ch. 1 first-order odes ordinary differential equations (odes) deriving them from physical or other...

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

Ordinary Differential Equation : An equation that contains one or several derivatives of an

unknown function of one independent variable

Ex.

Partial Differential Equation

: An equation involving partial derivatives of an unknown function of two or more variables

Ex.

Differential Equation : An equation containing derivatives of an unknown function

Differential EquationOrdinary Differential Equation

Partial Differential Equation

2 2 2' cos , '' 9 0, ''' ' 2 '' 2xy x y y x y y e y x y

2 2

2 2 0u ux y


Ex. (1) First order

(2) Second order

(3) Third order

Order : The highest derivative of the unknown function

2 2 2''' ' 2 '' 2 xx y y e y x y

First-order ODE : Equations contain only the first derivative and may contain y and any

given functions of x

, , ' 0F x y y

' ,y f x y• Explicit Form :

• Implicit Form :

'y

' cos y x

'' 9 0 y y


Solution : Functions that make the equation hold true

Solution

• General Solution

: a solution containing an arbitrary constant

• 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 :

General solution :

Particular solution :

Singular solution :

2' ' 0y xy y

2y cx c

2 / 4y x

2 4y x


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

0 0' , , y f x y y x y

Ex.4 Solve the initial value problem

Step 1 Find the general solution.

(see Example 3.) General solution :

Step 2 Apply the initial condition.


7.50 ,3' yydxdyy

3xy x ce

7.50 0 ccey

xexy 37.5



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



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 :

The initial condition :

Step 2 Mathematical solution.

General solution :


Always check your result :

Step 3 Interpretation of result.

The limit of as is zero.

dy dyy kydt dt

00.5 , 0 0.5 0.5ktdy ke ky y edt

0 0.5y

kty t ce

00 0.5 0.5 kty ce c y t e

y t


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


In this way, approximate sol is obtained. But it is sufficient.The exact solution can be obtained by the methods, in the following sections

Fig.7 CAS means computer algebra system (y(x)=1.213e^x^2/2)

Ch. 1 First-Order ODEs 1.3 Separable ODEs. Modeling

1.3 Separable ODEs. Modeling

Separable Equation :

Method of Separating Variables

dydx

dxdycdxxfdyxfyyg yg '

'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.

Ex. 1 Solve

2' 1y y

2 2 2

2

' /1 1 1 1 1

1 arctan tan 1

y dy dx dy dxy y y

dy dx c y x c y x cy

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 = 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

▶ The initial condition :

Step 2 Solution of the model.

▶ General solution :

▶ Particular solution :

'/ ydtdy

yyy 5000100

1100

50'

1000 y

1001 1 ln 5000 * 50005000 100 100

tdy dt y t c y cey

1000 49005000 4900 100500050000t

eycccey


Extended Method : Reduction to Separable Form

Certain first order equations that are not separable can be made separable by a simple

change of variables.

▶ A homogeneous ODE can be reduced to separable form by the substitution of y=ux

Ex. 6 Solve

' yy fx

' ' & ' ' 'y du dx yy f u x u f u y ux u y ux u x ux f u u x x

22'2 xyxyy

2 22

22 2 2

1 1 1 2 12 ' ' ' 2 2 1

1 1

y x uxyy y x y u x u u du dxx y u u x

c y cu x y cxx x x

1.3 Separable ODEs. Modeling

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 .

If ODE is an exact differential equation, then

Condition for exactness :

Solve the exact differential equation

• get

• get

, , 0 0 ,M x y dx N x y dy du u x y c

u udu dx dyx y

2

M N M u u u Ny x y y x x y x y x

, , , , u uM x y u x y M x y dx k y N x yx y

, , , , u uN x y u x y N x y dy l x M x yy x

& dk k ydy

& dl l xdx


Ex. 1 Solve

Step 1 Test for exactness.

Step 2 Implicit general solution.

Step 3 Checking an implicit solution.

2cos 3 2 cos 0x y dx y y x y dy

, cos sinMM x y x y x yy

2, 3 2 cos sinNN x y y y x y x yx

M Ny x

2 3 2

, , cos sin

cos , 3 2 *

u x y M x y dx k y x y dx k y x y k y

u dk dkx y N x y y y k y y cy dy dy

3 2 , sinu x y x y y y c

2 2

2

cos cos ' 3 ' 2 ' 0 cos cos 3 2 ' 0

cos 3 2 cos 0

u x y x y y y y yy x y x y y y yx

x y dx y y x y dy


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, ,

which is usually called the Integrating Factor.

, 0F x y

Ex. 3 Consider the equation

That equation is not exact.

If we multiply it by , we get an exact equation

0ydx xdy

1, 1 y xy x

2 2 2

1 1 10 y ydx dyx x y x x x x

2

1x


How to Find Integrating Factors

The exactness condition :

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)

Case 2)

0FPdx FQdy

F P F QFP FQ P F P Fy x y y x x

0F FF F x F', x y

1 1' where

exp

y xdF P QFP F Q FQ R x R x

F dx Q y x

F x R x dx

1 * 1* * * where * * exp **

dF Q PF F y R R F y R y dyF dx P x y



Ex. Find an integrating factor and solve the initial value problem

Step 1 Nonexactness.

Step 2 Integrating factor. General solution.

The general solution is

Step 3 Particular solution

1 0, 0 1x y y ye ye dx xe dy y

yyyxyyx yeeeyPyeeyxP

,

yy exQxeyxQ

1,xQ

yP

1 1 1 1 1

x y y y y x y yy y

P QR e e ye e e yeQ y x xe xe

Fails.

1 1* 1 *y x y y y yx y y

Q PR e e e ye F y eP x y e ye

0x ye y dx x e dy is the exact equation.

' ' , x x y y yuu e y dx e xy k y x k y x e k y e k y ey

cexyeyxu yx ,

00 1 0, 1 0 3.72 , 3.72x yy u e e u x y e xy e


Ch. 1 First-Order ODEs 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics

1.5 Linear ODEs. Bernoulli Equation. Population Dynamics

ODEs

Linear ODEs

Nonlinear ODEs

Homogeneous Linear ODEs

Nonhomogeneous Linear ODEs

Linear ODEs : ODEs which is linear in both the unknown function and its derivative.

Ex. : Linear differential equation

: Nonlinear differential equation

• Standard Form : ( r(x) : Input, y(x) : Output )

Homogeneous, Nonhomogeneous Linear ODE

: Homogeneous Linear ODE

: Nonhomogeneous Linear ODE

'y p x y r x

' 0y p x y r x

2'y p x y r x y

'y p x y r x

' 0y p x y


Homogeneous Linear ODE.(Apply the method of separating variables)

Nonhomogeneous Linear ODE.(Find integrating factor and solve )

is not exact

• Find integrating factor.

• Solve

' 0 ln * p x dxdyy p x y p x dx y p x dx c y ce

y

' 0y p x y r x py r dx dy 0 1py r py x

1 1 pdxP Q dFR p p F e

Q y x F dx

y 0pdx pdx

e py r dx e d

Ex. 1 Solve the linear ODE xeyy 2'

2 2 21, , x h h x x x x x x xp r e h pdx x y e e rdx c e e e dx c e e c e ce

' ' ,

pdx pdx pdx pdx pdx

pdx pdx pdx pdx pdx pdx

uu ye l x pye l x e py r l x re l x re dx cx

u ye re dx c ye re dx c y e re dx c


Bernoulli Equation :

We set

' 0 & 1 ay p x y g x y a

1 ' 1 ' 1 1 1

' 1 1

a a a au a y y a y gy py a g py a g pu

u a pu a g

: the linear ODE

1 au x y x

Ex. 4 Logistic Equation

Solve the following Bernoulli equation, known as the logistic equation (or Verhulst equation)

The general solution of the Verhulst equation is

2' ByAyy

2 2 1

2 2 2 1

' ' & 2

' ' '

, & h h Ax Ax

y Ay By y Ay By a u y

u y y y Ay By Ay B Au B u Au B

Bp A r B h pdx Ax u e e rdx c e eA

Ax Bc ceA

1 1

Axy

Bu ceA

ch. 1 first-order odes ordinary differential equations (odes) deriving them from physical or other...

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