution for this problem is
u(x, t) =
1
X
n=1
Bn sin nt sin nx = B2 sin 2t sin 2x = 2 sin 2t sin 2x.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
335
Example 242 (Summer 2014 Final Exam ).
The displacement u(x, t) of a string of length 8 cm with both ends clamped satisfies the differential equation
64uxx = utt .
a) If the initial displacement of the string is 5 sin 3⇡x and the initial velocity of the string
is 0. Find the displacement u(x, t) of the string at any position x at anytime t > 0.
Solution Here we have L = 8 and a = 8 since a2 = 64. This is the special case I wave
equation with u(x, 0) = 5 sin 3⇡x and ut (x, 0) = 0. The general solution is
u(x, t) =
=
=
1
X
n=1
1
X
n=1
1
X
An cos
an⇡t
n⇡x
sin
L
L
An cos
8n⇡t
n⇡x
sin
8
8
An cos n⇡t sin
n=1
n⇡x
8
Note here that, with t = 0, we have
u(x, 0) =
1
X
An sin
n=1
n⇡x
.
8
Compare this with an initial velocity of the string
u(x, 0) = 5 sin 3⇡x,
So we have
1
X
An sin
n=1
n⇡x
= 5 sin 3⇡x.
8
Compare the angle of sine function,
n⇡x
= 3⇡x
8
)
n = 24
)
A24 = 5.
Then the particular solution is
u(x, t) = A24 cos 24⇡t sin
24⇡x
= 5 cos 24⇡t sin 3⇡x.
8
b) Now assume that the initial displacement of the string in Part (a) is 0 and the initial
velocity is given by sin ⇡x, then what is the displacement of the string at any position x
at anytime t > 0?
Solution This is the special case II wave equation with u(x, 0) = 0 and ut (x, 0) = sin ⇡x.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
336
Its general solution is
u(x, t) =
=
=
1
X
n=1
1
X
n=1
1
X
Bn sin
an⇡t
n⇡x
sin
L
L
Bn sin
8n⇡t
n⇡x
sin
8
8
Bn sin n⇡t sin
n=1
Note here that
ut (x, t) =
1
X
n⇡x
8
n⇡Bn cos n⇡t sin
n=1
With t = 0, we have
ut (x, 0) =
1
X
n=1
n⇡Bn sin
n⇡x
.
8
n⇡x
.
8
Compare this with the given initial velocity ut (x, 0) = sin ⇡x,
1
X
n⇡x
= sin ⇡x.
8
n⇡Bn sin
n=1
Using the same strategy by comparing the angle of sine function,
n⇡x
= ⇡x
8
)
n=8
)
8⇡B8 = 1
)
B8 =
So the particular solution is
u(x, t) = B8 sin 8⇡t sin
8⇡x
1
=
sin 8⇡t sin ⇡x.
8
8⇡
1
.
8⇡
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.4.3
337
Summary of Wave Equation Problems
Wave Equation Problem
(Wave equation)
(Initial conditions)
a2 uxx = utt ,
u(x, 0) = f (x),
ut (x, 0) = g(x)
0  x  L, t > 0,
0 < x < L,
(Boundary conditions)
u(0, t) = 0, u(L, t) = 0,
t > 0.
✓
◆
1
X
an⇡t
an⇡t
n⇡x
The solution is u(x, t) =
An cos
+ Bn sin
sin
,
L
L
L
n=1
Z
2 L
n⇡x
where An =
f (x) sin
dx,
L 0
L
Z L
2
n⇡x
and Bn =
g(x) sin
dx.
an⇡ 0
L
( The steady state is 0 and lim u(x, t) doesn’t exist. )
t!1
Special case 1: If f (x) 6= 0, g(x) = 0, then all Bn = 0. Hence
Wave Equation Problem: Special Case 1 (f (x) 6= 0, g(x) = 0)
a2 uxx = utt ,
(Wave equation)
(Initial conditions)
u(x, 0) = f (x),
(Boundary conditions)
u(0, t) = 0,
The solution is u(x, t) =
1
X
ut (x, 0) = 0
u(L, t) = 0,
An cos
n=1
0  x  L, t > 0,
0 < x < L,
t > 0.
an⇡t
n⇡x
sin
,
L
L
Z
2 L
n⇡x
f (x) sin
dx.
L 0
L
( The steady state is 0 and lim u(x, t) doesn’t exist. )
where
An =
t!1
Special case 2: If f (x) = 0, g(x) 6= 0, then all An = 0. Hence
Wave Equation Problem: Special Case 2 (f (x) = 0, g(x) 6= 0)
a2 uxx = utt ,
(Wave equation)
(Initial conditions)
u(x, 0) = 0,
ut (x, 0) = g(x)
(Boundary conditions)
u(0, t) = 0,
u(L, t) = 0,
The solution is u(x, t) =
1
X
Bn sin
n=1
0  x  L,
0 < x < L,
t > 0.
an⇡t
n⇡x
sin
,
L
L
Z L
2
n⇡x
g(x) sin
dx.
an⇡ 0
L
( The steady state is 0 and lim u(x, t) doesn’t exist. )
where
Bn =
t!1
t>0
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
338
Exercises 8.4
1. Which of the following functions below is a solution of the wave equation
utt = 4uxx ?
4⇡ 2 t
(a) e
sin(⇡x)
(b) sin(x
2
(c) x + t
2t)
2
(d) 1 + 4 cos(t) + x2
2. Consider the initial-boundary value problems for the wave equation
9uxx = utt ,
u(0, t) = 0,
0 < x < 3,
t>0
u(3, t) = 0,
u(x, 0) = f (x),
ut (x, 0) = 0.
In what specific form will its general solution appear?
(a) u(x, t) =
(b) u(x, t) =
(c) u(x, t) =
(d) u(x, t) =
1
X
n=1
1
X
n=1
1
X
n=1
1
X
An cos(n⇡t) sin(
n⇡x
)
3
An cos(3n⇡t) sin(
Bn sin(n⇡t) sin(
n⇡x
)
3
Bn sin(3n⇡t) sin(
n=1
n⇡x
)
3
n⇡x
)
3
3. Suppose the displacement u(x, t) of a piece of flexing string is given by the initial-boundary
value problem
9uxx = utt ,
u(0, t) = 0,
0 < x < 2,
t>0
u(2, t) = 0,
u(x, 0) = 0
ut (x, 0) = 4
x2 .
(a) What is the physical meaning of the boundary conditions?
(b) What is the propagation speed of the standing waves?
(c) When t = 0, what is the velocity of the vibrating string at x = 12 ?
(d) In what specific form will its general solution appear?
1
X
3n⇡t
n⇡x
i. u(x, t) =
Cn sin
sin
2
2
n=1
ii. u(x, t) =
iii. u(x, t) =
iv. u(x, t) =
1
X
n=1
1
X
n=1
1
X
n=1
Cn sin
3n⇡t
n⇡x
cos
2
2
Cn cos
3n⇡t
n⇡x
cos
2
2
Cn cos
3n⇡t
n⇡x
sin
2
2
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
339
(e) True or False: The coefficients of the solution in part (d) above can be found using the
integral
Z 2
2
n⇡x
Cn =
(4 x2 ) sin
dx.
3n⇡ 0
2
(f) True or False: As t ! 1, the solution u(x, t) in (e) will reach a limit.
4. Suppose the displacement u(x, t) of a piece of flexing string that has both ends firmly fixed
in places is given by the initial-boundary value problem
9uxx = utt ,
u(0, t) = 0,
0 < x < 6,
t>0
u(6, t) = 0,
u(x, 0) = 0
ut (x, 0) = 0.5.
(a) State the general form of its solution. Then find the particular solution of the initialboundary value problem.
(b) True or False: As lim u(1, t) = 0.5. Explain your answer.
t!1
5. Suppose the displacement u(x, t) of a piece of flexing string that has both ends firmly fixed
in places is given by the initial-boundary value problem
16uxx = utt ,
0 < x < 4,
t>0
u(0, t) = 0, u(4, t) = 0,
p
p
5
u(x, 0) = 5 sin(⇡x) 2 7 sin( ⇡x),
2
ut (x, 0) = 0.
(a) State the general form of its solution. Then find the particular solution of the initialboundary value problem.
(b) True or False: As t ! 1, the solution u(x, t) will go to 0 for all x.
(c) True or False: If f (x) = 0, then u(x, t) = 0 is the unique solution.
Answers
1. (b)
2. (a)
3. (a) Two ends of the string are clamped in fixed positions at the horizontal level so they are
held motionless at all time. (b) a = 3, (c) 15
4 , (d) i, (e) T, (f) F
P1
n⇡x
4. (a) The general form of the solution is u(x, t) = n=1 Bn sin n⇡t
2 sin 6 . The particular
P1
solution is u(x, t) = n=1 (2n 41)2 ⇡2 sin (2n 21)⇡t sin (2n 61)⇡x , (b) F
P1
5. (a) The general form of the solution is u(x, t) = n=1 Cn cos n⇡t sin n⇡x
4 . The particular
p
p
5
solution is u(x, t) = 5 cos(4⇡t) sin(⇡x) 2 7 cos(10⇡t) sin( 2 ⇡x), (b) F, (c) T
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
8.5
340
Laplace’s Equation*
One of the most important of all partial di↵erential equations occurring in applied mathematics
is that associated with the name of Laplace: in two dimensions
uxx + uyy = 0,
and in three dimensions
uxx + uyy + uzz = 0.
• Laplace’s equation arises in the study of such diverse applications as steady state heat flow,
vibrating membranes, and electric and gravitational potentials. For this reason, Laplace’s
equation is often referred to as the potential equation.
• This equation does not contain the time t, so that we do not expect any “initial conditions”
to be imposed here.
• In the problems that arise in applications, we are given u, or its normal derivative, on the
boundary of a given region R, and we seek to determine u(x, y) inside R. The problem of
finding a solution of Laplace’s equation which takes on given boundary values is known as
a Dirichlet problem, while the problem of finding a solution of Laplace’s equation whose
normal derivative takes on given boundary values is known as a Neumann problem. Both
of these problems can be solved by the method of separation of variables if R is a rectangle.
Before we solve Dirichlet and Neumann problem, let’s consider the second order homogeneous
equation
X 00 µ2 X = 0.
We’ve learned in Chapter 3 that the solution of this equation is
X(x) = a1 eµx + a2 e
µx
since ±µ are roots of its characteristic equation r2 µ2 = 0. In this section, it’s more convenient
to express this in terms of hyperbolic cosine and sine. By definition,
cosh(x) :=
ex + e
2
x
,
sinh(x) :=
ex
e
2
x
.
Their properties are analogous but di↵erent to the ordinary cosine and sine function.
1. cosh 0 = 1,
sinh 0 = 0,
2. cosh( x) = cosh(x),
3.
sinh( x) =
d
cosh x = sinh x,
dx
4. cosh2 x
sinh(x),
d
sinh x = cosh x,
dx
sinh2 x = 1,
5. sinh 2x = 2 sinh x cosh x,
cosh 2x = cosh2 x + sinh2 x,
6. sinh(x ± y) = sinh x cosh y ± sinh y cosh x,
7. cosh(x ± y) = cosh x cosh y ± sinh x sinh y
By their definitions, we can show that
ex = cosh x + sinh x,
e
x
= cosh x
sinh x.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
341
Therefore
X(x) = a1 eµx + a2 e
µx
= a1 (cosh µx + sinh µx) + a2 (cosh µx
= (a1 + a2 ) cosh µx + (a1
Hence, the solution to X 00
sinh µx)
a2 ) sinh µx.
µ2 X = 0 can be rewritten as
X(x) = c1 cosh µx + c2 sinh µx.
8.5.1
Dirichlet Problem for a Rectangle*
Consider the problem of finding the function u satisfying
Dirichlet Problem for a Rectangle
(Laplace’s equation)
uxx + uyy = 0,
(Boundary conditions)
0 < x < a, 0 < y < b,
u(x, 0) = 0,
u(x, b) = 0,
0 < x < a,
u(0, y) = 0,
u(a, y) = f (y),
0 < y < b.
Figure 8.5: Dirichlet problem for a rectangle.
To solve this problem, we use again the method of separation of variables by assuming that
u(x, y) = X(x)Y (y).
Substitute u back into the Laplace’s equation,
X 00 Y + XY 00 = 0
where
)
X 00 Y = XY 00
)
X 00
Y 00
=
=
X
Y
,
is the separation constant. Two ordinary di↵erential equations are obtained as follow
Y 00 + Y = 0,
X 00
X = 0.
Now let’s change the boundary conditions,
u(x, 0) = 0
u(x, b) = 0
u(0, y) = 0
)
X(x)Y (0) = 0
)
X(0)Y (y) = 0
)
X(x)Y (b) = 0
)
)
)
X(x) = 0 or Y (0) = 0
X(x) = 0 or Y (b) = 0
X(0) = 0 or Y (y) = 0.
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
342
Since we’re looking for nontrivial solution, new boundary conditions are
Y (0) = 0,
Y (b) = 0,
X(0) = 0.
Until now we have transformed the problem to
Y 00 + Y = 0,
Y (0) = 0,
X 00
X(0) = 0.
X = 0,
Y (b) = 0,
(8.30)
(8.31)
The eigenvalue problem (8.30) gives
n
=
n2 ⇡ 2
,
b2
Yn (y) = sin
n⇡y
.
b
So Equation (8.31) becomes
X 00
X=0
n2 ⇡ 2
X=0
b2
X 00
)
)
X(x) = c1 cosh
n⇡x
n⇡x
+ c2 sinh
.
b
b
Apply initial condition X(0) = 0,
0 = c1 cosh 0 + c2 sinh 0
)
Xn (x) = sinh
n⇡y
.
b
So
c1 = 0.
Therefore
n⇡x
n⇡y
sin
b
b
is a solution for every positive integer n and they form the fundamental set of solutions. Thus the
function
1
X
n⇡x
n⇡y
u(x, y) =
cn sinh
sin
b
b
n=1
un (x, y) = Xn (x)Yn (y) = sinh
is a general solution for every choice of constant c1 , c2 , . . .. We evaluate at x = a,
u(a, y) =
1
X
cn sinh
n=1
n⇡a
n⇡y
sin
.
b
b
But it is given that u(a, y) = f (y). Therefore, we must choose the constants cn such that
f (y) =
1
X
cn sinh
n=1
n⇡a
n⇡y
sin
,
b
b
0 < y < b.
So we must expand f in a Fourier sine series on the interval 0 < y < b. Hence
Z b
n⇡a
1
2
n⇡y
cn sinh
= bn ) c n =
) cn =
f (y) sin
dy,
n⇡a bn
n⇡a
b
sinh b
b sinh b 0
b
for n = 1, 2, . . ..
In conclusion, we get the following result
Dirichlet Problem for a Rectangle
(Laplace’s equation)
uxx + uyy = 0,
(Boundary conditions)
The solution is u(x, y) =
u(x, 0) = 0,
u(x, b) = 0,
0 < x < a,
u(0, y) = 0,
u(a, y) = f (y),
0 < y < b.
1
X
cn sinh
n=1
where
cn =
0 < x < a, 0 < y < b,
2
b sinh n⇡a
b
n⇡x
n⇡y
sin
,
b
b
Z b
n⇡y
f (y) sin
dy,
b
0
n = 1, 2, . . . .
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
343
The method of separation of variables can always be used to solve the Dirichlet problem for a
rectangle R if u is zero on three sides of R. In other words, we can use the same strategy we did
above with these problems:
1.
(Laplace’s equation)
(Boundary conditions)
uxx + uyy = 0,
0 < x < a, 0 < y < b,
u(x, 0) = 0,
u(x, b) = 0,
u(0, y) = g1 (y),
u(a, y) = 0,
0 < x < a,
0 < y < b.
2.
(Laplace’s equation)
(Boundary conditions)
uxx + uyy = 0,
0 < x < a, 0 < y < b,
u(x, 0) = 0,
u(x, b) = f2 (x),
0 < x < a,
u(0, y) = 0,
u(a, y) = 0,
0 < y < b.
3.
(Laplace’s equation)
(Boundary conditions)
uxx + uyy = 0,
0 < x < a, 0 < y < b,
u(x, 0) = f1 (x),
u(0, y) = 0,
u(x, b) = 0,
u(a, y) = 0,
0 < x < a,
0 < y < b.
We can also solve an arbitrary Dirichlet problem for a rectangular R by splitting it up into four
problems where u is zero on three sides of R. In other words, if we would like to solve
(Laplace’s equation)
uxx + uyy = 0,
(Boundary conditions)
0 < x < a, 0 < y < b,
u(x, 0) = f1 (x),
u(x, b) = f2 (x),
0 < x < a,
u(0, y) = g1 (y),
u(a, y) = g2 (y),
0 < y < b,
we splits it up into four problems where each problem has only one nonhomogeneous boundary
condition while the remaining three are homogeneous ones as we mentioned earlier. Once we solve
all four problems, the solution to our original problem will be their sum
u(x, y) = u1 (x, y) + u2 (x, y) + u3 (x, y) + u4 (x, y).
This is true since the Laplace’s equation is linear and homogeneous, so the sum of solutions to
Laplace’s equation will also be a solution. Moreover, this solution will also satisfy each of the four
original boundary conditions. For example,
u(x, 0) = u1 (x, 0) + u2 (x, y) + u3 (x, 0) + u4 (x, 0) = f1 (x) + 0 + 0 + 0 = f1 (x).
8.5.2
Neumann Problem for a Rectangle*
Consider the problem of finding the function u satisfying
Neumann Problem for a Rectangle
(Laplace’s equation)
(Boundary conditions)
uxx + uyy = 0,
0 < x < a, 0 < y < b,
uy (x, 0) = 0,
uy (x, b) = 0,
0 < x < a,
ux (0, y) = 0,
ux (a, y) = f (y),
0 < y < b.
Assume that, by the method of separation of variables,
u(x, y) = X(x)Y (y).
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
344
Substitute u back into the Laplace’s equation,
X 00 Y + XY 00 = 0
where
X 00 Y = XY 00
)
)
X 00
Y 00
=
=
X
Y
,
is the separation constant. Then we get two ordinary di↵erential equations,
Y 00 + Y = 0,
X 00
X = 0.
Also the boundary conditions have changed to,
uy (x, 0) = 0
)
X(x)Y 0 (0) = 0
)
X 0 (0)Y (y) = 0
)
uy (x, b) = 0
ux (0, y) = 0
X(x) = 0 or Y 0 (0) = 0
)
X(x)Y 0 (b) = 0
X(x) = 0 or Y 0 (b) = 0
)
X 0 (0) = 0 or Y (y) = 0.
)
Since we’re looking for nontrivial solution, new boundary conditions are
Y 0 (0) = 0,
Y 0 (b) = 0,
X 0 (0) = 0.
Until now we have transformed the problem to
Y 00 + Y = 0,
X
00
X = 0,
Y 0 (0) = 0,
Y 0 (b) = 0,
(8.32)
0
X (0) = 0.
(8.33)
The eigenvalue problem (8.32) gives
n
=
n2 ⇡ 2
,
b2
Yn (y) = cos
n⇡y
,
b
n = 0, 1, 2, . . . .
So Equation (8.33) becomes
X 00
X=0
Note here that
)
X 00
n2 ⇡ 2
X=0
b2
X 0 (x) =
)
X(x) = c1 cosh
n⇡x
n⇡x
+ c2 sinh
.
b
b
n⇡
n⇡x n⇡
n⇡x
c1 sinh
+
c2 cosh
.
b
b
b
b
Apply initial condition X 0 (0) = 0,
0=
n⇡
n⇡
c1 sinh 0 +
c2 cosh 0
b
b
)
So
Xn (x) = cosh
n⇡
c2 = 0
b
)
c2 = 0.
n⇡x
.
b
Therefore
n⇡x
n⇡y
cos
b
b
is a solution for every nonnegative integer n and they form the fundamental set of solutions. Thus
the function
1
X
n⇡x
n⇡y
u(x, y) = c0 +
cn cosh
cos
b
b
n=1
un (x, y) = Xn (x)Yn (y) = cosh
is a general solution. Now we find ux (x, y) and evaluate it at x = a,
ux (a, y) =
1
X
n⇡
n⇡a
n⇡y
cn sinh
cos
.
b
b
b
n=1
CHAPTER 8. PARTIAL DIFFERENTIAL EQUATION
345
But it is given that u(a, y) = f (y). Therefore, we must choose the constants cn such that
f (y) =
1
X
n⇡
n⇡a
n⇡y
cn sinh
cos
,
b
b
b
n=1
0 < y < b.
(8.34)
So we must expand f in a Fourier cosine series on the interval 0 < y < b, which is
1
a0 X
n⇡y
+
an cos
2
b
n=1
Z b
1
X
1
2
=
f (y) dy +
b 0
b
n=1
f (y) =
Z
!
n⇡y
n⇡y
f (y) cos
dy cos
.
b
b
b
0
(8.35)
However, we cannot equate coefficients in (8.34) and (8.35) since the series (8.34) has no constant
term. Therefore, the condition
Z b
f (y) dy = 0
0
is necessary for this Neumann problem to have a solution. If this is the case, then by (8.34) and
(8.35),
!
!
Z b
Z b
n⇡
n⇡a
2
n⇡y
2
n⇡y
cn sinh
=
f (y) cos
dy
) cn =
f (y) cos
dy ,
b
b
b
b
n⇡ sinh n⇡a
b
0
0
b
where n 1. Lastly, note that c0 remains arbitrary, and thus the solution u(x, y) is only determined up to an additive constant. This is a property of all Neumann problems.
In conclusion, we get the following result
Neumann Problem for a Rectangle
(Laplace’s equation)
uxx + uyy = 0,
(Boundary conditions)
where
Z
0 < x < a, 0 < y < b,
uy (x, 0) = 0,
uy (x, b) = 0,
0 < x < a,
ux (0, y) = 0,
ux (a, y) = f (y),
0 < y < b,
b
f (y) dy = 0 is necessary.
0
The solution is u(x, y) = c0 +
1
X
cn cosh
n=1
n⇡x
n⇡y
cos
,
b
b
where c0 is arbitrary and,
2
cn =
n⇡ sinh n⇡a
b
Z
b
0
!
n⇡y
f (y) cos
dy ,
b
n = 1, 2, . . . .
Bibliography
[1] Shair Ahmad and Antonio Ambrosetti, A textbook on Ordinary Di↵erential Equations:
Springer International Publishing Switzerland, 2014.
[2] William E. Boyce and Richard C. Diprima, Elementary Di↵erential Equations and
Boundary Value Problems: John Wiley & Sons, Inc., 2012.
[3] M. Braun, Di↵erential Equations and Their Applications: Springer-Verlag New York,
Inc., 1983
[4] Christian Constanda, Di↵erential Equations: Springer Science+Business Media New York,
2013
[5] Morris Tenenbaum and Harry Pollard, Ordinary Di↵erential Equations: Dover Publications, Inc., New York, 1985
[6] Paul Dawkins, Di↵erential Equations: http://tutorial.math.lamar.edu/terms.aspx,
2007.
[7] Zachary S Tseng, Math 250/251 Notes: http://www.math.psu.edu/tseng/class/M251_
downloads.html, 2008, 2012.
[8] Morris Kline, Calculus: Dover Publications, Inc., New York, 1998
[9] E. T. Bell, Men of Mathematics: Simon & Schuster, Inc., 1986.
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Inc., 2010.
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//www-history.mcs.st-andrews.ac.uk/BiogIndex.html.
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346
Springer
Index
Abel’s theorem, 113
amplitude, 139
autonomous, 11, 268
Fourier series, 299
exact equation, 33
exactness, 33
explicit solution, 15
beat, 150
Bernoulli equation, 28
boundary conditions, 281
boundary value problem(BVP), 281
falling object, 74
first order di↵erential equation, 15
Fourier cosine series, 299
Fourier series, 292
even function, 299
odd function, 299
Fourier sine series, 299
fundamental peiod, 291
fundamental set of solutions
higher order, 163
fundamental set of solutions
second order, 83
canonical form, 22
center, 247
characteristic equation
matrix, 238
higher order, 168
second order, 82
characteristic polynomial, 238
column matrix, 236
column vector, 236
complementary solution, 120
convolution, 225
cosine series extension, 302
critical point, 60, 240, 268
critically damped, 142
gamma function, 182
general solution, 15
heat (conduction) equation, 309
heat conduction problem
bar with both ends insulated, 325
homogeneous boundary conditions, 310
nonhomogeneous boundary conditions,
320
type I, 310
solution, 315
type II, 320
solution, 321
type III, 325
solution, 326
Heaviside function, 203
higher order di↵erential equation, 162
homogeneous
system, 228
di↵erential equation, 3
higher order, 168
solution, 168
second order, 81
solution, 84
determinant, 237
Dirac Delta function, 215
Laplace transform, 216
direction field, 11
Dirichlet problem, 347
eigenfunction, 285
eigenvalue
boundary value problem, 285
matrix, 237
eigenvalues problem, 285
eigenvector, 237
equilibrium solution, 11, 60, 240
Euler equation, 99
Euler’s formula, 88
Euler-Fourier formulas, 292
even extension, 302
even function, 298
347
INDEX
homogeneous function, 160
homogeneous polar equation, 20
horizontal propagation speed, 335
hyperbolic function, 346
identity matrix, 235
improper node, 253
impulse function, 215
initial value problem(IVP), 15
initial-boundary value problem(IBVP), 310
integral curves, 11
integrating factor, 22
interval of validity, 54
inverse Laplace transform, 185
properties, 185
inverse matrix, 237
Jacobian matrix, 271
Laplace transform
definition, 174
properties, 175
table, 176
Laplace’s equation, 346
left-hand limit, 292
limiting velocity, 75
linear
ordinary di↵erential equation, 2
linearly dependent, 163
linearly independent, 163
main diagonal, 235
matrix, 235
multiplication, 236
addition, 236
dimension, 235
eigenvalue, 237
eigenvector, 237
inverse, 237
nonsingular, 237
scalar multiplication, 236
singular, 237
size, 235
subtraction, 236
mechanical vibrations
damped free vibrations, 142
undamped forced vibrations, 150
undamped free vibrations, 139
method of undetermined coefficients, 120
mixing problems, 67
modeling, 67
first order
falling object, 74
mixing problems, 67
348
motion of an object in a resistive
fluid medium, 78
second order
mechanical vibrations, 136
motion of an object in a resistive fluid
medium, 78
natural frequency, 139
natural period, 139
Neumann problem, 349
node, 244
nonhomogeneous
system, 228
di↵erential equation, 3
second order, 81
nonlinear
ordinary di↵erential equation, 2
odd extension, 302
odd function, 298
Fourier series, 299
order, 1
ordinary di↵erential equation(ODE), 1
overdamped, 142
partial di↵erential equation, 281
partial di↵erential equation(PDE), 1
partial fraction decomposition, 188
particular solution, 15, 120
period, 291
periodic function, 291
phase, 139
phase line, 60
phase portrait, 240
piecewise continuous function, 291
potential equation, 346
principle of superposition
higher order, 162
second order, 82
proper node, 251
quasi frequency, 143
quasi period, 143
reduction of order, 101
resonance, 151
Riccati equation, 30
right-hand limit, 292
row matrix, 236
row vector, 236
saddle point, 242
second order di↵erential equation, 81
separable equation, 15
separation of variables, 310
INDEX
sine series extension, 302
solution, 8
spiral point, 249
square matrix, 235
stability
first order, 60
system, 240
standard form, 22
star point, 251
steady-state solution, 320
systems of first order linear equations, 228
solution, 240
Taylor series, 271
the existence and uniqueness theorem
first order
linear, 54
nonlinear, 57
second order linear, 108
higher order linear, 162
349
the Fourier convergence theorem, 292
thermal di↵usivity, 309
trace, 235
trajectory, 240
transient part, 320
translation, 93
underdamped, 142
unit step function, 203
Laplace transform, 205
wave equation, 335
wave equation problem, 335
solution, 337
special cases, 338
Wronskian
second order, 83
higher order, 163
zero matrix, 235