mathematic 1st order differrentation

8/10/2019 mathematic 1st order differrentation

1/13Lesson 1_NCB 10303

1

DIFFERENTIAL

EQUATION(D.E.)

2

dy

dx

d y

dx,

2

2

INTRODUCTION

etc, is called a differential equation (D.E.).

What is a differential equation ?

An equation wh ich contains

04:Example xdt

dx

xdx

dyx

dx

yd2cos42

2

3

dy

dxx 2

If we are asked to solve the f irst order D.E.

, then what are we trying to find ?

We aret rying to f ind a function of y

in term o f x )(xfywhich satisfies the D.E.

dy

dxx 2

where x is the independent variable,

y is the dependent variable.

dy

dxx 2Given

Dependent variable

Independent variable

4

Example1 : Solve dy

dxx 2

Solution : Given

y x C 2

Solving a D.E. involves some form

of integration.

dy

dxx 2

integrate

5

THE CLASSIFICATION OF D.E.

The are many techniques for solvingdifferential equations

Different methods being applicable todifferent kinds of equation.

It is necessary to understand the variouscategories and classifications of differentialbefore apply the various method to solve

the equation.

6

1. Ordinary and Partial D.E

xdx

dfx

dx

fd2cos4:Example

2

2

Differential equation may involve either

ordinary or partial derivatives.

Thoseequations involving ordinaryderivatives are called ordinary differential

equations



7

While those equations involving

partial derivatives are called partial

differential equations

yxy

f

x

fExample 24:

2

8

2. The order of a D.E

yxy

f

x

f24

2

is a firstorderpartial D.E

The order of a differential equation is thedegree of the highest derivative that occurs

in the equation.

Examples :

9

yxy

fx

x

f

yx

f24 2

2

2

2

3

is a 3rdorder

partial D.E.

x

dx

dfx

dx

fd2cos4

2

2

is a 2ndorderordinary D.E.

04

2

dt

dx

dt

dx i s a 1st orderordinary D.E.

The order of an equation is not affected by

any powerto which the derivatives may be raised.

Example :

10

The variables with respect to which

differentiation occurs are called

independent variables

3. Independent and dependent variables

While those that are differentiated

are dependent variables.

11

DependentVariables

(upper)

IndependentVariables

(lower)

yxy

f

x

f24 2

tdt

dyt

dt

yd2cos4

2

2

f

y

yx,

t

12

The standard way of arranging terms in aD.E. so that all terms containing thedependent variable occur on the left-handside of the equality sign, and those termsthat involve only the independent variableand constant terms occur on theright-hand side.

4. Homogeneous &

non-homogeneous D.E.



13

When linear differential equation arearranged in this way, those in whichthe right hand side is zero are calledhomogeneous equations and those in

which i t i s n o n - z e r o a r en o n - h o m o g e n e o u s e q u a t i o n s

02

2

cydx

dyb

dx

yda

)(2

2

xfcydx

dyb

dx

yda

Homogeneous

Non-

homogeneous

14

Determine whether the following D.Eis homogeneous or non-homogeneous

04) xdt

dxa

Solution : Given 04 xdt

dx

Dependentvariable

Checking whether the given D.E is arrangedin standard way

*** All the derivatives and term

with dependent variable (x) must be at theleft hand side.

dt

dx

15

Given 04 xdt

dx

Dependentvariable

Right Hand side =0

Answer : Homogeneous D.E.

16

Determine whether the following D.Eis homogeneous or non-homogeneous

Solution : Rearrange in standard way

*** All the derivatives and termwith dependent variable (y) must be atthe left hand side.

dx

dy

dx

yd,

2

2

02cos42

2

xdxdyx

dxyd

17

02cos42

2

xdx

dyx

dx

yd

Dependentvariable

Independentvariable

xdx

dyx

dx

yd2cos4

2

2

Right Hand side

Answer : Non-homogeneous D.E.

0

18

Solving 1st Order

Differential Equation



19

ii) Exact product type

Types of 1st order D.E.

i) D.E. with variable separable (SOVA)

iii) The integrating factor type

iv) Solve by using substitution method

20

Type 1 : D.E. with Variable Separable(SOVA)

When a first order D.E is written as :

)().( ygxhdx

dy

Such differential equations of this formare separable i.e.

dxxhyg

dy

)()(

21

Next, integrate both sides

Thus, we get

dxxhdyyg

)()(

1

CxHyG )()((general solution of the D.E.)

Remark : Solving a 1st order D.E.involves some form of integration

22

EXAMPLE 1 : Solve the differential equation

3 5 2ydydx

x

STEP 1 : SEPARATE THE VARIABLES

In the form of

3 5 2ydy x dx

TYPE 1 : SOVA

Solution :

dxxhdyyg )()(

23

dxxydy 253

This is the general solution of the differential

equation because value C not yet determine.

STEP 2 : INTEGRATE BOTH SIDES

Cxy

3

5

2

3 32

24

Simplify answer :

Cxy

3

5

2

3 32

6

Cxy

35

236

32

cxy 6109 32

Axy 32 109 (where A =6C)



25

EXAMPLE 2 :

Given the following differential equation,solve for s .

s

dt

dst )3(

Solution :

26

27

29

NOTE :

If value of C is to be evaluated,i.e. the i n i t i a l c o n d i t i o n is given,then the solution of the D.E. is calledthe p a r t i c u l a r s o l u t i o n

Example 3 : Find the part icular solut ion

of the D.E.05 y

dx

dy

for the initial condition x=0, y=1

30

STEP 1 : Rearrange and separate the variables

Solution :



32

34

EXAMPLE 4

2 1 42 2x dy

dxx y

If y = 1 when x = 0, f ind the particular so lut ion

by expressing y in terms of x .

Find the general solution of the D.E.

35

dxyxdyx 22 412

)1(2)4( 22

x

dxx

y

dy

STEP 1 : SEPARATE THE VARIABLES

Solution :

22 412Given yxdx

dyx

dx

x

xdy

y )1(2)4(

122

STEP 2 : INTEGRATE BOTH SIDES



37

38

39 42



45

12ln4

12ln

4

1Cyy

Therefore, the general solution is :

dx

x

xdy

y )1(2)4(

1:From

22

partialfraction substitutionmethod

2

21ln

4

1Cx

46

Cxy

y

)1)(2(

2ln

2

DCBACD

AB

BAB

A

BAAB

lnlnlnlnln

lnlnln

lnln)ln(

Simplify Answer by usinglaw of logarithmic

2

2

1 1ln4

12ln

4

12ln

4

1CxCyy

12

2

1ln4

1

2ln4

1

2ln4

1

CCxyy

Let

)(412 CCC

Generalsolution

)(41ln2ln2ln12

2 CCxyy

47

Find value C to get particular solution,

substitute the initial condit ion into t he general solution

Cxy

y

)1)(2(

2ln

2

When 1,0 yx

C

)10)(12(

12ln

3lnC

48

3)1)(2(

22

xy

y

( substitute C=ln3 into general solution)

3ln)1)(2(

2ln

2

xy

y

Take exponential both sides

3ln)1)(2(

2ln

2

ee xy

y



49

To express y in terms of x

3)1)(2(

22

xy

y

)1)(2(32 2 xyy

yyxxy 33662 22

26633 22 xyxyy

Crossmultiplication

Rearrange y in theleft hand side

expand

50

2

2

34

46

x

xy

(particular solution

where y in terms of x)

26634 22 xyxy

266)34( 22 xyx

26633 22 xyxyy

Factorize y

51

FIRST ORDER D.E.

(Engineering Problems)

SOVA Method

52

EXAMPLE 1 : Materials

l

is given by : )( lxwdx

dM

where w is the constant load. Find an expression

M in term of x

.

The bending moment , M , of a beam of length

53

)( lxwdx

dM

Solution :

dxlxwdM )(

Step 1 : Separate the variables

)( lxwdx

dM

dependent variable (M)

independent variable ( x) constants (w, l)

Given

54

Step 2 : Integrate both sides

dxlxwdM )(

M

Clxx

wM

2

2

dxlxw )(

Take out coefficient

expression M in term of x


10/13Lesson 1_NCB 10303

55

The velocity, atuv of an object

dtdsv where

s is the displacement ,u is the initial velocity

a is the acceleration.

0t 0, s

show that the displacement ,2

2

1atuts

EXAMPLE 2 : Mechanics

is defined as :

Given that when

56

atuv

Solution :

-------equation (i)

dt

dsv -------equation (ii)

Substitute equation (ii) in equation (i)

atu

dt

ds

Given

Given

57 dtatuds

atudt

ds


Solution :

atudt

ds

dependent variable (s)

independent variable ( t) constants (u, a)

Given

58

dtatuds

Ct

auts

2

2

(general solutio n)


Find value of C by using the given initial condition

0C

Substitute ,0t 0s into the general solution

Cau

2

0)0(0

2

C 000

59

Ct

auts

2

2

02

2

at

uts

2

2atuts ANSWER :

Substitute C=0 into the general solution

(SHOWN !)

60

Consider a tank full of water which is being

drained out through an outlet. The height

of water in the tank at time is given by)(mH

Hdt

dH )108.2( 3

Given that when ,0t 4H

find an expression for H in terms of t

EXAMPLE 3 : Fluid Mechanics


11/13Lesson 1_NCB 10303

61

Solution :

H

dt

dH)108.2( 3

dtHdH 0028.0

dtH

dH0028.0


62


dtH

dH0028.0

dtdHH 0028.021

CtH

0028.0

21

12

1

CtH 0028.02 2

1

Rewrite

General

solution

63


Substitute 4H into the general solution,0t

C

C

042

)0(0028.042 21

CtH 0028.02 21

40028.02 21

tH

CtH 0028.02 21

4C (substitute into the general solution)

64

To get an expression for H in terms of t

.

2

40028.02

1

t

H

2

4

2

0028.02

1

t

H

220014.0 tH

40028.02 21

tH

Square both sides

20014.021

tH

65

mh

Area,

Further Engineering applications

at time t (seconds) is given by :

ghA

A

dt

dh20

0AWhere and A are the

cross-sectional areas of theoutlet and the tank respectively.

0A

1. Apply first order differential equations to

fluid mechanics. The differential equationdescribing the tank filled to a height

Area,0A

66

A cylindrical tank of diameter 3 m is filled with water to a

height of .The water is drained througha c i r c u l a r hole of diameter 0.1 m.

mh

i) Obtain the differential equation relating the height h

of water at time t.

,0t 2h

iii) How long (in minutes) does it take to empty

the tank which is 2 m full ?

EXAMPLE 4 : Fluid Mechanics

ii) Solve the differential equation f or the initial condition,


12/13Lesson 1_NCB 10303

67

SOLUTION (i) : From the formu la ghA

A

dt

dh20

The outlet area , is a circle of diameter 0.1 m0A

So, area of circle

0025.02

1.0 2

2

0

rA

Area,0A

Since we have a cylindrical tank of

diameter 3 m, therefore cross-sectional area of the tank:

25.22

3 2

2

rA

Step 1 : find 0A

Step 2 : find A

68

Substitute ,0025.00 A 25.2A and

81.9g into the formula :

ghA

A

dt

dh20

hdt

dh 81.92

25.2

0025.0

hdt

dh 62.190011.0

Answ er (i) : h

dt

dh00487.0

simplify

69

dthdh 00487.0

dth

dh00487.0

ii) Solve the differential equation for the initial condition,

,0t 2h

SOLUTION (ii) : From the D.E. hdt

dh00487.0

step 1 : Separate the variables

70


dth

dh00487.0

dtdhh 00487.021

Cth

00487.0

21

12

1

Cth 00487.02 21

Rewrite

General

solution

71


Substitute 2h into the general solution,0t

C

C

022

)0(00487.022 21

828.200487.02 2

1

th

Cth 00487.02 21

828.2C (substitute into the general solution)

72

To get an expression for h in terms of t

.

2

828.200487.02

1

th

2

828.2

2

00487.02

1

t

h

2414.1002435.0 th

828.200487.02 21

th

Square both sides

414.1002435.021

th


13/13

73

iii) How long (in minutes) does it take to empty

the tank which is 2 m full ?

Solution : If the tank empty, height of water, h=0

2414.1002435.0 th

t = ?h = 0

2414.1002435.00 t

414.1002435.00 t

From

74

0414.1002435.0 t

ondst sec581698.580002435.0

414.1

414.1002435.0 t

Time in minute : min60

581t min68.9

Answer : It take approximately 9.68minutes to empty the tank.

DISCUSSIONDISCUSSIONSolve the following 1st order differential equations :

75

22

22

)1xyx

xyy

dx

dy

2)0(thatgiven,2

3

2

1)2 yy

dx

dy

x

y

dx

dy 1)3

2

xydx

dyxy

22 sec4tan)4

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mathematic 1st order differrentation