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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
2
Ex.(Problem 16) ODE : y ' xy ' y 0
2
General solution : y cx c
Particular solution : y 2 x 4
Singular solution : y x2 / 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.
y0 ce0 c 5.7
Particular solution : yx 5.7e3x
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 0.5ke kt 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
gy 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
y0 100
Step 2 Solution of the model.
▶ General solution :
▶ Particular solution :
dy
1
dt
y 5000
100
ln y 5000
1
t c*
100
y0 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 2xyy' y 2 x2
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 xy 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 y2 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
PF
PF
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.
Px, y e x y ye y
Q x, y xe y 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
x y
e y e x y e y ye y 1
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 ux, y ex xy e y c
Step 3 Particular solution y 0 1
u 0, 1 e0 0 e 3.72
u x, y ex 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 e2 x
p 1, r e2 x , h pdx x
y e h eh 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 ya
a0
&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 py1a 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 By2
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 eh 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