Major topic-wise PPT · Major topic-wise PPT Subject Name : Mathematics-II Subject Code : KAS-203 Course Coordinator Dr. Dushyant Kumar (Associate Professor, Mathematics) Faculty

GALGOTIAS COLLEGE OF ENGINEERING & TECHNOLOGY

1, Knowledge Park-II, Greater Noida

Department of Applied Science (Mathematics)

Major topic-wise PPT

Subject Name : Mathematics-II Subject Code : KAS-203

Course Coordinator Dr. Dushyant Kumar

(Associate Professor, Mathematics)

Faculty Members

Dr. Padam Singh Tomar (Asst. Professor, Mathematics)

Dr. Vinay Gautam (Asst. Professor, Mathematics)

Galgotias College of Engineering and Technology1


Ordinary Linear Differential Equations of Second Order

• Linear differential equation of Second order with constant Coefficient

• Linear differential equation of Second orderwith variable coefficient

INTRODUCTION


General Form of Linear Differential Equation

Where P and Q both are constants

R is the function of (independent variable)or zero or constant

2

2

d y d yP Q y Rd x d x

x

• Solution of linear Differential Equation of SecondOrder with Constant Coefficient


2

2

d y dyP Qy Rdx dx

Now we can write as above D.E2D y PDy Qy R

2D PD Q y R


Complete (General) Solution of Differential Equation

General Sol.= Complementary Function + Particular Integral

Y = C.F. + P.I.

2

2

d y dyP Qy Rdx dx

Method of Complementary Function (C.F.)

Galgotias College of Engineering and Technology

6

2 .........(1)D PD Q y R

Replace D by m in Equation (1)

Auxiliary Equation (A.E.)2 0m Pm Q

Case-I when the roots of A.E. are real and distinct

Let

Let

1 2,m m m


ThenWhere C1 and C2 are arbitrary Constants.

Example: Solve the given differential equation

Solution: The given equation is

Auxiliary Equation

Where C1 and C2 are arbitrary Constants.


Here Particular Integral = 0 (because R=0)

Now General solution Y = C.F. +P.I.



Example: solve the given differential equation2

2 3 5 4 0d y d y yd x d x


A.E.Or

Now General Solution Y = C.F. + P.I.



Case-II When the roots of A.E. are real and equal

Let


Then

Example: Solve the given differential equation



Here Auxiliary Equation is

Or

General Solution Y = C.F. + P.I

Where C1 and C2 are arbitrary Constants


Case-III when the roots of A.E. are Imaginary (Complex)

Where C1 and C2 are arbitrary Constants

Let

Then

If

Then


Example: solve the given differential equation


Auxiliary Equation is

Now General Solution Y = C.F. + P.I.



Example-: Solve the given differential equation

Solution: The given equation isHere A.E. is

Now General Solution Y= C.F. + P.I.

Where C1 and C2 are arbitrary constants.

INTRODUCTIONThe series solution method can be defined in two ways:

• Power Series method

• General Series Solution method (Frobenius method)

General form of linear differential equations of Secondorder with variable coefficients:

We can reduce the equation (1) in the given form

Where P and Q both are variable

Ordinary and Singular Points

Let the differential equation

Compare and we get

Put

IfThen is an ordinary point of the differential equation (2)

Thus we can say both of the functions are analytic.

Then is not an ordinary point of the D.E. (2) Therefore is a singular point of the D.E.(2)

Types of Singular Points

• Regular Singular Points

• Irregular Singular Points

Regular Singular Points

Therefore x=a is called regular singular points.

Otherwise x=a is called irregular singular points.

Example-1 Prove thatPoints of Following Differential Equation.

Singular

Also show that is a regular singular point of D.E.

Solution: The given differential equation

is a regular singular point of the above D.E.

Example-2 Prove that is an ordinary point of

following differential equation.


is an ordinary point of above D.E.

Series solution of differential equation whenis an ordinary point of the given differential equation

Problem-1 Solve in Series the differential equation about


Hence x=0 is an ordinary point of above differential equation

Let the solution of the differential equation in the form

Put the value in D.E., now we get

Now equating the coefficient of

summation and


Put Summation &


Now we get recurrence relation

Substitute these values in equation (1) now get the solution

Where is called arbitrary constants

Periodic Functions and Fourier Series

Periodic Functions

A function f is periodic

if it is defined for all real and if there is some positive number,

T such that fTf .

Fourier Series

f be a periodic function with period 2The function can be represented by a trigonometric series as:

11

0 sincosn

nn

n nbnaaf

11

0 sincosn

nn

n nbnaaf

What kind of trigonometric (series) functions are we talking about?

32and32

sin,sin,sincos,cos,cos

We want to determine the coefficients,

na and nb .

Let us first remember some useful integrations.

dmndmn

dmn

cos21cos

21

coscos

0

dmn coscos mn

dmn coscos mn

dmndmn

dmn

sin21sin

21

cossin

0

dmn cossin

for all values of m.

dmndmn

dmn

cos21cos

21

sinsin

0

dmn sinsin mn

dmn sinsin mn

Determine 0aIntegrate both sides of (1) from

to

dnbnaa

df

nn

nn

110 sincos

dnb

dnada

df

nn

nn

1

10

sin

cos

000

dadf

002 0

adf

dfa

21

0

0a is the average (dc) value of the

function, f .

dnbnaa

df

nn

nn

2

011

0

2

0

sincos

It is alright as long as the integration is performed over one period.

You may integrate both sides of (1) from

0 to 2 instead.

dnb

dnada

df

nn

nn

2

01

2

01

2

0 0

2

0

sin

cos

002

0 0

2

0

dadf

002 0

2

0 adf

dfa

2

00 21

Determine naMultiply (1) by mcosand then Integrate both sides from

to

dmnbnaa

dmf

nn

nn

cossincos

cos

110

Let us do the integration on the right-hand-side one term at a time.

First term,

00 dma cos

Second term,

dmnan

n coscos1

m

nn admna

coscos1

Second term,

Third term,

01

dmcosnsinb

nn

Therefore,

madmf cos

,2,1cos1 mdmfam

Determine nbMultiply (1) by msinand then Integrate both sides from

to

dmsinnsinbncosaa

dmsinf

nn

nn

110

Let us do the integration on the right-hand-side one term at a time.

First term,

00 dmsina

Second term,

dmsinncosa

nn

1

01

dmsinncosa

nn

Second term,

Third term,

m

nn bdmnb

sinsin1

Therefore,

mbdmsinf

,,mdmsinfbm 211

dfa

21

0

The coefficients are:

,2,1cos1 mdmfam

,2,1sin1 mdmfbm

dfa

21

0

We can write n in place of m:

,,ndncosfan 211

,,ndnsinfbn 211

The integrations can be performed from

0 to 2 instead.

dfa

2

00 21

,,ndncosfan 211 2

0

,,ndnsinfbn 211 2

0

Example 1. Find the Fourier series of the following periodic function.

2

0whenA

whenAf

ff 2

021

2121

2

0

2

0

2

00

dAdA

dfdf

dfa

011

1

1

2

0

2

0

2

0

nnsinA

nnsinA

dncosAdncosA

dncosfan

ncosncoscosncosnA

nncosA

nncosA

dnsinAdnsinA

dnsinfbn

20

11

1

1

2

0

2

0

2

0

oddisnwhen4

1111

20

nA

nA

ncosncoscosncosnAbn

evenisnwhen0

1111

20

nA

ncosncoscosncosnAbn

Therefore, the corresponding Fourier series is

7sin

715sin

513sin

31sin4A

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Major topic-wise PPT · Major topic-wise PPT Subject Name : Mathematics-II Subject Code : KAS-203 Course Coordinator Dr. Dushyant Kumar (Associate Professor, Mathematics) Faculty