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Nonlinear Dierential Equations
and Chaos
(AM 3813 Course Notes)
Pei Yu and Christopher Essex
The University of Western Ontario
London, Ontario, Canada N6A 5B7
[email protected]
http:==publish.uwo.ca/pyu
January 2022

Chapter 1
INTRODUCTION
This course is aboutnonlinear dierential equations(primarily ordinary), but it is also
about a change in scientic perspective that has occurred in the last 30 years or so. Although
there are many interesting consequences and perspectives to this change, it primarily aects
our views about cause and eect (determinism) and ultimately what predicting the future
means in this area.
Had anyone predicted that new discoveries could be made in dynamics 300 years after the
publication of Newton's Principia, they would have been thought naive or foolish. Yet, in the
past three decades, new phenomena in nonlinear dynamics were discovered, principal among
these being chaotic and unpredictable behavior from apparently deterministic systems.
This course will touch on both classical (algebraic, e.g. perturbation) methods and
modern (geometric, e.g. phase-space) methods, but the emphasis will be put on the new
geometrical ideas that are revolutionizing dynamical systems theory. For being an applied
mathematician, an engineer, a scientist, an analyst or an experimentalist of all disciplines, we
are usually concerned with modeling and understanding the time evolution of real systems.
Most real-world problems confronting us, being neither linear nor even nearly linear, fall
outside the domain of traditional closed-form analysis, and must be tackled in the rst
instance on a computer. But numerical simulations, like physical experiments, typically
produce unwieldy masses of data, and for both of these, phase-space concepts must be
recognized as an essential guide to the structuring of the investigation and the interpretation
of the results.
At the present time we are indeed witnessing a spectacular blossoming of nonlinear
dynamics, made possible on the one hand by great theoretical strides in Poincare's qualitative
topological approach and on the other hand by the wide availability of powerful computers.
This has been stimulated and sustained by important and exciting new applications that have
1

AM3813B Chapter 1 2
multiplied throughout the biological, ecological, social and economic sciences, far beyond
the still vibrant traditional elds of mechanics, physics, chemistry and engineering. Indeed,
wherever the time evolution of a natural or human-made system needs to be modeled and
explored, the subtle and versatile techniques of dynamical systems theory are being invoked.
A signicant element of this major thrust, which nicely epitomizes its most advanced
theoretical, computational, and experimental features, is the discovery and delineation of
chaotic motions in remarkably simple deterministic models from a galaxy of disciplines rang-
ing from population dynamics and meteorology to lasers and particle accelerators. The recent
discovery of quite large regimes of chaos in the long-familiar, sinusoidally driven Dung and
Van der Pol oscillators emphasizes just how much was missed by the classical analysis of
ordinary dierential equations.
Before we start, we should agree about something: chaos is part of an even grander
subject known asdynamics. This is the subject which deals with change, with systems
which evolve in time. Whether the system in question settles down in equilibrium, keeps
repeating in cycles, or does something more complicated, it isdynamicsthat we use to
analyze the behavior.
1.1 Brief History of Dynamics
The subject of dynamics began in the mid-1600s, when Newton invented dierential
equations, discovered his laws of motion and universal gravitation, and combined them to
explain Kepler's laws of planetary motion. Specically, Newton solved the two-body problem
{ the problem of calculating the motion of the earth around the sun. However, it was
found later that Newton's analytical methods could not be extended to solve three-body
problems, in the sense of obtaining explicit formulas for the motions of three interacting
bodies. The breakthrough came with the work of Poincare in the late 1800s. He developed
a powerfulgeometricapproach to analyzing such questions, emphasizing qualitative rather
than quantitative properties.
In the rst half of this century, dynamics was largely concerned with nonlinear oscillators
and their applications in physics and engineering. Nonlinear oscillators played a vital role

AM3813B Chapter 1 3
in the development of such techniques as radio, radar, phase-locked loops, and lasers. The
invention of the high-speed computer in the 1950s nally led to the meteorologist { Lorenz's
discovery in 1963 of chaotic motion on a strange attractor. He studied a simplied model
of convection rolls in the atmosphere and found that the solutions to his equations never
settle down to equilibrium or to a periodic state { instead they continued to oscillate in
an irregular, aperiodic fashion. Moreover, if he started his simulations from two slightly
dierent initial conditions, the resulting behaviors would soon become totally dierent. This
is known as beingsensitiveto the initial conditions of chaos.
Table 1.1 Dynamics { A Brief History
1666 Newton Invention of calculus, explanation of planetary motion
1700s Flowering of calculus and classical mechanics
1800s Analytical studies of planetary motion
1890s Poincare Geometric approach, nightmares of chaos
1920{1950 Nonlinear oscillators in physics and engineering,
invention of radio, radar, laser
1920{1960 Birkho Complex behavior in Hamiltonian mechanics
Kolmogorov
Arnol'd
Moser
1963 Lorenz Strange attractors in simple model of convection
1970s Ruelle & Takens Turbulence and chaos
May Chaos in logistic map
Feigenbaum Universality and renormalization, connection between
chaos and phase transition
Experimental studies of chaos
Winfree Nonlinear oscillators in biology
Mandelbrot Fractals
1980s Widespread interest in chaos, fractals, oscillators,
and their applications

AM3813B Chapter 1 4
Lorenz's work had little impact until the 1970s, the boom years for chaos. Here are
some of the main developments of that glorious decade. In 1971 Ruelle and Takens proposed
a new theory for the onset of turbulence in uids, based on abstract considerations about
strange attractors. A few years later, May found examples of chaos in iterated mappings
arising in population biology. Next the physicist Feigenbaum surprisingly discovered that
there are certain universal laws governing the transition from regular to chaotic behavior;
roughly speaking, completely dierent systems can go chaotic in the same way. At the same
time, many experimentalists tested the new ideas about chaos in experiments on uids,
chemical reactions, electronic circuits, mechanical oscillators, and semiconductors. There
were two other major developments in dynamics in the 1970s. Mandelbrot codied and
popularized fractals, produced magnicent computer graphics of them, and showed how
they could be applied in a variety of subjects. And in the emerging area of mathematical
biology, Winfree applied the geometric methods of dynamics to biological oscillations. By
the 1980s many people were working on dynamics, with contributions too numerous to list.
Table 1.1 summarizes this history.
1.2 An Overview of Nonlinear Phenomena
There are two main types of dynamical systems:dierential equationsanditerated
maps(also known as dierence equations). Dierential equations describe the evolution of
systems in continuous time, whereas iterated maps arise in problems where time is discrete.
For dierential equations, the main distinction is between ordinary and partial dierential
equations. For instance, the equation describing the motion of a real pendulum is given by
d
2
x
dt
2
+
g
l
sinx= 0 (1.1)
wherexis the angle of the pendulum from vertical,gis the acceleration due to gravity, and
lis the length of the pendulum. This equation is an ordinary dierential equation, because
it involves only the ordinary derivative
d
2
x
dt
2
:That is, there is only one independent variable,
the timet :In contrast, the heat equation
@u
@t
=
@
2
u
@x
2
(1.2)

AM3813B Chapter 1 5
is a partial dierential equation { it has both timetand spacexas independent variables.
The main attention of this course will be focused on ordinary dierential equations and
dierence equations.
A general form for ordinary dierential equations can be described by the system
_x1=f1(x1; x2; ; xn)
_x2=f2(x1; x2; ; xn)
.
.
.
_xn=fn(x1; x2; ; xn)
where the overdots denote dierentiation with respect tot ;i.e. _xi
dxi
dt
:The variables
x1; x2; ; xnare called state variables, and the functionsf1; f2; ; fnare determined
by the properties of the problem. The oscillator (??) can be rewritten in form of (??) as
_x1=x2
_x2=
g
l
sinx1 (1.4)
by introducingx1=xandx2= _x :This equation isnonlinear. Nonlinearity makes the
pendulum equation very dicult to solve analytically. However, for smallx ;since sinxx ;
equation (??) can be approximated by
_x1=x2
_x2=
g
l
x1 (1.5)
which is alinearequation and can then be solved easily. But by restricting to smallx ;we
are throwing out some of the physics, like motions where the pendulum whirls over the top.
Ed Lorenz (1963) derived a three-dimensional chaotic system from a drastically simplied
model of convection rolls in the atmosphere. This well-known equation is now called Lorenz
equations, given by
_x=(yx)
_y= xyx z
_z=x y z (1.6)

AM3813B Chapter 1 6
where; ; >0 are parameters. The same equations also arise in models of lasers and
dynamos.
Lorenz discovered that this simple-looking deterministic system could have extremely
erratic dynamics: over a wide range of parameters, the solutions oscillate irregularly, never
exactly repeating but always remaining in a bounded region that they settled onto a com-
plicated set, now called astrange attractor(see Fig. 1.1).0
10
20
30
40
50
-20 -10 0 10 20
z
x
Fig. 1.1 Strange attractor of the Lorenz system.
1.3 A Dynamical View of the World
Having established the ideas of nonlinearity and phase space, we can now present a
framework for dynamics and its applications, which is shown in Fig. 1.2. The framework
has two axes. One axis tells us the number of variables needed to characterize the state
of the system, i.e. the dimension of the phase space. The other axis tells us whether the
system is linear or nonlinear. For example, the equation describing the exponential growth
of a population of organisms is given by the rst-order dierential equation,
_x= x
wherexis the population at timetandis the growth rate. We place this system in

AM3813B Chapter 1 7
the column labeled \n= 1" becauseonepiece of information { the current value of the
population { is sucient to predict the population at any later time. The system is also
classied as linear. As a second example, the equation of the pendulum (??) or (??) is
given bytwovariables: its current anglexand angular velocity _x :Because two variables
are needed to specify the state, the pendulum belongs in then= 2 column of Fig. 1.2.
Moreover, the system is nonlinear. Hence the pendulum is in the lower, nonlinear half of the
n= 2 column.
One can continue to classify systems in this way, and the result will be something like the
framework shown in Fig. 1.2. Admittedly, some aspects of the picture are debatable. You
might think that some topics should be added, or placed dierently, or even that more axes
are needed { the point is to think about classifying systems on the basis of their dynamics.
There are some striking patterns in Fig. 1.2. All of the simplest systems occur in the
upper left-hand corner. These are the small linear systems. Roughly speaking, these linear
systems exhibit growth, decay, or equilibrium whenn= 1;or oscillations whenn= 2:The
italicized phrases in this gure indicate that these broad classes of phenomena rst arise in
this part of the diagram.
The next most familiar part of the picture is the upper right-hand corner. This is the
domain of classical applied mathematics and mathematical physics where the linear partial
dierential equations live. These partial dierential equations involve an innite \contin-
uum" of variables because each point in space contributes additional degrees of freedom.
Even though these systems are large, they are tractable by using linear techniques such as
Fourier analysis and transform methods.
The lower half of Fig. 1.2 is the nonlinear part. As we increase the phase space dimension
fromn= 1 ton= 3;we encounter new phenomena at every step, from xed points and
bifurcations whenn= 1;to nonlinear oscillations whenn= 2;and nally chaos and
fractals whenn= 3:In all cases, a geometric approach proves to be very powerful, and
gives us most of the information we want, even though we usually can't solve the equations
in the traditional sense of nding a formula for the answer. Our journey will also take us
to some of the most exciting parts of modern science, such as mathematical biology and

AM3813B Chapter 1 8
condensed-matter physics.
It is noted that the framework also contains a region forbiddingly marked \The frontier".
These topics are not completely unexplored, of course, but it is fair to say that they lie at the
limits of current understanding. The problems are very hard, because they are both large
and nonlinear. The resulting behavior is typically complicated inboth space and time, as in
the motion of a turbulent uid or the patterns of electrical activity in a brillating heart.

AM3813B Chapter 1 9Elasticity Linear oscillator
Mass and spring
RLC circuit
2-body problem
(Kepler, Newton)
Pendulum
Anharmonic oscillators
Limit cycles
(neurons, heart cells)
Nonlinear electronics
(Van der Pol, Josephson)
Fixed points
Bifurcation
Overdamped systems
relaxational dynamics
Logistic equation
Wave equations
Electromagnetism (Maxwell)
Quantum mechanics
(Schrodinger, Heisenberg, Dirac)
Heat and diffusion
Acoustics
Spatio-temporal complexity
Nonlinear waves (shocks, solitons)
plasmas
Earthquakes
General relativity (Einstein)
Quantum field theory
Reaction-diffusion,
biological and chemical waves
Fibrillation
Epilepsy
Turbulent fluids (Navier-Stokes)
Life
Coupled nonlinear oscillators
Lasers, nonlinear optics
Nonequilibrium statistical
mechanics
Nonlinear solid-state physics
(semiconductors)
Josephson arrays
Heart cell synchronization
Economics
Neural networks
Immune system
Stranger attractors
Chaos
(Lorenz)
3-body problem (Poincare)
Chemical kinetics
Iterated maps (Feigenbaum)
Fractals
(Mandelbrot)
Forced nonlinear oscillators
(Levinson, Smale)
Practical uses of chaos
Quantum chaos ?
The frontier
Coupled harmonic oscillators
Solid-state physics
Equilibrium statistical
Civil engineering
structures
Electrical engineering
Nonlinear
Nonlinearity
Linear
Number of variables
n = 2 n = 1 n > 2n > > 1 Continuum
equilibrium
Growth, decay, or Oscillations Collective phenomena Waves and patters
Exponential growth
RC circuit
Radioactive decay
for single species
Biological oscillators
Predator-prey cycles
Molecular dynamics
Mechanics
Viscous fluids
Ecosystems
Fig. 1.2 A dynamical view of the world

Chapter 2
BRIEF REVIEW OF LINEAR ODE
In this chapter we will give a brief review of linear ordinary dierential equations (ODE). The
main attention will focus on the systems with constant coecients. (General theory of linear
dierential equations may be found in, for example, \Advanced Engineering Mathematics"
by Peter V. O'Neil, 1991.) These linear ODE are important because many physical and
engineering problems can be described by these kind of ODE. On one hand, we need to nd
solutions and study the properties of the solutions; on the other hand, many ODE can be
approximated by rst-order ODE.
2.1 Homogeneous Dierential Equations
We start with thenth-order dierential equation, given by
d
n
y
dt
n
+a1
d
n1
y
dt
n1
+a2
d
n2
y
dt
n2
+ +any= 0 (2.1)
whereai's are constants (real or complex). We will also use prime to denote dierentia-
tions, thus,y
0
,y
00
y(n), etc. represent the 1st, 2nd, n-th derivatives ofy. Use the
dierential operatorD=
d
dt
, one obtains
d
2
dt
2
=D
2
;
d
3
dt
3
=D
3
;
d
n
dt
n
=D
n
(2.2)
and dene the operator function
L(D) =D
n
+a1D
n1
+a2D
n2
+ +an; (2.3)
then,
L(D)y=D
n
y+a1D
n1
y+a2D
n2
y+ +any : (2.4)
Therefore, the linear DE (2.1) can be written as
L(D)y= 0: (2.5)
10

AM3813B Chapter 2 11
Suppose thatL(D) andM(D) are two polynomials ofD, then
L(D) (y1+y2) =L(D)y1+L(D)y2
[L(D) +M(D) ]y=L(D)y+M(D)y
L(D) [M(D)y] = [L(D)M(D) ]y : (2.6)
Moreover,
D e
t
=
d
dt

e
t

= e
t
D
2
e
t
=
2
e
t
.
.
.
D
n
e
t
=
n
e
t
(2.7)
which indicates that
L(D)e
t
=L()e
t
: (2.8)
Hence, ifis a root of the polynomialL(), then functione
t
is a solution of the DE (2.1).
Ife
t
is a solution of (2.1), thenmust be the root ofL(D). We call
L() = 0 (2.9)
thecharacteristic functionorcharacteristic polynomial(C.P.) of the DE (2.1).
2.1.1 Distinct roots.If the C.P. hasndierent roots1; 2; ; n, then it can
be shown that thenfunctions
yi(t) =e
it
; i= 1;2; n (2.10)
are independent and construct the basic solutions of the DE (2.1). Thus, the general solution
of (2.1) can be given as a linear combination of these basic solutions:
y(t) =
n
X
i=1
Cie
it
(2.11)
whereCi; i= 1;2; ; nare arbitrary constants.

AM3813B Chapter 2 12
When all coecientsaiin (2.1) are real and if the C.P. (2.9) has a complex root
=+i , then the complex conjugate

=i is also the root ofL(). Thus, both
e
t
=e
(+i)t
=e
t
e
it
=e
t
(cost+isint) (2.12)
(whereEuler's formulahas been used) and
e

t
=e
t
(costisint) (2.13)
are solutions of the DE (2.1). So, the real functions
y=
1
2

e
t
+e

t

=e
t
cost (2.14)
and
y=
1
2i

e
t
e

t

=e
t
sint (2.15)
are solutions of (2.1).
In general, supposeL() = 0 has 2kcomplex roots
1=

2=1+i 1
2=

4=2+i 2
.
.
.
2k1=

2k=k+i k; (2.16)
andn2kreal roots2k+1; 2k+2; n, then,e
1t
cos1t,e
1t
sin1t ,e
kt
coskt,
e
kt
sinkt,e
2k+1t
,e
nt
are basic solutions of the DE (2.1), each of which is a reals
function.
Example 2.1.Find all solutions of the DE
y
00
+ 3y
0
+ 2y= 0y2R: (2.17)
Solution:The C.P. of (2.17) is
L() =
2
+ 3+ 2 = (+ 1) (+ 2) (2.18)

AM3813B Chapter 2 13
which yields roots1=1 and2=2. So, the general solution of (2.17) is
y(t) =C1e
t
+C2e
2t
(2.19)
whereC1andC2are arbitrary real constants. (Ify2CthenC1andC2are complex
constants.)
Example 2.2.Find all solutions of the DE
d
3
y
dt
3
3
d
2
y
dt
2
+ 9
dy
dt
+ 13y= 0y2R: (2.20)
Solution:The characteristic polynomial is
L() =
3
3
2
+ 9+ 13 = (+ 1) (
2
4+ 13) (2.21)
which, in turn, results in1=1,2= 2+3iand2= 23i. Hence the general solution
of (2.20) is given by
y(t) =C1e
t
+C2e
2t
cos 3t+C3e
2t
sin 3t (2.22)
whereCi= 1;2;3 are arbitrary real constants.
The complex form solution of (2.20) is
y(t) =C1e
t
+C2e
(2+3i)t
+

C2e
(23i)t
(2.23)
whereC1is a real constant whileC2is a complex constant.
Example 2.3.Find all solutions of damped system
mx+k x= 0 (m; k >0): (2.24)
Solution:Note that we also frequently use dot to denote dierentiations with respect to
time. (If the independent variable is not time, then we usually use prime, not the dot, to
denote the dierentiations.) The C.P. of (2.24) is
L() =m
2
+k= 0 =) =
r
k
m
i : (2.25)
Thus,
x(t) =C1cos
r
k
m
t+C2sin
r
k
m
t (2.26)

AM3813B Chapter 2 14
whereC1andC2are arbitrary real constants. Suppose at timet= 0,
x(0) =x0 and _x(0) = _x0; (2.27)
then
x0=x(0) =C1 and _x0= _x(0) =C2
r
k
m
=)C2=
r
m
k
_x0: (2.28)
Hence,
x(t) =x0cos
r
k
m
t+ _x0
r
m
k
sin
r
k
m
t (2.29)
is a solution satisfying the initial condition (2.27).
Let!=
r
k
m
which is callednatural frequencyof system (2.24), and (2.24) becomes
x+!
2
x= 0: (2.30)
The solution of (2.26) can now be written as
x(t) =C1cos!t+C2sin!t=rcos(!t+) (2.31)
where
r=
q
C
2
1+C
2
2 and =tan
1

C2
C1

: (2.32)
2.1.2 Repeated roots.Next, let's discuss whenL() has repeated roots. We know
that if1and2are the roots ofL() = 0, then
1
12

e
1t
e
2t

is also the root of
L() = 0. But
lim
2!1
1
12

e
1t
e
2t

= lim
2!1
1
12
e
1t

1e
(21)t

=e
1t
lim
s!0
1e
st
s
(s=21)
=e
1t
tlim
s!0
e
st
(L
0
Hospital Rule)
=t e
1t
: (2.33)

AM3813B Chapter 2 15
This indicates that if1is a twice repeated root ofL() = 0, thene
1t
andt e
1t
are all solutions of (2.1). Similarly, if1is aktime repeated root ofL() = 0, then
e
1t
; t e
1t
; t
k1
e
1t
are all solutions of (2.1).
Suppose1; 2; mare distinct roots ofL() = 0, and they repeat, respectively,
k1; k2; kmtimes, such thatk1+k2+ km=n, then the functions
y1(t) =e
1t
; y 2(t) =t e
1t
; yk1(t) =t
k11
e
1t
;
yk1+1(t) =e
2t
; y k1+2(t) =t e
2t
; yk1+k2(t) =t
k21
e
2t
;
.
.
.
ynkm+1(t) =e
mt
; y nkm+2(t) =t e
mt
; yn(t) =t
km1
e
mt
are thebasic solutionsof the DE (2.1).
Example 2.4.Find all solutions of the DE
d
3
y
dt
3
+ 3
d
2
y
dt
2
+ 3
dy
dt
+y= 0: (2.34)
Solution:The C.P. of (2.34) is
L() =
3
+ 3
2
+ 3+ 1 = (+ 1)
3
: (2.35)
So=1 is a repeated roots (3 times) of (2.34). Hence,
y(t) = (C1+C2t+C3t
2
)e
t
(2.36)
is the general solution, in whichCiare arbitrary constants.
Example 2.5.Find all solutions of the DE
d
4
y
dt
4
+ 2
d
2
y
dt
2
+y= 0: (2.37)
Solution:
L() =
4
+ 2
2
+ 1 = (
2
+ 1)
2
(2.38)
which yields1=i(repeated twice) and2=i(repeated twice). Thus
y(t) = (A1+A2t)e
it
+ (A3+A4t)e
it
(2.39)

AM3813B Chapter 2 16
whereAiare complex constants. The real solution can be found following the procedure of
single roots. We conclude that
cost;sint; tcostandtsint
are the basic solutions of the DE (2.37). Hence,
y(t) = (C1+C2t) cost+ (C3+C4t) sint (2.40)
whereCiare real constants.
2.2 Non-homogeneous Dierential Equations
A homogeneous DE, as we discussed in the previous section, can be written in the general
form
L(D)y= 0: (2.41)
Non-homogeneous DEs are given by
L(D)y=f(t): (2.42)
The solution of (2.42) can be found using the followingprinciple of superposition:
(i)
p
(t) is a solution of (2.42) and
h
(t) is a solution of (2.41), then(t) =
h
(t)+
p
(t)
is a solution of (2.42).
(ii) 1(t) and2(t) are both solutions of (2.42), then(t) =1(t) +2(t) is also a
solution of (2.42).
(iii) (t) and (t) are solutions of the DEL(D)y=f1(t) andL(D)y=f2(t), respec-
tively, then(t) =a (t) +b (t) is a solution of the DEL(D)y=a f1(t) +b f2(t),
whereaandbare constants.
Now suppose that the basic solutions of the corresponding homogeneous DE of (2.42)
are1(t); 2(t); n(t). Then the general solution of (2.42) is given by
y(t) =
n
X
i=1
Ckk(t) +
p
(t) (2.43)

AM3813B Chapter 2 17
whereCkare arbitrary constants, and
p
(t) is aparticular solutionof (2.42).
Having found the basic solutions of the corresponding homogeneous DE, a particular
solution of (2.42) can be found by using methods such as themethod of undetermined coef-
cientsorvariation of parameters.
2.3 The Method of Undetermined Coecients
Suppose that the functionf(t) is given in so calledpseudo-polynomialform:
f(t) =
k
X
j=1
Pj(t)e
jt
; (2.44)
wherePj(t) are polynomials oft, andjare constants. Since
e
t
cost=
1
2

e
(+i)t
+e
(i)t

e
t
sint=
1
2i

e
(+i)t
e
(i)t

soe
t
[R(t) cost+S(t) sint] is also pseudo-polynomial, whereR(t) andS(t) are poly-
nomials oft.
According to the principle of superposition, if we nd a solution for each term off(t),
then adding them together results in a particular solution of the DE
L(D)y=f(t)
k
X
j=1
Pj(t)e
jt
:
Considerfj(t) =Pj(t)e
jt
then the DE
L(D)y=fj(t)Pj(t)e
jt
(2.45)
has a solution in the form of
yj(t) =Qj(t)e
jt
; (2.46)
whereQj(t) is a polynomial oft, whose degreem+s,mis the degree ofPj(t) while
s=
(
0 ifjis not the roots ofL() = 0 (Non-resonant Case);
lifjis the roots ofL() = 0;repeatedltimes (Resonant case):
(2.47)

AM3813B Chapter 2 18
Example 2.6.Find all solutions of the DE
dx
dt
=x+ 2ye
t
dy
dt
= 4x+ 3y+ 4e
t
: (2.48)
First, we can obtain a 2nd order DE forxoryas follows: rewrite (2.48) in the form
D xx2y=e
t
4x+D y3y= 4e
t
(2.49)
or
(D1)x2y=e
t
4x+ (D3)y= 4e
t
which yields (treatingDas a constant and then soloving the above two algebraic equations)
[(D1) (D3)8]x= 12e
t
[(D1) (D3)8]y=12e
t
by eliminating eitherxory. So we have
(D
2
4D5)x=
d
2
x
dt
2
4
dx
dt
5x= 12e
t
(2.50)
and so
L() =
2
45 =) 1=1 and2= 5: (2.51)
Hence, the basic solutions of (2.50) aree
t
ande
5t
. To nd a particular solution of (2.50),
notice thatf(t) = 12e
t
and=1 which is a single root of the C.P. Since the degree
ofP(t) = 12 is 0 andl= 1, som+s= 1. Therefore, the particular solution should be in
the form of
x
p
(t) =t a e
t
: (2.52)
Dierentiatingx
p
(t) once and twice yields
dx
p
dt
=a e
t
a t e
t
=(a ta)e
t
d
2
x
p
dt
2
=a e
t
+ (a ta)e
t
= (a t2a)e
t
: (2.53)

AM3813B Chapter 2 19
Substituting (2.52) and (2.53) into (2.50) gives
(a t+ 4a t5a t2a4a)e
t
= 12e
t
(2.54)
which, in turn, results ina=2 by balancing the coecients. Thus,x
p
(t) =2t e
t
and
so the general solution is
x(t) =C1e
t
+C2e
5t
2t e
t
: (2.55)
Having found the solution forx(t), the solution forycan be found from the rst equation
of the system as follows
y(t) =
1
2

dx
dt
x+e
t

=
1
2

C1e
t
+ 5C2e
5t
2e
t
+ 2te
t
C1e
t
C2e
5t
+ 2te
t
+e
t

=

C1+
1
2

e
t
+ 2C2e
5t
+ 2te
t
Example 2.7.Find all solutions of the DE
3 x+ 12x= 2 sin
2
t : (2.56)
Solution:Equation (2.56) can be rewritten as
x+ 4x=
1
3
(1cos 2t) =
1
3

1
3
cos 2t: (2.57)
The roots of the C.P.L() =
2
+ 4 are=2i. So the basic solutions are cos 2tand
sin 2t. Sincef(t) =
1
3

1
3
cos 2t, sof1(t) =
1
3
andf2(t) =
1
3
cos 2t. Forf1(t) =
1
3
, let
x
p
1(t) =a, then
x
p
1+ 4x
p
1=
1
3
=) 4a=
1
3
=) a=
1
12
: (2.58)
Hence,x
p
1(t) =
1
12
. Forf2(t) =
1
3
cos 2t(resonant case), let
x
p
2(t) =t(bcos 2t+csin 2t); (2.59)
then,
_x
p
2= (2c t+b) cos 2t(2b tc) sin 2t
x
p
2= (4b t+ 4c) cos 2t+ (4c t4b) sin 2t

AM3813B Chapter 2 20
and so
x
p
2+ 4x
p
2=
1
3
cos 2t =) 4ccos 2t4bsin 2t=
1
3
cos 2t (2.60)
which yields
b= 0 and c=
1
12
; (2.61)
and so we havex
p
2(t) =
1
12
tsin 2t. Therefore, the general solution of (2.56) is given by
x(t) =C1cos 2t+C2sin 2t+
1
12
(1tsin 2t): (2.62)
Example 2.8.Find all solutions of the DE
x+x=sin
3
t : (2.63)
Solution:Similar to the above example, we rewrite (2.63) as
x+x=sin
3
t=
3
4
sint+
1
4
sin 3t: (2.64)
It is clear to see that the rst term of the right-hand side gives the resonant case while the
second term yields non-resonant case. The roots of the C.P.L() =2+ 1 are=i.
So the basic solutions are costand sint. Forf1(t) =
3
4
sint, we have
x
p
1(t) =t(acost+bsint): (2.65)
Then,
_x
p
1= (b t+a) cost(a tb) sint
x
p
1=(a t2b) cost(b t+ 2a) sint
and so
x
p
1+x
p
1=
3
4
sint =) 2bcost2asint=
3
4
sint (2.66)
which results in
a=
3
8
and b= 0; (2.67)

AM3813B Chapter 2 21
Thus,x
p
1(t) =
3
8
tcost.
Forf2(t) =
1
4
sin 3t, we may let
x
p
2(t) =ccos 3t+dsin 3t: (2.68)
Then, x
p
2=9ccos 3t9dsin 3t, so
x
p
2+x
p
2=
1
4
sin 3t =) 8ccos 3t8dsin 3t=
1
4
sin 3t (2.69)
which yields
c= 0 and d=
1
32
: (2.70)
Thus,x
p
2(t) =
1
32
sin 3t. Hence, the general solution of the DE (2.63) is given by
x(t) =C1cost+C2sint+
3
8
tcost
1
32
sin 3t: (2.71)
Consider theEuler's equation:
t
n
d
n
y
dt
n
+a1t
n1
d
n1
y
dt
n1
+ +any= 0 (2.72)
whereai; i= 1;2; ; nare constants. Suppose 0< t <1, lett=e

, then
dy
dt
=
dy
d
d
dt
=
1
t
dy
d
ort
dy
dt
=
dy
d
d
2
y
dt
2
=
1
t
2

d
2
y
dt
2

dy
d

ort
2
d
2
y
dt
2
=
d
2
y
d
2

dy
d
.
.
.
Thus, (2.72) becomes an ODE with constant coecients, as discussed above.
Example 2.9.Find all solutions of the DE
(1 +t)
2
x+ (1 +t) _x+x= 4 cos ln(1 +t): (2.73)
Solution:First uset= 1 +tto transform (2.73) to
t
2
d
2
x
dt
2
+t
dx
dt
+x= 4 cos lnt (2.74)

AM3813B Chapter 2 22
which is in the form of Euler's equation. So lett=e

, we havet
dx
dt
=
dx
d
andt
2
d
2
x
dt
2
=
d
2
x
d
2

dx
d
, and thus (2.74) becomes
d
2
x
d
2
+x= 4 cos: (2.75)
The basic solutions of (2.75) are cosand sin. Here,f(t) = 4 coswhich is the resonant
case. So let
x
p
(t) =(acos+bsin) (2.76)
and so
dx
p
d
= (b +a) cos(a b) sin
d
2
x
p
d
2
=(a 2b) cos(b + 2a) sin:
Then,
x
p
+x
p
= 4 cos =) 2bcos2asin= 4 cos (2.77)
which yields
a= 0; b= 2: (2.78)
Thus,x
p
(t) = 2sin. Hence, the general solution of (2.75) is given by
x() =C1cos+C2sin+ 2sin: (2.79)
Returning to the original DE (2.73), one obtains
x(t) =C1cos [ln(1 +t)] +C2sin [ln(1 +t)] + 2 ln(1 +t) sin [ln(1 +t)]: (2.80)
2.4 Variation of Parameters
We will restrict to rst- and second-order ODE but with non-constant coecients. First
consider the rst-order linear ODE:
dy
dt
+P(t)y=F(t): (2.81)

AM3813B Chapter 2 23
WhenF(t)0, (2.81) is a homogeneous linear ODE. The general solution of (2.81) can be
found by rst nding the solution of the corresponding homogeneous DE
dy
dt
+P(t)y= 0 (2.82)
which isseparableand the solution is given by
y=k e

R
t
t
0
P()d
(2.83)
where the constantk6= 0. Now based on the solution (2.83) we use themethod of variation
of parametersto nd the solution of the DE (2.81). The idea is to suppose that the constant
kis now a functionk(t) to be determined such that
y=k(t)e

R
t
t
0
P()d
(2.84)
is the solution of (2.81). To determine the functionk(t), substitute (2.84) into (2.81) to
obtain
dk
dt
e

R
t
t
0
P()d
k(t)P(t)e

R
t
t
0
P()d
+P(t)k(t)e

R
t
t
0
P()d
=F(t) (2.85)
i.e.,
dk
dt
e

R
t
t
0
P()d
=F(t) (2.86)
or
dk
dt
=F(t)e
R
t
t
0
P()d
: (2.87)
Thus,
k(t) =C+
Z
t
t0
F(t1)e
R
t
t
0
P()d
dt1: (2.88)
Hence,
y(t) =e

R
t
t
0
P()d

C+
Z
t
t0
F(t1)e
R
t
t
0
P()d
dt1

(2.89)
is the general solution of the DE (2.81), whereCis an arbitrary constant determined by
the initial condition. The solution (2.89) is usually written as an indenite integral
y(t) =e

R
P(t)dt

C+
Z
F(t)e
R
P(t)dt
dt

: (2.90)

AM3813B Chapter 2 24R
E L
I
Fig. 2.1 An electric circuit.
Example 2.10.A simple electric circuit, shown in Fig. 2.1, consists of an electric source
E, a conductorLand a resistorRwhich are connected in series. Find the currentI(t) of
the circuit at timet.
Solution:According to the Kircho Law, we have the following DE
E=R I+L
dI
dt
; (2.91)
thus,
dI
dt
+
R
L
I=
E
L
: (2.92)
SupposeEis a constant,Iis direct current; and also assume that the currentIat time
tisI(0) =I0. Then, the formula (2.90) gives the solution
I(t) =e

R
R
L
dt

C+
Z
E
L
e
R
R
L
dt
dt

=e

R
L
t

C+
E
R
e
R
L
t

=C e

R
L
t
+
E
R
: (2.93)
Substituting the initial conditionI(0) =I0into (2.93) yields
C=I0
E
R
: (2.94)
Hence, the solution to the initial value problem (IVP) is
I(t) =

I0
E
R

e

R
L
t
+
E
R
: (2.95)

AM3813B Chapter 2 25
IfI0= 0, thenI(t) =
E
R

1e

R
L
t

.
Next, consider the quadratic DE
d
2
y
dt
2
+P(t)
dy
dt
+Q(t)y=F(t) (2.96)
withP(t); Q(t) andF(t) continuous on some intervalI. Assuming that we can nd two
linearly independent solutionsy1andy2of the corresponding homogeneous DE of (2.96),
then the variation of parameters method attempts to nd a particular solution of (2.96) of
the form
y
p
(t) =k1(t)y1(t) +k2(t)y2(t) (2.97)
in whichk1andk2are twice-dierentiable functions.
We need two equations to determine the unknown functionsk1andk2. In fact, we will
obtain equations from which we can solve for
_
k1and
_
k2, and then obtaink1andk2by
integration. In order to substitutek1(t)y1(t) +k2(t)y2(t) into the DE (2.96), rst compute
dy
p
dt
=k1(t)
dy1
dt
+k2(t)
dy2
dt
+
dk1
dt
y1(t) +
dk2
dt
y2(t): (2.98)
To simplify both
dy
p
dt
and
d
2
y
p
dt
2
, we will require thatk1andk2must satisfy
_
k1y1+
_
k2y2= 0 (2.99)
where the dot denotes the dierentiation with respect to timet, and the variabletis dropped
for simplicity. This gives us one equation for
_
k1and
_
k2. Now,
y
p
=
_
k1_y1+
_
k2_y2+k1y1+k2y2: (2.100)
The equation y
p
+P_y
p
+Q y
p
=Fnow becomes
_
k1_y1+
_
k2_y2+k1y1+k2y2+P(k1_y1+k2_y2) +Q(k1y1+k2y2) =F (2.101)
which can be rewritten as
k1(y1+P_y1+Q y1) +k2(y2+P_y2+Q y2) +
_
k1_y1+
_
k2_y2=F: (2.102)

AM3813B Chapter 2 26
Sincey1andy2are assumed to be solutions of the associated homogeneous equation y+
P_y+Q y= 0, the terms in brackets are zero, and we have
_
k1_y1+
_
k2_y2=F; (2.103)
a second equation for
_
k1and
_
k2.
Equations (2.99) and (2.103) give us two equations for
_
k1and
_
k2. For case of reference,
we rewrite these two equations together
y1
_
k1+k2
_
k2= 0
_y1
_
k1+ _y2
_
k2=F: (2.104)
This system of equation has a unique solution if and only if the determinant of the coecients
is nonzero. This determinant is
W=

y1y2
_y1_y2

; (2.105)
whereWrepresents theWronskianWofy1andy2. Sincey1andy2are assumed linearly
independent onI,W(t)6= 0 for allt2I. The solutions of (2.104) for
_
k1and
_
k2are
_
k1=

0y2
F_y2

y1y2
_y1_y2

=
y2F
W
and
_
k2=

y10
_y1F

y1y2
_y1_y2

=
y1F
W
: (2.106)
If we can carry out the necessary integrations, we have functionsk1andk2such that
y
p
=k1y1+k2y2
is a particular solution of the DE y+P_y+Q y=F. Usually we let the constants of
integration be zero in integrating to obtaink1andk2because we do not need more than
onek1andk2having the required properties.
Example 2.11.Find a solution of the DE
y+ 4y= tan(2t) (2.107)
on the interval

4
;

4

.

AM3813B Chapter 2 27
Solution:The associated homogeneous equation y+ 4y= 0 hasy1(t) = cos(2t) and
y2(t) = sin(2t) as a fundamental set of solutions. Their Wronskian is
W(t) =

cos(2t) sin(2t)
2 sin(2t) 2 cos(2t)

= 2 cos
2
(2t) + 2 sin
2
(2t) = 2: (2.108)
WithF(t) = tan(2t), equation (2.106) gives us
_
k1=
y2(t)F(t)
W(t)
=
1
2
sin(2t) tan(2t)
_
k2=
y1(t)F(t)
W(t)
=
1
2
cos(2t) tan(2t)
which yields
k1(t) =
Z

1
2
sin(2t) tan(2t)dt=
1
2
Z
sin
2
(2t)
cos(2t)
dt
=
1
2
Z
1cos
2
(2t)
cos(2t)
dt=
1
2
Z
sec(2t)dt+
1
2
Z
cos(2t)dt
=
1
4
lnjsec(2t) + tan(2t)j+
1
4
sin(2t)
and
k2(t) =
1
2
Z
cos(2t) tan(2t)dt=
1
2
Z
sin(2t)dt=
1
4
cos(2t): (2.111)
A particular solution of (2.107) is
y
p
(t) =k1(t)y1(t) +k2(t)y2(t)
=

1
4
lnjsec(2t) + tan(2t)j+
1
4
sin(2t)

cos(2t)
1
4
cos(2t) sin(2t)
=
1
4
lnjsec(2t) + tan(2t)jcos(2t): (2.112)
Hence, the general solution of (2.107) on the interval

4
;

4

is given by
y(t) =C1cos(2t) +C2sin(2t)
1
4
lnjsec(2t) + tan(2t)jcos(2t): (2.113)
Although the DE in Example 2. 11 has constant coecients, this is not a requirement for
the method of variation of parameters, as it was for the method of undetermined coecients.
Example 2.12.Find a solution of the DE
y
4
t
_y+
4
t
2
y=t
2
+ 1 fort >0: (2.114)

AM3813B Chapter 2 28
Solution:The associated homogeneous equation y
4
t
_y+
4
t
2
y= 0 is a Euler equation,
and we nd the fundamental set of solutionsy1(t) =tandy2(t) =t
4
fort >0, as follows:
Write the DE as
t
2
y4t_y+ 4y= 0:
Lety=t
m
. Then the DE becomes
t
2
m(m1)t
m2
4t m t
m1
+4t
m
= 0 =)m
2
m4m+4 = 0 =)m1= 1; m2= 4:
Hence, the fundamental solutions arey1=tandy2=t
4
.
The Wronskian ofy1andy2is
W(t) =

t t
4
1 4t
3

= 4t
4
t
4
= 3t
4
(2.115)
and this is nonzero fort >0. WithF(t) =t
2
+ 1, equation (2.106) gives us
_
k1=
t
4
(t
2
+ 1)
3t
4
=
t
2
+ 1
3
_
k2=
t(t
2
+ 1)
3t
4
=
1
3t
+
1
3t
3
(2.116)
which yields
k1(t) =
Z

t
2
+ 1
3
dt=
1
9
t
3

1
3
t
k2(t) =
Z
1
3t
+
1
3t
3

dt=
1
3
ln(t)
1
6t
2
: (2.117)
A particular solution of (2.114) is
y
p
(t) =

1
9
t
3

1
3
t

t+

1
3
ln(t)
1
6t
2

t
4
=
1
9
t
4

1
2
t
2
+
1
3
t
4
ln(t): (2.118)
Hence, the general solution of (2.114) is
y(t) =C1t+C2t
4

1
9
t
4

1
2
t
2
+
1
3
t
4
ln(t): (2.119)
We have considered the method of variation of parameters for the DE
y+P(t) _y+Q(t)y=F(t):

AM3813B Chapter 2 29
AlthoughPandQare not be constant to use the method, we have assumed that the
coecients of yis 1. If we encounter a DE in which the coecient of yis a nonconstant
functionR(t), we must divide the equation byR(t) (in an interval in whichR(t)6= 0)
before we can apply the method to the resulting equation.

Chapter 3
SOME \SOLVABLE" TYPES OF NONLINEAR ODE
In this chapter we will discuss some \solvable" nonlinear ordinary dierential equations
(NLODE). By \solvable" we mean that one can use some techniques to nd the analytical
solutions of the NLODE.
3.1 Classication
Before we can begin, in practice we must identify what makes a DE linear or nonlinear.
A DE (ordinary as opposed to ones involving partial derivatives) oforder (n)can generally
be written in the form
F

x; y; y
0
; y
00
; ; y
(n)

= 0 where y
(k)
=
d
k
y
dx
k
; k= 1;2; ; n (3.1)
which is a function ofx; yand all its derivatives. For example,
F(x; y; y
0
; y
00
) =y
00
+ 2 (y
0
)
2
+ sinx
3
= 0 (3.2)
wherexisindependent variableandyisdependent variable.
IfFis nonlinear in the dependent variable,y, or its derivatives, then the DE (3.1) is
said to be nonlinear. However, nonlinearity in the independent variable,x, is irrelevant.
Thus,
y
0
+x
2
y= 0 (3.3)
is linear { but not inx.
Sometimes for rst-order equations (considering one may choose which is independent
and dependent for rst-order equations), one can convert a seemingly nonlinear equation
into a linear equation in the other variable. For example,
y
0
+
y
2
x
= 0 (3.4)
30

AM3813B Chapter 3 31
is anonlinearDE ifxis chosen as independent variable whileyis chosen as dependent
variable. However, if we chooseyas independent variable whilexas dependent variable,
then the DE (3.4) islinearsince we can rewrite (3.4) in the form
dx
dy
+
x
y
2
= 0 (3.5)
which is alinearODE. Such possibilities are not normally available, of course, for higher
order equations.
ThedegreeofFis the degree (algebraic) ofy
(n)
inF. Thus,
y
00
+ 2 (y
0
)
2
+y+ sinx
3
= 0 (3.6)
is arst degree, second orderdierential equation; while
(y
0
)
2
+ 2x
2
y+x= 0 (3.7)
is arst order, second degreedierential equation.
Other nonlinear DEs may, of course, involvetranscendentalfunctions ofy, for example,
the simple (mathematical)pendulumdescribed by
y
00
+ siny= 0: (3.8)
Now let's consider an example of \solvable" types of nonlinear equations. Suppose the
equation is given by
F(x; y; y
0
) = 0: (3.9)
If one can treaty
0
as a variable to be solved algebraically one can often convert the equation
to a simple 1st-order ODE.
Example 3.1.Solve the DE
F(x; y; y
0
) =x y

dy
dx

2
+ (x+y)
dy
dx
+ 1 = 0: (3.10)
Solution:We can factor (3.10) in terms of
dy
dx
as

x
dy
dx
+ 1

y
dy
dx
+ 1

= 0 (3.11)

AM3813B Chapter 3 32
from which we havetwoequations actually

x
dy
dx
+ 1

= 0 or

y
dy
dx
+ 1

= 0; (3.12)
both of them are linear ODEs. From the 1st one, we have
dy
dx
=
1
x
and its solution is
y=lnjxj+C1
=lne
C1
jxj
=lnjC
0
xj(C
0
=e
C1
>0): (3.13)
From the 2nd equation, one obtains
y dy=dx =) y
2
=2x+C2: (3.14)
Thus, we have actuallytwodistinct families of solutions. In other words, for each point in
the (x; y) plane, there exist two dierent solution curves passing through this point. See
Fig. 3.1 for the case whenC1= 2 andC2= 6.6
1/2
6
1/2
-
(1, 2)
3
1
2 3 4 6 8
0
x
y
y = - ln |x| + 2
y = - 2 x + 6
Fig. 3.1 Two solution curves of DE (3.10).
Note that at the intersection point (1;2) the slopes,
dy
dx
, of the two curves are dierent.
If a solution is transited from one type to another type, the transition must occur at the

AM3813B Chapter 3 33
points where
dy
dx
of these two solutions are same, for this example, that is,

1
x
=
1
y
ory=x (3.15)
a straight line. For example, Fig. 3.2 shows the case that the two solution curves are tangent
at the point (1;1) at which they have the same slope1.1/2
3
1/2
3
-
x421
(1, 1)
0
y = - 2 x + 3
y = - ln |x| + 1
1
1-
y
Fig. 3.2 Two tangent solution curves of DE (3.10).
3.2 Singular Solutions
For some DEs there exist some particular solution which does not belong to family
of solutions of the DEs. This solution is calledsingular solution. Let's look an example.
Consider
y
2

dy
dx

2
a
2
+y
2
= 0 (ais a constant): (3.16)
This DE can be rewritten as

y
dy
dx

2
=a
2
y
2
(3.17)
and so
y dy
p
a
2
y
2
=dx (3.18)
y dy
p
a
2
y
2
=dx (3.19)

AM3813B Chapter 3 34
or (this should not be ignored)
a
2
y
2
= 0 (3.20)
which is an important equation. By applying direct antidierentiation to (3.18), we have
C1
p
a
2
y
2
=x (C1is an arbitrary constant) (3.21)
which represents left-hand semi-circles. Similarly, from (3.19) one obtains
C2+
p
a
2
y
2
=x (C2is an arbitrary constant) (3.22)
which denotes right-hand semi-circles. Equations (3.21) and (3.22) can be written as
(xC1)
2
+y
2
=a
2
and (xC2)
2
+y
2
=a
2
;
respectively, and these two equations can be combined to single equation
(xC)
2
+y
2
=a
2
: (3.23)
Another solution of (3.16) is solved from (3.20) as
y=a (3.24)
which can be veried by substituting it into (3.16) to show that
LHS =a
2
(0)a
2
+a
2
= 0 = RHS:
Hence, thegeneral solution(3.23) represents circles centered on thex-axis (depending
upon the choice ofC. See Fig. 3.3.) It is seen that any point on solutiony=aory=a
is on a particular circle and the particular curve (here, line) is tangent to the circle at this
point. However, the general solution, the family of circles (or family of semi-circles) does
not include the particular solutiony=a. From the viewpoint of geometry, this particular
curvey=aory=ais the envelope of the family of circles. While from the viewpoint
of dierential equations, at each point of the particular solution, theuniqueness of solution
is violated. (We shall discuss more about uniqueness of solution later.) These (y=a) are
special class of solutions known assingular solutionsthat only appear in nonlinear DEs.

AM3813B Chapter 3 35y = - a
y
x
y = a
Fig. 3.3 Solutions of the DE (3.16).
Denition 3.1:Asingular solutionis the one corresponding to an integral curve that
violates uniqueness of solutions at any point on the curve.
In rst-order DEs these solutions must be tangent to some element of the solutions from
the one-parameter family.
Next, let's discuss how to nd singular solutions for general rst-order nonlinear dier-
ential equations.
3.3 C-Discriminant Equation
Suppose that a general nonlinear ODE is given by
F(x; y;
dy
dx
) = 0 (3.25)
whereFis not necessarily a polynomial inx; yand
dy
dx
; and if it is a polynomial, it is
not necessary possible to factor it. We further assume that a solution of (3.25) exists (we'll
discuss existence of solutions later) which we can write in the implicit form
(x; y; C) = 0 (3.26)
whereCis an arbitrary constant. If a singular solution exists, it must be on the envelope of
the solutions as in the case of the example of circles discussed in the previous section. That
is because a singular solution must be tangent to the solutions somewhere. Each point of

AM3813B Chapter 3 36
the envelope must be a curve dependent parametrically only onC; i.e.,
x=x(C); y=y(C): (3.27)
Butxandyare still on the curves of family of solutions, so they must satisfy
(x(C); y(C); C) = 0: (3.28)
Thus, in general
d= 0 =)
@
@x
dx+
@
@y
dy+
@
@C
dC= 0 (3.29)
must hold. But on any member of the family of solutions,dC= 0. Hence,
@
@x
dx+
@
@y
dy= 0: (3.30)
This must be true even at points where a family member touches the envelope. Therefore,
on the envelope wheredC6= 0, we must have
@
@C
= 0 to satisfyd= 0. Hence, we have
two equations:
= 0 and
@
@C
= 0 (3.31)
which must be satised on the envelope. From these two equations, we can get so called
C-discriminant equationby eliminating the parameterC. This equation might represent a
singular solution. That is, thisC-discriminant equation needs not necessarily be a singular
solution. In other words, these two equations are necessary conditions, but not sucient! It
means thatif the family of solutions has any envelope, they must be included in the
C-discriminant equation, while a C-discriminant equation might not represent
an envelope.
Now let's use C-discriminant equation to consider the example given by equation (3.16).
Example 3.2.Find possible singular solutions of the DE
y
2

dy
dx

2
a
2
+y
2
= 0: (3.32)
Solution: From the discussion given in the previous section we know that the general solution
of the DE is
(x; y; C) = (xC)
2
+y
2
a
2
= 0: (3.33)

AM3813B Chapter 3 37
So, using
@
@C
=2 (xC) = 0 (3.34)
to eliminateCfrom (3.33) yields
y
2
a
2
= 0 =) y=a: (3.35)
We have seen thaty=aory=ais indeed the envelope of the family of solutions, i.e.,
circles
(xC)
2
+y
2
a
2
= 0:
Hence,y=aare singular solutions. Consider another example.
Example 3.3.Find possible singular solutions of the DE
y
2
3

dy
dx

3
5
= 0: (3.36)
Solution: It is easy to nd the solution by direct antidierentiation as
(x; y; C) =y
5
(x+C)
3
= 0 i:e: y= (x+C)
3
5: (3.37)
So, we have another equation
@
@C
= 3 (x+C)
2
= 0 (3.38)
EliminatingCfrom the above two equations results in the C-discriminant equation
y= 0: (3.39)
However,y= 0 is not the solution of the DE (3.36), therefore, it is not a singular solution
of (3.36). Indeed, Fig. 3.4. shows that the family of solutions has no envelopes.
Other family of solutions which do not have envelopes are circles having a same center
and parallel lines (Fig. 3.5).
Note: Something called thep-discriminant equationarises in the case of repeated roots of
the DE in
dy
dx
.

AM3813B Chapter 3 38x0
y
Fig. 3.4 Solutions for Example 3.3.(b)
y y
x x
(a)
Fig. 3.5 Examples having no envelopes.
3.4 Eliminating Dependent Variable
Again consider the DE
F(x; y;
dy
dx
) = 0:
Suppose we could write this equation in the form (of course, maybe we can't!):
y=g(x;
dy
dx
): (3.40)
Lettingp=
dy
dx
transforms (3.40) to
y=g(x; p); (3.41)

AM3813B Chapter 3 39
and then dierentiating this equation with respect toxyields
dy
dx
=
@g
@x
+
@g
@p
dp
dx
=h(x; p;
dp
dx
) (3.42)
in which the dependent variableyhas been eliminated. Thus, we have
p=h(x; p;
dp
dx
): (3.43)
Solving forpfrom (3.43) may be easier than solving foryfrom the original DE (3.40).
Example 3.4.Solve the DE
x

dy
dx

2
3y

dy
dx

+ 9x
2
= 0 for x >0: (3.44)
Solution: Letp=
dy
dx
, then (3.44) is transformed to
x p
2
3y p+ 9x
2
= 0 (3.45)
from which we can separateyout to obtain
3y=x p+
9x
2
p
: (3.46)
Next, dierentiating (3.46) with respect toxyields
3
dy
dx
=p+x
dp
dx
+
18x
p

9x
2
p
2
dp
dx
; (3.47)
which can be written, by using
dy
dx
=p, as
3p=p+
18x
p
+

x
9x
2
p
2

dp
dx
; (3.48)
or
2p

1
9x
p
2

=x

1
9x
p
2

dp
dx
: (3.49)
So, from (3.49) we have

2px
dp
dx

1
9x
p
2

= 0 (3.50)
which yields either
2px
dp
dx
= 0 (3.51)

AM3813B Chapter 3 40
or
1
9x
p
2
= 0: (3.52)
Equation (3.51) is separable and its solution can be obtained as follows:
dp
p
= 2
dx
x
=)
lnjpj= 2 lnjxj+C1= lnx
2
+C1
=)
jpj=e
C1
x
2
=C2x
2
(C2=e
C1
>0)
=)
p=C2x
2
=C3x
2
(C3=C26= 0): (3.53)
Substituting (3.53) into (3.46) gives us
y=
C3
3
x
3
+
3
C3
=C x
3
+
1
C
(C=
C3
3
6= 0) (3.54)
which is a one-parameter family of solutions.
Next, solving (3.52) yields
p=3x
1
2 (3.55)
and then substituting it into (3.46) produces
3y=x

3x
1=2

+
9x
2
(3x
1=2
)
=) y=2x
3
2: (3.56)
Now we want to show that (3.56) is actually the C-discriminant equation of (3.54). Thus,
consider the following two equations
(x; y; C) =yC x
3

1
C
= 0
@
@C
=x
3
+
1
C
2
= 0: (3.57)
The second equation gives usC=x
3=2
. Substitute it into the rst equation to obtain
y=x
3
x
3=2

1
x
3=2
=2x
3
2: (3.58)

AM3813B Chapter 3 41-8
-6
-4
-2
0
2
4
6
8
0 0.5 1 1.5 2 2.5
y
x
y = 2 x^(3/2)
y = - 2 x^(3/2)
Fig. 3.6 Solutions for Example 3.4.
Figure 3.6 shows that the C-discriminant equation is actually the envelope of the family
of solutions (3.54). Hence,y=2x
3=2
are singular solutions of the original DE (3.44).
A simple way to show that a C-discriminant equation is a singular solution is as follows:
Check if the C-discriminant equation satises the original DE; if it does, then it is a singular
solution, otherwise it is not. For Example 3.4, we know from (3.56) thaty=2x
3=2
are
solutions of (3.44), and indeed substitutingy=2x
3=2
into (3.44) results in
LHS =x

3x
1
2

2
3

2x
3
2

3x
1
2

+ 9x
2
= 9x
2
18x
2
+ 9x
2
= 0 = RHS;
so,y=2x
3=2
are singular solution of the original DE (3.44).
3.5 Eliminating Independent Variable
Similarly, the procedure discussed in the previous section can be applied to independent
variable. Suppose we can write the DEF(x; y;
dy
dx
) = 0 in the form
x=g(y; p) (3.59)
wherep=
dy
dx
. Then dierentiating (3.59) with respect toywith the aid
dx
dy
=
1
p
results
in
1
p
=
@g
@y
+
@g
@p
dp
dy
=h(y; p;
dp
dy
) (3.60)

AM3813B Chapter 3 42
which is a rst-order DE withyas independent variable andpas dependent variable,
and derivative
dp
dy
has been solved out. Then solving (3.59) and (3.60) together yields the
solution for the original DE.
Example 3.5.Solve the DE

dy
dx

3
4x y

dy
dx

+ 8y
2
= 0: (3.61)
Solution: Lettingp=
dy
dx
and solvingxfrom (3.61) give us
x=
p
2
4y
+
2y
p
: (3.62)
Then dierentiating (3.62) with respect toyyields
dx
dy
=
1
p
=

p
2y

2y
p
2

dp
dy

p
3
4y
2

2
p

; (3.63)
or
p
3
4y
4
2y p
2

dp
dy

p
2y

= 0: (3.64)
From the second factor of (3.64)
dp
dy

p
2y
= 0 we nd
p=cjyj
1
2: (3.65)
Substitute (3.65) (rememberingp=
dy
dx
) into the original DE (3.61) to obtain
c
3
jyj
3
24c x yjyj
1
2+ 8y
2
= 0 (3.66)
which is a one-parameter family of solutions of the DE (3.61). Wheny >0, canceling factor
y
3
2in (3.66) yieldsc
3
4c x+ 8y
1
2= 0; i.e.,
64y=c
2

4xc
2

2
or
y=C1(xC1)
2
(C1=
c
2
4
): (3.67)

AM3813B Chapter 3 43
Wheny <0, similarly canceling factor (y)
3
2in (3.66) leads toc
3
+ 4c x+ 8 (y)
1
2= 0;
i.e.,
64y=c
2

4x+c
2

2
or
y=C2(xC2)
2
(C2=
c
2
4
): (3.68)
It can be seen that (3.68) actually represents (3.67) ifC1takes negative values. Moreover,
y= 0 is obviously a solution of (3.61). Hence
y=C(xC)
2
(3.69)
is family of solutions of (3.61), whereCis an arbitrary constant.
Next, from the rst factor of (3.64)p
3
4y
2
= 0 one obtains
p=

4y
2
1
3
: (3.70)
Substituting (3.70) into (3.61) results in
y
h
3yx

4y
2

1=3
i
= 0:
So we have
y= 0 or y=
4
27
x
3
: (3.71)
On the other hand, from (3.69) we have
=yC(xC)
2
= 0
@
@C
= (xC) (3Cx) = 0
from which eliminatingCgives us
y= 0 or y=
4
27
x
3
:
Hence, bothy= 0 andy=
4
27
x
3
are singular solutions of the original DE (3.61).

AM3813B Chapter 3 44
3.6 First-Order Equations: General Strategies
In this section, we shall introduce several methods which can be used to solve for rst-
order DEs analytically.
3.6.1 Separate equations.Suppose algebraic manipulations of the DEF(x; y;
dy
dx
) =
0 leads to the form
f(x)dx+g(y)dy= 0: (3.73)
Then its solution can be obtained by just direct integration as
Z
f(x)dx+
Z
g(y)dy=C: (3.74)
Example 3.6.Solve the DE
(1 +y
2
)dx+ (1 +x
2
)dy= 0: (3.75)
Solution: (3.75) can be written as
dx
1 +x
2
+
dy
1 +y
2
= 0: (3.76)
So, its solution is
tan
1
x+ tan
1
y=C
0
(3.77)
from which we have
tan

tan
1
x+ tan
1
y

= tanC
0
=C
=)
x+y=C(1x y): (3.78)
Example 3.7.Find all the solutions of the DE
dy
dx
+
x
y
= 0: (3.79)
Solution: We need to discuss this equation in the two regions:
R1: 1< x <1;0< y <+1;
R2: 1< x <1; 1< y <0;

AM3813B Chapter 3 45
separately. In regionR1, from
Z
y dy=
Z
x dx+
C
2
2
; (3.80)
we have the solution
y
2
=C
2
x
2
ory=
p
C
2
x
2
;C < x < C (3.81)
whereCis an arbitrary positive constant. Similarly,
y=
p
C
2
x
2
;C < x < C (3.82)
is the solution of (3.79) in the regionR2.
Example 3.8.Find all the solutions of the DE
dy
dx
=
2
x
2
1
: (3.83)
Solution: We need to discuss this equation in the three regions:
1: 1< x <1; 1< y <+1;
2:1< x <+ 1; 1< y <+1;
3: 1< x <+1; 1< y <+1;
separately. For each point (x; y) of the three regions, there exists one and only one solution
curve passing the point. Equation (3.83) can be written as
dy
dx
=
1
x1

1
x+ 1
: (3.84)
Integrate the above equation to obtain
y=
Z
dx
x1

Z
dx
x+ 1
+C= ln

x1
x+ 1

+C: (3.85)
Thus,
y= ln
x1
x+ 1
+C for 1< x <1 and 1< x <+1;
y= ln
1x
1 +x
+C for1< x <1

AM3813B Chapter 3 460
x
y
-1 1
Fig. 3.7 Solutions for Example 3.8.
which implies that for a given constantC, there are three solution curves dened by (3.86)
(Fig. 3.8).
Example 3.9.Find all the solutions of the DE
dy
dx
=
y
2
1
2
: (3.87)
Solution: We need to discuss this equation in the three regions: 1< y <1,1< y <
1 and 1< x <+1. Integrating (3.87) yields
ln

y1
y+ 1

=x+C
0
: (3.88)
In region 1< y <1 and 1< y <+1, (3.88) is
y1
y+ 1
=e
C
0
e
x
=C1e
x
(C1=e
C
0
>0): (3.89)
So the solution is
y=
1 +C1e
x
1C1e
x
: (3.90)
When 1< x <lnC1,y >1; whenlnC1< x <+1,y <1. Note that none of
the domains of these two solutions is 1< x <+1.

AM3813B Chapter 3 47
In the region1< y <1, the solution of (3.87) is
y1
y+ 1
=C2e
x
(C2<0) (3.91)
which, in turn, gives
y=
1 +C2e
x
1C2e
x
(C2<0): (3.92)
Note that (3.92) is dened in 1< x <+1. The two solutions (3.90) and (3.92) can be
combined as
y=
1 +C e
x
1C e
x
(C6= 0): (3.93)
Buty=1 andy= 1 are obviously also the solutions of (3.87), the former of which can
be obtained from (3.93) by lettingC! 1while the latter can be got by settingC= 0
in (3.93). Therefore, letting the constantCvarying from 1to +1, one obtains all the
solutions of (3.87) from equation (3.93). See Fig. 3.8.y
-1
0
1
x
Fig. 3.8 Solutions for Example 3.9.
Example 3.10.Find all the solutions of the DE
dy
dx
=e
x
cos
2
y: (3.94)

AM3813B Chapter 3 48
Solution: First, it is easy to see that
y= (2n+ 1)

2
(n= 0;1;2; ) (3.95)
are solutions of (3.94). Wheny6= (2n+ 1)

2
(n= 0;1;2; ), for example, when

2
< y <

2
, the solution of (3.94) is determined by
Z
dy
cos
2
y
=
Z
e
x
dx+C i:e:;tany=e
x
+C:
Hence,
y= tan
1
(e
x
+C); 1< x <+1: (3.96)
Since the inverse trigonometric functions are multivalue functions, so for a given constant
C, (3.96) actually represents a set of innite solutions.
3.6.2 Exact rst-order equations.Iff(x; y) =C(Cis an arbitrary constant) is
the general solution of an ODE, then
df=
@f
@x
dx+
@f
@y
dy= 0 (3.97)
and
@
@y

@f
@x

=
@
@x

@f
@y

(3.98)
if functionfissmooth; or more generally iffisanalytic, then the conditions (3.97) and
(3.98) are satised. So if we have an ODE given by
g(x; y)dx+h(x; y)dy= 0 (3.99)
and
@g(x; y)
@y
=
@h(x; y)
@x
; (3.100)
then we expect there is a functionf(x; y) that produces the dierential; and
@f
@x
=g(x; y)
and
@f
@y
=h(x; y). The technique is to antidierentiategandhto ndf. Actually,
f(x; y) =
Z
@f
@x
dx+K(y) or f(x; y) =
Z
@f
@y
dy+L(x): (3.101)

AM3813B Chapter 3 49
We use the following example to illustrate the procedure of nding the solution of the DE
(3.99).
Example 3.11.Find all the solutions of the DE
(3x
2
y+y
2
+ 1)dx+ (x
3
+ 2x y1)dy= 0: (3.102)
Solution: Here,
g(x; y) = 3x
2
y+y
2
+ 1 and h(x; y) =x
3
+ 2x y1: (3.103)
Obviously,
@g
@y
= 3x
2
+ 2y=
@h
@x
:
Therefore,
f(x; y) =
Z
@f
@x
dx+K(y) =
Z
g(x; y)dx+K(y)
=
Z
(3x
2
y+y
2
+ 1)dx+K(y)
=x
3
y+x y
2
+x+K(y)
and thus
@f
@y
=x
3
+ 2x y+
dK
dy
(3.105)
which should be equal toh(x; y). So, we have
dK
dy
=1 =) K(y) =y: (3.106)
Hence,
f(x; y) =x
3
y+x y
2
+xy: (3.107)
Finally,f(x; y) =Cor
x
3
y+x y
2
+xy=C (3.108)
represents all solutions of (3.102).
Sometimes, however, a given DE
g(x; y)dx+h(x; y)dy= 0

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been unknown land to me. We Canadians had never gone much beyond a
little of Mendelssohn, which the teachers of music seemed to consider the
height of classical music, and the people were still singing the old
sentimental songs, not the ragtime the Americans love, but the deadly sweet
melodies that cloy and teach us nothing. Of course, no doubt, things have
changed there now; but it was that way when I was a girl in Montreal.
I did not want to leave New York even for two weeks. I had begun to
love my life here. There was something fine in the comradeship with the
boys in the old ramshackle studio building. I had been accepted as one of
the crowd, and I knew it was Bonnat’s influence that made them all treat me
as a sister. Fisher once said that a “fellow would think twice before he said
anything to me that wasn’t the straight goods,” and he added, “Bonnat’s so
darned big, you know.”
I had often cooked for all of the boys in the building. We would have
what they called a “spread” in Bonnat’s or Fisher’s studio, and they would
all come flocking in, and fall to greedily upon the good things I had cooked.
I felt a motherly impulse toward them all, and I wanted to care for and cook
for—yes—and wash them, too. Some of the artists in that building were
pretty dirty.
Paul had never spoken of love to me, and I was afraid to analyze my
feelings for him. Reggie’s letters were still pouring in upon me, and they
still harped upon one thing—my running away from him in Boston. He kept
urging me to come home, and lately he had even hinted that he was coming
again to fetch me; but he said he would not tell me when he would come, in
case I should run off again.
I used to sit reading Reggie’s letters with the queerest sort of feelings for,
as I read, I would not see Reggie in my mind at all, but Paul Bonnat. It did
seem as if all the things that Reggie said that once would have pierced and
hurt me cruelly had now lost their power. I had even a tolerant sort of pity
for Reggie, and wondered why he should trouble any longer to accuse me
of this or that, or even to write to me at all. I am sure I should not have
greatly cared if his letters had ceased to come. And now as I turned over in
my mind the question of leaving New York, I thought not of Reggie, but of
Paul. It is true, I might only be away for the two weeks in Providence; on
the other hand, I realized that should we succeed there, I would be foolish
not to go on with the troupe to Boston. I decided finally that I would go.

I went over especially to tell Paul about it. I said:
“Mr. Bonnat, I’m going away from New York, to do some more of that
—that living-picture work.” I waited a moment to see what he would say—
he had not turned around—and then I added, as I wanted to see if he really
cared—“Maybe I won’t come back at all.”
He stood up, and took me by the shoulders, making me look straight at
him.
“How long are you to be gone?” he demanded, as if he had penetrated
my ruse.
“Two weeks in Providence,” I said, “but if we succeed, we go on to
Boston and—”
“Promise me you’ll come back in two weeks. Promise me that,” he said.
He was looking straight down into my eyes, and I think I would have
promised him anything he asked me to; so I said in a little weak voice:
“I promise.”
“Good!” he replied. “I would not let you go, if it were in my power to
stop you, but I know you need the money, and I have no right to deprive
you of it. Oh, good God! it’s hell not to be able to—” He broke off, and
gently took my hands up in his:
“Look here, little mouse. There’s a chance of my being able to make a
big pot of money. I’ll know in a few days’ time. Then you shall not have to
worry about anything. But as I am now fixed, why I can’t stop you from
anything. I haven’t the right.”
I wanted to tell him that he could stop me from going if he wanted to;
but he had not told me he cared for me, and there was a possibility that I
was mistaken about him. He had that big, gentle way with every one, and it
might be that I had mistaken his kindly interest in me for something that he
did not really feel. So I laughed now lightly, and I said:
“Oh, I’ll be back soon, and if you like you can see me off on the train.”
When we were in the Grand Central the following night, I tried to appear
cheerful, but I could not prevent the tears running down my face, and when
finally he took my hand to say good-bye, I said:
“Oh, it’s dreadful for me to say this; b-but if I don’t see you soon again I
—t-think I will die.”

He bent down when I said that and kissed me right on my lips, and he
did not seem to care whether every one in the station saw us or not. Then I
knew that he did love me, and that knowledge sent me flying blindly down
the platform. After I was aboard, I found I had taken the wrong train to
Providence. I should have taken an earlier or a later one. Lil was already
there, and was to have met me at the station from the earlier train, but the
train I had taken would not get in till four in the morning.
When I arrived in Providence I did not know where to go. I had Lil’s
address, but she had written me she was living at a “very respectable house”
where the people would have been terribly shocked to know she was a
model, and I felt I could not go there at such an hour in the morning. The
rain was coming down in torrents. A colored boy was carrying my bag, and
he asked me where I wanted to go. Indeed, I did not know. When I
hesitated, he said that the hotels didn’t take ladies alone, but that he knew of
an all-night restaurant where I could get something hot to eat and I could
stay there till morning. So he took me over to Minks’. I had often eaten in
Minks’ restaurant in Boston, and the place looked quite familiar to me. I
had a cup of hot coffee and a sandwich, and then I asked the waitress if
there was some place where I could go and freshen or clean up a bit. She
whispered to the man at the desk, and he nodded, and then she beckoned to
me to follow her. We went upstairs to a sort of loft. It was bare, save of
packing cases, but she showed me to a little cracked looking-glass where
she said I could do my hair. I told her I had been on the train all night, and
she said sympathetically:
“Sure, you look it.”
I went over to Lil’s boarding-house about seven in the morning. She was
right near Minks’, and said I was foolish not to have come right over.
Well, we played every night in the theatre in Providence, and we made
what theatrical people call a “hit.” The whole town turned out to see us. The
girls were all as pleased as could be, and so was Mr. Hirsch, and they made
all kinds of plans for the road tour, but I could think of nothing but New
York, and I was so lonely, in spite of the noisy company of the girls, that I
used to go over and look at the railway tracks that I knew ran clear to New
York. And I thought of Paul! I thought of Paul every single minute. The
little maid would slip his letters every morning under my door, and I used to

cry and laugh before I even opened them and I held them to my lips and
face, and I kept them all in the bosom of my dress, right next to me.
We had finished our engagement. Lil and I were coming out of the
dressing-room the last night when somebody slapped me on the back. I
turned around, and there was Mr. Davis. He was so glad to see me that he
nearly wrung my hand off, and he insisted on walking home with us. He
told me he was now manager of a theatrical company, and that he had been
looking around for me ever since Lil told him I was in New York.
“Now, Marion,” he said, “you are going to begin where you left off in
Montreal, and it’s up to you to make good. You’ve got it in you, and I want
to be the man to prove it.”
I asked him what he meant, and he said he was starting a new “show” in
Boston that week, and that he had a part for me that would give me an
opportunity.
I said faintly:
“I was going back to New York to-morrow.”
Lil exclaimed:
“What’re you talking about? Aren’t you going along with Mr. Hirsch?”
“Instead of going to New York,” said Mr. Davis, “you come along with
me to Boston. Cut out this living-picture stuff. It’s not worthy of you. I
always said there was the right stuff in you, Marion, and now I’m going to
give you the chance to prove it.”
For a moment an old vision came back to me. I saw myself as “Camille,”
the part I had so loved when little more than a child in Montreal, and I felt
again the sway of old ambitions. I said to Mr. Davis:
“Oh, yes, I think I will go with you!”
But when I got back to my room, I took out Paul’s last letter. How
confident he was of my keeping my promise to return! He wrote of all the
preparations he was making, and he said he had a stroke of luck, and that I
should share it with him. We should have dinner at Mouquin’s, and then we
would see some show, or the opera. Whatever we did, or wherever we went
we would be together.
I got out my little writing pad, and I wrote a letter hurriedly to Mr.
Davis:

“Dear Mr. Davis:
“Will you please excuse me, but I have to go to New York. I’ll let you
know later about acting.”
I sent the note to Mr. Davis by the little maid in the house, and he sent
back a sheet with this laconic message upon it:
“Now or never—Give me till morning.”
Lil talked and talked and talked to me all night about it, and she seemed
to think I was crazy not to grab this chance that had come to me, and she
said any one of the other girls would have gone clean daft about it. She said
I was a little fool, and never knew when opportunity came in my way. “Just
look,” she said, “how you turned down that chance you had to be a show
girl, and all of us other girls weren’t even asked, and I’ll bet our legs are as
pretty as yours. It’s just because you’ve got a sort of—of—well, I heard a
man call it ‘sex-appeal’ about you, but you’re foolish to throw away your
good chances, and by and by they won’t come to you. You’ll be fat and
ugly.”
I said:
“Oh, Lil, stop it. I guess I know my business better than you do.”
“Well, then, answer me this,” said Lil, sitting up in bed, “are you
engaged to that fellow who sends you letters every day?”
I could not answer her.
“Well, what about Reggie Bertie?”
“For heaven’s sakes, go to sleep,” I entreated her, and with a grunt of
disgust she at last turned over.
Next morning Paul’s letter fully decided me. It said that he would be at
the station to meet me! He was expecting me, and I must not, on any
account, fail him.
“Lil, wake up! Wake up!” I cried, shaking her by the arm. “I’m going to
take the first train back to New York.”
Lil answered sleepily:
“Marion, you always were crazy.”
All of a sudden the room turned red on all sides of us, and I realized that
it was on fire. The little stove had a pipe with an elbow in the wall, and

when I put a match to the kindling, the flames must have crept up to the thin
wooden walls from the elbow, and in an instant the wall had ignited. I had
on only a nightdress. I seized the quilt off the bed, and threw it on the
flames, but it seemed only to serve as fresh fuel. Lil was
And both shrieking we ran out into the hall.
crouched back on the bed, petrified with terror, and literally unable to move.
Desperately screaming, “Fire, fire!” I seized the pitcher and flung it at the
flames, and then somehow I grabbed hold of Lil by the hand, and both
shrieking, we ran out into the hall. Then I fainted. When I came to, the fire
was out, and the landlady and her son and husband and Lil were all standing
over me, laughing and crying.
“Well,” said the man, “did you try to burn us out?” He turned to his wife,
and said: “It’s a good job I got that insurance, eh?”

My clothes were not burned, but soaking wet, and so I missed my train
—the train that Paul was going to meet.

O
XLIX
H, how good it was to enter New York once more! I remembered how
ugly the city had looked to me that first time when I had come from
Boston. Now even the rows of flat houses and dingy tall buildings
seemed to take on a sturdy and friendly beauty.
Paul was walking up and down the station, and he came rushing up to
me, as I came through the gates. He was pale, and even seemed to tremble,
as he caught me by the arm and cried:
“When you did not come on that train, I was afraid you had changed
your mind, and were not coming back to me. I’ve been waiting here all day,
watching each train that arrived from Providence. Oh, sweetheart, I’ve been
nearly crazy!”
I told him about the fire, and he seized hold of my hands, and examined
them.
“Don’t tell me you hurt yourself!” he cried. And when I reassured him, it
was all I could do to keep him from hugging me right there in the station.
All the way on the car he held my hand, and although he did not say
anything at all to me, I knew just what was in his heart. He loved me, and
nothing else in all the whole wide world mattered.
He had helped me out at the studio building, and now as I went up the
old rickety stairs, I realized that this was my home!
It was a ramshackle, very old, neglected, rickety sort of place, and I do
not know why they called it Paresis Row. The name did not sound ugly to
me, somehow. I loved everything about the place, even the queer business
carried on on the lower floors, and old Mary, the slatternly caretaker, who
scolded the boys alternately and then did little kindnesses for them. I
remember how once she kept a creditor away from poor Fisher, by waving
her broom at him, till he fled in fear.
I laughed as we went by the door of that crazy old artist that the boys
used to tease by dropping a piece of iron on the floor after holding it up
high. They would wait a few minutes, and then he would come hobbling up
the stairs. There would be three regular taps, and then he would put his head
in and say:

“Gentlemen, methinks I heard a noise!”
On the first floor back a man taught singing, and he had gotten up a class
of policemen. It seemed as if they sang forever the chorus of a song that
went like this:
“Don’t be afraid, don’t be afraid, don’t be a-f-rai-d!”

Several artists had committed suicide in the building. I am not sure of
the causes, and we never dwelt upon the reasons. There was nothing pretty
about the place; it was cold and not even very clean; but—it was my home!
Paul opened the door of his studio. The place was all cleaned up and new
paper on the walls. He showed me behind the screen a little gas stove, pots
and pans hanging at the back of it, and dishes in a little closet. Then, taking
me by the hand, he opened a door, and showed me a little room adjoining
his studio. It seemed to me lovely. It was prepared in soft gray, and the
curtains of yellow cheesecloth gave an appearance of sunlight to it. There
were several pieces of new furniture in the room, and a little mission
dresser. Paul opened the drawers, and rather shyly showed me some sheets,
pillow slips and towels, which he said he had purchased for me, and added:
“I hope they are all right. I don’t know much about such things.”
I knew then that Paul intended the room to be for me. He had only the
one studio room before.
“Well, little mouse,” he said, “are you afraid to live with a poor beggar,
or do you love me enough to take the chance?”
Thoughts were rushing through my mind. Memories of conversations
and stories among the artists, on the marriage question, by some considered
unnecessary and somehow with Paul it seemed right and natural, and the
primitive woman in me answered: “Why not? Others have lived with the
man they loved without marriage. Why should not I?” He was waiting for
me to speak, and I put my hands up on his shoulders, and said:
“Oh, yes, Paul, I will come to you! I will!”
A little later, I said:
“Now I must go over to my old room and have my trunk and some other
things I left there brought over, and I must tell Mrs. Whitehouse, the
landlady, as she expects me back to-day.”

“Well, don’t be long,” said Paul. “I’m afraid you will slip through my
arms just as I have found you.”
Mrs. Whitehouse, the landlady, met me at the door. I told her I was going
to move over to Fourteenth Street, to Paresis Row. She threw up her hands
and exclaimed:
“Lands sakes! That is no place for a girl to live, and I have no use for
them artists. They are a half-crazy lot, and never have a cent to bless
themselves with. If I were a young and pretty girl like you, Miss Ascough, I
would not waste my time on the likes of them. Now there’s been a fine-
looking gent calling for you the last two days, and I told him you’d be back
to-day. He’s a real swell, and if you’d take my advice, you’d get right next
to him.”
Even as she spoke the front doorbell rang. She opened the door, and
there was Reggie! I was standing at the bottom of the stairs, but when I saw
him, I fled into the parlor. He came after me, with his arms outstretched. I
found myself staring across at him, as if I were looking at a stranger.
“Marion,” he cried, “I’ve come to bring you home.”
I backed away from him.
“No, no, Reggie, I don’t want you to touch me,” I said. “Go away! I tell
you go away!”
“You don’t understand,” said Reggie. “I’ve come to take you home.
You’ve won out. I’m going to marry you!”
He looked as if he were conferring a kingdom on me.
“Listen to me, Reggie,” I said. “I can never, never be your wife now.”
“Why not? What have you done?” His old anger and suspicion were
mounting. He was looking at me lovingly, yet furiously.
“I’ve done nothing—nothing—but I cannot be your wife.”
“If you mean because of Boston—I’ve forgiven everything. I fought it
all out in Montreal and I made up my mind that I had to have you. So I’m
going to marry you, darling. You don’t seem to understand.”
Further and further away I had backed from him, but now he was right
before me. I looked up at Reggie, but a vision arose between us— Paul
Bonnat’s face. Paul who was waiting for me, who had offered to share his
all with me, and somehow it seemed to me more immoral to marry Reggie
than to live with the man I loved.

“Reggie Bertie,” I said, “it’s you who don’t understand. I can never be
your wife because—because—” Oh, it was very hard to drive that look of
love and longing from Reggie’s face. Once I had loved him, and although
he had hurt me so cruelly in the past, in that moment I longed to spare him
the pain that was to be his now.
“Well? What is it, Marion? What have you done?”
“Reggie, it’s this: I no longer love you!” I said.
There was silence, and then he said with an uneasy laugh:
“You don’t mean that. You are angry with me. I’ll soon make you love
me again as you did once, Marion. You’ll do it when you are my wife.”
“No—no—I never will,” I said steadily, “because—because—there’s
another reason, Reggie. There’s some one else, some one who loves me,
and whom I adore!”
I hope I may never see a man look like Reggie did then. He had turned
gray, even to his lips. He just stared at me, and I think the truth of what I
had said slowly sank in upon him. He drew back.
“I hope you’ll be happy!” he said, and I replied:
“Oh, and I hope you will be, too.”
I followed him to the door and he kept on staring at me with that dazed
and incredulous look upon his face. Then he went out and I closed the door
forever on Reggie Bertie.
* * * * * *
The expressman had just put my trunk in the studio. I opened the door of
the little room that Paul had fixed up for me.
“Are you afraid, darling?” he asked. “Are you going to regret giving
yourself to a poor devil like me?”
I answered him as steadily as my voice would let me, for I was
trembling.
“I am yours as long as you love me, Paul.”
I had started to remove my hat.
“Not yet, darling,” said Paul, and he took me by the arm and guided me
toward the door. “First we have to go to the ‘Little Church Around the
Corner.’”

THE END.

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Nonlinear Differential Equations And Chaos Pei Yu Christopher Essex

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