ChewMA1506-09Ch1

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MA1506

Mathematics IIGroup B

1

Thank Dr Toh Pee Choonfor providing me

most of the PowerPoint slides

Chew T S MA1506-09 Chapter 1

Chapter 1

2


1.1. Introduction

In this Chapter, we study

1st order ordinary differential equation

and its a lications

3

2nd order ordinary differential equations

However its applications will be given in

Chapter TWO


1.2 Separable equations

WestudyODEofthefollowingform

( )dy M x

4

( )dx N y=

Weshalllearnhowto

solve separableequations

byexamples


Example 1


5

Thenintegratebothsides,

2

1

1

xe dx dy

y=

+


Example1(cont)

get


6Chew T S MA1506-09 Chapter 1

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Example 2

A radioactive substance decomposes

(so the amount of substance is decreasing)


7

.

Amount of substance = 2 mg at t=0.

what is the amount at time t ?


Let x(t) be the amount of substance at

time t.

Then represents the rate of change

dx

dt

Example 2 (cont)


8

of the amount


Why x in

Example 2 (cont)


9

Since the amount is decreasing


Example 2 (cont)


10


Example 3

A copper ball heated to 100C.

At t=0, it is placed in water maintained at 30C.

At the end of 3 mins tem erature of the ball is


11

,reduced to 70C.

Find the time at which the temperature of the ballis 31C.


Example 3 (cont)

Physical information:

Rate of change dT/dt of the temperature T

Newtons Law of Cooling


12

of the ball is proportional to the difference between

Tand the temp T0of the surrounding medium.


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Example 3 (cont)

100C

30C

70C

3 min


13

whent=0,T=100, whent=3,T=70

WhenT=31, findt

T0=30


Example 3 (cont)

Initial condition



Example 3 (cont)

2nd condition


15

Solve


Example 4

A sky diver falls from rest .

When the sky diver's speed is 10m/s,

the parachute opens .

Assume t=0 the arachute o ens


16

Find the speed of the sky diver at time t.


Example 4 (cont)

Physical assumptions and laws:

weight of the man + equipment = 712N

air resistance = bv2, where b=30 kg/m=30kg/meter


17

ew on s

2nd Law

712N

g=accelerationduetogravity

=9.8m/s2


Example 4 (cont)

Newtons

2nd Law


18

v > k then vdecreases

v < k then vincreases


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Example 4 (cont)



Example 4 (cont)



removing abs | |

get

Example 4 (cont)


21

WHY? See Example 6(a)

Clarification on removing abs | |


How to find c

Recall when t=0, v=10m/s , subst. this

into

Example 4 (cont)


22

to get c


Example 4 (cont)


23

lim ( ) 4.87t

v t

=


Example 5

A 2000m3 room contains air with 0.002%

CO at time t=0

The ventilation system blowing in air which


24

con a ns

The system blowing in and out air at a rate

of0.2m3/min

When the air in the room containing

0.015% CO?


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Example 5 (cont)

Let x(t) = vol of CO in the room at time t

x(t) of CO

0.2 m3 /min

3% CO 0.2 m3 /min

Room 2000 m3


25

COperm3


x(t) of CO

0.2 m3 /min

3% CO 0.2 m3 /min

Vol: 2000 m3

Example 5 (cont)1.2 Separable equations


Example 5 (cont)


27

0.015% CO means x(t1) = 0.00015 X 2000 = 0.3


What happens when ODE is not separable?

For examples,


28

2 tricks:

reduction to separable

linear change of variables


Reduction to separable form

Set


29

' ( )dv

y g v v xdx

= = +


Example 6a : Reduction to separable form


30

1

2 2

d v vx

d x v=


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Example 6a (cont)



Clarification on removing abs | |


32

positive negative

Can be positive

or negative


Example 6b : Reduction to separable form

yv

x=Let ' ' y v xv= + andThen


33

Nowfindvthen

findy

' ( )v xv g v+ =

where22 cos

'x x

v

v

=Then


Linear Change of Variable

ODE of the form

y'=f(ax+by+c),


34

by setting u=ax+by+c

where b 0


Example 7: Linear Change

( 2 ) 3'

2 2 5

x yy

x

=

+


35

set


Example 7 (cont)

2 51

4 11

u du

u dx

+=

+

4 102

4 11

udu dx

u

+=

+1

1 2du dx =


36

4 11u+


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Integrating factor

d v kv g

d t m+ =(cont)

1.3 Linear 1st order d.e


Example 10

At time t = 0 a tank contains 20 lbs of salt dissolved

in 100 gal of water.

Assume that water containing 0.25 lb of salt pergallon is entering the tank at a rate of3 gal/min


44

.

Find the amount of salt at time t.


amt of water = constant3 gal/sec.25 lb/gal

Example 10 (cont)

dQ=inflow outflow


45

3 gal/sec100 gal

3 0.25 3100

dQ Q

dt=


amt of water = constant3 gal/sec

.25 lb/gal

Example 10 (cont)1.3 Linear 1st order d.e

46

3 gal/secga


Reduction to linear form: Bernoulli Equations

Set

Nonlinear if

n > 1 or n< 0


47

Giveneqmultipliedby(n1)yn get


Example (ii): Bernoulli Equations

Set


48

Integrating factor:


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Review: First Order ODE

Separable

Linear

Use integrating factor


49

What if neither applies? Use some clever substitution

a) Reduction to separable, v = y/x

b) Linear change, u = ax+by +c

c) Bernoulli eq: z= y1-n


hasmanysolutions.

Howeverifaninitialconditiony(x0)=y0,(very

oftenx0=0)isgiven,

(cont)1.3 Linear 1st order d.e

50

thenthereisoneandonlyonesolution,

i.e.,thesolutionisunique.

Intheabove,Q(x)maybezerofunction.

HereweassumethatPandQarecontinuous


1.4 Second-order linear ODE withconstant coefficients

The general form is

2

2( )

d y dy A By R x

dx+ + =

51

where A, B are constants.When R(x) is zero function, we have

This equation is called homogeneous.

2

20

d y dy A By

dxdx+ + =


When R(x) is not zero function,

is called nonhomogenous.

2

2( )

d y dy A By R x

dxdx+ + =

1.4 Second-order linear ODE

52

We shall consider homogeneous case

first


1.4.1 Second-order homogeneouslinear ODE

2

20

d y dy A By

dxdx+ + =


53

It is clear that

zero function is a solution,

which is called a trivial solution


Now we shall look for nontrivial solutions.

Recall that the general solution of first-order

linear homogeneous ODE


54

is

( ) 0dy

p x ydx

+ =

( )P x dx y Ce

=

1( )dy p x dx

y=


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Consider a special case: when p(x) is

constant , say B. Then the general

solution isBx

y Ce

=From this solution , we may guess thatFrom this solution , we may guess that


55

2

20

d y dy A By

dxdx+ + =

a nontrivial solution

of

is of the form

a nontrivial solution

of

xy e

=Chew T S MA1506-09 Chapter 1

Then and

Subst.these into the given ODE, get

xdye

dx

=

22

2

xd ye

dx

=

2 0 x x xe A e Be + + =


56

Which is called characteristic equation

or auxiliary equation

2 0A B + + =


Thus

When solving

There are three cases:

Two distinct real roots

Only one real root

2 0A B + + =


57

Two distinct complex roots


Two distinct real roots

Suppose that two distinct real roots

are and

Then we have two distinct (linearly

independent, see Appendix 1)

1 2


58

solutions1xy e

= 2xy e=

1 2

1 2

x xy c e c e

= +Chew T S MA1506-09 Chapter 1

General soln is

The above property is called

superposition principle (see Appendix 2)

In fact , we can prove that every solution

is of the form


59

Here and are any constants.1C 2C

1 2

1 2x x

y c e c e = +


Example: Solve '' ' 6 0 y y y =

Solution: Let xy e=


60

Subst this y into the given ODE , get

2 6 0 =


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(cont)

We have two distinct real roots,

.

Thus the general solution of the equation

1 22, 3

= =


61

is

2 31 2

x x y c e c e= +


(b)Only one real root

Suppose that the only one real root is

Then we have a solution

1

1xe=


62

For 2nd order ODE, we can prove

that we should have two distinct (linearly

indep.) solutions.

What is the 2nd solution?


The 2nd solution is

We can verify that

1x y xe=

1 1

1 2x x

y c e c xe = +


63

is also a solution (superposition principle)

In fact , we can prove that every solution is

of the form 1 11 2

x x y c e c xe

= +


Example: Solve

Solution

The auxiliary equation is

2

24 4 0

d y dyy

dxdx + =

24 4 0 + =


64

We have only one solution

.

Hence the general solution is

2 2

1 2

x x

y c e c xe= +

1 2 =


Two distinct complex roots

Suppose that we have two distinct

complex roots, namely and

Then we have two distinct (l inearly indep)

complex-valued solutions

1 2


65

and

Suppose that

Then

1xy e= 2

xy e

=

1 a ib = +

2 a ib = Chew T S MA1506-09 Chapter 1

Note that these two solutions are

Complex-valued . However we want

real-valued solutions. How to get

real-valued solutions ?

We shall look at the real art and ima inar


66

part of the solution

(cos sin )ax

e bx i bx= +1x ax ibx y e e e

= =


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We can verify that the real part

and the imaginary part

are two (real-valued)

solutions

we can prove that every solution is of the

cosax y e bx=sinax y e bx=


67

form

1 2

1 2

cos sin

( cos sin )

ax ax

ax

y c e bx c e bx

e c bx c bx

= +

= +


Do we need to consider

2x

y e

=


68

ANS: NO, since it induces the same

general solution.


Example: Solve

Solution

The complex roots of the auxiliary

e uation

2

22 2 0

d y dyy

dxdx + =

2 2 2 0 + =


69

are and

Hence the general solution is.

1 2( cos sin )

x

y e c x c x= +

1 1 i = + 2 1 i =


Remark: As in the case for 1st order ODE,

2nd order ODE

has many solutions.

2

20

d y dy A By

dxdx+ + =


70

If initial conditions are given, then there is

ONLY one solution,

see the following example.


Example

are two linearly indep solutions

Initial value conditions

Initial value problem (IVP)1.4 Second-order linear ODE

71

is the general solution


1.4.2 Second-order nonhomogenouslinear ODE

The general form is

Solving this equation can be reduced to

2

2( )

d y dy A By R x

dxdx+ + =


72

1.Find the general solution to

the homogrneous equation

, say the solution is

2

20

d y dy A By

dxdx+ + =

hyChew T S MA1506-09 Chapter 1

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2. Find a particular solution to the

nonhomogeneous equation

3. Add the solutions from step 1

py

2

2

( )d y dy

A By R xdxdx

+ + =


73

,

,which is the general solution to

(see Appendix 3)

h p

2

2( )

d y dy A By R x

dxdx+ + =


We have learnt step 1. There are two

methods for step 2.

Method 1.

The method of undetermined coefficients.

Method 2.


74

The method of variation of parameters.


Example 1 solve

Can we guess a solution?

2'' ' 2 4 y y y x =

or

2p y Ax=

2 A Bx Cx= + +


75

We can verify that

is NOT a solution

2p y Ax=


(cont)

Now we can verify that

is a solution

2p y A Bx Cx= + +

First ( ) ' 2p y B Cx= +

( ) '' 2py C=


76

Subst above into 2'' ' 2 4 y y y x =

get

So

2 22 2 2 2 2 4C B Cx A Bx Cx x =

2, 2, 3C B A= = =


(cont)

Hence

is a particular solution of

23 2 2p y x x= +

2

'' ' 2 4 y y y x =


77

On the other hand

is the general solution of

21 2

x xh y C e C e

= +

'' ' 2 y y y =Chew T S MA1506-09 Chapter 1

Therefore

is the general solution of the

nonhomogeneous ODE

2 21 2 3 2 2

x xh p y y C e C e x x

+ = + +


78

Here and can be any constant

2'' ' 2 4 y y y x =

1C 2C


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Example 2 Solve

Guess a particular solution

We guessAs in Example 1, we can find the values of A

'' 3 ' 4 2sin y y y x =

cos sinpy A x B x= +


79

and B

Hence a particular solution is

3

17A =

17B

=

3 5cos sin

17 17p y x x

= +


(cont)

On the other hand , the general solution

of

is

'' 3 ' 4 0 y y y =

41 2

x xh y C e C e

= +


80

So the general solution of

is

'' 3 ' 4 2sin y y y x =

h py y+Chew T S MA1506-09 Chapter 1

Example 3

Consider

We guess a particular solution is

'' ' ax y py qy e+ + =

axp y Ae=


81

u s n o e g ven , ge

Hence

p y Ae=

2( ) ax ax A a pa q e e+ + =

2

1A

a pa q=

+ +


(cont)

We have to assume that

i.e., is NOT a solution of the

2 0a pa q+ +

axe

Case1


82

equation

'' ' 0 y py qy+ + =


Case 2

Suppose that is a solution of

Then we guess a particular solution is

axe

'' ' 0 y py q+ + =


83

Subst into get'' 'ax y py qy e+ + =

axp y xAe=

axp y xAe=


2( ) ax A a pa q xe+ + (2 ) ax ax A a p e e+ + =

Hence

So

(2 ) 1 A a p+ =

1

2A

a p=

+


84

i.e., is NOT a double root ofa2 0p q + + =


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Double root?

If a is a double root of

2 0p q + + =


85

then2 4

2 2

p p q pa

= =

i.e., a is a double root iff a is the only

root.Chew T S MA1506-09 Chapter 1

Case 3

Suppose that is a double root

Then we guess a particular solution is

a

2 ax x Ae=


86

Subst this solution into

get

Hence

'' ' ax y py qy e+ + =1

2A =

21

2

axp y x e=


Summary

(A)The general solution of

is

(1)A particular solution of2

'' 3 ' 4x

y y y e =

'' 3 ' 4 0 y y y =4

1 2x x

C e C e+


87

is of the form

(2)A particular solution of

is of the form

2xAe

4'' 3 ' 4

x y y y e =

4xxAe


Summary (cont)

(B)The general solution of

is

'' 2 ' 1 0y y+ + =

1 2x x

C e C xe +


88

So a particular soln of

is of the form

'' 2 ' 1x

y y e+ + =

2 x x Ae


Example 4 Find a particular soln of

First the general soln of

'' sin y y x+ =

''


89

Is

As in the summary, a particular soln is of

the form

y y+ =

1 2sin cosC x C x+

( sin cos )py x A x B x= +Chew T S MA1506-09 Chapter 1

(cont)

We can check that

Hence a particular soln is

10,

2A B

= =

1( cos ) y x x

=


90

The general soln of

Is

'' sin y y x+ =

1 21

sin cos cos2

C x C x x x+


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Example 5 Consider

We can guess that a particular soln is

By the method used in previous

3 2 2( ) x Ax Bx Cx D e+ + +


91

Examples, we can find A, B, C, D.

However the computation is very

involved. We will use the following

method to simplify the computation


(cont)

Let

Then a particular soln is

3 2( )u x Ax Bx cx D= + + +

We have


92

Subst the above into the given ODE, get,


3 3u x x x=

Subst

into

We can find A,B, C, D. We get

3 2( )u x Ax Bx Cx D= + + +3

'' 2 2u u x =


93

3 2( 3 ) xp y x x e=

Thus a particular soln is


Example 6

Consider

First note that

has only one root 2 (double root)

3 2'' 4 ' 4 20 x y y y x e + =

24 4 0 + =


94

the general soln of

is

'' 4 ' 4 0 y y y + =

2 2

1 2

x xC e C xe+


Example 6 (cont)

So a particular soln of

is of the form2 3 2 2

( )x

x Ax Bx Cx D e+ + +

3 2'' 4 ' 4 2 0 x y y y x e + =


95

Note that we have extra term above

By method used in Example 5 , we can

get A=1,B=C=D=0

2x


Why we have extra term x in the above?

Example 7

We may guess that a particular soln is of the form

[( )sin 2 ( )sin 2 ] x Ax B x Cx D x+ + +


96

Again, it is not easy to find A,B,C, D.

We shall use the method in Examples

5,6 to find a particular soln.

'' 4 y y o+ =

Since sin2x and cos2x are solns of


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first notice that

(cont)

To find a particular soln of

2

cos(2 ) sin(2 )ix

e x i x= +So now we consider the following ODE


97

where z is a complex valued function ,

say z (x)=w(x)+iy(x)


(cont)

Remark (*) :

The imaginary part of

is

So if z(x) is a complex soln of

216 xixe

16 sin 2x x


98

Then the imaginary part Im(z) of z is a

soln of

2'' 4 16 xi z z xe+ =

'' 4 16 sin 2 y y x x+ =


As in Example 5, we assume a particular soln is

We have


99

where

A, B are complex numbers

Subst z,z,z into the z- equation, get

( ) ( )u x x Ax B= +

extra term x in u


(cont)

Subst u(x)=x(Ax+B) into the above u-equation,

we get


100

2

2 22

iA i

i i= = = 1B =

So


Now look at what we have done


101

From the remark (*), we have


Example 8

As in Example 7, we consider

2'' 2 ' 5 16 x ix z z z xe e+ + =


102

( 1 2 )( ) ( )

i xz x u x e

+=

where ( ) ( )u x x Ax B= +As in example 6, we can find A and B. The

real part of z(x) is a particular soln of y-

equation . Why we have extra term x in u(x)?Chew T S MA1506-09 Chapter 1

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Example 9

We known that a particular soln of

is

A particular soln of

'' 3 ' 4 2sin y y y x =

1 (3cos 5sin )17

p y x x= '' 3 ' 4 4 y y y =


is

Then a particular soln of

is103

*1py =

'' 3 ' 4 2sin 4 y y y x = +*

p py y+Chew T S MA1506-09 Chapter 1

Remark: Method of undetermined coeff only works

for the following

Conditions: constants


104

Polynomials

Exponentials

Sine/Cosine


Method 2: Method of Variation of

Parameters

Conditions: 1) Continuous functions,


105

mainly constant functions

2) Homogeneous solutions known

Try 1 2( ) ( ) ( ) ( )py u x y x v x y x= +


Method of Variation of Parameters


106

Set


Subst the above into the given ODE, we

get

So we have two equations


107

Solving these two eqs , we get u and v

integrate u and v , we will get u and v

For details, see following


Method of Variation of Parameters



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2

Example 3



Example 3 (cont)



Example 3(cont)



Summary: 2nd Order Linear D.E.


118

If pand qare constants,

use charac. equation

Method of undetermined coeff

Variation of parametersChew T S MA1506-09 Chapter 1

Appendix 1 Optional

Linearly independent

Two solutions u(x) and v(x) are said to

be linearly dependent if we can find a

constant c such that u x =cv x , for all

119

x, otherwise they are linearly

independent

For examples, sinx and cosx are linearly

indep; sinx and 2sinx are linearly dep.


Appendix 2 Optional

Proof: Superposition principle

If y1 and y2are solutions then so is c y1 + d y2


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Appendix 2 (cont) Caution

Does not hold for nonhomogeneous ode

121

and 1 are solutions, but

1 1 cosx+ +is NOT a solution


Appendix 3 Optional

General soln of nonhomogeneous ODE

Particular, i.e.no arbitrary

constants

GeneralSolution

122

We can

check that


Appendix 4


Appendix 4 (cont)


Appendix 5


Appendix 5 (cont)


Chapter 1

END

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