Download 슬라이드 제목 없음

Ch. 1 First-Order ODEs
Ch. 1 First-Order ODEs
 Ordinary differential equations (ODEs)
• Deriving them from physical or other problems (modeling)
• Solving them by standard methods
• Interpreting solutions and their graphs in terms of a given problem
Ch. 1 First-Order ODEs
1.1 Basic Concepts. Modeling
1.1 Basic Concepts. Modeling
 Differential Equation : An equation containing derivatives of an unknown function
Ordinary Differential Equation
Differential Equation
Partial Differential Equation
 Ordinary Differential Equation : An equation that contains one or several derivatives of an
unknown function of one independent variable
Ex. y '  cos x,
y '' 9 y  0,
x 2 y ''' y ' 2e x y ''   x 2  2  y 2
 Partial Differential Equation
: An equation involving partial derivatives of an unknown function of two or more variables
Ex.
 2u  2u

0
x 2 y 2
Ch. 1 First-Order ODEs
1.1 Basic Concepts. Modeling
 Order : The highest derivative of the unknown function
Ex. (1) y '  cos x
(2) y '' 9 y  0
 First order
 Second order
(3) x 2 y ''' y ' 2e x y ''   x 2  2  y 2

Third order
 First-order ODE : Equations contain only the first derivative y ' and may contain y and any
given functions of x
• Explicit Form : y '  f  x, y 
• Implicit Form :
F  x, y, y '  0
Ch. 1 First-Order ODEs
1.1 Basic Concepts. Modeling
 Solution : Functions that make the equation hold true
• General Solution
: a solution containing an arbitrary constant
Solution
• Particular Solution
: a solution that we choose a specific constant
• Singular Solution
: an additional solution that cannot be obtained from the general solution
 Ex.(Problem 16) ODE :  y '  xy ' y  0
2
2
General solution : y  cx  c
Particular solution : y  2 x  4
Singular solution : y  x 2 / 4
Ch. 1 First-Order ODEs
1.1 Basic Concepts. Modeling
 Initial Value Problems : An ordinary differential equation together with specified value
of the unknown function at a given point in the domain of the solution
y '  f  x, y  ,
y  x0   y0
 Ex.4 Solve the initial value problem
y' 
dy
 3 y,
dx
y 0  5.7
Step 1 Find the general solution.
(see Example 3.) General solution : y  x   ce3x
Step 2 Apply the initial condition.
y 0  ce0  c  5.7
Particular solution : y  x   5.7e3 x
Ch. 1 First-Order ODEs
1.1 Basic Concepts. Modeling
 Modeling
 The typical steps of modeling in detail
Step 1. The transition from the physical situation to its mathematical formulation
Step 2. The solution by a mathematical method
Step 3. The physical interpretation of differential equations and their applications
Ch. 1 First-Order ODEs
1.1 Basic Concepts. Modeling
Ex. 5 Given an amount of a radioactive substance, say 0.5 g(gram), find the amount present at any later time.
Physical Information.
Experiments show that at each instant a radioactive substance decays at a rate proportional to the
amount present.
Step 1 Setting up a mathematical model(a differential equation) of the physical process.
By the physical law :
dy
y
dt

dy
 ky
dt
The initial condition : y  0  0.5
Step 2 Mathematical solution.
General solution : y  t   cekt
0
Particular solution : y  0  ce  c  0.5

Always check your result : dy  0.5kekt  ky,
dt
Step 3 Interpretation of result.
The limit of y as t   is zero.
y  t   0.5ekt
y  0   0.5e0  0.5
Ch. 1 First-Order ODEs
1.2 Geometric Meaning of y` = f ( x , y ). Direction Fields
1.2 Geometric Meaning of y` = f ( x , y ). Direction Fields
 Direction Field , y’=f(x,y) represents the slope of y(x)
For example, y’ = xy
- short straight line segments, lineal elements, can be drawn in xy-plane
- An approximate solution by connecting lineal elements, Fig.7(a)
 Reason of importance of the direction field
• You do not have to solve the ODE to find y(x).
• The method shows the whole family of solutions and their typical properties., but its accuracy is limited
Ch. 1 First-Order ODEs
Fig.7 CAS means computer algebra system (y(x)=1.213e^x^2/2)
In this way, approximate sol is obtained. But it is sufficient.
The exact solution can be obtained by the methods, in the following sections
Ch. 1 First-Order ODEs
1.3 Separable ODEs. Modeling
1.3 Separable ODEs. Modeling
 Separable Equation : g  y  y '  f  x 
 A differential equation to be separable all the y ’s in the differential equation is on the one side and all the x ’s
is on the differential equation is on the other side of the equal sign.
 Method of Separating Variables
g  y  y'  f x 
 Ex. 1 Solve
y'
1
1 y2

 gy dy   f x  dx  c
 dy

 dx  dy 
 dx

y '  1 y2


dy / dx
1
1 y2
1
 1 y
2

dy   dx  c
dy
 dx
1 y2

arctan y  x  c

y  tan  x  c 
Ch. 1 First-Order ODEs
1.3 Separable ODEs. Modeling
 Modeling
 Ex. 3 Mixing problems occur frequently in chemical industry. We explain here how to solve the basic model
involving a single tank. The tank in Fig.9 contains 1000gal of water in which initially 100lb of salt is dissolved.
Brine runs in at a rate of 10gal/min, and each gallon contains 5lb of dissolved salt. The mixture in the tank is
kept uniform by stirring. Brine runs out at 10 gal/min. Find the amount of salt in the tank at any time t.
Step 1 Setting up a model.
▶ Salt’s time rate of change dy / dt  y' = Salt inflow rate – Salt outflow rate
“Balance law”
Salt inflow rate = 10 gal/min × 5 lb/gal = 50 lb/min
Salt outflow rate = 10 gal/min × y/1000 lb/gal = y/100 lb/min

y '  50 
▶ The initial condition :
y
1
5000  y 

100 100
y0  100
Step 2 Solution of the model.
▶ General solution :
▶ Particular solution :
dy
1

dt
y  5000
100

ln y  5000  
1
t  c*
100
y0  5000  ce  5000  c  100  c  4900
0

y  5000  ce
y  5000  4900e


t
100
t
100
Ch. 1 First-Order ODEs
1.3 Separable ODEs. Modeling
 Extended Method : Reduction to Separable Form
Certain first order equations that are not separable can be made separable by a simple
change of variables.
y
▶ A homogeneous ODE y '  f   can be reduced to separable form by the substitution of y=ux
x
 
 y
y '  f    u ' x  u  f u  
x
du
dx 
y


& y '   ux  '  u ' x  u 
 y  ux  u 
f u   u x 
x

 Ex. 6 Solve 2 xyy'  y 2  x 2
2 xyy '  y 2  x 2

1 y x
y'    
2 x y

c
u 1 
x
2

1
1
u ' x  u  u  
2
u

2

c
 y
  1 
x
x

x 2  y 2  cx
2u
1
du   dx
u 1
x
2
Ch. 1 First-Order ODEs
1.4 Exact ODEs, Integrating Factors
1.4 Exact ODEs, Integrating Factors
 Exact Differential Equation : The ODE M(x , y)dx +N(x , y)dy =0 whose the differential form
M(x , y)dx +N(x , y)dy is exact, that is, this form is the differential du 
u
u
dx  dy .
x
y
 If ODE is an exact differential equation, then
M  x, y  dx  N  x, y  dy  0
 Condition for exactness :

du  0
M N

y
x




u  x, y   c
M   u   2u
  u  N 
  
  

y y  x  xy x  y  x 
 Solve the exact differential equation
• M  x, y  
u
x

u  x, y    M  x, y  dx  k  y 
• N  x, y  
u
y

u  x, y    N  x, y  dy  l  x 


u
 N  x, y 
y
u
 M  x, y 
x


get
dk
& k  y
dy
get
dl
& l  x
dx
Ch. 1 First-Order ODEs
 Ex. 1 Solve
1.4 Exact ODEs, Integrating Factors
cos  x  y  dx   3 y 2  2 y  cos  x  y   dy  0
Step 1 Test for exactness.
M
  sin  x  y 
y
N
N  x, y   3 y 2  2 y  cos  x  y  
  sin  x  y 
x
M  x, y   cos  x  y 

M N

y
x
Step 2 Implicit general solution.
u  x, y    M  x, y  dx  k  y    cos  x  y  dx  k  y   sin  x  y   k  y 


u
dk
 cos  x  y  
 N  x, y 
y
dy
dk
 3y2  2 y
dy


k  y3  y 2  c *
u  x, y   sin  x  y   y3  y 2  c
Step 3 Checking an implicit solution.
u
 cos  x  y   cos  x  y  y ' 3 y 2 y ' 2 yy '  0
x


cos  x  y    cos  x  y   3 y 2 2 y  y '  0
cos  x  y  dx   3 y 2  2 y  cos  x  y   dy  0
Ch. 1 First-Order ODEs
1.4 Exact ODEs, Integrating Factors
 Reduction to Exact Form, Integrating Factors
Some equations can be made exact by multiplication by some function, F  x, y   0,
which is usually called the Integrating Factor.
 Ex. 3 Consider the equation  ydx  xdy  0

  y   1,
y
If we multiply it by 1

 x  1
x
x
2
 That equation is not exact.
, we get an exact equation 

y
1
dx  dy  0 
2
x
x

  y
1
  1 
 2    2   
y  x 
x
x  x  
Ch. 1 First-Order ODEs
1.4 Exact ODEs, Integrating Factors
 How to Find Integrating Factors
FPdx  FQdy  0
The exactness condition :


 FP    FQ 
y
x

F
P F
Q
PF

PF
y
y x
x
Golden Rule : If you cannot solve your problem, try to solve a simpler one.
Hence we look for an integrating factor depending only on one variable.
Case 1) F  F  x 

FPy  F ' Q  FQx
Case 2) F *  F *  y  
F
F
 F',
0
x
y

1 dF
1  P Q 
 R  x  where R  x   


F dx
Q  y x 

 F  x   exp
  R  x  dx 
1 dF *
1  Q P 
 R * where R*  

  F *  y   exp
F * dx
P  x y 
  R *  y  dy 
Ch. 1 First-Order ODEs
1.4 Exact ODEs, Integrating Factors
 Ex. Find an integrating factor and solve the initial value problem
e
x y
 ye y  dx   xe y  1 dy  0,
y  0   1
Step 1 Nonexactness.
Px, y   e x  y  ye y
Qx, y   xey  1 

P
 e x  y  e y  ye y
y
Q
 ey
x
P Q

y x
Step 2 Integrating factor. General solution.
R
1  P Q 
1
1
x y
y
y
y
x y
y


  y  e  e  ye  e   y  e  ye 
Q  y x  xe  1
xe  1
R* 
1  Q P 
1

e y  e x  y  e y  ye y   1

  x y
y 
P  x y  e  ye

e
x

 Fails.
F *  y   e y
 y  dx   x  e y  dy  0 is the exact equation.
u    e x  y  dx  e x  xy  k  y 

u
 x  k '  y   x  e y
y

k '  y   e  y , k  y   e  y
The general solution is u x, y   e x  xy  e  y  c
Step 3 Particular solution y  0  1

u  0, 1  e0  0  e  3.72
 u  x, y   e x  xy  e y  3.72
Ch. 1 First-Order ODEs
1.5 Linear ODEs. Bernoulli Equation. Population Dynamics
1.5 Linear ODEs. Bernoulli Equation. Population Dynamics
Homogeneous Linear ODEs
Linear ODEs
ODEs
Nonhomogeneous Linear ODEs
Nonlinear ODEs
 Linear ODEs : ODEs which is linear in both the unknown function and its derivative.
Ex.
y ' p  x  y  r  x  : Linear differential equation
y ' p  x  y  r  x  y 2: Nonlinear differential equation
• Standard Form : y ' p  x  y  r  x  ( r(x) : Input, y(x) : Output )
 Homogeneous, Nonhomogeneous Linear ODE
y ' p  x  y  0 : Homogeneous Linear ODE
y ' p  x  y  r  x   0 : Nonhomogeneous Linear ODE
Ch. 1 First-Order ODEs
1.5 Linear ODEs. Bernoulli Equation. Population Dynamics
 Homogeneous Linear ODE.(Apply the method of separating variables)
y ' p  x  y  0
dy
  p  x  dx
y


ln y    p  x  dx  c *

 p  x  dx
y  ce 
 Nonhomogeneous Linear ODE.(Find integrating factor and solve )
y ' p  x  y  r  x 
• Find integrating factor.
• Solve e 
R
 py  r  dx  e
pdx
1  P Q 


 p
Q  y x 

is not exact 





 py  r   p  0  1 
y
x 
1 dF
p
F dx

 F  e
pdx
dy  0
pdx
pdx
pdx
pdx
u
 pye   l '  x   e   py  r   l '  x   re  , l  x     re  dx  c
x
pdx
pdx
pdx
pdx
 pdx
pdx
u  ye    re  dx  c

ye    re  dx  c

 y  e    re  dx  c 


u  ye 

pdx
 py  r  dx  dy  0

pdx
 l  x 
 Ex. 1 Solve the linear ODE
y ' y  e 2 x
p  1, r  e2 x , h   pdx   x

 y  e  h   e h rdx  c   e x   e  xe 2 x dx  c   e x e x  c   e 2 x  ce x
Ch. 1 First-Order ODEs
1.5 Linear ODEs. Bernoulli Equation. Population Dynamics
 Bernoulli Equation : y ' p  x  y  g  x  y a
 a0
&1
1 a
We set u  x    y  x  

u '  1  a  y  a y '  1  a  y  a  gy a  py   1  a   g  py1a   1  a  g  pu 

u ' 1  a  pu  1  a  g : the linear ODE
 Ex. 4 Logistic Equation
Solve the following Bernoulli equation, known as the logistic equation (or Verhulst equation)
y '  Ay  By 2
y '  Ay  By 2
p  A, r  B

y ' Ay   By 2 & a  2  u  y 1 

u '   y 2 y '   y 2  Ay  By 2    Ay 1  B   Au  B

h   pdx  Ax

u ' Au  B
B
B

& u  e  h   e h rdx  c   e  Ax  e Ax  c   ce Ax 
A
A

The general solution of the Verhulst equation is y 
1
1

B  ce  Ax
u
A

