Introduction to Differential Equations CHAPTER 1
Introduction to Differential Equations
CHAPTER 1
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Chapter Contents
1.1 Definitions and Terminology1.2 Initial-Value Problems1.3 Differential Equations as Mathematical Methods
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1.1 Definitions and Terminology
Introduction: differential equations means that equations contain derivatives, eg:
dy/dx = 0.2xy (1)
Ordinary DE: An equation contains only ordinary derivates of one or more dependent variables of a single independent variable. eg: dy/dx + 5y = ex, (dx/dt) + (dy/dt) = 2x + y (2)
An equation contains the derivates of one or more dependent variables with respect to one or more independent variables (DE).
Definition 1.1.1 Differential equation
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Partial DE: An equation contains partial derivates of one or more dependent variables of tow or more independent variables.
(3)
Notations: Leibniz notation dy/dx, d2y/ dx2
prime notation y’, y”, ….. subscript notation ux, uy, uxx, uyy, uxy , ….
Order: highest order of derivatives
tu
tu
xu
yu
xu
2 ,0 2
2
2
2
2
2
2
2
xeydydx
dxyd
45
3
2
2
second order first order
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General form of n-th order ODE: (4)
Normal form of (4) (5)
eg: normal form of 4xy’ + y = x, is y’ = (x – y)/4x
Linearity: An n-th order ODE is linear if F is linear in y, y’, y”, …, y(n). It means when (4) is linear, we have
(6)
0) , ,' , ,( )( nyyyxF
) , ,' , ,( )1( nn
n
yyyxfdx
yd
)()()()( 011
1
1 xgyxadxdy
xadx
yda
dxyd
xa n
n
nn
n
n
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The following cases are for n = 1 and n = 2 and
(7)
Two properties of a linear ODE:(1) y, y’, y”, … are of the first degree.(2) Coefficients a0, a1, …, are at most on x
Nonlinear examples:
)()()( 01 xgyxadxdy
xa
)()()()( 012
2
2 xgyxadxdy
xadx
ydxa
')1( yy ysin 2y
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That is, a solution of (4) is a function possesses at least n derivatives and
F(x, (x), ’(x), …, (n)(x)) = 0 for all x in I, where I is the interval is defined on.
Any function , defined on an interval I, possessing at least n derivatives that are continuous on I, when replaced into an n-th order ODE, reduces the equation into an identity, it said to be a solution of the equation on the interval.
Definition 1.1.2 Solution of an ODE
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Example 1 Verification of a Solution
Verify the indicated function is a solution of the given ODE on (- , ) (a) dy/dx = xy1/2; y = x4/16 (b)
Solution:(a) Left-hand side:
Right-hand side: then left = right
(b) Left-hand side:
Right-hand side: 0 then left = right
xxeyyyy ;02
4164
33 xxdxdy
4416
322/142/1 xx
xx
xxy
0)(2)2(2 xxxxx xeexeexeyyy
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Note: y = 0 is also the solution of example 1, called trivial solution
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Example 2 Function vs. Solution
y = 1/x, is the solution of xy’ + y = 0, however, this function is not differentiable at x = 0. So, the interval of definition I is (- , 0), (0, ).
Fig 1.1.1 Ex 2 illustrates the difference between the function y = 1/x and the solution y = 1/x
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Explicit solution: dependent variable is expressed solely in terms of independent variable and constants.Eg: solution is y = (x).
G(x, y) = 0 is said to be an implicit solution of (4) on I,
provided there exists at least one function y = (x)
satisfying the relationship as well as the DE on I.
Definition 1.3 Implicit solution of an ODE
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Example 3 Verification of an Implicit Solution
x2 + y2 = 25 is an implicit solution of dy/dx = −x/y (8)
on the interval -5 < x < 5.
Since dx2/dx + dy2/dx = (d/dx)(25)
then 2x + 2y(dy/dx) = 0 and dy/dx = -x/ysolution curve is shown in Fig 1.1.2
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Fig 1.1.2 An Implicit solution and two explicit solutions in Ex 3
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Families of solutions: A solution containing an arbitrary constant represents a set G(x, y) = 0 of solutions is called a one-parameter family of solutions. A set G(x, y, c1, c2, …, cn) = 0 of solutions is called a n-parameter family of solutions.
Particular solution: A solution free of arbitrary parameters. eg: y = cx – x cos x is a solution of xy’ – y = x2 sin x on (-, ), y = x cos x is a particular solution according to c = 0. See Fig 1.1.3.
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Fig 1.1.3 Some solution of xy’-y=x2 sinx
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Example 4 Using Different Symbols
x = c1cos 4t and x = c2 sin 4t are solutions of x + 16x = 0we can easily verify that x = c1cos 4t + c2 sin 4t is also a solution.
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Example 5 A piecewise-Defined Solution
We can verify y = cx4 is a solution of xy – 4y = 0 on (-, ). See Fig 1.1.4(a).
The piecewise-defined function
is a particular solution where we choose c = −1 for x < 0 and c = 1 for x 0. See Fig 1.1.4(b).
0 ,
0 ,4
4
xx
xxy
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Fig 1.1.4 Some solution of xy’-4y=0 in Ex 5
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Singular solution: A solution can not be obtained by particularly setting any parameters. y = (x2/4 + c)2 is the family solution of dy/dx = xy1/2 , however, y = 0 is a solution of the above DE.
We cannot set any value of c to obtain the solution y = 0, so we call y = 0 is a singular solution.
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System of DEs: two or more equations involving of two or more unknown functions of a single independent variable.
dx/dt = f(t, x, y) dy/dt = g(t, x, y)(9)
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1.2 Initial-value Problems
IntroductionA solution y(x) of a DE satisfies an initial condition.
ExampleOn some interval I containing xo, solve:
subject to:
(1)This is called an Initial-Value Problem (IVP).y(xo) = yo , y(xo) = y1 ,are called initial conditions.
) , ,' , ,( )1( nn
n
yyyxfdx
yd
10)1(
1000 )( , ,)(' ,)( n
n yxyyxyyxy
10)1( )(
nn yxy
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First and Second IVPs
(2)
and
(3)
are first and second order initial-value problems, respectively. See Fig 1.2.1 and 1.2.2.
00 )( :
) ,( :
yxytosubject
yxfdxdy
solve
1000
2
2
)(' ,)( :
)' , ,( :
yxyyxytosubject
yyxfdx
ydsolve
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Example 1 First-Order IVPs
We know y = cex is the solutions of y’ = y on (-, ). If y(0) = 3, then 3 = ce0 = c. Thus y = 3ex is a solution of this initial-value problem
y’ = y, y(0) = 3.If we want a solution pass through (1, -2), that is y(1) = -2, -2 = ce, or c = -2e-1. The function y = -2ex-1 is a solution of the initial-value problem
y’ = y, y(1) = -2.
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In Problem 6 of Exercise 2.2, we have the solution of y’ + 2xy2 = 0 is y = 1/(x2 + c). If we impose y(0) = -1, it gives c = -1. Consider the following distinctions.
1) As a function, the domain of y = 1/(x2 - 1) is the set of all real numbers except x = -1 and 1. See Fig 1.2.4(a).
2) As a solution, the intervals of definition are (-, 1), (-1, 1), (1, )
3) As a initial-value problem, y(0) = -1, the interval of definition is (-1, 1). See Fig 1.2.4(b).
Example 2 Interval / of Definition of a Solution
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Fig 1.2.4 Graph of the function and solution of IVP in Ex 2
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Example 3 Second-Order IVP
In Example 4 of Sec 1.1, we saw x = c1cos 4t + c2sin 4t is a solution of x + 16x = 0Find a solution of the following IVP:
x + 16x = 0, x(/2) = −2, x(/2) = 1 (4)
Solution:
Substitute x(/2) = − 2 into x = c1cos 4t + c2sin 4t, we find c1 = −2. In the same manner, from x(/2) = 1 we have c2 = ¼.
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Existence and Uniqueness:
Does a solution of the IVP exist?If a solution exists, is it unique?
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Example 4 An IVP Can Have Several Solution
Since y = x4/16 and y = 0 satisfy the DEdy/dx = xy1/2 , and also initial-value y(0) = 0, this DE has at least two solutions, See Fig 1.2.5.
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Let R be the region defined by a x b, c y d that contains the point (x0, y0) in its interior. If f(x, y) and f/y are continuous in R, then there exists some interval I0: (x0- h, x0 + h), h > 0, contained in [a, b] and a unique function y(x) defined on I0 that is a solution of the IVP (2).
Theorem 1.2.1 Existence of a Unique Solution
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Fig 1.2.6 Rectangular region R
The geometry of Theorem 1.2.1 shows in Fig 1.2.6.
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Example 5 Example 3 Revisited
For the DE: dy/dx = xy1/2 , inspection of the functions
we find they are continuous in y > 0. From Theorem 1.2.1, we conclude that through any point (x0, y0), y0 > 0, there is some interval centered at x0 on which this DE has a unique solution.
2/12/1
2 and ) ,(
yx
yf
xyyxf
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Interval of Existence / UniquenessSuppose y(x) is a solution of IVP (2), the following sets may not be the same:
the domain of y(x), the interval of definition of y(x) as a solution, the interval I0 of existence and uniqueness.
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1.3 DEs as Mathematical Models
Introduction Mathematical models are mathematical descriptions of something.
Level of resolutionMake some reasonable assumptions about the system.
The steps of modeling process are as following.
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AssumptionsMathematicsformulation
Check model Predictionswith know
facts
Obtainsolution
Express assumptions in terms of different equations
If necessary,alter assumptions
or increase resolutionof the model
Solve the DEs
Display model predictions,e.g., graphically
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Population DynamicsIf P(t) denotes the total population at time t, then
dP/dt P or dP/dt = kP(1)where k is a constant of proportionality, and k > 0.
Radioactive DecayIf A(t) denotes the substance remaining at time t, then
dA/dt A or dA/dt = kA(2)where k is a constant of proportionality, and k < 0.
A single DE can serve as a mathematical model for many different phenomena.
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Newton’s Law of Cooling/WarmingIf T(t) denotes the temperature of a body at time t, Tm the temperature of surrounding medium, then
dT/dt T - Tm or dT/dt = k(T - Tm)(3)
where k is a constant of proportionality.
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Spread of a DiseaseIf x(t) denotes the number of people who have got the disease and y(t) the number of people who have not yet, then
dx/dt = kxy (4)where k is a constant of proportionality.From the above description, suppose a small community has a fixed population on n, If one inflected person is introduced into this community, we have x + y = n +1, and
dx/dt = kx(n+1-x) (5)
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Chemical ReactionsInspect the following equation
CH3Cl + NaOH CH3OH + NaClAssume X is the amount of CH3OH, and are the amount of the first two chemicals, then the rate of formation is
dx/dt = k( - x)( - x)(6)
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MixturesSee Fig 1.3.1. If A(t) denotes the amount of salt in the tank at time t, then
dA/dt = (input rate) – (output rate) = Rin - Rout
(7)
We have Rin = 6 lb/min, Rout = A(t)/100 (lb/min), then dA/dt = 6 – A/100 or dA/dt + A/100 = 6 (8)
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Fig 1.3.1 Mixing tank
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Draining a TankReferring to Fig 1.3.2 and from Torricelli’s Law, if V(t) denotes the volume of water in the tank at time t,
(9)
From (9), since we have V(t) = Awh, then
(10)
ghAdtdV
h 2
ghA
A
dtdh
w
h 2
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Fig 1.3.2 Water draining from a tank
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Series CircuitsLook at Fig 1.3.3.From Kirchhoff’s second law, we have
(11)
where q(t) is the charge and dq(t)/dt = i(t), which is the current.
)(1
2
2
tEqCdt
dqR
dtqd
L
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Fig 1.3.3 Current i(t) and charge q(t) are measured in amperes (A) and coulumbs (C)
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Falling BodiesLook at Fig1.22.From Newton’s law, we have
(12)
Initial value problem
(13)
or 2
2
mgdt
sdm g
dt
sd 2
2
002
2
)0(' ,)0( , vsssgdt
sd
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Fig 1.3.4 Position of rock measured from ground level
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Falling Bodies and Air ResistanceFrom Fig 1.3.5.We have the DE
(14)
and can be written as
or (15)
kvmgdtdv
m
dtds
kmgdt
sdm 2
2
mgdtds
kdt
sdm 2
2
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Fig 1.3.5 Falling body of mass m
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A Slipping ChainFrom Fig 1.3.6.We have
(16)or 2
32 2
2
xdt
xdL 064
2
2
xLdt
xd
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Fig 1.3.6 Chain slipping from frictionless peg
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Suspended Cables From Fig1.25.We have
dy/dx = W/T1 (17)
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Fig 1.3.7 Cables suspended between vertical supports
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Fig 1.3.8 Element of cable
Fig 1.3.8 explains the Element of cable.