FIRST ORDER ODE.ppt

8/10/2019 FIRST ORDER ODE.ppt

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ORDINARY DIFFERENTIALEQUATIONS

BUM2133LECTURER: ROZIEANA BT KHAIRUDDIN

019-9661379


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Lesson Objective :

Classify the ordinary and partial differentialequationsDetermine the order of differential equationsDistinguish and determine

* the independent and dependent variables,

*linear and nonlinear differential equations and*homogeneous and non-homogeneousequations.


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1.2 The Classification of

Differential Equations

y x y f

x f

242

is a PDE, why?

2

2 4 cos 2d f df

x xdx dx

ORDINARY AND PARTIAL DIFFERENTIAL EQUATIONS( ODE & PDE )

is a ODE, why?


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Cont..

INDEPENDENT AND DEPENDENT VARIABLES( IV & DV )

y x y f

x f

242

xdxdg xdxdf cos46

xdxdf

xdx

f d 2cos42

2

is a PDE, iv are x and y.dv is f .

Now, do this exercises


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Cont..

THE ORDER OF A DIFFERENTIAL EQUATION

y x y f

x f

24 2

xdxdf

xdx

f d 2cos42

2

xdxdf

xdx

f d 2cos42

2

xdx

df x

dx f d

2cos42

3

3

is a first-order PDE

is a second-order ODE

The order of an equation is not affected by any power to which thederivatives may be raised.

Now, do this exercises.


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Cont..

LINEAR AND NONLINEAR DIFFERENTIAL EQUATIONS

y x y f

x f

24 2 y x y x f

24 223

are linear PDE

2

4 0dx dxdt dt

are all nonlinearODE

2

2 4sind x dx

x t dt dt

4 sin 0dx xdt

Linear equation as those in which the dependent variable orvariables and their derivatives do not (occur as products,raised to power or in nonlinear function).


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Now, do this exercises.2

25 4 9 2cos3d x dx x t dt dt

2 3

1 3

y xdydx x y

2

1 , where is a constantdy

y C C dx

2

2

1, where is a constant

N N N kN k

t r r r

2

2 0d y dy

x xydx dx


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Cont..

HOMOGENEOUS AND NONHOMOGENEOUS EQUATIONS

02

y x f 04 x

dt dx

and 0)(sin4 xt dt dx

are all homogeneous equations,

y x y f

x f

24 2 , t dt dx

t dt

xd sin42

2

and xdxdf

xdx

f d 2cos42

2

are all nonhomogeneous equations,

,


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Objectives :Distinguish terms between general andparticular solutions, boundary and initialconditions, analytical and numerical solutionand problems that under-determined and fullydetermined.

Solve the initial value problems of differentialequations.

1.3 Solving Differential Equations


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Cont..

Integrate the differential equation xdxdy

2

On integrating, we obtain a general solution

C x y 2

Now do this exercise

Integrate twice the differential equation )(22

2

2 xlxw

dx yd

here w and l are constants, to find a general solution for y

General and Particular Solutions


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Cont..

Find the unique solution of DE23 x

dxdy

which satisfies the condition

4)1( y

The general solution of the DE is C x y 3

Applying the boundary condition

C C x y 33 14 3C

and the particular solution is 33 x y

General and Particular Solutions


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Cont..

Find the unique solution of DE23 x

dxdy

which satisfies the

condition 4)1( y

Additional conditions on the solution of a DE are called boundary conditions. In the special case in which all the boundary conditions are given at the same value of the

independent variables the boundary conditions are calledinitial conditions.

Boundary and Initial Condition


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Cont..Boundary and Initial Condition

2 23 2

3 2 0 0 0, 0 1, 2 0d x d x dx dx

t x x xdt dt dt dt

2 1/ 23 2 2

2 23 2 2

14 0 0 0, 0 , 0 0

d x d x dx dx d yt x t x x U

t dt dt dt dt dx

BC

IC


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Objectives :

Determine and find the solutions(for case initial value problems)of separable equations.

Determine and find the solutions(for case non initial value problems)of separable equations.

1.4 First-Order OrdinaryDifferential Equation


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Cont..

Separation of variables is a technique commonly used to solve first-orderODE. It is so-called because we rearrange the equation to be solved such that all

terms involving the dependent variable appear on one side of the equation, andall terms involving the independent variable appear on the other. Integrationcompletes the solution.

Examples of such equations are

32 y xdxdy

4 , 0

dx xt x

dt

Elementary Analytical Solution Methods :Separable Equations


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Cont..

1. Use the method of separation of variables to solve the differential equation

a. y

x

dx

dy 23

b. xt dt dx

4

Solution

a. C x y 32

21

b. 22 22

22lnt ct Aee x

C t x



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Cont..

2. Find the solution of the initial-value problem

a. 1)1(,0)1()1(22 y x y

dxdy

y x

b. 2)0(,212 x x

t dt dx

3. Find the general solution of the equation

a. ye

dxdy x

b. ye x

dxdy 23

c. y x

dxdy sin6

Elementary Analytical Solution Methods :

Separable Equations


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Cont..

But, some DE, while not being in separable form, can betransformed, by means of a substitution, into separableequations.For example :

a)2 2 , 0, 0dxt x xt t x

dt

b)23

dx t xdt t x

c)2 24 4 2

dxt xt x

dt



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Objectives :

Determine and find the solutions(for case initial or non initial valueproblems) of exact equations.

Cont.. 1.4 First-Order Ordinary

Differential Equation


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Cont..

Some first-order DE are of a form (or can be manipulated into a form) that is

called EXACT .

How to define an EXACT EQUATIONS ?

o Let

,

,

M x ydydx N x y

(1)

o By definition ; Equation , , 0 M x y dx N x y dy (2)is said to be an EXACT EQUATION if there are

exist a continuous function ,u x y , so then , ,du M x y dx N x y dy (3)


Exact Equations


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Cont..

o Since ,u x y is a continuous function,

then

2 2u u

y x x y

(6)Or

u u y x x y

that is represent

M N y x

(7)

o M N y x

will proved that

, , 0 M x y dx N x y dy is an EXACT EQUATION .


Exact Equations


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Cont..

Solution : 1.(a)

2

2

2 2 0

2 2 0

dx xt x t

dt

xtdx x t dt

Since x is the dependent variable and t is the independent variable,

Then 2 2u

M x t t

and 2u

N xt x

,

2

2u u M x x t x t x

and 2

2u u N xt x t x t

Since M N x t

, therefore this DE is an EXACT EQUATION .


Exact Equations


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Cont..

How to solve this EXACT EQUATION ?

o Remember thatu

M x (8)

andu

N y (9)

o Integrate equation (8) with respect to x, so then

u Mdx y (10)o Differentiate equation (10), which is u with respect to y and compare

the result with equation (9) to get the unknown y .


Exact Equations


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Cont..

Example 2:

Find the solution of 0222 t x

dt dx

xt by using

EXACT EQUATION since it is proved that the DE is an exact.


Exact Equations


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Cont..

Solution : 2

2 2u M x t t

(2.1)

and

2u N xt x

(2.2)

1. Integrate equation (2.1) with respect to t to get u,

2 2u x t t 2 2u x t t x (2.3)


Exact Equations

C


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Cont..

2. Differentiate equation (2.3) with respect to x, to getu x ,

2 2 x t t xu x x

2u xt x x

(2.4)

3. Compare equation (2.4) with equation (2.2).

Since 2u

xt x x

2u

xt x , then 0 x and x C .


Exact Equations

C


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Cont..

Therefore,2 2u x t t C .

4. Since the general solution for an EXACT EQUATION

in the form of ,u x y k , then 2 2 x t t k C

2 2

x t t D where D k C

or

2 D t x

t


Exact Equations


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Objective :

Determine and find the solutions(for case initial or non initial value

problems) of linear equations.

1.4 First-Order Ordinary

Differential Equation


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Cont..

The most general first-order linear DE must have theform

)()( t q xt pdt

dx


Linear Equations

How to solve first-order linear DE?


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Cont..

1. Rearrange the equation to be in the form of

)()( t q xt pdt dx

.

(1)

2. Get ( ) p t and solve p dt .

3. Calculate the integrating factor

pdt

t e (2)

4. Rearrange the equation in the form

d x qdt

(3)5. Integrate equation (3) with respect to t , which is the solution

x q dt (4)

Elementary Analytical Solution Methods :Linear Equations


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Cont..

Note : Remember that

ln

, for 0 f x

e f x f x

For example : ln tan tan xe x

ln t e t


Linear Equations


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Cont..

Example :

Solve the differential equation

4 0t dx

t x edt

.


Linear Equations


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Cont..

Solution :

1. Rearrange the DE,

4 0t dx

t x edt

4 t dx x edt t t

2. 4 p t

t and4

p dt dt t

4

4ln lnt t


Linear Equations


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C


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Cont..

Exercises :

Find the solutions of the following initial-value problems:

1. 2 1 0, 2 2

dxt x x

dt

2. 2 2 1 0, 0 0dx

t x xdt

Elementary Analytical Solution Methods :Linear Equations

Solution:


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Solution: 2

2 2

( ) ( )

12

2

1

question 1

1 0, 2 2

1) Rearrange to be in the form of linear DE, ( ) ( )

1 1

1 12) Find the

1

3) Evaluate the integrating factor, =

4)

p t q t

pdt

t

dxt x x

dt dx

p t x q t dt

dx xdt t t

t pdt dt t dt

t t

e

e

1 1

2

1 1

2

1 1

2

1 1 1

drearrange to be in the form,dt

d 1dt

5) solve the integration,

1

we know that, '

1Therefore,

1

applying the condit

t t

t t

f x f x

t t

t t t

x q

e x et

x qdt

e x e dt t

f x e dx e c

e dt e ct

e x e c x e c

1 12 2

12

1 1 11 22

ion (2) 2

12 1

1 1t t

x

e c c ee

x e e x e

2

( ) ( )

22

2

question 2

2 2 1 0, 0 0

1) Rearrange to be in the form of linear DE, ( ) ( )

4 2

2) Find the

44 2

2

3) Evaluate the integrating factor, =

4) rearra

p t q t

pdt

t

dxt x x

dt dx

p t x q t dt

dx t x t dt

pdt

t tdt t

e

e

2 2

2 2

2

2

2

2 2

2 2

2 2

22

22 2

dnge to be in the form,

dtd

2dt

5) solve the integration,

2

we know that, '

Therefore, 22

12 2

applying the

t t

t t

f x f x

t t

t t t

x q

e x e t

x qdt

e x e t dt

f x e dx e c

ee t dt c

ee x c x e c

2 2

0

2 2

condition (0) 0

1 10

2 2

1 1 1 12 2 2

t t

x

e c c

x e x e


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Objective :

Apply ordinary differential equations insolving engineering problems.

1.5 Applications of OrdinaryDifferential Equations


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Cont..

A chemical reaction is governed by thedifferential equation

25dx K xdt

where x t is the concentration of thechemical at time t . The initial concentrationis zero and the concentration at time 5s is

found to be 2. Determine the reaction rateconstant K and find the concentration at time10s and 50s. What is the ultimate value ofthe concentration?

A

Answer: x (10)=2.857 and x (50)=4.348Ultimate value, x =5


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Cont..

A skydivers vertical velocity is governed bythe differential equation

2d m mg K dt

where K is the skydivers coefficient of drag.If the skydiver leaves her aeroplane at time

0t

with zero vertical velocity find at whattime she reaches half her final velocity.

B

1 1Answer: tanh2

mt

gK


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Cont..

A chemical A is formed by an irreversiblereaction from chemicals B and C . Assumingthat the amounts of B and C are adequate to

sustain the reaction, the amount of A formedat time t is governed by the differentialequation

71dA K Adt

If no A is present at time 0t find anexpression for the amount of A present attime t .

C

1

6

1 1Answer: A 1

1 6

t

Kt


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Cont..

Water is heated to the boiling pointtemperature 100C. It is then removedfrom heat and kept in a room which is at a

constant temperature of 60C. After 3minutes, the temperature of the water is90C. Find the temperature of the waterafter 6 minutes.

D

Answer: T =82.5C


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Cont..

A 12V battery is connected to a simpleseries circuit in which the inductance isH and the resistance is 10 . Determine

the current i if i(0)=0. Hint: Use

E

di L Ri E t dt

20

1.2Answer: 1.2 t i e

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FIRST ORDER ODE.ppt