Transcript Chapter 1

Chapter 1
Differential
Equations
Chapter 1
1
Introduction
• In sciences and engineering, mathematical
models are developed to aid in the
understanding of physical phenomena.
• This model often yield an equation that
contains some derivatives of an unknown
function which is called differential equation.
• Examples:
free fall of object, radioactive decay, electric
circuit, management, rate of change in
temperature, mixing problem in tank and many
more
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Chapter 1
Example (DE)
n
• Any equation containing
• Example:
dy
x  y2  0
dx
d y
n
dx
xy  y 2  0
d2y
xy 2  y 2 sin x  0
dx
xyy  y 2 sin x  0
Chapter 1
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Definition
• An equation containing the derivatives of
one or more dependent variables, with
respect to one or more independent
variables, is said to be differential
equation (DE).
• In order to talk further about differential
equation we shall classify differential
equations by type, order and linearity.
Chapter 1
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Types of Differential Equations
Types of
differential
equations (DE)
Ordinary Differential
Equation (ODE)
Partial Differential
Equation (PDE)
Our Focus
Chapter 1
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Ordinary Differential Equations
A differential equations involving only ordinary
derivatives with respect to a single independent
variable.
d 2x
dx
 a  kx  0
2
dt
dt
Our Focus
Single independent variable, t
Chapter 1
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Example (ODE)
dy
 2 x  y
dx
x = independent variable
y = dependant variable
d 2x
dx

4
 2x  0
2
dt
dt
t = independent variable
x = dependant variable
Chapter 1
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Partial Differential Equations
A differential equations involving only partial
derivatives with respect to more than one
independent variable.
u u

 x  2y
x y
Multiple independent variable, x and y
Chapter 1
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Order of Differential Equations
Order of
differential
equations (DE)
First Order
Second Order
Our Focus
Chapter 1
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Order of Differential Equations
 Largest derivatives present in the differential
equations
 Example:
dy
x  y2  0
dx
Our Focus
xy  y 2  0
first order
d2y
xy 2  y 2 sin x  0
dx
xyy  y 2 sin x  0
second order
Chapter 1
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Example
• Classify each as an ordinary differential equation
(ODE) or a partial differential equations (PDE) and
give the order:
(i) Competition between two species, ecology
dy y (2  3x)

dx x(1  3 y )
ODE, First Order
(ii) Laplace’s equation, heat, aerodynamics
d 2u d 2u
 2 0
2
dx
dy
Chapter 1
PDE, Second Order
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Linearity of Differential Equations
Linearity of
differential
Only for ODE
equations (DE)
Linear
Nonlinear
Chapter 1
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Linear Differential Equations
A linear differential equations is any differential
equation that can be written in the following form
dny
d n 1 y
dny
an  x  n  an1  x  n 1  ...  a1  x  n  a0  x  y  F  x 
dx
dx
dx
an  x  , an1  x  ,..., a0  x  = depend on independent variable, it
can be zero or non-zero functions,
constant or non-constant functions,
linear or non-linear functions
Chapter 1
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Criteria of Linear Differential
Equations
i) the dependent variable and its derivatives occur to
the first power only
2 3
d y
dx2
ii) no products involving the dependent variable with
its derivatives
dy
y
dx
iii) no nonlinear functions of the dependent variable
such as sin, quadratic, exponential, etc
sin y
y
2
Chapter 1
e
y
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Nonlinear Differential Equations
A nonlinear ordinary differential equation is simply
one that is not linear. Nonlinear functions of the
dependent variable or its derivatives, such as sin y
or ey, cannot appear in linear equation.
Chapter 1
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Example
d2y
3
1.

y
0
2
dx
Order : Second order
Linear : Nonlinear because the dependent
d2y
3
2.

y

x
dx 2
Order : Second order
Linear : Linear because the dependent
variable occur to the power of three, y3
variable, y occur to the first
power
2
d
y
dy
3.
y
 cos x
2
dx
dx
Order : second order
Linear : nonlinear because y dy/dx
Chapter 1
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Exercises
Decide whether or not the following equations are linear
and determine the order of each equations.
a)
dy
sin x  y  x
dx
b)
dx
 x  t3
dt
c)
d2y
2

y
0
2
dx
d)
dy
 sin y  0
dx
Chapter 1
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Answers
a) Order = first order ; Linearity = linear
b) Order = first order ; Linearity = linear
c) Order = second order ; Linearity = nonlinear
d) Order = first order ; Linearity = nonlinear
Chapter 1
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General solution and particular
solution of a differential equation
The solution of a differential equation is a
relationship
between
the
dependent
and
independent variables such that the differential
equation is satisfied for all values of the
independent variable over a specified domain.
Chapter 1
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Example
Verify that y  e is a solution of the differential equation
x
dy
y
dx
Solution:
dy
If y  e , then
 ex
dx
dy
 y , for all values of x
dx
x
Therefore;
dy
y  e is the solution for
y
dx
x
Chapter 1
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Example
dy
 y , where C is
Verify that y  Ce is a solution of
dx
any constant.
x
Solution:
dy
If y  Ce , then
 Ce x
dx
dy
 y , for all values of x with C for any constant
dx
x
C is called an arbitrary constant
Therefore;
dy
y  Ce is the general solution of
y
dx
x
Chapter 1
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continue…
If x  0, y  4;
Condition
y  Ce x
4  Ce0
4  C (1)
C4
Therefore;
y  4e x
This is called a particular solution.
Chapter 1
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Conditions
Conditions
Initial
Conditions
Boundary
Conditions
Chapter 1
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Initial Conditions
Solution to the differential equation on an interval I
that satisfies at x0 the n initial condition
y ( xo )  y 0
dy
( x o )  y1
dx
.
.
d n 1 y
( x0 )  y n 1
n 1
dx
or
y' ( xo )  y1
x I
Where 0
Chapter 1
and y1,y2,…,yn-1 are given constant
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Initial Value Problem
A differential equation together with an initial condition is
called an Initial Value Problem (IVP)
y  f ( x, y), y( x0 )  y0
• With x as independent variable (instead of t)
• x0 and y0 are given values
• The initial condition y(x0) = y0 is used to determine the value of
A and B in the general solution
• Values of a function and its derivative at the same point
• Number of initial conditions for a given differential equation
depend upon the order of the differential equation.
Chapter 1
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Example
The following is an Initial Value Problem (IVP)
y   xy ,
y(0)  1
Here’s another IVP
4 x y  12 xy  3 y  0 ,
2
1
y (4)  ,
8
Chapter 1
y(4)  
3
64
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Boundary Conditions
- solution to the differential equation on an interval I
that specified at two distinct points x j and xk
y( x j )  y j ,
y ( xk )  yk


Chapter 1
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Boundary Value Problem
A differential equation together with a boundary condition
is called an Boundary Value Problem (BVP)
y  f ( x, y), y( x j )  y j ,
y( xk )  yk
- Values of a function and its derivative not at the same point
- Number of boundary conditions for a given differential
equation does not depend upon the order of the DE
Chapter 1
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Example
The following is a Boundary Value Problem (BVP)
y  2 y  2 y  0 ,
y(0)  1,
y( 2)  0
Here’s another BVP
3 y  8 y  3 y  0 ,
y(3)  1,
Chapter 1
y(3)  1 e2
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First Order Differential
Equations
• In this section, we will learn how to solve
the first order differential equations. But in
order to do that we need to understand
what type of equations we are dealing
with. The types of equations that we are
going to discuss are :
Separable Equations
Chapter 1
Integral Factor
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Separable Equations
A first order differential equation of the form ;
dy
 g ( x ) h( y )
dx
(1)
is said to be separable or to have separable
variables.
Chapter 1
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Example : Separable Equations
• For example the equation
dy
 y 2 xe 3 x  4 y
dx
• is said to be separable because we can factor it
as;
f ( x, y)  ( xe )( y e )
3x
g(x)
2 4y
h(y)
Chapter 1
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Solution : Separable Equations
Equation (1) can be rewritten to isolate the variables x
and y on opposite side of the equation as in ;
1
dy  g ( x)dx
h( y )
(2)
Then we integrate both sides ;
1
 h( y)dy   g ( x)dx
(3)
And we obtain ;
H ( y )  C1  G ( x)  C 2
Where C = C1 + C2
H ( y )  G ( x)  C
(4) is called the general solution for (1)
(4)
Chapter 1
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Example 1
• Solve the nonlinear equation
dy x  5
 2
dx
y
• Solution :
• Step 1 : Separate equation
y dy  ( x  5)dx
2
• Step 2 : Integrate equation
2
y
 dy   ( x  5)dx
Chapter 1
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Example 1
• Step 3 : The general solution is
y3 x2

 5x  c
3
2
#
Chapter 1
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Example 2
• Solve the equation
(1  x)dy  ydx  0
• Solution :
• Step 1 : Separate equation
(1  x)dy  ydx
1
1
dy 
dx
y
1 x
• Step 2 : Integrate equation
1
1
 y dy   1  x dx
ln y  ln(1  x)  cChapter 1
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Example 2
• Step 3 : The general solution is
y
 ec
1 x
y  (1  x )e c #
Chapter 1
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Integral Factor
The linear first order differential equations can be
expressed in the form ;
dy
(1)
a1 ( x)
dx
 a 0 ( x ) y  b( x )
Where a1(x), a0(x) and b(x) depend only on the
independent variable x. For example, the equation;
dy
(sin x)  (cos x) y  x 2 sin x
dx
a1(x)
a0(x)
Chapter 1
b(x)
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Solution : Linear Equations
First, we have to divide all terms in equation (1) with
a1(x);
a1 ( x)dy a0 ( x)
b( x )

y
a1 ( x)dx a1 ( x)
a1 ( x)
(2)
Equation (2) can be rewritten as ;
dy
 P( x) y  Q( x)
dx
(3)
Standard
Form
Where ;
a 0 ( x)
P( x) 
a1 ( x)
Chapter 1
b( x )
Q( x) 
a1 ( x)
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Solution : Linear Equations
Second, we need to determine the integrating factor
v(x);
P ( x ) dx

v( x)  e
Third, multiply v(x) to both sides of equation (3) will give
us;
 dy

v( x)   P( x) y   v( x)Q( x)
 dx

(4)
From left-hand side of equation (4) will be ;
Product rule;
d
(uv )  udv  vdu
dx
d
v( x) y   v( x)Q( x)
dx
Chapter 1
(5)
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Solution : Linear Equations
Then integrate both sides of equation (5) we obtain;
d
 dx v( x) y    v( x)Q( x)
Finally we will get the general solution;

1
y ( x) 
v( x)Q( x)dx  C

v( x)
Chapter 1

General
Solution
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Example 1 : Linear Equation
• Solve the linear first ODE
dy
2x
 xy  3 x
dx
• Solution :
• Step 1 : Change the equation to standard form (1)
by dividing with 2x
dy 1
3
 y
dx 2
2
• Step 2 : To find the integral factor let P(x) = 1/2
v( x)  e

1
dx
2
e
Chapter 1
1
x
2
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Example 1 : Linear Equation
• Step 3 : Multiply v(x) = e1/2x to both sides of
standard form equation;
 dy 1  3 2 x
e   y  e
 dx 2  2
1
x
2
1
1
1
d 2x
3 2x
(e y )  e
dx
2
• Step 4 : Integrate equation in Step 3
1
x
2
1
x
2
d
3
 dx(e y)dx   2e dx
1
x
2
e y  3e
1
x
2
C
Chapter 1
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Example 1 : Linear Equation
• Step 5 : The general solution
y  3
C
e
1
x
2
#
Chapter 1
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Example 2 : Linear Equation
• Solve the linear first ODE
dy
 y  e3x
dx
• Solution :
• Step 1 : Since the above equation is in the
standard form then, you don’t have to change
anything.
• Step 2 : To find the integral factor let P(x) = 1
1dx
x

v( x)  e
e
Chapter 1
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Example 2 : Linear Equation
• Step 3 : Multiply v(x) = ex to both sides of standard
form equation;
 dy

e   y   e x .e3 x
 dx

x
d x
(e y )  e 4 x
dx
• Step 4 : Integrate equation in Step 3
d x
4x
(
e
y
)
dx

e
 dx
 dx
1 4x
x
e y  e  C Chapter 1
4
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Example 2 : Linear Equation
• Step 5 : The general solution
1 3x C
y e  x
4
e
#
Chapter 1
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Mathematical Modeling Involving First
Order Differential Equations
• Mathematical modeling is the technique of
representing real world problem which is complex,
involving multiple variables and some interrelated
processes. This method can be used in the study
of
growth
population,
radioactive
decay,
economics problems, changes in temperature,
mixtures, chemical reactions, biological reactions,
mechanics, velocity of a falling object, electric
circuits and etc.
Chapter 1
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References
• Zill, D.G. (2001). A First Course in Differential
Equations with Modeling Applications, 7th ed.
Brooks/Cole.
• Zill, D.G. and Cullen, M.R. (2005). Differential
Equations
with
Boundary-Value
Problems.
Brooks/Cole.
Chapter 1
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