chapter 1 - differential equations

8/20/2019 Chapter 1 - Differential Equations

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CHAPTER 1 DIFFERENTIAL EQUATIONS

1.1 INTRODUCTION

1.1.1 DIFFERENTIAL EQUATIONS

• What is the meaning of differential equations?

• In sciences and engineering, mathematical models are developed to aid in

understanding of physical phenomena.

EXA!LE 1

Any equations containingn

n

d y

dx, we called differential equations. The differential equations

can be written as (i 2 0dy

x ydx

− = 2 0 xy y′ − =

(ii2

2

2 sin 0

d y xy y x

dx− = 2 sin 0 xyy y x′′ − =

1 | P a g e

So"e of a##li$ations in%ol%ed in t&is 'orld(

free fall of ob!ect

radioactive decay

electric circuit

management

rate of change in temperature

mi"ing problem in tan#

DEFINITION

An equation containing the derivatives of one or more

dependent variables, with respect to one or more independent

variables, is said to be Differential Equations )DE*.

$athematical models often yield an equation that

contains some derivatives of an un#nown function

which is called differential equations.


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• In order to tal# further about differential equations, we shall classify differential

equations by t+#e, order and linearit+.

EXA!LE ,

%rdinary &ifferential 'quations (%&' An equation that involve onl+ ordinar+ deri%ati%es

of one or more dependent variable with respect to a sin-le inde#endent %ariale. )uch

that

EXA!LE /

*artial &ifferential 'quations (*&' An equation that involve #artial deri%ati%es of one or

more dependent variable with respect to t'o or "ore t&an one inde#endent %ariale.

)uch that

1.1., ORDER OF DIFFERENTIAL EQUATIONS

%rder of &ifferential 'quations is determined by the highest derivative.

2 | P a g e

T+#es of differential equations(%rdinary &ifferential 'quations (%&'

*artial &ifferential 'quations (*&'

(i2

2 0

d x dxa kx

dt dt + + = + )ingle independent variable, t

(ii 2dy

x ydx

= − + + )ingle independent variable, x

(i) 2u u

x y x y

∂ ∂− = −

∂ ∂+ $ultiple independent variable, x

and y

ii 3V V

r ∂ ∂

+ = + $ulti le inde endent variable, r and h

Order of differential equations(

irst %rder

)econd %rder


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EXA!LE 0

Order Ea"#le

irst 2 0dy

x ydx − =

2 0 xy y

′ − =

)econd

2

2

2

ydx

yd x −

0" 2

=− y xy

EXA!LE 2

-lassify each of the following as an ordinary differential equation (%&' or a partial

differential equations (*&' and give the order

(i -ompetition between two species, ecology+(2 3 )

(1 3 )

dy y x

dx x y

−=

−.

(ii aplace/s equation, heat, aerodynamics+

2 2

2 2 0

d u d u

dx dy+ = .

)olution

T+#es of differential equations Order of differential equations

(i %&' irst order

(ii *&' )econd %rder

1.1./ LINEARIT3 OF DIFFERENTIAL EQUATIONS

A linear differential equations is any differential equation that can be written in the

following form ( ) ( ) ( ) ( ) ( )1 1

1 1 01 1...

n n

n nn n

d y d y d ya x a x a x a x y F x

dx dx dx

−

− −+ + + + = .

Where ( ) ( ) ( ) ( )1 1 0, ,..., ,n na x a x a x a x− are depending on independent variable, it can be 0ero

or non10ero functions, constant or non1constant functions, linear or non1linear functions.

3 | P a g e

Order of differential equations(inear 2onlinear


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• Criteria of linear differential equations(

i the dependent variable and its derivatives occur to the first #o'er onl+

'"ample2 3

2

d y

dx

ii no #rodu$ts in%ol%in- t&e de#endent %ariale with its derivatives (in other word

each coefficient depends only on the independent variable, "

'"ampledy

ydx

iii no nonlinear fun$tions of t&e de#endent %ariale as trigonometric, quadratic,

e"ponential, etc

'"ample sin y (trigonometric functions, 2 y (quadratic equations, ye (e"ponential

A nonlinear ordinar+ differential equation is simply one that is not linear. 2onlinear

functions of the dependent variable or its derivatives, such as sin y or ye , cannot appear in

the linear equation.

EXA!LE 4&etermine each of the following equation with order and linearity

(i2

3

2 0

d y y

dx+ =

(ii2

3

2

d y y x

dx+ =

(iii2

2 cos

d y dy y x

dx dx− =

(iv sin dy

x y xdx

+ =

(v 3dx x t

dt + =

(vi2

2

2 0

d y y

dx+ =

(vii sin 0dy ydx

+ =

Solution :

Order of differential equations Linearit+

(i)econd order 2onlinear because the dependent

variable occur to the power of three,3 y

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(ii)econd order inear because the dependent

variable, y occur to the first power

(iii )econd order 2onlinear because

dy

y dx

(iv

(v

(vi

(vii

1., T5E SOLUTION OF DIFFERENTIAL EQUATIONS

• What is t&e solution of differential equations6

EXA!LE 7

3erify that x y e= is a solution of the differential equation

dy y

dx= .

Solution :

If x y e= , then xdy

edx

= .

dy y

dx= , for all values of x .

Therefore, x y e= is the solution for

dy y

dx= .

5 | P a g e

The solution of a differential equation is a relationship

between the deendent and indeendent variables

such that the differential equation is satisfied for all

values of the independent variable over a specified

domain.


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EXA!LE 8

3erify that x y Ce= is a solution of dy ydx =, where C is any constant.

Solution:

If x y Ce= , then x

dyCe

dx= .

dy y

dx= , for all values of x with C for any constant.

C is called an arbitrary constant.

Therefore, x y Ce= is the -eneral solution of

dy y

dx= .

If 0, 4; x y= =

04

4 (1)

4

x y Ce

Ce

C

C

=

===

Therefore, 4 x y e= . This is called a #arti$ular solution.

1.,.1 CONDITION

1.,.1.1 Initial Condition

)olution to the differential equation on an interval, I that satisfies at 0 x the n initial

condition.

0( )o y x y=

1( )ody

x ydx

= or 1'( )o y x y=

6 | P a g e

Conditions

Initial-onditions

4oundary-onditions


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1

0 11 ( )

n

nn

d y x y

dx

−

−− = Where 0 x I ∈ and 1 2 -1, , , n y y y… are given constant.

A differential equation together with an initial condition is called an Initial 9alue !role"

)I9!* . )uch that 0 0( , ) , ( ) y f x y y x y′ = = ;

• With x as independent variable (instead of t

• 0 x and 0 y are given values

• The initial condition ( )0 0 y x y= is used to determine the value of A and 4 in the

general solution• 3alues of a function and its derivative at the same point• 2umber of initial conditions for a given differential equation depends upon the order

of the differential equation.

EXA!LE :

The following e"ample are the Initial 3alue *roblem (I3*

(i , (0) 1 y xy y′ = − =

(ii 2 1 34 12 3 0 , (4) , (4)

8 64 x y xy y y y′′ ′ ′+ + = = = −

1.,.1., ;oundar+ Condition

)olution to the differential equations on an interval, I that specified at two distinct points j x

and k x where ( ) , ( ) j j k k y x y y x y= = .

igure 5

A differential equation together with a boundary condition is called an ;oundar+ 9alue

!role" );9!*. )uch that ( , ) , ( ) , ( ) j j k k y f x y y x y y x y′ = = = ;

• 3alues of a function and its derivative not at the same point

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• 2umber of boundary conditions for a given differential equations does not depends

upon the order of the differential equations

EXA!LE 1<

The following e"amples are the 4oundary 3alue *roblem (43*(i 2 2 0 , (0) 1, ( 2) 0 y y y y y π ′′ ′+ + = = =

(ii 23 8 3 0 , ( 3) 1, (3) 1 y y y y y e′′ ′− − = − = =

1.,., FIRST ORDER DIFFERENTIAL EQUATIONS

In this section, we will learn how to solve the first order differential equations. 4ut in order to

do that, we need to understand what type of equations we are dealing with.

T&e t+#es of "et&od that we are going to discuss are

1* Se#arale equation

,* Inte-ral Fa$tor

1.,.,.1 Se#arale Equation

A first order differential equations of the form, ( ) ( )dy

g x h ydx = is said to be seara!le or have

to searate t"e #aria!les.

EXA!LE 11

6ow to separate the variables of2 3 4 x ydy y xe

dx

+= ?

Solution:

irstly, we #now that ( , )dy f x ydx = .

Then, factori0e2 3 4 x ydy y xe

dx

+= to be 3 2 4( , ) ( )( ) x y f x y xe y e=

8 | P a g e

( )h y( ) g x

6ow to solve the differential equations by using

se#arale equation "et&od? '"plain generally.


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7iven that ( ) ( )dy

g x h ydx

= can be rewritten to isolate the variables x and y on the opposite

side of the equation. It is such as,

1( )

( )dy g x dx

h y=

Then we integrate both sides from the above equation+

1( )

( )dy g x dx

h y=∫ ∫

And we obtain+

1 2( ) ( )

( ) ( )

H y C G x C

H y G x C

+ = +

= + Where 1 2 C C C = + .

Therefore, ( ) ( ) H y G x C = + is the -eneral solution for ( ) ( )dy

g x h ydx

= .

EXA!LE 1,

)olve the nonlinear equation 25dy x

dx y

−= .

Solution:• Step 1 )eparate the variables of equation

2 ( 5) y dy x dx= −

• Step 2 Integrate both sides of the equation2 ( 5) y dy x dx= −∫ ∫

• Step 3 The general solution is3 2

53 2

y x x c= − + .

EXA!LE 1/

)olve the equation (1 ) 0 x dy ydx+ − = .

Solution:• Step 1 )eparate the variables of equation

9 | P a g e


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(1 )

1 1

1

x dy ydx

dy dx y x

+ =

=+

• Step 2 Integrate both sides of the equation

1 1

1

ln ln(1 )

dy dx y x

y x c

=+

= + +

∫ ∫

• Step 3 The general solution is

1

(1 )

c

c

ye

x

y x e

=+= +

1.,.,., Inte-ral fa$tor

The linear first order differential equations can be e"pressed in the form of

1 0( ) ( ) ( )dy

a x a x y b xdx

+ = , Where ( )1a x , ( )0a x and ( )b x depends only on the independent

variable x .

or e"ample, the equation2(sin ) (cos ) sin

dy x x y x x

dx+ =

irst, we have to divide all terms in equation 1 0( ) ( ) ( )dy

a x a x y b xdx

+ = with ( )1a x ;

01

1 1 1

( )( ) ( )( ) ( ) ( )

a xa x dy b x ya x dx a x a x

+ =

)econd, from the above, it can be rewritten as+

( ) ( )dy

P x y Q xdx

+ =

Where0

1

( )( )

( )

a x P x

a x= and

1

( )( )

( )

b xQ x

a x= .

Third, we need to determine the integrating factor, ( )v x ; ( )( ) P x dx

v x e∫ = .

10 | P a g e

( )1a x

6ow to solve the differential equations by using inte-ral

fa$tor "et&od? '"plain generally.

( )0a x ( )b x

Standard Form


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1 1

2 2( )dx x

v x e e∫ = =

• Step 3 $ultiply ( )1

2 x

v x e= to both sides of standard form equation+

1 1 1

2 2 21 32 2

x x x

dye e y edx + =

1 1

2 23

( )2

x xd e y e

dx=

• Step 4 Integrate both sides from equation in Step 31 1

2 2

1 1

2 2

3( )

2

3

x x

x x

d e y dx e dx

dx

e y e C

=

= +

∫ ∫

• Step 5 The general solution1

2

3 x

C y

e

= +.

EXA!LE 12

)olve the linear first %&'3 xdy y e

dx+ =

)olution• Step 1 )ince the above equation is in the standard form then, you don=t &a%e to

$&an-e an+t&in-.

• Step 2 To find the integral factor let ( ) 1 P x = into ( )( ) P x dx

v x e∫ = . Then,1

( ) dx xv x e e∫ = =

• Step 3 $ultiply ( ) xv x e= to both sides of standard form equation+

3. x x x xdy

e e y e edx

+ =

4( ) x xd

e y edx

=

• Step 4 Integrate both sides of the equation in Step 3

12 | P a g e


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4

4

( )

1

4

x x

x x

d e y dx e dx

dx

e y e C

=

= +

∫ ∫

Step 5 The general solution is31

4

x

x

C y ee

= + .

1.,.,./ Con$lusion $athematical modeling is the technique of representing real world problem which is

comple", involving multiple variables and some interrelated processes. This method can be

used in the study of growth population, radioactive decay, economics problems, changes in

temperature, mi"tures, chemical reactions, biological reactions, mechanics, velocity of a

falling ob!ect, electric circuits etc.

Referen$es• 8ill, &.7. (9::5. A Firt !oure in "i##erential $%uation &ith 'odelin Appliation, ;th

ed. 4roo#s. "i##erential $%uation &ith *oundary+alue -roble,

4roo#s


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a 0sin2

2

=+ ydx

yd

Solution :

• 2onlinear • )econd order

b xe y y y =+− 2')1(

Solution :

c 0'2'' =+− y y y

Solution :

d 04)( =+− xdydx x y

Solution :

e y xdx

yd 4)12(

2

2

=+

Solution :

f 22t x

dt

dx=+

Solution :

g 2

2

2

xudx

ud =+

Solution :

h 32

2

2

2 x ydx

yd =+

Solution :

. ind the solution to given initial>%alue #role" by using separation of variablesmethod.

a 24( 1), ( ) 1

4dx x xdt

π = + = b2

21 ; (1) 21

dy x ydx y

−= =−

15 | P a g e


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Solution :

Ste# 1 dt dx x

41

12

=+

Ste# ,

C t x

dt dx x

+=

=+

− ∫ ∫ 4)(tan41

1

1

2

To find t&e %alue of C?

π

π π

π

4

3

4

)4

(4)1(tan

4)(tan

1

1

−=

+=

+=

+=

−

−

C

C

C

C t x

Ste# / )

4

34tan(

)4tan(

π −=

+=

t x

C t x

Solution :

c ; (3) 4dy x

ydx y

= − =

Solution :

d 1)1(;2 −=−−= y xy ydx

dy x

Solution :

e5

2 1; (0)2

dy y y

dt + = =

Solution :

f ; (0) 2dy

xy ydx

= = −

Solution :

16 | P a g e


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B. ind the general solution for the following %&' by searation of #aria!les.

a 2)1( += x

dx

dy

Solution :

Step 1 dx x xdy )12(1 2 ++=

Step 2 ∫ ∫ ++= dx x xdy )12(1 2

Step 3 C x x x y +++= 23

3

b 0)1( 2 =−− dx ydy

Solution :

c 02 =+ xydx

dy

Solution :

d

++

=54

32

x

y

dx

dy

Solution :

17 | P a g e


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e y xe

dx

dy 23 +=

Solution :

f ydx

dy x 4=

Solution :

g 03cos23sin =+ xdy y xdx

Solution :

h )70( −= Qk dt

dQ

Solution :

>. ind the general solution for the following %&' by using the inte-ral fa$tor "et&od.

a

x

e ydx

dy 3

=+

Solution :

Ste# 1( )ince the equation is already inthe standard form then, no need to doanything.

Ste# , ( xdx ee x

x P

==

=

∫ 1)(

1)(

µ

Ste# / (

[ ] x x

x x x x

e yedxd

ee yedx

dye

4

3.

=

=+

b

xe ydx

dy363

=+

Solution :

18 | P a g e


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Ste# 0 (

[ ]

C e ye

dxedx yedx

d

x x

x x

+=

= ∫ ∫

4

4

4

1

Ste# 2 ( x x

eC e y += 3

41

c xe y y =+'

Solution :

d 0cos2 3 =−− x x ydxdy

x

Solution :

e 32 =+ ydx

dy x

Solution :

f xy ydx

dy x 485)2( 2 −−=+

Solution :

19 | P a g e


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g 4123 =+ ydxdy

Solution :

h x xdx

dy

412 =+

C. &etermine the general solution of this non linear first %&' using inte(ral fa$tor)

xe ydx

dy5

3

13 =+

!"en0

= x

an#1−= y

20 | P a g e


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Solution :

;. In a chemical reaction, hydrogen pero"ide is continuously converted into water and

o"ygen. At time t minutes after the reaction started, the quantity of hydrogen pero"ide

that has not been converted is x cm and the rate at which 9 cm is decreasing

proportional to x .

The problem above can be translated by mathematical word, 2dx

xdt

= − .

(i )olve the above equation by using separation of variables. 7iven that initial

amount of x is 5 where ( )1 0t = . (ii &etermine the solution when it too# minutes for the hydrogen pero"ide to be

reduced to half the original amount where1

32

t = ÷

.

Solution :

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)@ul+ ,


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)@ul+ ,


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)Se#te"er ,


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)@anuar+ ,


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)Se#te"er ,

chapter 1 - differential equations

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