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Nod) OF APPLIED
MIT EMATICS-AII
PROBLEM SOLVING APPROACH

Eo Engincering, Science and Technology Students

Method of Variation of Parameters:
tation of Parameters:

1) j f(x)
= oo dx
LL une: "I Woy)

|

Author /Begashaw/Moetot;
¡Gual ifications/MEDEMSC

A HAND BOOK OF
APPLIED
MATHEMATICS-IH

Revised Edition
By Begashaw Moltot Zemedhun |
2009 E.C.

©By Begashaw Moltot

All Rights Reserved.

Contents

CHAPTER-1
ORDINARY DIFFERENTIAL EQUATIONS

1.1 Definition and Classifications of Differential Equations
1.2 Linear and Non-linear Differential Equations.
1.3 Solutions of Differential Equations.
1.4 Initial Value Problems (1VP) .
1.5 Solving First Order Differential Equations.
1.5.1 Separable Differential Equations: Method of Separation...
1.5.2 Homogeneous Differential Equatio
1.5.3 Exact Differential Equations: Method of Exactness
1.6 Non-Exact ODEs: Method of Integrating Factors
1.7 Bemoulli’s Differential Equations:
1.8 Second Order Linear Differential Equations) with Constant Coefficients. 36
1.8.1 Solutions and their Properties.
1.8.2 Second Order Homogeneous Linear DE with constant coefficients
1.8.3 Initial and Boundary Value Problems (IVP and BVP).
1.8.4 Solving Non-homogeneous Linear DE with Constant Coefficient
1.8.4.1 Method of Undetermined Coefficients (MUCs)..
1.8.4.2 Method of Variation of Parameters (VPs).
1.9 System of First Order Linear Differential Equation
1.9.1 Homogeneous Systems with Constant Coefficients.
1.9.2 Non-homogeneous Systems with constant coefficients.
Review Problems on Chapter-1

CHAPTER-2

Laplace Transform and its Applications

2.1 Definition and Examples of Laplace Transform 97
Properties of Laplace Transform. ica er renee 9B
Unit-Step (Heaviside) Function and its Laplace Transform... 102

2.4 Inverse Laplace Transform and Their Pr a 107

2.5 Convolution Product and Laplace Transforms .. „11

2.6 Laplace Transform of Derivatives and Integra „112

2.7 Applications of Laplace Transform „115

CHAPTER-3: CALCULUS OF VECTOR VALUED FUNCTIONS

3.1 Definition and Examples of VVF … 127

32 Limit and Contnuty of Vector Valued Functions 128

3.3 Derivatives and Integrals of Vector Valued Functions 131

. 133

3.4 Space Curves and Length of Space Curves...
3.5 Tangent and Normal Vectors and Curvatures.
3.6 The concepts of Scalar and Vector Fields
3.7 Line Integrals of Scalar and Vector Field
3.7.1 Line Integrals of Scalar Fields.
2 Line Integrals of Vector Fiel
3,73 Line Integral in Differential Form...
3.7.4 Fundamental Theorem of Line Integral.
3.7.4 Green's Theorem for Line Integral
3.8 Surface Integrals.
3.9 Surface Integrals over Oriented Surfaces.
3.10 Stokes's and Gauss's Divergence Theorems
3.10.1 Stokes's Theorem (Relates Line and Surface integrals
3.10.2 Gauss’s Divergence Theorem..

$ nd Bok of ppd ae sn. Fu our commen an Sizes e ML

CHAPTER-4
FOURIER-ANALYS

4.1 Fourier-Series
4.1.1 Periodic Functions... pa =
4.1.2 Fourier Series and Euler’s Formule." | sas 20
4.1.3 Fourier Series of Discontinuous Functions u
4.1.4 Fourier Series of Functions with Arbitrary Periods
4.1.5 Fourier Series of Even and Odd Functions...
4.1.6 Half-Range Expansions
4.1.7 Parseval's Identity (PI)

4.2 Fourier Integrals and Transforms.
4.2.1 Fourier Integrals un
42.2 Sine and cosine Fourier Integrals
4.2.3 Fourier Transforms un
4.2.4 Fourier Sine and Cosine Transforms.

IS

CHAPTER-5

CALCULUS OF COMPLEX FUNCTIONS

5.1 Functions of Complex Variables.
5.2 Limits and Continuity of Complex Functions.
5.3 Differentiability and Cauchy-Riemann Equations.
5.5 Analytic and Harmonic Functions .
5.6 Complex Integrations
5.7 Power Series Representations of Complex Functions
5.8 Calculation of Residues and Residue Theorem
4.9 Applications of Residue Theorem.

.. 246

251

.. 264
268

¡land Book of Applied Mathematics Il by Fra"

CHAPTER-1

Ordinary Differential Equations (ODE)

1.1 Definition and Classifications of Differential Equations

An equation containing (involving) derivative of the dependent variable with
respect to independent variables is known as differential equation (DE),

‘There are two basic types of differential equations.

2) Ordinary Differential Equation (ODE): A differential equation involving
derivatives of the dependent variable with respect to one independent variable.
b) Partial Differential Equation (PD! differential equation containing the
derivatives of the dependent variable with respect to more than one

independent variables.

sone: 2 0 à
Notations: © =’, =)! in ODE.
stations: À = y" yin

Examples:
2
a) Examples of ODE: i) 2. 2x iDy'dy=xe" ii) E 7

oz
a
Order and Degree of Differential Equations:
3) Order of a Differential Equation: It refers the order of the term with the
highest derivative in the Differential Equation.
In a differential equation, y' represents first order, y" represents second order,

b) Examples of PDE: i) £a ii)

y" represents third order and in general, y” represents n" order derivative.
So, if we get derivatives of different orders in a single differential equation, We
take the highest order as the order of that differential equation.

b) Degree of a Differential Equation: It refers the highest power or exponent
(positive integer only) of the term with the highest order derivative.

1) The degree of a DE is identificd.afer the exponents of all the terms with any
order derivatives are made to havgpositive idlgger exponents. This may need
squaring or cubing both sides of thelequation. |.”
Besides, if the term with the highest order is part of an operation like
multiplication, you have to simplify the: before reading the degree.
2) Degree of a differential equation is defined gply if the DE is a polynomial in
all the terms with the derivatives y, y", "sage y". In general, the degree of
a differential equation that contains any of the following exceptions need not be
defined. Because of this, every DE bas order but may not have degree.
Exceptions: ¢# or e” in or in yysiny/) or sigh ).c08(y) sin" 7,008
Examples: Identify the order amos of following DEs.
a) y""+4y" 2)! = 2°. For this WB, order m and degree d=5.
Why the dogs? Base htm wit te ide order drive y"
and its exponent is 5.

Na LY = -
»(& ) a ee E 0. Hor, onder wand ge d = 1.
hg ear 1 is because the term with the highest order derivative

inthe given DEis E 2 and its power is 1. $

e) y“-yl=0. Here, order = Zend deg d=8. How? In this DE, we
cannot read the degree directly because the exponent of y' is not integer.

So, frst square both sides to make integer. +

Thatisy"-yta0= teint} GA Sym des.

d) y” +sin4y=x. Here, order m2 but is gre is not defined because it is
not a polynomial with respect 103 MThis is cof the termsin 4y ).

€) y”+sin4y=x. Here, order = 2 and ite) is also defined which is
d =3 even though it is not polynomial in y.

Py" emails E (How?)

A fod on Apr te 0 HAGEN
1.2 Linear and Non-linear Differential Equations

Definition: A differential equation is said to be linear if and only if the

foliow two conditions are satisfied:

) The dependent variable and its derivatives in all terms have first degree.

ii) There is no term involving product of the dependent variable and any order

of its derivatives

A differential equation which violates either of these two conditions is said to

be non-linear differential equation. In general, any differential equation that

contains atleast one ofthe following terms is automatically non-linear.
sin(y"),cos(y"),see()"),In »’,e” sin y!,cos”! y.

Examples:

dy

Ko has order n
& ds

b) The DE y"*+3y'-y =Ohas order = 2and degree m = 4. It violates the first
condition of the definition. Thus, itis non-linear,

©) The DE y"'+yy'=1has order =2and degree n=1 but it contains the term
99° which is the product of the dependent variable and its derivative. It violates

the second condition of the definition and thus it is non-linear.

a) The DE

+ Thus, it is linear.

and degree n=

d) The DE y"=V1+x" has order m= 2 and degree n=1. ‘Thus, it is linear:
e) The DE "= i+)" has order n=2and degree = 2. It violates the first
condition and thus itis non-linear.

N) Consider the DEs: "ex = y and Jy"+y =x. Which equation is linear and
which equation is non-linear. Why?

nd Bk Api thematic y ee Y or pocos a eens e 24262
Solutions of Differential Equations

Defin

on: Any function (involving the independent and dependent variables)
h satisfies the given DE whenever substituted is called solution of the DE.
Types of solutions: There are two forms of solutions for a given DE (if it has).
2) General solution: The solution of a DE which contains arbitrary constants in
its expression is called general solution (primitive).

b) Particular solution: The solution of a DE free from arbitrary constants (that
does not contain arbitrary constants) is called a particular solution. Usually,
Particular solutions are solutions obtained from the general solution by
assigning particular values to the arbitrary constants.

Examples:

4) =12e** -12e* +8=8

Fo Ie 260”
This means y=3e™—4x satisfies the given DE when substituted.

b) Consider the DE: y""=8y. Can both y=3e™ and y =ce”*be solutions?
Here, y =. Ge, y"=12e™, y"=24e™ . So, y""=24e*" =8(3e*)=8y.
Again, y = ce", y'=2ce™, y"=4ce*", y""=80e*. So, y"=8ce" =8(ce™*) = By.
‘Therefore, both y =3e* and y=ce™are solutions.

Besides, y =ce*" is general solution because it contains arbitrary constant ¢ in

3e** is particular solution obtained by assigning h
+c,e™ +.’ is the solution of the DE

its expression. But
©) For arbitrary constants cjand €, y =
)x*. Therefore, it is a general solution. But if we give c, = Qand

yy -6x
2, we obtain y=2e"* +x which is a particular solution.
> isa solution of y""-29+6y = 6but y=x* is not.

ELT ANIME AE HAN TUTO ARMES ¢

9 Ity=

is the solution of the DE Baro , then find k.

Solution: Since is y "is given to be a solution, it must saisy the DE.
2 5
Thatis ye => ya 2e ya 40 > Ty-h=0
=> 467 +66" ké" =0>k-10=0>k=10
©) For what value of adoes y =" isthe solution ofthe differential equation
Ñ a

given by A ray 480?
Solution: Since y = x7is a solution, it must satisfy he given DE.
Thatis Dar 2 pere 2e ayy = 20 +8

tra? 6 = 27 Ho la-ó=230=4

14 Initial Value Problems (IVP)

“The problem of finding a function solution satisfying a differential equation and

an initial condition is called am initial value problem (IVP).

Consider a differential equation = $6). ‘Then the initial value problem
Df)

for such first order DES is of the form de «Then, the problem of
Lx) = Yo

finding a function which satisfies this differential equation with the given

additional condition is sid to be Initial Value Problem.

The condition (x)= yp is called initial condition.

Examples: 0) x/42y=4x", HD=3 b) y=2y+6, HO=2

Note: To solve IVPs, frst find the general solution of the given DE using any

method and finally find the constants using the given initial conditions.

ECS

1.5 Solving First Order Differential Equations

Solving a differential equation means finding the unknown function which

satisfies the given differential equation,

Forms of First Order Differential Equation:

Any DE of the form M(x,y)de+N(x,y)dy=0 or 2

first order differential ‘equation. Here, under we will discuss how to find the

general solution for such form of DE.

Since there is no general method to solve all forms of DEs, we will see

different methods for different forms of DES.

Methods of Solutions for First Order Differential Equations:
1) Separable Differential Equations: Method of Separation
2) Homogeneous
3) Exact Different
4) Non-Exact Differential Equations: Integrating Factor Method
5) Bemoulli's Differential Equations: Transformation (Reduction)

(x,y) is said to be

1.5.1 Separable Differential Equations: Method of ‘Separation

Separable Differential Equation:

Any first order differential equation MQs yJde+N(x,y)dy =0 which can be
expressible in the form ¿0d = f(a)deis said to be separable differential
equation. Here, the form g(y)4) = (HE is sid tobe the separable form.
Method of solution: The general solution of such separable DE is obtained
by integrating the separable form g(y)dy = FCO both sides.

Thatisf gra = [ Sr.

EAS +

| nd ck of pl Mahal Bes Fr your comments a seston we
Example:

the following DES

eparable and solve them,

und
(yy (y 5) = 0 y

©) dy (y de yay -dy)de=0 dede,

€) tanxsin* yx + cos’ xeot ydy = 0 fre’ sinxas—

8) (e" +3)c05.xdx—e” sin xdy =0

DOF +97 dx (6 yy =0 30 )dy = 20 Dir
Solution:

E ot : 7
A 220 205 di

Hence, the DE is separable.

Thus,2ydy = [290 = [aes y = Lin +I +e

Yat

vide ~ (2 3) = 09 “(DJs 7 ab =

DADES AP y +? -Ay)dr=0= 24d) =P ~4y) + 3° ay Jer
> Udy = (9? ya +de dy =(x? +1)de

‘Hence, the DE is separable and by PFD,
y en Fe

d) ca Ga arder fe =[er>-e
24
sof oe (E

y ==
ESE

naa

2° ya
tanx

= de cos ysin” y ==sin xcos” x. Hence it is separable

So, [eos ysin” yay =—[sin xeos” xde=> ———te

{ok es Mie Tk cl TE 6020
y. Sinx sin

y= Sard fer = IE

Seidel +23)de

Se t2ndeoe ze +2 40

(i-y)dy=0

- y. de —— ya
O

2. Solve the following IVPs using separable method

a) 2e” der? De) =0, (0) =2 om” 0)=0

0) +3 co 2x =0,y0)=

4) xD scary=0, 1D

Solution: Very Important: To solve IVPS with »(x%) = Yo,
First: Find the general solution using appropriate method
Second: Find the constant by putting x = x, and y = y, in the general solution.

a) 2xe” dx + (= dy =0> Py = 2x8" de
= Joo? -Day=f-2xe" de>

Now, find the arbitrary constant by using x =

y=2:

rs yet Dies ense
a

Le A EOL
Equations Reducible to Separable: Any DE of the forn 2

Slac+by+e)

de
can be reduced to separable form sing the substitutions = ax + by +e

at
Then, —=a+bf (1)

a Ud ari fé ET
Examples: By reducing into separable a solve the following DEs:

2 dy_2x+2y+3
Ze YA 2 rra
»_ dy _2xs3yt5 a. 5

Vay Va axeby-3 Dn

Solution: We use the above substitution rule

Dee 3 But from the given 2

de 2 de
‘by Lat yt 1+- ead ak 1
Ta Lar LENS
de eel
[Ed fan [io jan farsintre see
Pal

ryan (+ y) exo pa tan (+ y +) le

a dd
Ola r=y Ho El
ma. ev2y+3_2ety)+3_ de RUDA
ya yl de 1

1 1
Ga [orha=[aeoaeiy see
SI y eee y += 0402043) +In)e+ y +1] =c

A function /(x,y) is suid to be homogenous of degree 1 if for any parameter
150, (9) = f(xy). In general, Fis homogenous function of degree nif
and only if for any parameter #0, fa, =" f(x,y).

Homogenous Differential Equation:

A differential equation of the fm = /(&,y) is said to be homogenous if

4(+,y) is homogeneous function. Otherwise, itis sid to be non homogeneous.
Test of Homogeneity:

A differential equation 2. (x, )is homogenous of degree 1 if and only if

for any parameters 20, f(y) = f(xy).

Method of Solutions: By Reducing into Separable DEs

First: Use Test of Homogeneity to check whether it is homogeneous or no.
Second: Change the homogeneous DE into separable DE.

‘To change into separable DE, use the substitutions: vente,

This will change the given DE into separable DE in the variables >1nd y. Then,
integrating the separable form gives the general solution in terms of xandv.
Finally, by solving y =vxfor v, express the answer in terms of x and y.
Examples:
1. Check whether the differential equations are homogenous or not. For these
which are homogeneous, solve using homogeneous method.

a Bate rr

FE

NEL?
N -y'X= 298
Aw ylny-tnz)

tt

e)x'dy=(y}-x)&
DEA (+
DO? -2ay)dr=(x* -29)

i yea wm
Solution:
Firsts Use Test of Homogeneity: To use this KE for Di

given DE in the form a. Flog) Then, apply te test on /

2) Then the DE is the form = = f(x,y) where Sl

Here, forany 1#0, /(@50) ghee.
ES

‘Thus, fis homogeneous which implies the DE itself is homogenous.

lan Banner
de x = de de x
ofa tex y=xImj+er
dy #

yt Foy
à

B= 10,9) vit nein is FC »=

le in the form

ery? EAN

Here, forany 120, SD),
ey ar

“Thus, fis homogeneous which implies the DE itself s homogenous

1 3
Fl =l|==Inp|+e

2 -i[+inp|=e=2 Info? ~Ds|=e= 6" Deed

EEE om A EL
Y Y sese% Here, forany 1#0,

o Here, xdy = (y+ ess E
1 € re

»

fe. Dei sel sy) Zach Arc = S29)

Thus, f is homogeneous which implies the DE itself is homogenous

Second: Use the substitution y = vr=> = v+ x to change into separable
de ae

dla

rro seer
ra tar

> Jsinvay= [+ 40 cosy =Inf]+0= eos? = bte

Me y(t indide => &

an eft

Ez
0%
1 1 1
ia EE
de VRR o ht vie, = ils
DAA AA AA e Pal

an = eo =| As main?
> focver= [de [sen cot =p ro 100% -asint

AAA 12

(ERE pi aT TO een WS a wi alsa eae PD
ad yes u vel
xty ae ity Lev

But Teckel, u=v +2v-1= du = oie

v+l = 4 ni
(3 A de | hdi=ml+c>> 1 inla| = Inf +

> In? + 2v-I]+-2Inff = c= nf +2v-1):

(Er)
E

Bay}

Sy +2 sem xp =dyte

tr
Fa 2 2

aros
de

mor] =mpjre=e0-34)=0

ole -3y)=0

DILO y e VEDA
d Er”

art

E api
> Emp-1J- Ippo c= tfp—Ii+ n= 3h +e => (9-2) =

Bye yiny-inx)=> dein) Y 2 (in?
Ax/=yiny-nz) > x7 n= Ai

1 1
Sg id lay =n

>hv-I=&>h

dv

=vinv

loa Bok pl Mahe ge or meal el eins we EGO

2. Solve the following homogenous DES
at) bp = f+ y

Wy=(vtysin Dd da = +

ay
de

e) xsin?,

Wirt N Ayer =0

gy 2-2 + tant) ayy ®
Da PRE

D det (yee =0 Dx + y de =0
Solution:

Now using tig subio, = m0>d=
SEO. 49 fsec@10 =tnscco stand =IoVi+ e +]

lee “Wino |
sj mabel ofabhiseif liz oe

at

tty" +
=

ann
Fi
«E

FON yal de vel x

bols y=xhfo
lo}

y y dy y
e) xsin2 dy = (x+ ysin2)de= sin? © 2 14 Fine),
x E x dx $ y
dv. dv
Ssinv(r+r E) =1+vsinv> vsinvsxsiny® 21 à sino
dx de

>si?
&

>sinvd=Ldr> [sinva»
x

= ~cosv=In)y|+e= cos? = Ins|+.c

4) xdy =(y+ y(in2)*)de D 4X my
rages

Sd nain Hasty!

‘Wand Boo of pied Mathematics y Begushue M. For yar comen and Ris we OSLER.

cosy
D [ds i
x

vet
wey
vel ¡de
Wav Wave Y al
[e ar

y deal

But by Partial Fraction decomposition, 222

=v-hb=hhites 2 =Inpl+e

la fp iki by Baha M Vo oi comics sad sas td OEA

3. Solve the following problems using the Methods of homogenous DES
a) (sec + y)de—xdy=0, (I) =0 6) xdyy~(2xe* + y)de=0,3(1) =0
y
a) xdy = 0e stn}

dy = (x? + y? )dx f) xdy = (yt xeos hdr

Solution: First, find the general solution and then use the initial conditions.

secy=> cos val

+scev= x

&

= [lacs sinv= bte sin 2 = Inde
z x

E
Besides,y()=0>e*=Inl+c>c=1>e* =Inx "+1
1

Parr der

, 2
> cre dv Lar fredy Ha 26° =Inkl+c»2e”

Br Evans tony
Da ar art tan

= codos de foutue [Lars [pe
sv

Line or sin

0-2)"
x
paz san

her
Dea er EY 17

> nkinfeinhf+c> hing

Jav=- fd de 2 +t
FH

dr = Infeov-+tanvf= nf+e = Il

id ik i
1.5.3 Exact Differential Equations: Metho

id of Exactness

Definition: A differential equation of the form M(x, par + N(x, y)dy = Ois
said to be exact differential equation if there exists a function u(x, y) such that
u,(%, 9) = M(x, y) & u,(x, y) =M(x,y)-

The function u(x, y) with such properties is called potential function.

But determining exactness by finding the potential fugctionu(x, y) is a difficult
task. So, we have the following test for exactness.

Test for exactness: Any DE of the form M(x,y)dx+ N(x, y)dy = 0 is exact if

and only if M,

Solving exact DE: Method of Exactness

Suppose M(x, yee + M(x, y)dy = 0 is exact. If u(x, y)is its potential function,
then, the general solution is u(x,y) = where cis arbitrary constant.

This means that once the DE is exact its general solution is obtained from the
knowledge of the potential functionu(x, y) that satisfies the definition.

So, the basic task is how to determine the potential function.

Procedures to find the general solution of exact DEs:

Step-1: Integrate 1,(x,y)= M(x») watt x to get u(x,y) = | M(x, pds + 80)
Here, assume y as constant and g(y) as constant of integration.

Step-2: To find g(y), equate u, (x, y) = N(x, y) from step-1.

That is u,(x,y) > Jue, pet Fg N=NG, »

Step-3: Integrate the result in step-2 with respect to y to obtain g(y).
Therefore, the general solution is u(x,y) = | Mix, y)dr+g(y) =e where g())
is to be substituted by the result in step-3.

Short-Cut Formula:

By collecting and rearranging the results from step-1 to step-3 in the above
procedures, we get the following short-cut formula for the potential function.
Potential function: u(x,y) = [Max + [IN-[M,dDd.

ie wf ped Nai 9 Bg ory ens Sug EBD
1. Verify thatthe following DEs are exact and solve them.
a) (Pet +200 x)dr + Bye" -Ay)dr =0
e) (Sr+-dy)de+(4r-8y")dy =0
e) -2aysin(x" jé + cos dy =0 D) ye™de + (2y +30” My
hi) 2xsin pde +x? cos yeh

b) 31 yde +(6y + x )dy =0
d) (2y?x-3)de+ (2)? +4)dy =0

g) le’ siny+3x*)dr +e" cos pdy = 0
Solution: For your understanding, first le' follow the procedures and then use
the short-cut formula, Please compare the results in each problem.
ye ay > M, =3y"e" Ny x

ye

a)M = ye" +2c0sx,N
Since M,
To clarify the above steps, let's solve step by step only this part.
Step-l: u,(x,y)= pe" +2cosx => u(x,y) = ye" +2sinx +80)
Step-2: Equate u,(x,y)=N(x,y)-
From step-1, u(x,y)=y'e" +2sinx+g() >, (x,y) =3y"e" +8 0).
Then, 1,(%,9) = (x,y) = 3y7e" +2") =3y%e" -4y>8'0)=-4y
Step-3: Integrate the result in step-2 with respect toy.
That is g'(y) =—4y= [e'Q)4y= [-4ycdy=> g(y) =-2y".
So, using g(y)=~2y’, the potential function is u(x,y)= ye +2sinx—2y*.
Using the short-cut formula: Here, M=e*y? +2cosx, N=3y"e"—4y.
uy) = [Me + [IN - [Malay

= fc» +200s 2 + [[By"e"—4y— [33404

=ey +2sinx+[0ye =4y- 3 e y

=e y +2sinx+ [—4ydy =e'y +2sinx—2y?
‘Therefore, the general solution is ey? +2sinx—2y? =e,
DHere, M(x,y)=3r'y, N=6y+0° SM =337 an,
Hence, the equation is exact.

DENE

pe", the differential equation is exact.

kk Le No ek AA

Using the short-cut formula: Here, M=3x"y,N = 6y+x.Then
13,3) = [Mar + [[N- [M dé
* yee + Îl6v+» fard = ry +3y"
Therefore, the general solution is u(x,y) = x y+3y° =c.
€) M=5x+4y,M, =4, N=4x-8y, N,=45 M,=N,.
Hence, the equation is exact.
Using the short-cut formula: Here, M=5x+4y,N=4x-8)” .Then
(x,y) = [Má + [IN [M ae

= [6x + apres fl4x-8) - [Acc = 0 y +3y*

=f sty [aay Hype

‘Therefore, the general solution is E say —2y*

d) Here, M=2y*x-3,N=2yx?+4= M, =43
Hence, the equation is exact.
Using the short-cut formula: Here, M=2y*x-3,N=2yx* +4 Then
u(x,y) = [Max + JIN— [Marley

= J@y*x-3)de+ [lan +4- anule) =x"y* 31 +4y
Hence, (x, y)= y’x’-3x+4y=cis the general solution.
e) M=-2xysin(x*),M, =-2xsin(x?), N=cos(#*), N, =—2xsin(x?)
Using the short-cut formula: Here, M = -2xysin(x"), N = cos(x").Then
u(x,9)= [-29ysint + [eos = [-2xsing ya} ay = yo)
Hence, u(x,y)=yoos(x*) = is the general solution.
DM =” +e”, N, =e” +92” => M, =N,
Hence, the equation is exact.

A AS

wack pe tea. sh MF amen and grs ase EET?
Using the short-cut formula: Here, M = ye" ,N=2y+xe” ‚Then

u(x.y) = [Max + [[N-[M,deldy

= Dear + [Ry + xe” -fer +9e”)dr]dy
sxte™+[2ydy=xte™+y?

Hence, the general solution is_u(x,y)=x+)" +e" =0.
DM, cos y=> M, = N,

Hence, the equation is exact.

Using the short-cut formula:

u(x, y) = [Mar + [[N - J M,dr]dy

= (e sin y+3x*)de + fle" cos y Jet cos yarlay =e" sin y+ x’
‘Thus, the solution is e* sin y+ x =c.
h) M=2xsin y, N = x" cos y=>M, = 2xcos y, N,

2xcosy=> M, =N,.
Using the short-cut formula: Here, M =2xsin y,N =x* cos y.Then
u(x,y) = [Max + [IN - [M,dx]dy
= [2xsin par + fx? cos y - [2xcos yededy = x’ sin y

Therefore, the general solution is U(x,y)=c=>x'siny =e.
2. Check the exactness and solve the following DEs

a) (x? —4xy—2y? de + (y? — day -2x*)dy = 0, (0) =3

5) (e? — pe )dx + (xe” —e")dy = 0, (3) =0

©) Gx? + yoosx)de + (sinx—4y")dy =0,y(2)=0

d) (e" sin y—2ysin x)dx+(e” cos y+ 2ycosx)dy =0, y(0) = 4

e) 2xydx +(x? +008 y)dy =0

SP) (xy +327 de +x7dy = 0

Using the short-cut formula:

(8,2) = [Mar + [[N— JM dd

= (x? 4929 Jae 4p? — ay 20 - [Ar dy
>

ya
Ay

carry +

‘The general solution is À 2x'y-2"

But (=
b)M
The DE is exact. So, by themethod of exactness,

uy) = te’ ye jae + fe” et - [le eddy = x0” 3e"
The nel sohtion se” — ye" =o. Buy) 02e

©) M, =cosx, N,.=0osx=>M, =N,. Hence, the equati

Fy Oxy? + y
=M,=e -

ye.

‘Thus, the general solution is x? — y' + ysinx=0.
mens 0 —y +ysinx=8.

‘Thus, the general solution is u(x, y) =e" sin y+2ycosx = €
Besides, (0)=2 => ¢=2n=>e" siny+2ycosx=27.

D M, =2x, N, =2x=> M, = N,. Hence, the equation is exact.
Using the short-cut formula: Here, M=2xy,N =x? +cos y .Then
Zope + fx? +008 y- | 2xde]dy =x y +sin y

Hence, the general solution is x’y+siny=c.

8) M, =2x, N,=2x= M, = N,. Hence, the equation is exact.
=2y +31 Nox.
ye

Using the short-cut formula: Here,
Then u(x, y) = f 2x9 +31 )dr+ Ste = fard

Hence, the general solution is x°y+x =c.

Sea 22

A
A A a OEE

ar wf ped ata Far. or or came a sai eZ

1.6 Non-Exact ODEs: Method of Integrating Factors

So far, we discussed how to solve DES when they are separable, homogencous
or exact, However, there are many situations that do not fit to either of such
cases. So, our next discussion focuses on how to solve DEs that are neither of
the above forms. Consider the non-exact DE P(x,y)de+ OC, y)dy = 0. Now,
multiply this equation by a nonzero function, say 4 (it will be a function oF x,
y, or both) such that the resulting equation Pa + Od)
‘The function jz which is used to change the non-exact differential equation

P(x,y)de+ OCs, y)dy=0 into an equivalent exact DE of Pde + uOdy = Dis
known as Integrating Factor. For example, the equation 2ydx + xdy = 0 is not
exact but if we multiply it by (x) =x, it becomes - 29h —a "dy = 0 such
that M(x,y)=-2xy, Ma) ==> M, =-2x = N, which is exact and its
solution is obtained easily. But, here the main problem is how to choose or
select the function jz which is used as multiplier to change the non-exact DEs
into exact DE, Even though there is no hard and fast rule on how to find
integrating factor, any way let's see the general procedure to find such function.
From the condition of exactness, the DE Pde + uOdy = 0 will be exact if and

only if zum hon. Then, by product rule, 4, P+ uP, = 1,0+ 40, -

Now, solving this equation for y is too complicated. So, to simplify the
‘complication let's consider different cases for y.

Case=1: Suppose y is a function of x only. Then, using the relation between
partial and ordinary derivatives, we have

UP + BP, = 10+ 12, = = OHR. Cp, =0, ty

is exact.

di

de

‘a Raf pd NA Eg For your comments: nt
1.6 Non-Exact ODEs: Method of Integrating Factors

lo far, we discussed how to solve DES when they are separable, homogeneous
ions that do not fit to either of such
Ive DEs that are neither of
Now,

or exact. However, there are many situati
cases, So, our next discussion focuses on how to sol
the above forms. Consider the non-exact DE P(x, pd + O(a»)
multiply this equation by a nonzero function, say zu (it will be a function of x,
, or both) such thatthe resulting equation uPde + Ody =0 is exact

The function je which is used to change the non-exact differential equation
P(x,y)de+O(x,y)dy = 0 into an equivalent exact DE of uPde + wOdy
known as Integrating Factor. For example, the equation 234 + ach = 0 is not
exact but if we multiply it by (x) ==, it becomes — 2xpedx —x "dy = 0 such
that M(x, y) =-2ay, Nu) == = M, =-2x=N, which is exact and its
solution is obtained easily. But, here the main problem is how to choose or
select the function y which is used as multiplier to change the non-exact DES
into exact DE. Even though there is no hard and fast rule on how to find
integrating factor, any way let’ see the general procedure to find such function.
From the condition of exactness, the DE uPdr + Ody = 0 will be exact if and

a
dy it 2
only it LU)

is

Zum Then, by product rule, 4,P4 uP, = 2,0 40, +

Now, solving this equation for yz is too complicated. So, to simplify the
complication lets consider different cases for zu.

Case-1: Suppose ya is a function of x only. Then, using the relation between
partial and ordinary derivatives, we have

PQ, WP,

‘Again, dividing this result by 4 gives
du 8-0. 28,
va Q #0

NITRO RAAT 23

and Bok pd a stg a ee are}
Now, integrating this equation with respect rly
Ir
From this explanation, we can state the following theorem
‘Theorem: [Integrating Factor of the form (x)]:
If P(x, y)de+ O(x, y)dy =0 is non-exact such that UPdx + Ody =

=
to B-
JAP y em, where fa) Be

is exact
Le
depends only on x, then the: ‘integrating factor is (x)=e" ©

Case-2: Suppose y is a function of y only.
‘Theorem: [Integrating Factor of the form u(y)}:
MPG, y)de+ OCs, y)dy=0 is non-exact such that uPde + uOdy =0 is exact

B 2,
+ depends only on y, then the integrating factor is p(y)=. el 7.

and

P

Examples:
1. Verify that the following DES are not exact and solve by finding the
appropriate Integrating Factor:

a) Gr +3y*)de+ 4aydy: b) Epa +(4y+9x7\dy =0
erde rg | AY ma
IH dry =0 D yte+Bx-y+3)dy=0

Ey +2y?)de + Qay + 5y")dy=0 | MAR + y)de+(x"y—2)dy=0
Solution:

a) Here, P=8x +3y', O=4xy’ => P =12y’, Q, =4y’,

Since P, + Q,, the DE is not exact or it non-exact.

So, to solve this DE, first find the integrating factor that changes it into exact.

But, 22 _ 12949? _ 89” „2, (2 which depends only on x.
17) sy 4m x

pr meh à

Hence, the integrating factor is ua) = M9 =
Now, multiply by (x) = x? to get the new exact DE.
Hence, the new exact DE is (8x7 +3x"y")dr+ 4x°y'dy=0.

Er AT A Gs CORTES 24

BRS pi wake Behl Fo jour gen We BESTS?
Next, solve by method of exactness. Let M = 8x" +3x'y", N=40y
Then integrating u, (x,9) = M(x, y) and integrate with respect 10x
That is fu, (x y)de = [6 +3x° yde ule, »)
Differentiating u(x, y)

dy +80).

5 +22 pt + g(y) with respect to y and equating with N

gives u, y +20) =40 y > 809-0804

Therefore, the solution is u(x, y) = c = x" + y
b) P=6xy,0=4y+9x" = P, =6x#18x= 0, . This means it is not exact.
besides, LME 2, py ypc” ay?

P oy y
Hence, the exact DE is 6xy"dr+(4y" +9x"y*)dy =0 %
Next, solve by method of exactness. Let M =6x9°, N=4y? +9x7x7.
‘Then integrating u,(x,y) = M(x, y) and integrate with respect to x.
Thatis Ju,(x,ybde= [op > us, )=3 y +80)
Differentiating u(x,y)=3x"y’ + g()) with respect to y and equating with N
gives u,(x,y)=9 y +8 0)=4y +9 y => g'O) =4y?.
Integrating with respect to y gives [e'(y)dy=[4y’dy=> g0)=y"-
‘Therefore, the general solution is 31")? + y* =c.

©) M,=x,N,=2x+y=M, #N,. Hence the equation is not exact

o My, dy (+) = LAG
Besides, te = Henn, the integrating factor
isulx)=e!* =1 and the new exact DE is (++ pdt + (x+y)dp=0.

x
Next, solve by method of exactness. Let M= x’ + y, N=x+y.
‘Then integrating #,(x, ») = M(x, y) and integrate with respect to x.

Thatis [u,(m, de f(x? + Su )= +2940).

id fi Nabe Wg

279 +29) with respect 10 y and equating with N

reg O)=xty=gy)=y.

Differentiating u(x, y)

gives a, (x,

Therefore, the general solution is

d) Here, P=x°y*+y, => P.=29*+1,0,
Since P, # Q,, the DE is not exact or it non-exact.
To solve this DE, first change it into exact using integrating factor.
Alta")
A SU) depends on y.
YA

any ¿2
Yao?

Now, multiply by 4(»)= > to get the new exact DE.

A 1
Hence, the new exact DE is (x? +0.

1
Next, solve by method o exaciness. Let Max" +,

‘Then integrating u, (x, y)= M(x, y) and integrate with respect to x.

1 >
Thats fu (a pr = J? +4)" = +5 +80).
Differentiating ut ro respect to y and equating with N

>80).
y

Integrating g’(y)=1 with respect to y gives us g0)=)-

es
Therefore, the general solution is u(x,y)=e=> + +726:

TE 2:

Ta cua le ME fo omens an iene Ape)

e Pex! +y?,0=-1 > P, =2y#—-y=Q,.. This means itis not exact

BO,

Besides, WARE

x

Hence, the exact DE is (x+2,)ar-%
27

exactness, we get the solution to be x? - À
Y

DP=3,0=3x-y+3>P, =1#3=Q,.. This means it is not exact

Hence, the exact DE is y"dr+(3xy* — y +3y")dy=0.

Next, solve by method of exactness. Let M=y*, N =39? y +3?
Then integrating u,(x, »)= M(x, y) and integrate with respect to x.
Thats Ju, (23e = [de = ul, ») = 9" +80).

Differentiating u(x, y)

? +g(y) with respect to y and equating with N gives
5d +8 0)=39 -y +3y? > g'Q) =-y +39".
Integrate g'(y)=-2y’ +3y? with respect to y

Thatis [Wr +3y*)dy = gy) =

Therefore, the general solution is xy? "+ ys

8) Here, P=4ay+2y?, O=2xy+5y* >P,=4x+4y, Q, =
Since B, # Q,, the DE is not exact or it non-exact.

Q.-P, _2y-(x+4y)_-(4x+2)
But, Le AAN, (42 429)
P DAR) y FC) depends on y.

Hence, the integrating factor is u(y) = el = ¿[39 _ ur o 1

er nenne

Hence, the new exact DE

Gx+2y)de+(2x+5y))dy.

and Book of Applied ae gs MF your nant DE)
Next, solve by method of exactness. Let M=4x+2y, N=2x+5y” .
‘Then integrating u,(x, y) = M(x, y) and integrate with respect to x.

Thatis u(x y)de= [ax + 2p)de = u(x, y) = 20° +29 +80).

Differentiating u(x,y)= 2x" +2ay+ g(») with respect to y and equating with N

gives u(x) =2r+g¢)=2445y >=.

y
-.
Therefore, the general solution is u(x,y)=¢=> 22° +2ay ae

Integrating g'(y)=Sy? with respect to y gives us g(y) =

h)P=2x" +y,0

B-Q, _1-Q9-D_-29-D__2_ Br
Zt rea D > Abe

Hence, the new exact DE is er @-Dér=0.

2y-x> P, =1#2xy-1=Q,. This means it is not exact.

1
Besides,

‘Then integrating ,(x, y) = M(x, y) and integrate with respect to x.

That is [u,(x, y =[0+ au) = 2x-%4 g(y) and
E x

CS

2,920)
Ñ

Therefore, the general solution is 2?

2

2. Verify whether the following DES are not exact and solve by finding the
appropriate Integrating Factor: =
a) (y? +2ay)de+ (Ax? +5ay+6)dy=0 b)pde+(2xy-e")d=0

0) 3xtytde+ (2 y+x y) =0 d) 2mdx+ y dy=0

e) (y-e™” de —(1+ xe”? dy =0 N G8 y-x?)de+dy=0
8) (y+ x*)de—xdy =0 A) (I+ pd +(1-x)d =0
Sol

2y+2x, Q,=8x+5y= P, #0.. Hence, the equation is not exact.

EAT +

re Te

Hence, the integrating factor is u(y) = Si? om zer =y.
Then, the new exact DE is (y* + 21y*)dr+ (4x? y? +5xy* +6y")dy =0.
Here, M, =Sy' +81, N, = 8x +5y* M,

‘Thus, the general solution is obtained as follow:
Using the short-cut formula:

u(x, y)= [Mdr + [IN - [Malay
= J 0% +2ay' e+ [laxty? +50 +69 - fy! +89 hy

3
sa +9 0 2 try toy"

So, itis exact.

‘Therefore, the general solution is u(x, y)==y* +x°y* dy =.

b) M, =1, N, =2y= M, # N, . Hence, the equation is not exact. Besides,
N,-M,
3 „a 2h20). Hence, the integrating factor is

we ae ee and the new exact DE is

dre =0 . Thus, the solution is
1, (9) = M(x,y) =e” => u(e,y) = 22” +80)
ENS ree Ls aly

MA

‘Thus, the solution is xe” -Iny[=c.

e) M, =62"y, N, en # N,. Hence, the equation is not

Hence, the integrating factor is (y)

Grp he det (2 +2 y )e dy =0.

DE

& 2
u(y) =@x yr ye? +2'Q)=2rytxy'Je? = gO) =k
d) M, =2x, N, =0=> M, #N,. Hence, the equation is not exact. Besides,

57 £0) Henes the integrating factor is

a zei = and the new exact DE is 2xdr+)uy =0
$

Thus, the solution is x? + y?/2=c.
e) Since P, =1-e" #-{1+2)e"” =Q, the DE is not exact. But,

1. Hence, p(x) = ef“ = e* and the new

exact DE is (ye -e”)dr—(ae” +6") = 0 .

Here, let M=ye"—e7, N =-(xe’ +e"*). Therefore, integrating

(x,y) = M(x,y) with respect to x gives u(x,y)=—xe” —ye™* + HO). Then,
differentiating u(x, y) =-xe” ye + O) wr y and equating with N gives
aloe - EHH) =e” +e) HY) =O) =e

‘Therefore, the solution is u(x,y) =e=>.e” + ye" =0

) Since P, =3x* +0=0,, the DE is not exact.

mb. act = f(a) Hence, he integrating air is) =e me?

and the new exact DE is (3xy-27)}? dire” dy=0-
Now, let M(x.) Gym 00”, May) =e

A +

| Begin MEF yout commit and ogg
Method of Integrating Factor for First Order Linear DEs;
Suppose y'+p(x)y = /(@)is first order linear DE.
Then, B+ pay = $0) = Indy - FCO dy = 0
Here, P(x,y)= p@y-S@), Q05y)=1>P,=p(3), 0,=0=>P #0,
-2.

P,
Thus, the DE is not exact. Besides, —- 2

p(x) is the function of x.

Hence, the integrating factor is p(x) =e 9.

Furthermore, multiplying the DE with this integrating factor gives
ebay fare" dy = which is exact.

Then, the general solution becomes (x) = POr oc).
Examples:

1. Find the general solution ofthe following first order linear DEs

a) xy-2y=xre" 5) ssl ©) y+y=sinx
d)y#6xy=x e)(2y-3x)de+xdy=0 f}xy+y=xt+l
Solution: First change in the standard form y'+p(x)y = f(x)

a à
Dry te sde 9 dre) axe sex?

I sxsystf las’ c =H. E

Doyle

x

dad” ne = y= fe sinxdr+ Co” iaa
dig e syse™ [re dt ce zeal
6

(2y-33 =
©) (y Inlet cdi, PEA, a pee +5

2. Solve the following IVPs,

a) y+5y=3e"-1, y(0)=1 6) y+ytanx=sin2x, y(0)=1

e) xe" de+(y?-Ndy=0, y(0)=0 dd) yy »0=3
Solution: Use integrating factor method for first order linear DEs

fsa

a) The integrating factor is s(x) =e’ ‘= e™ Then, the general solution is

A (LL ae) abet ng,

Now, find c using the initial condition. That is y(0)=1=>¢= i.

Hence, y(x) =: je + a ee + 3
1

b) The integrating factor is (x) =e!""™" sig Then the general solution is

LE" 2
= |—sin 2xdx +c] = cos x(-2008x+c)=cc08x-2c05" x
Yer) = cos ff sin 2ade +c] = cost )
‘Now, find e using the initial condition. That is y(0)=1=>¢=3.
Hence, y(x)=3c0sx-2c0s*x.

€) xe der (y —Ndy = 0 = (y? -Ddy=-xe" de
>[0 -Ndy=-fxe"dr>

y

Hence, =
ar)

2 Je
yy = dx! = y+ y = 4x. Hence, M(t) = E
Then, reac ner 1a (fard e)er ++

2
Now, find ©. That is »(1)=3>0=2. Hence, et:

EEE TEN

E ppd tlic Beste lor your comments and gt PTE 2 y
1.7 Bernoulli’s Differential Equations

Definition: Differential equations of the form y+p(a)y = f(y". me Rare
known as Bernoulli's Differential Equations. If» # 0,1, the DE is nonlinear.
‘Such non linear differential equations are transformed into lincar as follow,

First, multiply both sides by + That is y*y'+p(xdy" = f(x)
y

Using the substitution z = y", 2' =(-2)y" 2 »y

(-my"

Putting this in the equation y” wer

Fix), we have

YY" = SES". coy" PQ» = f(x)

races I(x) 22 HI-Mp()2=(1-Mf0)
Therefore, we get firs ms near DE 2'H1-n)p(x)z =(1-m) f(x).
‘Simple steps to solve Bernoullÿs Equations y'+p(x)y = f(x)y", ne R

First: Identify p(x), /(x),n from the given problem.
Then, use the following formula to get the general solution.

Integrating Factor: (x) =".

General solution: y?" [En der where € is constant.

Examples:
1. Solve the following Bemoulli's Differential Equations.
ayyeZaxty!ix>0 D y'-y= ay, (0) =-1 Ly”

cm
nm?

=? ,.
DIES, D) 2y'=103'y + y

O NI
Fey th
ly

IEA IAE ©

son EEE A A i ee EER
Solution:

fax, n=2

a) Here p(x) =

Integrating Factor: zu) e PO

General sol

oF e
=—|(1-m)(x) —
= Jamas ca ==

A RS

>
ay joder jay

xy? Here p(x)=-l, f(x) =x, n=2.Then,
ray loro li oy

Iwan Letter mé jeta y>
z

1

I-xtee”

Besides, y(0)=-12L-=-130=2>
Tee

e) Here p(x)=-=, f(x) =",
Integrating Factor: (x)= e (“90% = DR
A
General solution: y? = x Le do x +a =

d) First rearrange in Bernoulli's Sats BE Px), bi er

Integrating Factors ee

General xfdrtorax ter.

OA +

sand en 02

LI

Juno

Integrating Factor: u(x) =
General solution:

ey
ca.
AG)

OSOS

D Here p(x) =-2, (a) = 6,

Integrating Factor: (x) =e
General solution:

yt ae [126% de+ce

Then, we have p(x)=

Integrating Factor: p(x) =e!" _
General solution:
Lars

„A à
yt == f-20 +
= Li

1.8 Second Order Linear Differential Equations) with
Constant Coefficients (SOLDE

Second-Order-Linear-Differential Equations with constant coefficients are
equations of the form ay!"+by"+cy = f(x (e)
Here, if f(x) =0, then the differential equation is known as homogeneous

and if f(x) #0 , itis known as non-homogeneous.

i) y"3y-4y=0

ii) Ty"46y'45y =0
iti)
iv) y”

1.8.1 Solutions and their Properties

} are homogeneous differential equations.

$ }enm-honoennus ine equations.

Particular and Complementary solutions:
Consider second -order linear non homogeneous differential equation and the
corresponding (reduced) homogeneous differential equation of the form

¡EA AS

Then, any function free of arbitrary constants that satisfies the DE in (i) is said
to be particular solution and denoted by y, .

The general solution of the corresponding homogeneous DE given
to be complementary solution and denoted by y, .

Fundamental Set of Solutions and Superposition Principle
Any set F=(y,,y,)of linearly independent solutions of the homogenous
differential equation ay"+by'+ey =0 is said to be fundamental set of solutions.

A Mei NTS min sin gu,

‘Superposition Principle:
Let y,and_y,be any two solutions of the equation ay"*by+cy=0.Then, the

linear combination y = ci +c,y, is also a solution.
In particular, if y, and y, are fundamental (linearly independent) solutions, then
y =) +6,39, is its general solution and this is called complementary solution
denoted by y, = ay +¢2)2-

So, once we have the fundamental solutions to ay"+by'+cy =0, we can easily
determine its general solution from these fundamental solutions.

Here, to determine the general solution of ay""+by'+cy = 0, it is sufficient to get
any two we fundamental or linearly independent solutions y, and y,.
Question:

How to check whether any two solutions are fundamental or not?

Wronskian Test and Fundamental Solutions:
Wronskian: Let fand gbe differentiable functions. Then, the determinant

defined by W(f, g) + 4 = f.8'—f gis known as Wronskian of f andg .

Linearly Independent Functions: Any two functions are said to be linearly
independent if and only if their Wronskian is non-zero,
Example: Verify that f(x)=e™and g(x) = e* are linearly independent.

a
Solution: The Wronskian of f and g is W(f,8Xx) E e e
22 se

Since W(f,gXx)=7e* #0,VxeR, then fand g are linearly independent.
Wronskian Test (For Fundamental Solutions):

Any two solutions y, and y, of the equation ay"+by"+ey =Oare fundamental or
linearly independent solutions if and only if W(y,,y,Xx) # 0 for all x.

AAA Y

[iad Wo of Appi Mi by BrgeshaeW ¿For yo comments ad grins ar PELIS
1.8.2 Second Order Homogeneous Linear Differential
Equations with constant coefficients (SOHLDE)

where aandb are constants.
(Oh! How?)

General Form: ay"+by +
Form of Solution: The solutions are of the form y

Question: What is the basic task to get the solution? As you see in y =e the

constant rin the exponent is arbitrary. If we know the value of, then the
solution is known. Therefore, the basic task is to determiner
Method to determiner: Assume y = eis the solution of ay"+by'+0/=0-

"is assumed to be the solution of ay""+by'+cy =0, it must satisfy

Since y.
this equation whenever substituted. Here, y = €”, = re", y"
‘Then, substitute these in the equation ay/"+by'+<y

ayy = 0 are +bre™ +e" =0
De" (ar +br+c)=0
>e” =0or ar? +br+c=0

Since e” +0,we must have ar? +br+c=0.

Note:
à) The equation ar? +br+c=0is called Auxiliary (Characteristics) equation
of the differential equation ay"+by'*y=0.

ii) The function y=e is a solution of ay"+by'+cy=0 if and only if the
constant r is the solution of the quadratic equation ar? +br+c=0,

—b+ VE -4ac
a

Here, we may get two distinct real roots, single real root, or complex roots
depending on the sign of the expression b? —4ac under the radical sign. Since
the type of the root determines the form of the solution of the DE, let’s see the
three different cases based on the type of the roots. The forms of the solutions
based on the natures of the roots are summarized using table as follow.

Thatis a? +br+c=0>

NA a on sens x,
[Forms of Fundamental and General Solution foray"=by'=<y=9
T Fundamental | Complementary

Solutions y;. y; | Solution y,

Cases | Roots of the
‘Auxiliary equation

ar’ +br+c=0
Two Real Roots Woh rey
-b+ Vb" le
= 0e" +02]
Db*-4ac>0 2a Y Y
=b= vb -4ac
2a
Single Root san ren
mét-4ac=0| y =} le
fe Ye = +
2 =p ted
Ib? -4ac< fr
yurce"cosfk
+
ce sin A
Notice about case-III: For the two complex roots =a+ Al, =a A.

Here, y,=e™ and y, =e? are the fundamental solutions. But these
are complex solutions while our problem is real. So, we have to change these
solutions in to their real forms. This is possible by using Euler’s formula.

Uy +y,) = 6 0h

Pate Où -p)=e" sind

Hence, the corresponding real solutions are y, =e* cos fx, y, =e” sin ft

2 lillies GAMING RAINS Cems ”

ead IG Api Mala Boshi Yor our oma and ge CIO

Examples:
1. Solve the differential equations using the above procedures.

a) y'-5y+6y=0 b) y"+8y'+16y=0 c) y'-4y+13y=0
d)2ÿ"-1ÿ4y=0 e)y#Iy=0 D y#9y=0
g)4y"-4y'ty=0 A) y"4y'4Ty =0 i) y"#2V2y'2y =0
Solution:

a) Here, the characteristics equation is r? -5r+6=0.
Solving this gives us r? —Sr+6=0=>(r-2Xr-3)=0>x =2,7, =3.
Hence, the fundamental solutions are y, =e**, y, =e”.

Therefore, by case-I, the solution is y
b) Here, the characteristics equation is r? +8r+16=0.
Solving this gives us r? +8r+16=0= (+4)
‘Thus, the fundamental solutions are y,
Hence, by case-II, the general solution is y = ce +c,xe*

©) Here, the characteristics equ: is r?-4r+13=0.

Solve this using quadratic formula give the following complex roots.

4+V-36 _4+6i
2 2
From these complex roots, we have @=2 and B=3. Thus, the fundamental

solutions are y, =e* eos fic =e™ cos3x, y, =e" sin fic=e™ sin3x

je" + ce’

r-4r+13=0>r=

+3.

Hence, by case-IIl, the general solution is y =e**(c, cos3x+c, sin3x).
4) Here, the characteristics equation is 2r* - r-4=0.
74V8i _749

A =

4
Hence, the fundamental solutions are y, =

Therefore, by case-l, the general solution is y = je" + ce Ÿ
€) Here, r? +77 =0=>r(r+7)=0=>7=0,r=-7.

Hence, the fundamental solutions are y, = 1, y, =e"
Therefore, the general soluti Wi HEY, =O + Ge

A A e <c

U Book of Applied Mühe U Rgishae M. ee

9=0=> 7? =-9=>r=4V-9 7, =3i, 1 =

Hence, by case-Ill, y =e (c, cos Ar +¢, sin fie”")=c,0052x+c,sin2y,
2. Find the values of the constantk for which the DE y+ky!+hy gy
general solution of the form y =e (c, cos / +c, sin A). 3
Solution: The characteristics equation is r? +kr+k=0.Then the DE will
a general solution of the form y =e* (e, cos Ax +c, sin ft) if and only if
7? +kr+k=0 has complex roots. But the roots will be complex if and only
dk <0>k(k-4)<050<k<4,

Obtaining HUDE ‘from its General solutions:

Once we understand the forms of solution of second order homogeneous DE
we can also determine the reduced DE from its general solution as follow.
First: Identify the fundamental solutions y, and y, from the given gener
solution and obtain the roots 7, and r, from these fundamental solutions.
Second: Form the characteristics equation using the roots obtained.
That is (r-n)r-n)=0>r°-(n+n)rtnn=0-

Third: Deduce the DE. It is y'-{n +r,)y'+1.7,9=0.

Examples: Find second order LHDE whose general solution is given.

ay=ce+oe* b)y=csinV2x+e,cosv2x od) y= qe +1
Solution: First identify the roots of the characteristics equation.

a) Using case, pe y => ln
So, the characteristics equation is determined as follow.

(r= Xr=r,)=0>(r-Ir-3)=0=>r*-4r+3=0.
Therefore, the required homogeneous DE is y"-4y43y =0.
b) Here, using case-IIl, y, =sin/2x, y, =cosy2x=> = mn
So, (r-5Xr-r,)=0= (r-V2i)r+ V2) =0= 7? +4=0
Therefore, the required Homogeneous DE is y"+4y =0..
©) Using case-M, y = ae" +c,xe™ => y =e", y, = x8" =H =
So, (r-nXr-n)=0>(r-2r-2)=0>r'-4r+4=0. 0°
Therefore, the required Homogeneous DE is y'4y+4y=0:

Las WHET Mid Matec WBN or oa comme od ge PPEHZE
1.83 Initial and Boundary Value Problems (IVP and BVP)
Ñ ‘A differential equation together with some specific conditions on the dependent
Variable and its derivatives which are given at the same value of the
independent variable is known as Initial Value Problems (IVPs). The specific
‘conditions are said to be initial conditions. On the other hand, if the specific
‘conditions are given at different values of the independent variable, the
problem is known as Boundary Value Problem (BVPs) and the specific
conditions are said to be boundary con
à Forms of Initial Value Problems (IVPs)
ay" +by'tey = f(X)-DE
(0) = Yor Y (50) == ICS
Y%o) = Yor Y (60) = M1 um
+
(Both y and y'at thesame value x = x,)
ji) Forms of Boundary Value Problems (BVPs)
(ay"+by'+ey = f(x)-DE
0) = Yor YU) = 71 — BCS
> Ce) re Y BVP

(Here y and y'tare given at diiferent values x=, = x)

Examples:

1. Solve the following IVPs
O)y"3y'-4y=0,9(0) =4,'@)=14 _B)y"-3y'=0,9(0) = 7,0) = =
a)y"+y=0,y(0)=2,y'(0)=0 d)y'Ay+4y=0,y(0)=1,y(0)=4
eyutriy=0,0M=3,Y'0=4 Ny Ry=0,=L y (m=0
Solution:

a) Here, the characteristics equation is r+3r-4=0>r=14.

‘Thus, the general solution is y= ce" +66".

Now, let's determine the constantsc,, c, using the initial conditions.

b) Here, the characteristics equation is r* 3
‘Thus, the general solution is y =c, +c,e". Now, determine c,, c,.

y +0
Be, =
‘Therefore, the solution is y=10-3e”.

©) Here, the characteristics equation is r?+1=O=>r =: Thus, the
solution is y = c, cos x+c, sin x. Now, let's determine c,, cz.

That is y(0)=2= ¢, =2, y (0) =0>c, = 0.Therefore, y= 2cos x,

d) Here, r?-4r+4=0=>7=2.Thus, the general solution is y=ce™

>

That is y(0) =7,y"(0) = sal 5 =-3,¢,=10.

and y(0)=1,y"(0)= a MU , = 2 So, 0

2, Solve the following BVPs
a) y'+4y=0, (0) =2,y'(z)=-6 5) y'+4y'4+4=0, 10) =6, 3):
Solution:

9) Here, the characterises equations 7? +420.
Solving this gives us r? +:
Hence, the general solution is y =c, 0052x+C, sin 2x.

Now, let's determine the constantsc,, c, using the boundary

e, cos0+c,sin0=2
027026 2 sin 2) +2, 008(28) =-6 > si
Hence, the solution ofthe BVP is y =2c052x—3sin2x
b) Here, r? +4r+4=0>r=-2.. The solution isy
Now, lets determine the constants;, c, using the
Therefore, the solution of the BVP is y = 6e* ~2xe

ia fp aay gh Wi oa oi sag we PA
1.8.4 Solving Non-homogeneous Linear Differential
Equations (SONHLDE) with Constant Coefficients

Form of SONHLDE with constant coefficients: ay'+by'+c = f(x).
Particular solution of ay"+by'+cy = f(x):

Any function y, free of arbitrary constants that satisfies this SOHLDE is said
10 be a particular solution.

‘Theorem (The General Solution Theorem, GST):

If y, is any particular solution of the non-homogeneous DE ay"+by"+ey = f(x)

and y. is the complementary solution of the homogeneous part ay "+by'+ey =
then the general solution of ay"+by'+ey= f(x)is given by y =y,+),-

In short, this theorem says that the general solution of ay"+by'+cy = f(x)is the
sum of the general solution of the corresponding HLDE ay'"+by'+cy=0 and
any particular solution of ay""+by"tey = f(x). That is y= y, +y,-

So far, we have seen how to find y, of ay"+by'+0y =Obut how to get y,?
Procedures to solve ay"+by"tey = f(x).

First: Solve ay"+by'+cy =0 and obtained the solution y, =
Second: Find any particular solution y, of ay"+by'+ey = f(x).
Third: Form the general solution y =c,y, +¢,y, + y, using the GST.
There are different methods to find y, .

‘Method of Undetermined Coefficients (MUCs)

Variation of Parameters (VPs)

‘The Operator Method (OM)

Diagonalization Method (DM)

Laplace Transform Method

Power Series Method

Foy

Sean

Tid y pt al ES ea ra | /
1.11.1 Method of Undetermined Coefficients (I (
Suppose we want to solve ay"+by"+ay=f(2) where the coefficients a,b and | y
€ are constants using the Method of Undetermined coefficient.

Main principles to notice about MUCs: o
Assumption: The method of undetermined coefficients assumes that the | {

solution to the DE equation isthe same form as f(x).
ii) Starts with trial form: Making educated guess .
Once we assume, has he same form as /(2), the method proceeds with an | |

educated guess by expressing y, using undetermined coefficients. |
iil) Coefficient Determination: From the tral form obtain, y,, y, and y”, and
substitute in ay"+by'+cy=J(x) to determine the coefficients. With this it is
possible to determine the undetermined coefficients to be determined.

For this reason, the method is named as Method of Undetermined Coefficients.
Conditions to use the method: When do we use the method?

The general method is limited to non-homogeneous linear DE of
Y'+by +) = FG) with the assumption that the coefficients are constants,
and f(2)is only of the form p,(x), "sin ax, cos fis, p,(x)e", and their
combinations like p,(3)sinax,p,(2)00s rp, (3)e" sinax+ p, (xe cos fe.
Exceptions: When does the method fails?

‘The Method of Undetermined Coefficient (MUCs)

is not applicable if the

m 1
functi 1 cost
tion f(a) is of the form In x; VE, tan x cos * 3,500 5 ct x, cc or any

‘other transcendental functions.

Hand Bok of Applied Mathematics by Bgher Fr pun comments and sugestins ase 09924242

ABLI The basic Trial Forms of Particular Solution
The form of F(x) Trial form of y,

x)= ae de”

DETTES) 3, = 4x8

Sn=artsbi+e At Bx+C

fasin ix or beoskx
SO" a sin tx +bcoskr Asin kx + Beoskx

asinh kx or
HORS coshix Asinh kx + B.coshkx

asin Ax +bcosmx y, = Asin Ax + Beoskx + Csin mx

SOHN a kam |?
asin Ax + bsin mx +Dcosmx
F(x) =ae® +be”,k=m y, = de" + Be™
For Product forms
[9 = (ars De*™ y,= (ax +BJe"
(ax + b)sin Ax or Y, = (Ax +B)sin kx + (Cx+D)coskt|

=

fo (ax + b)cosix

P(e) = ae" sin orbe™coskk | y, = de™ sinkx + Be” coskx

PACS y, = (Ax+B)e”sinkr
ax + bye”” coskx +(Cx+ De" coskt

SG = (ae +br+c)e® (Ax? +Bx+ Ch

Cautions:
The table will give you only hints on how to guess y, based on f(x).
Always, ASK yourself the following questions about y, :

Does what we guess for y, always work? No!

How do we know when it does not work? From analysis of roots!
How do we correct if it does not work? Use Modification Rules!

Le se

Find the complem
+) Step-A: Find th

acieristics equation

Hence, the complementary solution is

he same formas f(x

Sup: particular solution y, havi

From the tabl

seems ofthe form y, =a2*. Then, y, = 2ae™, y",=

ine the co
3y'-4y

tant a by substiniting these values in the given DE.

Deia > a

> 120%

Thus,
b) Step-

Here, r? ~2r+

26

: Find the complementary solution y, of y"-2y'+y =0.

(1? =0=>r=1
Hence, the complementary solution is y, = ce" +c,xe".
Step-2: Find the particular solution y, of the form f(x) =x" -x+3.

Thatis y, is ofthe form y, =ax? +bx+c

Then, substitutey, =ax* +öt+c, y',=2ar+-b, y", =2a in the DE.

2y'ty

~¥4+3=>2a-2(Q2ar +b) ax? +bx+c= x =x43
= ax? 4+(6~4o)x42a~-2b+e=x? -x43
= a=l,b-4

So, the particular solution is y,,

P3547,

Hence, the general solution is y= yg + y, =

©) Step-1: Find the complementary solution y.of y'-3y—4y=0.

{tis the same as in part (a). That is y, = ce +

hand Bako pled theme Bash. or nor om ad upon we IKT

Find the particular solution ;», of the form f(x) = 34sin.x

ne
seems ofthe form y, =asinx+beesx,

-asin x-beosx

cosx—bsinx, y

Then, Y,
low, determine a,b by substituting these values in the given DE.

yy ay =34sinx
2 -asinx=bcosx-3(acosa

bs
> (Sa +3h)sinx+(3a-5b)cosx=34sinx

ax) A(asin x + bcosx) = 34sin x

Equation the coefficients of sin x, cos x on both sides, we have

us, y, =asinx+beosx=-Ssinx+3cosx.

fore, Y= * +c,e" —Ssinx+3cosx.

First, find the complementary solution y, of y"+y'-2y=0.

P4r-220> (7-2) = 0297 Shh =-2.

fence, the complementary solution is y. =ce™ +:

lext, find the particular solution y, of the form f(x) =2xe™.
seems of the form y, =(ar-+b)e".

en, ÿ,= ae — (ax be, y", =—Dae™ + (ae +
us, substitute these values to determine the coefficients.

yity'-2y =2xe*
> las” +(ax+b)e” +40" -(ax+b)e” Uat ble = 2xe*

> Dare -(a+2b)e™ = 2x6"

S-a-2-a-2=02 42-127)

fore, the general solution is y = Je + Y,

ka pid Malema y Be Fr ur coments and
9) '42y= 0-77 +2=0=97 =4V2i = y, =c, cos /2x+e, sin Be

Now, lets’ find the particular solution y, of the form f(x) = 3xsinx

Here, y, is of the form y,

(ar+b)cosx+(cx+d)sinx. Then, wehave |

= 400$ x —(ar+b)sin x + esin x + (ex + d)cos x

2asinx - (ax + b)cos x + 2ccosx ~ (ce +d)sin.x
So, y"42y

> Ge +bcosx + (d- 2a)sin x +axcos x + exsin
=2%+b=0.d-2a=0,a
> y, =3xsinx=6c05x

3xsinx
= 3=9 a=0,b=-6,d=0

Hence, the general solution is

D ri+2r+s=0>

e, cos ¥2x +c, sin 2x + 3xsin x - 6cosx |

142i

—2i. Hence, the fundamental solutions ae
“sin 2x. Now, find y, ofthe form f(x) =e" cos2.
Itis of the form y, = ae" sin2x+be" cos 2x

cos2x, y

(a~2b)e" sin2x+(2a+b)e" cos2x,
"p= Ba +4b)e" sin2x+ (4a—36)e" cos 2x
yy

€ cos 2x

> Ga~8b)e" sin 2x + (Ba + 4b)e" cos 2x = e* cos 2x
> 4a-8b=0, 8a +46

=1S a 224, 8a+4b=1
D 144-1561 ol

‚as

20°" 10

>»,

Ls 1
jp Sin2x+ et cos 2x == (2sin 2x + 00521)

‘Lio Book of Applied them by Fc Far our comen an igestins ise 2945-6242
Remark (Sum Rule for Trial forms):

Inthe DE ay"+by+cy= f(x), the function (x) may be the sum (difference) of
functions like f(x)=g(x)+h(x),in such case, guess the trial forms y, for
g(x)and y,, for h(x) separately as the particular solutions.

Then, their sum y, = y, +y,,i8.a trail form for f(x) = g(x) +h(x)..

Examples: Solve the following DEs using MUC.

a) y"4y"s5y = 30e" +10x-7 b) y"+4y =8x+9sinx

c) y'-y'-2y = 2x? +e" d) y"4Sy'H6y =4e™ +Ssinx
Solution:

Step-1: Find the complementary solution y, of y"+4y'+5y=0.

Here, r° +4r+5= +i Then, y, =e cosx+c,e sinx

Step-2: Find y, which is the same form as f(x) =30e" +10x-7.
Here, f(x) = g(x) +(x) where g(x) =30e", h(x)=10x-7.
So, guess y,, =Ae" for g(x)=30e" and y,, = Bx+C for h(x)=9sinx.
Then, y, for f(x) = 300" +10x—Tbecomes y, = y, +3 pn = Ae + Br +C.
That is y, =ae" +br+e=9 y',=a0" +6, y",=ae".
So, y"+4y'+5y = 300" +10x-7
> 4e +4(4e" +B) +5(4e" +Bx+C)=300' +10x-7
> Ae" +44e* +4B+54e" +5Bx+5C=30e" +10x-7
1046 +5Br+4B+5C=30e* +10x-7
104=30= 4=3
=45B=10=3B=2 =y,=3e"+2x-3
4B+5C=-1>C==3

Therefore, y=y, +y, = ce cosx+cye™" sinx+3e" +2x—3.
b) Step-1: Find the complementary solution y, of y"+4y=0.
0=r=#2i.

cos 2x +c. sin2x

Here, the characteristics equation #° +
So, the complementary solution is y, =

od nf ped ae ge. fos men Ein o
ind the particular solution y, of the form f(x) %

Step.
Guess y,, = 4x + B for g(x)
+ ya = Ax + B+Ceosx+ Dsinx for f(x

x and yg = Coosx+Dsinx for h(x) 9,

Then, y, *+9sinx

Ccos x Dsinx

Hence, y',= 4—Csinx+ Deos x, y",
Now, determine the constants À, 8,C, D.
"shy =Bx+9sinx

>-Coosx- Dsinx + [dx + B+ Ceosx+ Dsin.x]=8x+9sinx
244=8,4B=

Hence, the particular solution is y, = 2x +3sin x

Therefore, y=c,00521+c,sin2x+2x+3sinx.
©) Step-1: Find the complementary solution y, of y"=y/=

Solving this gives us

Here, the characteristics equation is +7 =

e

.So, y,

Step-2: Find the particular solution y, having the same form as

®+br+e for g(x) =4xand

Sla)=4x? +86". Here, we guess =

bx+c+de*for

Yor = de for h(x)=8e. Then, Y, = YA + Y po
I(x) = 4x? +8e" Hence, y,=2ar+b+de", y",=20+de".
Now, determine the constants a,b,c,d.
Y'-ÿ-2y = 46 +80"
2 2a+de ~[2ar+b+de"]-2ar? +bx+c+de")=4x" +8e"
> Lar" -(2a+2b}x+2a-b-2c-2de = 4x" +8e"
>-u ,2a-b-20=0,-2d=8
,0=3,d=-4
Hence, the particular solution is y, =-2x* +2x-3-4e*

Therefore, y= ce +c,e* ~2x? +2r-3-40".

A ALAND ove 02H)

A Hand Bok of Applied the mais iyBgesh Far our comments ad sozgsions use PRESS
Modification Rules for Trial Forms: Generalization on MUCs
Now, let's analyze the three questi

ns about », that we have posed earlier.
Y Does what we guess for y, always work? No!
Y” How do we know when it does not work? From analysis of roots!
Y How do we correct if it does not work? Use Modification Rules!

As we have discussed for the DE ay"tby'tqy= f(x), we used an educated

guess of the particular solution y, based on the form of f(x). But this is not
always true. There are cases where the form of the particular solution y, that
we guessed may not work. In what follow, let's discuss the cases where the
form of y, is determined based on the relation between the roots and 7, of
the characteristics equation and some part of f(x).
Modification Rule-1: For the form f(x) =(A,x" +...+ Ax+ Ae
In such case, the form of the particular solution y,of ay"+by'+<y=0
depends on 7,,r, and the exponent k.

DIT 4k,r, 2h, then y, =(0,1"+...+0,1+0,)e%

5) If =k or r,=kbutr #7, then y, =x(a,x" +..+a,x-+a,)e"

iii) If 7, =n =k, then y, = ax" ++ ax + ae"
Examples: Solve the following DES using undetermined coefficients.
a) y"-3y-4y=10e" b)y"-3yAy=xe" 0) p-2ÿ+y=6e
yx yl ef) y A yy axe
Solution:
3) Step-1: Find the complementary solution y, of y"-3y-4y = 0
0. Solving this gives us

Here, the characteristics equation of is 7° -3r-
r-3r-4=0>(r+IXr-3)
Hence, the complementary solution is y, = ce" +00 % =qe* +c,e".

Step-2: Find y, Here, f(x) =10e",with k=-1, *rjand 7 =-L,1, =4 but

here 1, =k. Thus, y, is ofthe form y, =

AS 52

y'-By-4y = 2e* > -2ae” +axe”* - (ae - axe”*)-4ae”
=> Sar” = 10e >-5a=10>a=-2

Thus, y,=-2xe”. Therefore, Y=y, +), =00* +06" 2e,
b) Here, r?-3r-4=0>1 =-1,1, =4and f(x)=xe™, with.
So, by the second part ofrule-1, y, = (ar +b)e” =(ar +bx)
Then, y,=(b+2ax-bx-ar)e”, y", = (ax? -dar+br+2a-2b)e”.

So, ay dy mae”

=> (a -dar+bx+2a-2b)e” -3(6+2ax-bx- ax le” -4ar +bre* ze

> (-l0ax+2a-Sb)e™ =xe* >-100=1, 2a-5b=0= a:

Here, the solution of the correspondi

7 =-2r+1=0>(r-Ir-1) = 05, =

Hence, the complementary solution is y,

Step-2: Find the particular solution y, .

Here, f(x)=6e",with k=! =

‘Then, y,=Qartar)e', (2a+4ax-+ax*)e* So,

y'-2y+y = 6e = (a+ dar + ar Je" -Udartar')e' +ade =6e"
= 26 =6e => 20=630=3

Thus, , =3x7e". Therefore, =}. +), = Ge tape’ +3x7e",

d) Here, 7? +6r +9 =0=>7, =1, =-3and f(x)=xe™, withk =-3.

So, by third part of rule-1, y, ="(ar+5)e™. (Complete it!)

©) Step-1: Find the complementary solution y, of y"+y-2y=0.

BAPE Ec aire mn nen THANE
=

PE CE ET, IIS

A and Bock of Applied Mathematics by Bash M For your comments and suggestions use 093923424)
Here, the characteristics equation is 0. Solving this gives us

Or =1r, =-2. So,

= (r-Ir+2)

A

r ae +

Step-2: Find y,having the same form as f(x)=e" +e". Here, we guess

yp = axe” for g(x) =e" and yy, =be” for h(x)=

«Then,

= are +be™ for f(x) =e" +e”*. Now, determine aand b.

ya

1
Therefore, y=ce" tee thie’ es;

Modification Rule-2: For the form f(x) =4,x" +..+ Ax+ 4
In such case, the form of the particular solution y,of ay"+by'tey= f(x)
depends on the coefficients.a,b.
1) 100 #0,then y, = a,x"
ii) If b=0,but a # 0, then y,
ii) fa=b=
ind the particular solution of the following DEs.
18x? +2 b) y"=24x ©) }'=9x +2x-6

ax+a,

(a,x" ++ x +0)

0, then y, (ax +..+ax+@)

Solution:
2) Here, the characteristics equation is r? —3r=0=> a= -3,b=0 and
J (x)= 18x" +2. So, y, is of the form y, (a +br+c)= ax’ ++ er.
Then, y', =3ax? +2bx+c, y",=60x+2b.S0,
y'Ay = 26" => 6ax+2b-3Bax? + 2bx-+c)=18x7 +2
> Jar + (6a-6b)x+2b-3e = 18x? +2
=>-9a=18,6a-6b = 0,2b-3¢=2 = a=-2,b=-2,c=-2
| Thus, y, = 212) =2 - 2x? Bu.

NTT MAMTA OAT sa

iB pie Way Bias. or our commons a ei e ey
b) Here, the characteristics equation is r°=0. So, the coefficients of the
characteristics equations are a=b=0 and f(x)=24x. So, y, is of the form
Y,=2*(av+b)= ax? +bx?, Then, y',= 30 +2bx, y",=6ax+2b.
But y"=24x= 6ax +26 = 24x= 6a=24,2b=0=>a=4,b=0
Thus, y, =4x and the general solution is J=J. +}, =Cı +0,x+4x°,
=Owhich is a single root,

©) Here, the characteristics equation r?=0=>% =";
‘Then, the fundamental solutions are y, =e"=1,), =39, =x-

Next, let’s determine the particular solution y,. The direct trial form is

y, = ax" +bx +c but it does not work because the coefficients of the
characteristics equations r? = Oare a=b = 0. Besides, /(x)=9x" +2x-6.
‘Thus, by the above modification rule, the direct trial form of y, must be

modified as y, =" (ax? +bx+c)= ex +bx? +a,

Then, y',=4ar? +3bx? +2cx, y", =12ax* +6bx+2c.

So, y'"=9x? +2x-6>12ar? +6bx+2c=9x* 42x-6
>120=9,65=2,20--6>a=

3 losen 34,1
+ ‘3x7, Therefore, y: atastzx +e Ir,

Modification Rule-3: For the form (x)= (acos A +bsin Age".
In such cases, the form of the particular solution y, of ay"+by'sey = f(x)
depends on the relation between the characteristics roots 7,7 and @,ß .
Mfr, *a+ fir, #a- Bi, then y, =(Acos fic + Bsin Be”
ii) If, =a+ fir, =a fi, then y, = x(Acos fir+ Bsin fe”
Examples: Find the particular solution of the following DES.
a '#y=sinx B)y"Hy+Sy=e"sin2x c)y"Hy=xcos2x
dYHy=Bcos2x e) y"-2y+y=e"sinx Ny+16y=4cos4r

he 55

y!

ad of pes Wehr Bega Fi vant adsense ESDP
Solutio
a) Here, f(x) =si

But 7? +1=

x, with B=1,@=0.

a- pi
dacosx +bsinx)

So, by part (ii) of modification rule-5, y

yi,=acosx+bsin x+x(boos.x—asinx), y", = (-2a~Bx)sinx + (2b—ax)oos +

So, yry =sinx= (-2a—bx)sin x + (2b—ar)oosx+ a(acos.x+bsin3

> -2asinx+ 2beosx=sine>-2a=12b=0>a=

Thus, y, =x(acos x +bsinx)

DIRYSy=03 1 +2r+S=037 =
es A

Now, lets” find y, which is the same form as f(x)

Here, in f(x) =e" sin2x, we have @ =-1, 8 =2. But we have

=-1-Asnzatf,n

n=-1+2i,n = fi . Thus, y, is of the form
y, =x(Acos2x+ Bsin2x)e™ . (Complete the solution!)

Now, lets’ find y, which is the sam > arm as

©) y'#4y=0=r =2,
f(x) = xc0s2x. Here, in f(x) = xcos2x, we have a =0, 5 = 2. But we have
2i>r, = a+ fir, = a fi . Thus, y, is of the form

y, = (ds + B)cos 2x +(Cx+ D)sin2x]. (Complete the solutiont).
Miscellaneous Examples on MUCs:

The following problems will help you to test yourself whether you understood
all the main concepts about MUCS. First try by yourself and then see the hints,
1. Solve the following DES using MUCs.

a) y-y-2y= 4x? b) y"By't2y ax? tet
e) y"-Sy't6y = ne” d) y "by w9y=x+ e"
y '2y3y=12xe% +6x-11 DIIYRy=8 +1

‘Solution:
4) Step-1: Find the complementary solution y. of y"-y-2y=0.

OAT 56

si Wate by Degas M Fr your comments and suggestions us 0

Kind oo

=0>(r+I(r-2)

Step-2: Find the particular solution y,,of the

x +he+c. Hence, Y,

Here, we guess
Now, determine the constants a,b,c.
2y=4 »2a-[2ar+b}- ax +br+e)

ya

Hence, the particular solution is y, =-2x° +2x—3 .

Therefore, y= Ge" +e,e% -2x? +2x-3
b) Here, r? ~3r+2=0=9(r-I(r-2)=0=9 7, =1,0=23 y, ce" +0".
br +6, yy = det

Here, y,scems of the form y, = +, Where ÿ =
but not because e*is already in the solution y, = ce" +e,e*
Thus yy, = dee" such that y, =ax? +bx+c+dre
Then, Y,=2ar+b+(d+d9e",y",=2a+(2d+dJe.
So, y"-3y't2y=x' +e"
>2a+ (2d+dı)e' -Har+b+(d+dr)e‘]+ far’ +bx+e+ dre") =x +e"
= 2ax! +(-d -2)xe" + (2b~6a)x—de" + 2a-3b+ 202 x +e"

=> 2a=1, ,2b-6a =120-35+20=-0>a=

1239.
lo y Degen one ee
©) Here, the characteristics equation is
PF ~$r+6=0=9(r-2)r-3)=0-7,=2,7, =3and fQ)=
So y, is of the form y, =x(ar+bJe* = (ae +bx)e™,

Then, ¥,=(2ar? +2ar+2br+b)e*, y (far? +8ax+4br+20+4b)0"

**, withk=2.

57

Und Dock of pp them A M Ft jour commen ad agen se OED

So, y"=Sy'46y = xe" > (~2ar+2a~b)e™ = xe™

d)Here, r? -6r+9=0>(--3) =0>r=3.Then, y,

Now, lets’ find y, which is the same form as f(x) = x+e*-
Here, y, seems ofthe form y, = Ja + Ya Where yy =ar-+b, y o =e"
Then, Y
So, y'-6y'49y = x40" = dee +0ax-6a+9b=x+e"

121, 2, ee 2)
AN "27
©) Step-1: Find the complementary solution y,of y"-2y/-3y=0.

a+cet,y",= ce".

æ4c=1,9a=1,9b-6a=0=c=

Here, the characteristics equation isr? —2r-3=0. Solving this gives us
r-27-3=0>(r+1(r-)=0>n =-L,1, =3. 50, y, = + ce"

Step-2: Guess y, having the same form as S(x)=12xe* +6x-11

Here,, Ya =(ar+bje” for h(x)=12xe "and y, =cx+d for g(x)=6x—11.
Then, Y, = Ya typ = (ax + be +er+d for f(x)=12xe* +6x-11
(4ax+4a+4b)e™.

Hence, y',=(2ar+2b+a)e™ +6, y",
y"-2y'-y = 12xe™ +6x-11

> (Bar+2a-3b)e™ ~3ex—2e~3d =12xe +6x-11
=6, -2e-3d=-11

>30=12,20-3b=
=5>), ae 2545

1) Step-1: Find the complementary solution y,of y ayy =0

3/7 Y.

Here, *+3r+4=031= I >

IE SBE HEN. nem
eihod of Undetermined Coffee E :
8) y"=y=0082x

DIayay=d —h) yty'=3rede'

Solution:
2) Here, +4 =
Hence, y, = ce" +c,e”. Now, y, =artb+ce

But the term ce™ alrengy

arth see

exists in y, So, it must be multiplied by x. That is y,
Bee" — 30e + 9x0 %

Therefore, y=y, +, =

b) Here, 7
J'y=0052x= -Sacos2x-Sbsin2r= cos 2x => -Sa=1,-50=0
V5,b=059 y, =-1/50052x> y=ce" +e,
©) Here, the characteristics equation r(r—1) =0>7, =0,r,

= =Lr =-1 and y, =a00s2x+bsin2r.

Sosa ta", Now, le’ findy, having the same form as f(s) =1+¢

Since 6=0(as in rule-2) and ri =1as in rule-1, we guess y,
*. Then,
IV Sa tbe cre" for f(x) = 4er Hence,

Platt, Y 2204200 +00 Ths,

dar +) fie

B(x) ce? for h(x)

Paba 4e Ja 6°
Arabe dal, 0

Aad oof pled ha Degas 2 Far our comment and sggetions we 442
3. So,y, =¢,+c,e". Now, let's find y,

d) Here, 7? +3r=0=>r=0,

+4x. Since b=0( as in form 2) and

having the same form as /(x) = 6e
, =-3as in form I, we guess y,,
for h(x) = 66"

Then, Yp=Yp +
y,=lar+b+e” —3ere™, y", = 2a-6oe™ + 9ere™ Thus,

x(ax+b) for g(x)=4xand y, =e

x? +bx+ere” for f(x) =4x+6e™ . Hence,

y3y=4x+ 6e = 2a - 6ce”” +90xe 4 3(2ax + b+ ce Sere) = 4x + 607

= 2a-3ce * + 6ax +3b = 4x +66"

>6a=4,-30=6,2043=0=a= 4
a, Fame

‘Therefore, y=c, +c,e* erase”.

y 2y=0>r -2r=0>r(r-2)=0>5 =0,7 =2>y,=0 +00”
Now, lets’ find the particular solution y, which is the same form as
f(x)=eF sinx. Here, y, is ofthe form y, = (acosx+bsinx)e” and thus
Y ,=(boosx—asinx+acosx+bsinx)e",y", (-2asinx+2bcosx)e*
So, y"2y=e*
= (2asinx+2bcosx)e* -2(bcosx—asinx-+acosx-+bsinx)e" =e" sinx

sinx

Rae! cose" sin =e" sins=9-2a=0,-2b=1=>a=0,5=-5

Hence, y, =-Le"sinx= y=0 +00 je sinx.
9) Here, f(x) = 4x +e* indicates y, =ar+b+ce”,

Then, using y',=a+3ce™, y",=9ce™, we have

EEE e 60

lod ok a pied aie by Bega Fr yur comms ge a

d) Here, 7’ +3r=0>r=0,r=-3. So,

=0 +00

having the same form as _f(x)=6e" +4x . Since

3as in form 1, we guess y,, =x(ar+b) for gC
for h(x) =6e"

Then, Y, = Yi + Yo = ax? +br+oe” for f(x) =4x+6e™ « Henco,
2a-6ce”" +9exe™ Thus,

ya 41466" 29 20~ bee + Dore + Dar bee” dere) =A 66

Y= dart btoe™* —3cre™, y

= 2a-3ee™ + 6ax +3

re

Therefore, y

Oy"-2y'=0=9 7? -2r=03r(r-2)=0>5=0,1 =23y,=0 +00"
Now, lets’ find the particular solution y, which is the same form as
S@)=
¥,,=(boosx—asinx-+ac0sx+bsinx)e",y", =(-2asinx + 2bo0s.x)e"
So, y"2y=e*sinx
> (-2asin.x+ 2beosx)e' -Abcosx-asinx+acosx+bsinx)e

inx. Here, y is ofthe form y, =(acos.x+bsin.x)e" and thus

sinx
> -2ae cosx-be'sinx=e'sine>-2a=0,-25=1>a=0,

as
Hence, y, =—Letsinx=>y=a +0" -ze'sinz.

2
8) Here, f(x) =4x+e™ indicates ), ax+b+ce™.

‘Then, using y',=a+3ce™, y", 9ce™*, we have

LS

yyy =4x+e" = Ice” +Xa+ 300”)
> 2ar+da+20+2000" =4x+e”

>) +Uar+b reo

>2a=4,30+2b=0,200=1>a=

PR
Sy mo

h) Here, +r=0=ær=0r=
f(x) = 4x +e indicates Y, =ac+b+ce”.

Then, using Y,=a+3ce”, y",=9ce”, we have

3. Using Method of Undetermined Coefficients, solve the IVPs and BVPs.
a)y"4y=16x,y(0)=1,y(0)=3 b)y'-2y+y =sinh x, (0) = Lyo se

Solution:
2) Here, r°-4r=0=7=04.Thus, y, =¢, +c,e"". Besides, as f(x)=16x,
y,isof the form y, = x(ax +b). Then, y ,=2ax+b, y",=2a.

So, y"~4y'=2x=>2a-4(2ar+b)=16x=>a=-2,5=-1

Hence, the general solution is y=

re 2x? x

Here, 70) =1,y,0=3> {07 =!
y'0) lisas

Therefore, the solution is y =e ~2x?—x à

>c,=1,0 =0

4. Find a DE whose general solution is y =c, +c,0 + Lx
2

Solution: Here, from y, =c, +c,6*, we can infer the roots 7, =0, 7, =
So, the characteristics equation is (r-0)(r-+1)=0=>r?+r=0,

Hence, the corresponding homogeneous DE is y'ty'=0.
2

Now, puting y, ==> x in y"4y/= 2) we get y'a x.

LN ©

Kia oko Spe Nabe Dees I at ar ae a ga EOD
1.8.4 Method of Variation of Parameters (VPs)

How the method is developed? Why it is so named?

‘The particular solution y, is a pseudo-linear combination of the homogeneous
equation. By a pseudo-linear combination we mean an expression that has the
same form as a linear combination, but the constants in the linear combination
are allowed to depend on x: y,(x)=14(x)y, +u(x)y>.

In the combination y, (x)=,(x)y, +u,(x)y,, the parameters (the constants in
the linear combination) u,and 1, are assumed to be variables. That means the
particular solution is the combination of the fundamental solutions with
variable coefficients. That is why the method is so named.

Suppose y, and y, are fundamental solutions of ay"+by'+y=0.

“Then, we look for a pair of functions u, and 1, that will make the combination
given by y, (x) =u, (x)y, +42(x)y28 solution of ay"+by+cy = F(x)

Putting the values y,(2),Y',(0,y", (0 in ay"+by'+ey = f(x) gives the linear

ui +H mic b >]
un are’ masse Ly, y Ju

Since y, and y, are linearly independent, won]

Hence, solving the above linear system by Cramer's rule, we have

, 0 |
bes | SER, Qe b, HOMO

Oo) WO FO) FO»)
Ñ 10) Ion 4
Henes, by integration, u) =], = Faz st

‘Therefore, the particular solution y, (x) =24(1)y, +4,(9)y, that we are

[fas Pas +}, JO à. LON à

looking for becomes y,(x)=—J1, WO)

CEA <>

(i ik pe ee. ro a
Summary of Procedures to find y, using VPs

(3)
Step-1: Find y, and y, of part ay!"+by'tey'=0 and compute their wr

Objective: To solve the DE ay"+hy+cy =

Skin
, Yo
Thatissolve or +br+c=0 and compute Winys)=[% À
à i yf LOL a
Si Fi ' Ey)
Finally: The General solution ¡sy = y¢ + y, by superposition principle.
Examples:
1. Solve the following NHLDES using Variation of Parameters
a) y"ty=seex DA ©) y'42y4y=e"lnx

x
d) y"#3y'42y=e* cosx e) y"-Sy'iy=x
Solution:

4) Step-1: Find the fundamental solutions y;,
Here, 4103 (—D(r+)=03 = ir,

Hence, the fundamental solutions are y, = cos x,

Dy'2y+2y=e"tanx

20 y "ty

»=sinx.

Then the Wronskian becomes (y, y,) =| 905% sinx] =1
inx cos.

Step-2: Find the particular solution y, using the formula

= cos xInfoos x/+ xsin x
Therefore, the solution is YC Cosx+e, sinx+.c0s.xInoos {+ xsinx

0>r=1. Hence, the fundamental solutions are
ronskian becomes (y, y,)

DET] e:

b) Step-1: Here, r? = 2741 =

+ Ya =xe*and their Wr

Ad pp thema 1 by Bras For yor comme ad sagan we A 3.6262
Step-2: Find y,
y Loe
WO.

yp) =

e fade set] Seide

= far [lac

* Inf] xe"

‘Therefore, the general solution is y = ce" +e,xe" + xe" Infa|—xe"

e) Step-1: Find the fundamental solutions y), y¿0f y"+2y'+y = 0

Thatis r? +2r+1=0>(r+1)

="

Hence, the fundamental solutions are y, =

Then, the wronskian becomes W(y,, y») 4 >

Step-2: Find the particular solution y,

em LO a
LE
[ae er nai de se [CI a
€ TAS

€” [in adr + xe" finxde

Now using by parts, [ln xd = —Inx-

2 4

Hence, y, =e" [nade +xe™ fin ae Homo

and finxde = xinx—x

à
"cede,
2 4

Therefore, the solution is y

d) Step-1: Here, r? +3r+2=0>r,
Hence, the fundamental solutions are y,

‘Then, their Wronskian becomes (y, y)

DRE

Step: Find Yor

(sinx+cosx)

dr - oF
Therefore, the general solution is y= ge" + (in +6052)

e) Step-1: Find the fundamental solutions y,, y, of y"-Sy'#6y =0
‘As we did in part (a), y, =e, y, =e"and Wa".
Step-2: Find the particular solution y,

LOW ey [LOM ef
ee re ee

Therefore, the general solution is y= ce" cet +245,

6°36
D Here, 1? -2r+2=031 =1+i,n,=1-7. Henee, the fundamental solutions
are y, = €" cos x, y, =e" sin and their Wronskian becomes

esx e'sinx
ro gee eine ne
Now, find the particular solution y,

‘at
yy are co) dete! sins snade

2-6" cosx| (sex -co8 dr + sin [sind =e 008 inc x + tan: eh
“Therefore, y= Ce" 0s3-+¢,e" sine" cos nec x + tana}

SS cs

EE Yo CREE TS

Step-2: Find y,

en osx:
A (© dr
LOS ote e >

ef Po ij =

2-07 fxs adc" fe osado

eo

sinx+ (6° sin xe" cos x) = —— (sin x +008 x)

sina cosa)

Therefore, the general solution is y = a+

9 Step-1: Find the fundamental solutions y, y2of y"-5y'+6y =0
‘As we did in part (a), y, =e", y, =e"and WO, »:)= €". |
Step-2: Find the particular solution y,,

y LO tee yf LOM eee i dere j= LE

Le

ee 70er
i

[xe dec [ede EZ

3
ERWERBEN
31792476 36

Therefore, the general solution is y = ce " +c,0?* +.

D Here, F-24202 =I+ in

1. Hence, the fundamental solutions
are y, = 6" 006, y, =e" sin.xand their Wronskian becomes
ef cosx e'sinz
WO) =|
vn Be ee] e.
Now, find the particular solution y, |
A |
ye] de sinzfsinzde |

=-e"cosxf(secx-cos-mjdr-+e" sin Sin ad =e" cos xInboc + tan x}
Therefore, y= ce" c0sx+c,¢" sinx—e" cos xinfsecx+ tanz]

\ En Ga at Lt

2, Solve the following DEs using Variation of Parameters

esc3x 6) y"-2y"42y=8e"sinx c)y"-9)
tanx DN yy = cote

yidy=xte’ i) ty =axsinx

sinh2x e)y"+y

Solution:
a) Step-1: Find the fundamental solutions y,, y¿0f y"+9y=0

Here, the characteristics equation is +* +9=0. Solving this gives us

7492025 ="
Hence, the fundamental solutions are y, =

cos3x in 5,
|-3sin3x 3c0s3x| ”

(cos 3x, y, =sin 3x and their

Wronskian becomes HO, y:

Step-2: Find the particular solution y,

¡LO e+ yf LO
WO) 7057)

i on

BAG

=~ feos def sin foot 3aae= roos3+ sin 3rInfsin3x]

os ressens anden]

Therefore, y
b) Here, the characteristics equation is r? —2r+2= 0. Solving this gives us
P-2r+2=0>1 =1+i,n, =1-i. Hence, the fundamental solutions are

=e" cos, y, =e" sin xand their Wronskian becomes

TE a eat |,
* cosx—e*sinx e” sinx+e* cos.

Now, find the particular solution y,

»

A A ES

AAA METAS DA cio e PER
yyy LOL a LW: as y, [La SOM qe
AT?) HO)

= 6" cosx[Bsin? xd +e" sinx{ sin xc0s xd

=e" cosx (l-cos2x)dr+-4e sin [sin 2ade

F cos x(4x—2sin 22) 2e" sin x00s 2x

Therefore, y= ce” cos.x-+c,e" sin xe cos x(4x—2sin 2x) —2e" sin.xcos 2x
01, =3,1,=-3. Hence, the fundamental solutions are
and W(y,, y) =-6. Now, find y,.

9 Here, r-
wre y=

Sey, Sen
sn ,
jo

=a tere en

8) Step-l: Here, r?-1=0>n =1,7, =
Hence, the fundamental solutions are y =e", y=e™and their Wronskian

anf

Step-2: Find y,. Observe that f(x) =sinh 2x =

Sy,
tr
== "i MAGO)

Eis ok ped Nahe hig, Toy aan a ap RE
Therefore, the general solution is y= ce" +0,” +sinh2x
Step-1: Find the fundamental solutions yı,y,0f y"4y =0

Her, the characteristics equation is r? +7=0. Solving this gives us
Par=0>2 (Mt) =037 =>

Hence, Y, = 005 x, y, =sinxand (y, y») ese me.
Step-2: Find the particular solution y,

LO ey LO ae
7e) wo.) ‘Ti, Word”

= nn cd

| d+ sin sin d= cos x

de-sinxcosx

cots ons esse
= cos x( {cos ade — sec xd) — sin cos x = cos xInlsecx + tan x
‘Therefore, the general solution is y= 6 C05x +c, sin x—cos x nsec x + tan a|
f) Here, r? +1=0=>7, =-i,n, =i. Hence, the fundamental solutions are
R osx sinz|
LA Ten
y

cos x cot xsin xx + sin x cot x 006 ae

= 008 xf cos xde + sin xf

sinx
=—cos sin x +sin x{ (esex—sin.x)dr, cos? x=1-sin’x

== cos xsin x sin xinfesex + ot | +sin x cos x = —sin xinfesex+ cot x
Therefore, y=c,05x+0,sinz-sinzinlsex+eota]

D Step-1: Here, 7? + 4r +4 = 02 (+2) =0>n=n=-2.

Hence, the fundamental solutions are y, =e", y, = xe

En ee PERERA AO]

Step-2: Find the particular solution y,,
e) fee
¡[A dre I

(xe! et) +20 O) =e

=P fred ae fede
Therefore, the general solution is y= ¥. +), = GE
h) Step-1: Find the fundamental solutions y,, y; of y'-4y#y = 0.
Here, r* -4r+4=0>(r-2) =0>r = 2(we have repeated roots).
So. the fundamental solutions are y, , ya = xe™*and their Wronskian

= a

becomes wo re

Step-2: Find the particular solution y,
IO LO Ea x
=», 2d = Pa

WOM e, be OO he
= a [werd = (x7 +4x46)e*

‘Therefore, the general solution is y = ce” +c,xe™ +(x° +4x+6)e".

3. Solve the following IVPs using Variation of parameters.

à) y'-Sy+4y =e, (0) =1, (0) =0

b) y"-2y'+y =e" sin x, y(0) =3, y'(0)=0

©) Y"Ry+Sy = €" sin x, (0) =0,y'(0)

Solution:

ae Find te fundamental solutions Ir Of y"-Sy4y =O

à characteristics equation is 1? -5r+4=0. Solving this gives us

POS (44-2027 24, =1,

Hence, y, =e", y, =e"and Po e|

+ +e”,

hig Ma atu ed TED

Step-2: Find the particular solution y,

“Therefore, the general solution is y= y, + y, =0,0% +c,e' -—
Now, let's determine the constants cjand c, using the initial conditions.

1 3
O= 134 +0-3=130 +0 =5

(0) =0>4c, +c, -1=0>3 4c, +0, =1 623

Hence, the solution of the IVP becomes y

b) Step-1: Here, r? -2r+1=0>(r- Ve

Hence, y, =e", y, =xe"and WO,y2)=|,

€ +e"

Step-2: Find the particular solution y,
in x{xe*) spe’ sinx(e"
arte (RO

ef;

aoe" [sin de + xe" [sin xd

=" (sinx—x00sx)- xe" cosx=-e"sinx
Therefore, the general solution is y=y,+y, =e" +cp1e* -e'sinx.
Now, let's determine the constants c,and c,using the initial conditions.,

()=3>< =

edad mí Aia

Hence, the solution of the IVP becomes y =3e* -2xe* ~e" sin x.
©) Step-1: Find the fundamental solutions y,, of y"+2y+5y =0
Here, À 42r45= 027 =-1+2, 1-2.

Hence, the fundamental solutions are y,

Step-2: Find the particular solution y,
y... cos2xf a sin 28) de 4 e° sin2xf

sin xsin 2e + £58 sin xe0s2ade (Completeit)

0s 2

2 -

4. Solve the following NHLDES using Variation of parameters |
2e nr: i e
Dry DI €) y'-3y#2y; Dr
Solutior
a) Step-1: Here, r?-4r+4=0=(r-2) =0>r=

. Hence, the fundamental

an
solutions are y, =e", y, = xe*and W(y,y:) LE

Step-2: Find the particular solution y,
ae ([ 26% Y x0* aff 20% er
noe (Saree ae
wf 2x ap_2 =
e tre E +1) + 2xe* tan tx
‘Therefore, the solution is y = cie +.c,xe™ —e? In(x? +1) + 2xe™ tan” x.
D Step-t: Here, r*-1=0=>7 ml, 1, =-1. Hence, the fundamental solutions

e, y =e*and nf E
-e
Step-2: Find the particular solution y,

[Gare (55) La

are y,

E O
«Ej pe
let je

nd bi of Spied Matematica by rs M, Forja tment ad a CESS
1.9 System of First Order Linear Differential Equations

Revision on Matrix:
i A art+by=k,
System of Linear Equations: Consider a 2x2system:
xt dy=k,

We know that such system of linear equations can be written or expressed in

a b k,
\B=| "|.
ea) lk
Eigen-values and Eigen-vectors:
From the matrix representation, we have also seen how to determine the eigen=
values and the corresponding eigen-vectors associated with the coefficient

matrix A by forming its characteristics equation.
Characteristics Equation: det(A — AI) = 0 where I is identity matrix.

Eigen-values: The solution of the characteristics equation det(A — Al
Eigen-vectors: The vector with the property (A — AI)v =0.
These algebraic concepts are the basis for analysis of systems of DE.

matrix form as AX=B where X e Ja
=

0.

1.12.1 Homogeneous Systems with Constant Coefficients

A system of linear differential equation is a-system of equation involving the
derivative of two or more variables about the same input parameter, usually
denoted by r. Here under, we are going to see a system of two variables x and
y about the same parameter £.

Notations: Derivatives wit respect to 1: &
#=ax+by
or
Heard

In matrix Form or in vector:

X'=AX or I) X=!

Form of the system: {

CELA i E
Fundamental solutions: In general, like that of second order DE, such systems
have two fundamental solutions X,and Xy. But the form of the fundamen]
solutions X,and X, depends on the nature of the roots of the characteristic,
equation det(A = AT) =0 of the system. So, our next task is how to find such
fundamental solutions using eigenvector method.

Since the eigenvalues are the roots of det(A~AI)=0, they could be two
distinct real roots, single root or complex roots. That is ifthe coefficient matrix

of the system is af, der

Oe ee à à )-9=#-@ra1ra1-%=0

d-À

= A -trace(A)1+det A =0 (Note:trace(A)=a+d)
Now, consider the three cases for this quadratic equation.
Case-I: Two distinct real roots À # 4,.
‘Then, the fundamental solutions are X, = ve”, X, = v,e#.

X, mv +, )e% where v, is
a vector to be determined from the condition(A—AI)y, = v,.
(Case-III: Complex conjugate roots 4 =a AA.

M nr ar E

0
el
Now, identify the real and Imaginary parts of v= v'e|
añ { X, =Re(v)=Re[v'e* (cos A+ sin A)]
X = Im(v)=Imy‘e" (eos e+ isin A))
Complementary Solution: Therefore, X, =c,X, +c,X,.

‘That is (A-2D)v" -0>-a(¿)-

(A +isin A).

or LE he a ani REED
Eugenvalue Approach to solve system of DE:

The process of solving system of linear differential equations using the
characteristics equation is known as eigenvalue or matrix approach:
Procedures:

First: Find the eigenvalues of the coeffici

1 matrix A using|A~ Al] =0

Second: Find the eigenvectors corresponding to each eigenvalue. But be careful
to analyze the three cases of the forms of the eigenvalues.

Third: Find the fundamental solutions X, and X, and write the complementary
solution X, Thatis X, =0,X, +c,X,.

Examples:

1. Find the general solution for the systems of differential equations.

‘a3x+2y ¥o3rty 2x43y 1-1
b) d) X'=| IX
“fe fie, OE, al;

Solution:
a) This is an example with two distinct real roots,
First, identify the coefficient matrix A and find its exigent-values.

3 2
‘The coefficient matrix is af} ?}:

analzo=f 7 Joao =0

> (12-32-44+2)-2=0
>32 -74+10=0>(2-24-5)=0
>24=2,4=5

a
Now, let's determine the corresponding eigen vectors. Let Y «(2

1 2Ya)_(0)_ fa+2=0
Drea=2,(4-ane=0= DO aan 2

2 (-2
Hence, letting b = 1, we get the basis vector to be el » > 1 ).

A 0 (-2a+2b=0
Drwa=5, (4-02 E CES >a=b
AA

E pd Waa he WF ae ad
1 à

Hence, letting a=1, we get the basis vector to be v,
‘Therefore, the complementary solution for the homogeneous system js |
of?) ae (eral |

X, zone tw =cCl "Je tee

b) This is an example with single repeated real root.

301
The coefficient matrix is a{ 1 ) 1

0=(3-2X1-2)+1=

B-a
la-aj=o>)| Ñ

>0-32-4+4')+1=0
A -44+4=0= (4-2) =0>4=2

Now, let's determine the corresponding eigen vectors. Let V -()
al” 11210). [jat2=0 5
1 alo) lo)” La-ó=0>*
Z -b -T
+ we get the basis vector to be v, ={ y |=| |.

We need with the condition that (A—Al)v, = v,. Let y, el

1 1Ye E

A Of ı )Perd=-1>e=-1-0

=0, we get the basis vector, (596)
d vr

‘Therefore, the complementary solution for the homogeneous system is

a wel bt eel bal N
same" elegi we =al fe reall y b+ y Je
EER 75

Ford =2, (4A

Hence, letting b

Question: How do we get the second basis vector v, ?

Hence, letting

x,

a ped Mat che For or omens ios RA
¿y This is an example with complex conjugate roots
p-a 3|
3 2-4
4-24-2048 +9=0
SR -4413=0>4=

|A-al]=0> 0>(2-AX2-4)+9=0

Ex}
Here, we need how to form fundamental solutions for such complex roots.

Now, let's determine the corresponding eigen vectors. Let y

0) _ [-3ai+36=0_,
Lo) \-32-35i=0

Hence, letting a=-i, we getb=ai

So, the complex basis vector becomes v" 6-0)

‘Now, identify the real and imaginary parts of v=v’e” (cos ft +isin fi).
Here, 2=243/30=2, B=3.Then, v=v'e™(cos3+isin3¢).

reve (csr eins ("ris
El ie” cos3r+e"sin3r))_(e” sin3r) E —ie* cos 3
e* cos3t+ie” sin3r) ) (e%cos3r) \ ie” sin3r)
* sin3t)) [—e* cos3r
FA >.
e*cos3t) Le” sin3r)
EE el
pt ET)
(e* sind) —e cos3r
= Im(v) =
Gay nn e*sin3r)
‘Therefore, the complementary solution for the homogeneous system is

x e sin) e? cos3r
o oa oe | ot gin)

Here, we got v, = Re(v):

ERA 76

nd sugar

ii ik ped a za cm
ate roots.

4) This is an example with complex conjug
First, identify the coefficient matrix A and find its €

la-all=0>) 5 „se MIA

>(1-2-24+4)+1=

> -2042

Here, we need how to form fundamental solutions for such complex roots.
(a)

Now, let's determine the corresponding eigen vectors. Let V=| , |

) Ford =1+i,(A—Alv=0 fee „feinem,
Ford =14i,(A-Av=0=>/ pe Lo)" la-bi=0 7"

Hence, letting à

‚we getb = ai

So, the complex basis vector becomes v” = {) = ( )
)
Now, identify the real and imaginary parts of v=v'e (cos Ar +isin A).

That is 4=1£i=>a=1, B=1.Then, v=ve (cost +ising).

y veto ssn [csr sising

fe! cost—e'sint) (-e'sin) fi ,
0 cost=e' sind) | „fie'cose)_f-e’ sing)
'cost-+ie'sint) | e' cos sine

e'cost
ET

sing a
eh Im mmor(: 2)

\

nd if pid Mathew she Ne K
1.12.1 Non-homogeneous Systems with constant coefficients

A system of differential equation of the form X'= AX +g(0)is said to be non-

function g(t)#0 is said to be forcing or

homogeneous when g(t)#0. The
AXis said to be the

AX+git), the part X

input funcion. In the system X
corresponding homogeneous part

+ by + (0
x+dy+g2(0)

General form of non-homogeneous system: | E

In matrix or in vector Form:
he above non homogeneous system can be written in matrix oF in vector form

x ab E
2 AX+ g(t) where X'=|*, LA x=[" Lee
as follow: X'= AX+B(0) where X () (: al E ) (a
6)
e dy) (al,
How to solve such non-homogeneous system?
Given: Suppose you a system is given in one of the notations or forms.

{ Hy q FAIRE ox, ( Y „ao
yeatdtg) lar e dhy) la,

Objective: The main objective is to solve this system.
Fundamental solutions: The solutions of the corresponding homogeneous part

which is xo axen(” A ll
y) Ne ayy,

Then, the fundamental solutions are X, =we*, X.

It can also be expressed in the form X'=|

vie.
‘Complementary (General) Solution: X, = ¢,X, +c,X,
General Solution of the non-homogeneous system:

The general solution of the system X'= AX+g()is X = X, + X, . Here, X, is the

complementary solution of the homogeneous part and X, is any particular
solution of the non-homogeneous.

ECR] 7e

ssp a eT ONE FE |
tation using X =X. + N,?

the challenge is how tc

asd Bo fr!

is the challenge to express the sol
ne Kis

sed how to determi
= AX+8(0)

Since we have discu
¢ Eigenvalue APF

In general, to solve the system x
First: Find X, of X'=AX How? Us
second: Find X, of X'=AX+8Ú)
sion: How to determine X, ?
0 fin

ach,

Quest

There ar een approaches o Find panicuar solution X, +

1. Method of Undetermined Coeiciens

Variation of Parameter

Laplace Transform method

Fourier Transform method

‘The Operator Method

Diagonalizaion method.

7. Power series Method

1. Method of Undetermined Coefficients

nd mois in the sume way as we used 10 solve non-homogeneolf

second order differential equation.

e csemprio: Assume te for ofthe particular solution X, based onthe

form of ge) That means make educated guess forX,

Second: Substitute X,and X; inthe system to determine constants.

Hints: To make educated guess, always notice the folowing points.

à) Since the form of g(is 800 = 0 ;

) of Wis fü rte frm af X,isalo X, = 6)
2

Pap

That means guess x, for g,(£)and y, for 8,()-
Some ral forms: As example consider the following tral forms.
D PME the guess will be X, -(“)
be leal
m (2) then the guess will be X, =| At+B
a+d) x (a pel

Pe. a 0]

Cai Wf pied Wales U by Boise fo ot ra ad gg we VEZ
(asinks + bcoskr) (

el o ke Bcoskr
L #0 | sinks decora) Po Ko" Conte + Deosker}
be à

men | D° |, the guess will be x, =| (4°+ De
(ct+d)e' "(Cr+ Dje*

sal Ta ky)

vw) g(t) =| 2%, | then the guess will be x, =| 42°” +20
bet "car 4 De

M (Ae® \
ve) so ] sen es ite x rente
bt+c PAD EnF

Notice: IN addition to these hints, modification rule also works here in a more

generalized way. That means the trial form of X,may not work directly. In

such case, we use the rule depending on the situation.

Examples:

1. Find the general solution for the following systems of differential equations.
vaxty+l0cost

{ y=3x-y-10sint

Solution:
a) First: Determine the complementary solution using eigenvalue approach.

The coefficient matrix is A (i >

la-a1=0>

I? ec ¿roza=aci-a-
>(-1-4+4+2)-8=0
> P-9=0>2=921=33
Here, we need how to form fundamental solutions for such complex roots.

a
Now, let's determine the corresponding eigen vectors. Let V = (})

ES

ind Bok ppl eet ari "oc ns 04

. a4 Ge
prora=3,(4-ar=02[ > aly Lo

24

a-45=0

Hence, leting b=1, we geta=

a) (4
So, the complex basis vector becomes Y, = - ( i}

48 0)_, féa+8=0 ,,
El = =%
i)ForA=-3,(4-Av=0=|, 5 4 lazo

2.

Hence, letting b= I, nn
2

So, the complex basis vector becomes v, =| GH )

‘Therefore, the complementary solution for the homogeneous system is

ee

Second: Determine the particular solution X, using MUCs.

x
X,=quje¥ ton,

or
Since g() (i a linear polynomial in both components, X, is also a linear

at+b a
I; jial. That is X, =| A =
voran =) x).

vext8y+9_ [x=ar+b+8(a+d)+9r=a
yy +9 U+b=c=d+9=0

(a+8c+9N+b+8d=a_ [ 0+80+9=0,b+8d=a
(a-c+9)+b-d=c ~ | a-c+9=0,b-d=e

jane
> =
a-c+9
brsd=a_ [b+8d=9_[b+8d=9
>) b-d=e |b-d= beds u
91-1
Hence, X,=[~_) |

of the non-homogeneous system is

the general solut

a

INES
1 a
ty First Determine the complementary solution using eigenvalue approach.

. Li 0 2
‘The coefficient matrix is A= 2 0)

A 2
paja, Joma =021=4>1=22
Here, we need how to form fundamental solutions for such complex roots.

Now, let’s determine the corresponding eigen vectors. Let Y -())

-2 2Ya)] (0 -2a+2b=0
paa? A

Hence, letting a =1, we getb =.

So, the complex basis vector becomes v, =|
2 2Ya\_(0\_ {2a+2b=0

ii =-2,(A-. = = =

ii) Ford =-2, (AAD sf À) (Races a

Hence, letting a=-1, we getb=-a=1.

E
So, the complex basis vector becomes v, (') -( : )
Therefore, the complementary solution for the homogeneous system is
i 1 <P
= pa
CG}

Second: Determine the particular solution X, using MUCs.

X,=0 vet +c,v,e

. 4-5
Since so. a Je a linear polynomial in both components, X,is also a

a at +b
linear polynomial, That isX, a =) Then, x,-(").

niki ted ea Pe N" For your commen and supgesins Dr
yes fa=2a+2d+di-5=4 (2c+4)+2d-5=a
, (2a-6)r+2b=

|
le

Hence, the particular solution becomes X, (3 2)
‘Therefor, the general solution of the non-homogeneous system iS
X=X,+X were ve" sof (el a i
+X, =0 e +evie* sal; +E y ah
«e complementary solution using eigenvalue approach.

e) First: Determine th

The coefficient matrix is af aj

la-a]= 07 1 pore -4%-1-4)-3=0

Prom anaes
28-4202 7 =4>45:
Here, we need how to form fundamental solutions for such complex roots.

Now, let's determine the corresponding eigen vectors. Let V -())
5

-1 1Y2 0 (=,
D Ford =2, (A-Alv=0: A a+b=0
dE (5 e) 3»

Hence, letting a=1, we getb=a=1.
So, the complex basis vector becomes y

3 1a’
Ford. =-2,(A-Al)v=0: nl
area carol Je
1, wegetó=-3a=3.

Hence, leting a
So, the complex basis vector becomes Y, = -()- Gi
3

Tia Bf ped Nena eg Ft ar oma nl a ED

‘Therefore, the complementary solution for the homogeneous system is

eer

‘Second: Determine the particular solution X,,.

Xx = amet +ev.e"

linear polynomial. That is X, cl

10cost
Since so-( ii one a linear polynomial in both components, X, is also a
esint +d cost,

(x= x+y+l0cost

en, X, 0
B
paix y-10sint

> acost—bsint = asin + bcost +csint +dcost +10cost
ccost—dsint = (asin! + bcost)—csint -deost -10sinr
+d+10

(-b=b+d+10+35-d (-55=10
-d=3b+d+10)-@b-d)- 107 |-4=44+20

=3b-d e=-2
> > >
la=b+d+10" la=4

ee |

Hence, the particular solution becomes X, ( ndaradoner)'

Therefore, the general solution of the non-homogeneous system is
u 1 Ma, ,( 4sint=20088

x=x.+x,=0( ref) a(t ace

4) First: Determine the complementary solution using eigenvalue

The coefficient matrix is “|, At

AAA ee

(Cid Woo ofp Mesa Weges M = ForFour kommenis and sages ae y
7

IS

la-a=0>| a.
ai

= (-24+#)-4=0 (
> -24-3=02(2+ 0-3) =0>34=-1,3
Here, we need how to form fundamental solutions for such complex roots ,
Now, let's determine the corresponding eigen vectors. Let Y = le |
)

i) Fora

2 IYa 0 2a+b=
a |

Hence, letting a=-1, we geta=-2b=2.
a)_f-1
os bé rome) (2!)
of? *Y2)-(°)> a ee
a 2/5) lo) [4a Bo 7
Hence, letting a=1, we getb = 2a=2.
a)_(1
Se. comporte veo tcomers =(2)=(*).

. complementary solution for the homogeneous system is
nn fils
Er-()

Determine the particular solution X,.

i) Fora =3, (4-AN»=
“

Therefore, th

x, =0 ve reve

Second:

85

OR

=0>0-200-4)-4=0

1-2
ja-aj=o=| 4

i >(1-24+2)-4=0
2-21-3202 (A+I(A-3)= 0 4=-13
Here, we need how to form fundamental solutions for such complex roots.

a)
Now, let's determine the corresponding cigen vectors. Let V { )

2 1Ya) (0) f2a+b=0 _,_,
Drwa== dam 0(, IS PARE 2a

Hence, letting a=-1, we geta=-2b=2.
6

e m
-2 1 Ya) (0). [2
i Ford =3, (4-4N)y=0 =|
al RES
Hence, letting a=1, we getb=20=2.
Se compl veer onen, +2) (7)
2

So, the complex basis vector becomes v,

s
‘Therefore, the complementary solution for the homogeneous system is

ane ae

Second: Determine the particular solution X,

2. Variation of Parameters

ax+by+e\(,

Considera non-homogeneous linear system: {
Poets

In matrix Form:
‘The above system of equations can be written in matrix or in vector form as

[ al? }-(J}a0 (60)

It can also be expressed in the form X'=! (3) = () El (: a (s 2)
2) AY) le dhy) "lao,

What are we going to do?

Given: Suppose you a system is given in one of the notations or forms.

{ thy +a) { Fractby+e (0, ee xl LE)
arts) | fscr+dy+g,) e dA») (0

Objective: The main objective is to solve this system.

Fundamental solutions: The solutions of the corresponding homogeneous part

Which is xaaxer(?)-(2 DB
y) Ke dA»,

Then, the fundamental solutions are X, = vie”, X, =v,e*.

Complementary (General) Solution: X, =c,X, +0,X,

Particular solution: The general solution of the non-homogeneous is of the
fom X=X,+X,. Here, X, is the complementary solution and X, is any

follow: X= AX+g(t) where

=

Particular solution to the non-homogencous system.

Since we have discussed how to get X, , the challenge is how to get X,.
Let's see how we can apply variation of Parameters.

Procedures to use Variation of Parameters effectively:

First: Determination of fundamental solutions

Using eigenvalue method, determine X, and X, of the homogeneous part.

weal} al)

That is solve the homogeneous part: X'

a AT A aS
2, Variation of Parameters

Consider a non homogeneous linear system:

E ax+by+e,()
yeardy+s,)

In matrix Form:
“The above system of equations canbe writen in matrix or in vector form as

follow: X= AX+ g(t) where x-(F}a- ¢ ‘x (ro. (+)
uoinabobeerresdinte mx) -(*)-[* ach (60)

What are we going to do?
Given: Suppose you a system is given in one of the notations or forms.

{ eaxtby+gi(t) «($ Lay fa 19

+” | gar MO)
Objectives The main objective is to solve this system.
‘Fundamental solutions: The solutions of the corresponding homogeneous part

pel

‘Then, the fundamental solutions are X, = vie*, X, = ze”.

Complementary (General) Solution: X, =0,X, +0,X,

Particular solution: The general solution of the non-homogeneous is of the
form X=X,+X,. Here, X, is the complementary solution and X, is any
particular solution to the non-homogeneous system.

Since we have discussed how to get X, the challenge is how to get X,.

Let's see how we can apply variation of Parameters.

Procedures to use Variation of Parameters effectively:

First: Determination of fundamental solutions

Using eigenvalue method, determine X, and X, of the homogencous part.

‘That is solve the homogeneous part: X'= AXor| e Lh

AAA i

solution matrix and computing its inverse

Second: Formation

Form the solution matrix M by using X,25 first column and X25 say

column of M and compute its inverse M*. Use the formula for the inverse
x a |” x

2x2 matrix. That is M = (X, xo Jan “alo a

‘Third: Determination of the particular solution.
“Apply the formula X, = M[M* 804 where Ut) is the forcing function inte

given system of differential equations.

#
#
#
ss
ya
#
1
nr
Hen

Examples:
1. Find the general solution for the systems of differential equations. =
E a ,
a à pxe(! Tx [ es
ya=r+ayme 11 le sins
ox-(} 82e Pete
1-1)" late! yout!
Solution:

2) First: Determination of fundamental solutions using eigenvalue method.

E

Thecoefcient matixis 4=[ 9. 7
13

poentonayeactos atraco

= (A-I\A-2)=0 4, =1,4 =2
Now, let's determine the corresponding eigen vectors. Let Y = OF
5
Dre, Dr 05 (Ts 2V2)_(0).., {-2+2b=0
a AJ lose

Hence, letting b=1 in a= 25 = 2, we get the basis vector v, -() E
vi).

ford =2a-ano=0>(—; 1-0 | {-2+25=0
-1 1 0, 0 >a=b

a+

solution isX,
Second: Formation of solution matrix and computing its inverse.

Using, = fe = first column and X, (

. od: E gr
matrix is M = (X, se E maaan nae É ) e.
€

)» second column, the solution

ee, e
the, ml © El e -#) fe -e
deM\-y x e de) ler 2%)

Third: Determination of X, using the formula X, = M.[M".g(0dt.

Here, g()isthe forcing function in the given system which g() { e ):
-e
un fe elf et E
Hence, X, = M[M aoa-f Al “hf 3 EXC
fe af 2 V2 A fase
Le elle Te à fe) Lee
‘ate! +3e")_(4) (3
‘Thus, X, =| =| 4 .
’ (ut) ( -(}
‘Therefore, the general solution ofthe non-homogencous system is

rox ota Foo (C4

+) First: Determination of fundamental solutions using eigenvalue method.
x par armonia
T ie

yas Fok oa on a ar ag
a)_ ae ai-b=0
») lo)” la-bi=o ts
Hence, letting a =, we getb = -ai

._fa)_fi
So, the complex basis vector becomes Y -( ) «() h

i) ForA=1+i, (4-41)

Now, identify the real and imaginary parts of v= vie” (cos ft +isin pr).
That is A=lti>a=l, ‚Then, v= v'e' (cost + isint)
le! cost-e' sin »)

i à
ve, isint)= | t+isint) =

’sint)) fie’ cost) _(—e'sint) (oe)
e cose ) lie'sino)) \ e’cose ) Le sine),
”

Hence, we have X, VE wi = Im(v)= a

Second: Formation of solution matrix and computing its inverse.
—e sint e'cost

ele
cost e'sint

sinn —e'cost)_(—e"sint cost
lee cost -e'sint) (e*coss e*sine)
Third: Determination of X, using the formula X, = M.[M‘.g(0)dr.

Here, g()is the forcing function in the given system which u a

‘That is M=(X, X, land det M =-e*.

Thus, M' =

é'cost e'sint Pl e* cost e”sins

(a econ, _(—e'sint e'cosıYo
ecos e'sint LIS | ercosı esi it

sine At,
Therefore, the general solution of the non-homogeneous system is
X=X,+X,=

9) ef e ser) (cose
cost "e sinty +

x, emo [ns ze elas dei

|
e) First: Determination of fundamental solutions using eigenvalue method.
1-2 Les

Now, let's determirie Ihe corresponding eigen vectors. Let Y = 8)
N y 6,

a -2 8Ya 0 -2a+8b=0
y, Ford, =3,(4-AN)v=0= E =
DForA ) E >.) (del sant

¥Hence, letting 6 =1 in

a 4 8Ya) (0 4a+8b=0
yaa {fa

Hence, letting b=1, we get the basis vector v, = 04)

4 A
So, the fundamental solutions are X, = ve (ik. ¿Xy = vet «( qe

Second: Formation of solution matrix and computing its inverse.
pe a

ex (1 Ju first column and (E Je second ecians, the
€

, We get the basis vector v,

(de 5

De" he
solution matrix is M { ee oe ) seas = ae A
e,

CA (te*
aM u à] (o ae Jo

Third: Determination of X, using he formula X, =M.[M*g(Qát.

de’

| 4 a
sent CC
be EC)
e e poe +168"

Es Ale te >)

ee” eX "+4" et
(res) | -16te' -2e' )

je" +te' je! +(-e'+e'-e)) (de +Ste'-3e
‘Therefore, the general solution of the non-homogeneous system is

4 -2) , —16te! -28
ed)

4) First: Determination of fundamental solutions using eigenvalue method.
la-aj=o=[+ ¿Joni nmomama

X,=

Here, we need how to form fundamental solutions for such complex roots.
Now, let's determine the corresponding eigen vectors. Let V =| 7

| D Ford =i, (4 ADv= 02(7; NI Ea

Hence, leting a=-i, we gotó=-ai=-P =1,

So, the complex basis vector becomes v' ()-( )

Now, identify the real and imaginary parts of v= v'e (cos + isin 2).
That is 4=i>a=0, Bel.

y o Hoost+sinf) (sine) (-oosr
vo[fowrstenn- (. an)" HE)
Hence, we have X, =. 20 ON da J encuen

solutions.

\

[ied Bek f ppd Mathematics Boga TE
Second: Formation of solution matrix and compu

. sing cos
Thatis M=(X, e Justa.
D

sint cost
cost sine

Determination of X, using the formula X, = M.[M* g(*)dt.

CRT

Ti

Here, g(()is the forcing function in the given system which 8() = 0

x, 2 ai
cost sing J?\-coss sins

sine cos” \{ sint+teost

Lin re)

sint -cosi\f tsint F

de ol
‘Therefore, the general solution of the non-homogeneous system is

use)

2. Find the general solution for the systems of differential equations.
fee N [ae
y =5x+6y-6é Y =x+ y -cost+sint
Solution;
a) Firs, find the complementary solution X,of the homogeneous system.

314)
Thats ind the sohution of | "7
y=5x+6y

f A onen

5 6.

> 18-34+64-4')+20=0
> -34+2=0>(2-IX4-2)=0
>4=14=2

fg Bas Ty Fab. pr comments is |
Review Problems on Chapter-1

1. Determine the order nand degreed (if defined) of the following! DEs.

| ey

a

| a 2) ==

atyy"ayy =
| pyy?
d
i (8) +008(3y)=1
Answer:a)n=4,d=6
d)n=3,d=38
g)n=3, h)n=2,Nodegree i)
2*. Show that y=tan”' xis a solution of the DE y"+2sin ycos? y=0.
3. In each of the following, find the constants k anda.

2) Ify=e* is the solution of the DE Loy =0, then find #

,

b) Ify= x is the solution of the DE sd -ay=x?,then find a.

©) If y =x" +kx+1 is the solution of the DE y"+xy"-2y’=0, then find k.
4) Ify= x" is the solution of the DE 16x*y"424:y"+y = 0, then find 4.
Answer:a) k=9 Ba=IS ©) k=3 d) k=/4
4*. Verify thatthe following DES are homogeneous and solve them.

a) dy = (y+ stan), 6) dy = (y+ xe0t2)ae y
E ‘

in? dy =(x+ ysinZ mL rt 2
a) có + yla yde e) = y ind 4

Pr

ser

wt kf pid Males Bega Heya cor a nis i OE
5, Verify that the following DES are homogencous and solve the IVPs il

Or tj + (2x7 + )dy = 0, y(2)= -3
pro) EU

Answer:a) 2tan (9/3) =In( +3) b)y
ayy =3e 1h € xytexty=24 —f)sin(y/x)=Inls

6. Verify that the following DES are exact and solve them.
Array D) ade + yey = (+)
€) ede + xe" dy =0 d) 2xpdx +(1+ x )d =0
e) (es y +3x*)de +e" cos dy = 0 N (et +2x)dx+edy = 0
Dr Mader = 0

Answer: a) 3xy+2x° =c byin(x?+y*)-2yse c)e™ =e
diyry=e ee sinyp+x =c. Pré+é=c grrr
2. Verfy that the DEs are not exact and solve them.

a) (0 -2y? )dx+2xydy = D) (xy + 4ay + 2y)de +(x? + x)dy = 0
= Day ++ (ay? -x)d = 0
90) +2x*)de + dy = 0 No + )de+O-xy)d = 0
DO +2x*)de—x(1+ ay)dy=0 h) 2 +7 - ay =0
Dil =0 DIE +327 HD =0

2
Agwer:a)x+2-=c b)(xtHelry=e
x

e

2
dxy -3y=0 +

x

2
2

92m: LG nc pre Daze Deora

A 5

NETA AA

miso that the Following DES are exact
> Lees ial

$, Find the constants a,
D pes tar +42) 20 yd + (xy
€) area
Answer:a)a=? bym=4n=3 D] d)m
9. Solve the following Bernoulli's Diferential Equations
yyy mA yy 7073
Jay ay nna!

+770 nia
Bat D Zeya" Dy+y=7 70 1

1

1
ye De”
za Dr
10. Solve the following DES using Methods ‘af Undetermined coefficients
a) y"Ty'#l0y = 24e" D) y "ay ray = 2e™ +4212

d) y"-2y'ty =e" sin

ap ze sin 2x
Bee NIAYy= e sinx

gry= 1120 D y"-2ÿ+7=2e"

Dry +7 = 2xsinx DIA

Da m) yy 2y=2xe"
Dynqet eye ext 44-2

raya He” +66

Answe
a eer
pala tenet tern: Dylan eis
yet re” prete ein
1 hy y=qet +c,xe' +xX’e

grat rate

or you oem ad sario we (9088342
11, Solve the following DEs using Variation of Parameters

b) y"-4y'edy = = ©) 4y"+36y = oschr

ay" ayy

pyty=snx e) Ay y =(e+ e” f) y"ey=scextanx

e

py'nyreoshs yy

Answer:a) y(x)=c/0* +c,xe" —(In]xf+ Ne” 5) y(x)=(c, + gear

9 y=420s3+0,sin38— cos + sin ofi df

se . per
d) yrccosxte,sinx—Fe0sx e) y= rc E +

y =3008x+c,sinx+xcosx—sin.rinfosx|

Dre ta E

12. Find the value(s) of the constantk for which the DE y"+6y+ky=Ohas a
genera solution of the form y=(c, +c¿2)e””. Answer :k =9

13%. Fnd a and k for which the DE y"+ay+ky=0has a general solution of
thefom y= (c, cos2x+c, sin2x)e™. Answer:a=6, k=
14. Ling the appropriate methods, find the particular solution
aytytyne’+4 —b)y"-y=sinhx 0) y"42y'ty=e™Inx

4 ray dy 2yHy=etmx f) ya
2"2y+y=4x 7-3 h) y"43y+2y=sine”

aswer:a) y, =e" 8) y, =Foosh Oy, = doit

2

de nr dy, =e" cos.xin(secx + tan 2)

dy, = Het int+e*)-e in +e")=1]
Dy,=42 416x421 hy, =-e™ sine?

fee 1, he SS |

CHAPTER-2

Laplace Transform and its Applications

2.1 Definition and Examples of Laplace Transform

function defined for all 1>0. Then, te Laplace

Definition: Suppose f isa
F(s) defined by

Transform of denoted by L{/()} is a function
F(s)= Lf} = -je +f (at where sis a parameter (real or conplex),

Note: The interval of integration is infinite which is an improper intgral and

thus it is evaluated by the rule F(s) = U} = Jerrod = lim je fe.

Examples: Find the Laplace Transform L{f(O} of the following functns.

asombi20 SOE" 950-1120
1,051<2
alas profes

Solution: Let’s apply the definition.

n= jerodr=imjea= ==] 4 „= “li
à y sl |

» rokuer-jerra-fea-in] Fe

ET LE en) JE MES
sa
s-a

D er ir tc. Femme i ed
y FO)=LS0}= fe “fOd= je dt = lim infe "dt

1
=

a N
DUO fe" far = Je Sell "Sat = [erario Yerdı
;

s sos

ı . : is
NLgD\=[e"fleodrs joa [a-yera-E,:>0
° 1 4 E

2.2 Poperties of Laplace Transform

Propety-I: Linearity Property:

If fnd_g are functions whose Laplace Transforms exist, then for a € R
D Leaf} =al{fO}
i) LUO+s0)=LUO+ LEO)

Exple: Find the Laplace transform of /() =2e* ~3¢+4

Sition: In the above example, we got Z{} = Le = Zu

ing linearity properties, we have
(2e -31+4)=L(20*) + L(30+L(4)
3,4

AS

os NT ANT OR LE Tenor Amey ot ache? Ame

Sega)
The Laplace Transform 1 of f Trigonometric Functions:

The Laplace Transform of /()=sinal and _f()=cosatem be
directly by the definition using Integration by Parts which is tco demanding,
But it can be done easily using the Euler’s formula with linearitypropertes,
Recall: Euler’s formula: e* =cosar +isinat-
Using linearity properties, we have
ee cosa! tisinat > L{e")= L{cosat +isinat}

= L{e}= L{cosat} +iL{sinat}..
1 1

I Se) —.
s-a s-ia

But as we discussed above, L{e“

‘Now rationalizing, L{e*'} using the conjugate of s—ia, whave

1 sti s

1 __ sig $4 = 545%
(Gays +ia) Sea Sta Sa

Le

}

s

Equating (i) and (ji), we have L{cos af} + iL{sin at} =

sta sa
But we know that two complex numbers are equal if their correspoding real
parts atthe same time imaginary pan equal. Using this property,

L{cosat} + L{sin ar). =- + + i
aa” io a fsa tal

Rd ree SS
ele ginny pat

> L{cosat} Lísina)=2>

Sa a

Therefore, L(cosat)==+ Fag sina) = a ris >0
‘Examples:

Ant
Least) = sine) = Lcos22)

,L{sin3}= 2, 5
4 Np?

| The Lap
Consider the functions (1) =sinhaf , Jae = coshat .

Recall: By definition, we know hatsinhar = ©”

Then, by linearity property, we have

i) acetal len en nn
But =H 13 Le") Le") = ar

ere ea a
2 jure 3-9 TE a) Yet) PA

, Lisinh30)= usar E

s-2

a Moorh2t}= 5, feos) = 55 Leosh Vn =

Property-Il: Power property: u = den

Examples: L{?}=—r A >= = iso
Property-III: satin Properties (First Shifting Rule):
Suppose L{f()}= F(s). Then,
D Le" fO}=Fls-a) i) Lle“fO}=Fls+a)
Examples:
1. Find the following Laplace transforms
Que DUES c)Le‘coshé) d)Lte“”})
Solution: Identify fand compute L{f()}=F(s) Then apply property-IL.

3) Here, let f(() =00s2t. Then FO= LOS Lee

Therefore, by S-shifting property, L{e 00s24) = F(s-3)=

Fan erde NE Ar ar ui nm ATAR ove 02909 AMEN 100

‘The Laplace Transform of Hyperbolic Functions:
Consider the functions f(t) =sinhat , f(t) =coshat .

Recall: By definition, we know thatsinh at

‘Then, by linearity property, we have

)L(sinhar) le jte em.

= ETC

ohne JE 14 ä pan 1, 1
Mba] { 3 ju PE Ter) Ts 26+0)

Examples:

D L{sinh21} =

Property-Il: Power property: Li"
Examples: L(0}= = =
Property-III: S-Shifting Properties (First Shifting Rule):

Suppose L{f()} = F(s). Then,

D Lle"fQ}= Fls-a) ii) Le“ SO} =F(s+a)

Examples:

1. Find the following Laplace transforms

a) Le” cost} b) Le sinSt}—c) Le“ cosh6t} Let”)
Solution: Identify and compute L((1)) =F (s).Then apply property-IlL.

a) Here, let f(t) =00s2t. Then F(s) =L {RQ} = Les) =,

‘Therefore, by S-shifting property, Lie! os} = Fl = TE yea

oyna Pann Ia PRAT BIRNEN 100

aT, rns
Here let 0)=sinst.Then, F6) LU) =L/sinsy = >

Therefore, by S-shifling property, L{e™™ sin St} = F(s+2)=

(542) 425
©) Let f(t) = coshét, F(s) = Lfcosh6t} "7%
Therefore, L{e” cosh 61} =F(s+1)= me
d) Le f=, F(s)= LA} + = = = Lie") = F(s—4)= 20

6-9
2. Find the Laplace Transform of

a) f=sin3rcos 4) f(t) =cos4tcos% c) f(t) =cosı

€) f()=sin2rcos2e DSO =4e sin’ı
Solution: Recall: Product to sum formula of trigonometric functions.

sinxeos y= sine = y)+infe+ y) coso y = [reste y) +costr+y)

a) L(sin3rc0s21) = uso) + Las =

1 1 los ©
ad. + Lcosér} = LL,
8) Leos 4100521) = Lfcos21)+ L{cos 61} Aer ae)

cos2 el eo
—= = L}+=L{cos2r}=— 4 5 _
2 Los = Zube Lfos2e} 22559

SN
D 0 sa sina 2") Aina,

Then, LUTO) "le sin ae sina =

41
e) sin21cos21= hoy

1 1
6-34 Capa
2
+16 |

nd Book of pple Mathematics by Regus N Fa
property-IV: t-Shifting Property:

If LO} =

Examples: Find the following Laplace Trama using tshifting property.
a) L{tcos21} LS) Lt") ad) L{tcosh3i)
Solution:

a) Let (1) =c0s21. Then, LSO) =Leos21) =.
F+

rimes dre MERE

(s),then Lf" f()}= (1) Erro)

Therefore, by tshifting property, L{rcos 2s} =-2 (y 2 24
ow
Using Lin» E and sshifting, F(s)= fe" sina => —
6 > +9
Therefore, byt-shifing, L(te” sins =L 3) =D)

ds Ge +9 (5-2) +9)

Ha ft) =e", F(s) = Le" atea cnt £(4)- 24

as) +
F pid si) 49
4) Here, L{cosh3t) == L{reosh3ÿ = (D 5 5): ES

23 Unit-Step (Heaviside) Function and its Laplace Transform

Definition: The unit-step function (also known as Heaviside function), denoted
Lt>a
by U,()=U(-a) or H,()=H(t-a) is defined as U,()= de te

For any unit step function U, (1) = H(t—a), L{U, (1)} =L(H(=0)= €*
This is obtained using the definition of transform as follow.
LU.) = je, (dt = fe EU, sone "UY ()dt = ja

“jean tim ET =| =
md e ai ur

2

Examples: L{U,(0)=

,L{H(t-3)}= E sie »)

s
Theorem (Second-Shifting Rule): Let L/@}= F(s). Then,

D LUG aU} = LU a)H(t-a)} =e LY} =e "F(s)

i) LOU, ()}= LU OH(- @)} =e Lf +0)
Examples: Find the following Laplace transforms
a) LM-DH(-2) 5) L{2At-3- H(-3)} o) AFH(-4)}
AD) LASt+3HE-) e) Llos21-mH(-x)) f) Lte*HU-4)
Solution: To use the above shifting rule, first identify /(r) from the problem.

a) Here, f(t-2)=4(t-2) > f() = 4=> LF) = L{4n +

Therefore, L{4(1-2)H(1-2)}=e LM) = e

Her f-3)= 2-3)? = f)=2P = LUO} = La? =
F

Therefore, Lit DHL DSL ee LE) = Se
E
€) In this case, let’ use the second property with a=4, (U) =p",
Here, f()= 0 > /(+4)=(1+4Y arrasar 48 418

Therefore, LPH (t-4)} =e“ LUft+4)} = e* (3 +>

ey Here, f(t) = 008 211) = f(t) = c0s21= L{f()} = L{cos2} =?
Ber]

se”

‘Therefore, L{cos (1 -MHL-M}=e"L/W)=

Pina f() =e" > f+4) =e nee wen
=

e
2

Therefore, ECHA) = LSO eE
FE
Laplace Transform of Piecewise Functions

First, let's see how any piecewise function is expressible in terms of step
PAONETET]
AO, t>a

functions. Suppose (0) = { is any piecewise continuous

function.

Then, f(t) can be expressed in terms of step function as
SO=LO+40-KOW.O=FON-U.0)* FAQ AO]

Laplace Transform:

LO) = LUO HO -AOU O1 LAON LL, O- AOU}

Examples: Express using step functions and find their Laplace Transforms.

3,0<r<1 10<1<2 o0sr<l
oe role area
opsr<l
ii oro
sint +cos(¢~/2),02 4/2 7,128

A *

sk ip ii I. CA
Solution: 4u,(0)
a) f= SOHO OR =3+ 1-1-3} (0 = 3-4

Therefore, LI) = LB—4u N} = LES) Lt (0) = LE
D O0=LO0HRO- HOW 0 = +30 = 4312

Therefore, LIO) = LA +3ua(0))= LIN + Ltd}

LE

OL > LO = LD OSLO} =
a) SO = f+ AOU, = sint +eoste- BE

se"?
+
à f0= AOL -VOl+ OO -U + LOWO

= (6-0, -U, + 7U,(0 = ¢- DU) - C- 8+ TU, 0)

Therefore, L{/(D}= (DU (}~L{(¢-8)U,()} =(0* e “Ys?

Laplace Transform of functions of the form 2 a 0]

Therefore, L{f(0) = Lisinr) + L{costt—7/2)u,,.()} =

If LEf()}= F(s), then, de] = [072

Examples: Find the Laplace Transform of

sin 2t e-1 2sin? 24 Cul
D

2 ) E = ao
Solution: i

a) Here, JOnAS FO =140) = Laon == FO)
7

sin21] _? en 2 x s
majo nj tania

à {Had oo of Applied Mothemati il by egos Y PR EEE
D f= -1 F(S)=LEf(O} = Lie! - Us tore =
r-lr

nef 1. fr = jh ti y

“(D

à 102621 Eon cot Roya

mes er 2).¡ je E aye Fitri? vf

“tale FT 46. arr:

d) f()=e" > F(s)=Lle*} oh

host
7

ens
fe = [4e- 0. This means that Z

Ir

Po

Dir
2.4 Inverse Laplace Transform and Their Properties

If L(J(0) =F(s), then the function f is said to be the inverse Laplace
transform of F(s) and is given by /()=L"{F(s)}-

Examples: a) L{l}= tert) =1 b)L{e*} + =L"t
5

Properties of Inverse Laplace Transforms
Let LU/(0) = F(s), L{g()} =G(s)and let a be any constant. Then,
i) LaF(s)} = al-"{F(s)) =0f(0)
i) LF (s)+G(9)} = L'{F()}+L'(G()}= +8)
ill) L*{F(s-a)} =e"L{F(s)} =e" f(), L'{F(s+a)} =e fl)

im (Sula =H(-a)

WL e“F(s)} =U(-a)f(t-a)
How to find Inverse Laplace Transforms?
Simple Cases: Using manipulations or rearrangements and then properties of
Laplace Transforms and basic tables of transforms.
Examples-1: Observe the following inverse computations,
44s

ae yt) a =o" ci
BE ae ate nt

°F eB ad cost

D ae ara D nn

af 4s I
= fc en a ont

af +7 Hl 6) be
PE (23) ee (ata) 008214307 sin 2

RS mn 07

‘etd ook pi ha a Fo a
General Cases: Inverse Laplace Transform by Partial Fractions

The tricky procedures used in the above examples to find inverse transforms are
not useful for many cases. So, le

see the procedures to find inverse transform.

The procedure to find the inverse Laplace Transform of Fs) =P)

Qs)
Step-L:Factriz the denominator Q(s) completly
Pl
Step-2: Decompose F(s) = an sum of rational functions.
5

Step-3: Determine the constants in the decomposition.

Step-4: Take the Laplace Transform of each term separately.

In doing so, we may get different cases based on the factors of Q(s)

i) For each non-repeated linear factors like s-aof Q(s)assign a rational

function of the form — where Ais a constant tobe determined.

sa
ii For each repeated linear factors like (s—a)" assign a rational function of
weten Mi A, 4 A
D = sma

iii) For each non-repeated quadratic factors like as’ +bs +c assign a rational
function ofthe form —# +2

as +bs+e
Examples: Find the inverse Laplace Transforms.

2-7 y ger 85+10

arr: rer" (s-2)(s7 +1) (s+IXs +2)
Solution:

a) Here, O(s)=s* -5s+6=(s-2Xs-3).Then,
2-7 __ 2-7 A, A A6D46D
5546 (s-245-3) 5-2 5-3 (-2\(s-3)
> A(s-3)+4)(s-2)=25-6>(4, + 4,)8-34, -24,=25-7
24 +4,=2,34,-24,=-7>4 =3, 4, =-1

AA 10:

AAA
25-7 3 1 2 qn
Hence, LÀ = 13) = es ae
_ E =} vi) 53
> — 65? +95 =s(s-3)'.Then,
18 AB), By _A(-3Ys-3)+Bs(s-3)+Bs

56-362) tg ” 6-39
= A(s-3Xs-3)+Bs(s-3)+ Bs =18

>4=2,B=-2, B,=6
) Here, we have mixed factors. So, the partial fraction decomposition is of the form

b) Here, Q(s),

18 2 2 6 ”
18 AS A 2 A 6 EA
x) E) e +

5 -65+2]_
(s-2Xs* +1)
8+0 4

te

d)Har,

= 4,=2,B, =-2,B,
85410 af 2 af_2 a
ER Gh: ar ts)

= 20" 2% Verse

Inverse using Second -Shfting Rule: Suppose L{/()} = FC).
Then, SUD 0) = TESEO) = faut —a).We use
this property to computo nese Laplace Transforms ofthe form Le (5)
«To use this rule, first identify F(s)and e”” from the given problem, second

find f(using f()=1"(F(s)), finally use the rule,
Examples: Find the inverse Laplace Transform of

PS A pe
aa ai “aa ?

As 43070

ONO pled Wari by Babe For your commen an sages me O 3
2 fi) a

ot
s(5-3X8-3) 5
> A(s-3(5—3)+8,5(8-3)+ Bas =18

>24=2,B,=-2, B,=6

ME

©) Here, wehave mixed factors.So, the partial fraction decompositi
556542 _ 4, Bs+C
Ge +) 5-2 +

sf 55? -65+2 ¿[2 af 3s
Hence, SEE pd pa
= (55) [he E
a) Here, 5 +10 2,2,

Sn MON
>4=28,=2,B,=2,8,=6

m { 85410 pul 2 Jul 2 Heil 2 ef 6 }
(+642) s+l s+2 (+2) +2)

207 Vet He”

Inverse using Second -Shfting Rule: Suppose L{f()} = F(s).

Then, L{f(t—a)u, (0) =e FO) LE" F(s)} = f(t-a)u(t-a).We use

this property to compute inerse Laplace Transforms ofthe form _L*(e""F(s))

“To use this rule, first identify F(s)and e*“ from the given problem, second

find fusing = L'{F(), finally use the rule.

s(s-3Xs-3)

n is of the form

= 4=2,B=3,C=0

Examples: Find the inverse Laplace Transform of

= ng =

€ etme € 454302
a) das “ma D #9

s(5+35+2)

——

id Bk pled Maen eh i al IO EEL
Solution:

2 1 En
a a Flo) where
tota
rr rr ON

1/2 af 1 | 1/2 1
meso eit

Thus, Le" F(s)}= f(r -2)u(t-2) =(

D

fer" {tsk

Thus, De F(s)-e "F(s)} = Efe" F(s)}-L fe" F(s))
= SC-U(-3)- f(t-Mut-7),

er

Et
SGD Gaal
F( Li e BE oer
OL 3/0= ar + su

Thus, L' En = f(t-3)u(t-3) =e” sin(t -3)u(t-3)
D Her, £ a

=e F(s) where

=4G(s) +e" F(s) where

sos ar) Fa en. vor =) sinh3r

Thus, LAG) +" F(s)} =4L UG) +L (eFC)
= 4cosh3t+sinh3(¢=20)u(t-20)

AMR 220

|

2.5 (Cenvelution Product and Laplace Transforms
Defiai ties: Suppose f and g are continuous functions on [a,b]
Then tor 1 e[a,5], the convolution product of fand g, denoted by /*g, i,
defined us Weejo= [senator

>
Example:
La 10=3,80=0. Then, (1*8X0= | fl-a)a(a\de= [30 d=30 3

> 2

Properties of convolution product
Dftess*f MSMeth=Sietseh SE MN=U A
Convolution Theorem: If L{/()} = F(s)and L{g()} = G(s),then
DA O) =LUOMEO= FIG) DEIN = U

Essmgies:

Find L{{f * gX¢)} using Convolution Theorem where

DIOR IO 0 DIO Os
Solution: First find LE/()) and L{g(¢)} so as to apply Convolution Theorem.

amoo lod
A Lt) À
uo) lee)
E ile EE
oral Le ls

1 Lt0)= Lin Jah
>= Le eso Ll Juli) ae

+4

ut

2.6 Laplace Transform of Derivai es and Tntegrats

2.6.1 Laplace Transform of Derivatives

If F(s) is the Laplace transform of f and if S has a value f(0), when t=0,
then HPO) =5F(s)- 0), LE O}=5°F()-0- $0).
inge, A/C] = UNO] LO (0). rnb
LF (O)is the value ofthe n* derivative of {at 120.
Example: Given f with f*()-6/()+5/(9=0,£(0)=3,50)=7. Find
the Laplace Transform of fand deduce /{itself.
Solution: Take transform both sides.
LS"-6FO+5f(O} =0
LF) - SL O) +SLUW) =0
SLU) HO -F0)- ALSO) - FORSLYO) =0
FPLYO)-3s-7-GLf()+18+SLYO) =0
Dl 64 LY) = 3-11

ao eh

Now let's deduce S (itself by using inverse transform. That is.
3s-11

WO as

3s-11
A ett}

pepe

TOBA 101

2.6.2 Laplace Transform of Integrals

Suppose f(t)is piece wise continuous for all #2 Oand Lf O}= Els).

Then, if Der FO) and thus E (En DE j Nod.
i ;

Examples:
1. Find the following Laplace Transforms

of java iljera) aufn) ajena)
aus) :

D] yes ==

s-l
LA [et cos dr} =—L{e! cost =
2 {fe Y ans, iw Gy {6-1 +1)
2. Find te Laplace Transform of the following functions.

DfO=et[rerd HSO= na 9 10=["E&
Lx

Solution: Transform step by stp.

à) First using shine, Lf? er 2 (f)- Ha >

Second, using Laplace Transform of integrals, find Mes:
d

Thatis Fs)= djéra)-! Le

Se

(Cia Book pl Nahen by egos Wot sa ames ad EEE

wal

That is F(s)= Uf sind = = lu {rsin 21) =

Finally, by shin,
LSO) = Le] xsin2xdx) =

So, LEAS
x

sinar y

Therefore, L{f()} = ‚ja

se“

3. Gh —
O er

.[sexe=8, 80 = for. Find Lie)
Solution: ° :
sta} = 1 fee = rod | rod

= re fé

=Lvo-18-uo-i-

rag ee ERNE SRE ie ay
2.6 Applications of Laplace Transform
Laplace Transform is widely applicable to solve many real life problems
Especially it is useful to solve Initial Value Problems, Integral-Equations,
System of Differential Equations. Here, let's see some applications.
1) Initial value problems: Here, lets see how Laplace transform can be
used to solve initial value problems for linear differential equations.
Procedures to solve IVPs using Laplace Transform:
Given the IVP: ay"+by'tey = f(x), 9(0) = (0) =
‘Then, to find (+) satisfying the IVP, we use the following procedures.
Step-1: Take Laplace Transform of the DE both sides.
That is L{ay"+by'+cy} = LE f(x)} = aL{y"} +bL{y’}+cL{y} = Hs)
At this step use Laplace transform of derivatives and solve for y(s) by
substituting the given initial conditions y(0) =m, y'(0)=7.
Step-2: Solve for y(t) by taking inverse Laplace Transform of ¥(s)
Y)
[At this step, apply inverse finding techniques to solve L"{¥(s)} for y(1)
Examples:
1. Solve the following initial value problem using Laplace Transforms.
a) y+3y=e', (0)
D) y'2y-3y=3" , y(0)=0,y'(0)=5
©) y'4y=24e”, y(0)=0,y'(0)=0

Take inverse Laplace Transform of ¥(s). That is y(t) =

a) y+y-2y=31e” „0-70
e y'-3y#2y=6e" , (0)
DRY ye sine, y

Soluti

a) Step-1t Determine the Laplace Transform Y(s) of »(1). That is take Laplace
‘Transform both sides and solve for ¥(s) by substituting i

ial conditions.

Ly) = Le} > Ly} +L By} =sy{s)- y(0)+3y(s) i
otr

= (5)-143y(5) =

s-1 s-1
5

= y(sXs+3)=

s-1 GT EP)

Step-2: Solve for y(t) from inverse Laplace Transform of ¥(s).
ct D
That is y()=L'{y(s)} = (= >)
Apply inverse finding techniques to solve £*(Y(s)) for X)
That is find the inverse transform L”| an)
+
To determine the inverse, ie li Fraction Decomposition.

He — =. 2 und determine A, Band C.
er, = FG wt 5-1

Now, determine the constants A, Band C.
By using Cover-Up Method, at s=-3, we have as andats=1, Bet.

sea sit.
So, the decomposition is (8) = aye +

Therefore, the function y()is determined as

AA En

b) Step-1: Determine the Laplace Transform Y(s) of (1). That is take Lape,
‘Transform both sides and solve for ¥(s) by substituting initial conditions

3
Ly) ==
SS =

Liy"-2y-3y} = LB} > Lp)

2

2190-90-10 - Ast) YON -39(5) =

52 25-265)

Ss-7 E
IO | EDG

Ss-7
(5+D6-D6-2)"
Step-2: Solve for (0) by inverse Laplace Transform of Y(s).

Hence, the transform of y(t) is found to be y(s)=

7
a)

To evaluate the inverse let's use Partial Fraction Decomposition.

5s-7
Here, 9 erg
15-7 8

By using cover-up method, at s=3, we have A=——
QU]

That is (0) =1"p1s) =

Ats=-1, B

TTL Nae EA
‘Therefore, y(t) = 1 ke

=2e* e

L{24e") > L{y"}+ Ltay}= 24L{e")

O Liy"4y

FO ae
= =

20 Lo os

— E
+96)

2

oa as

ll af 3]
Zr: { salt {|
2 af 1

ee

Therefore, y(t) =. vf;

an

=-300821-3sin21+3e%
DL +y2y)=L(80")
>LY")+LL)-2Lp)=8L(te”)

=>5°y(s)-5y(0)-y'(0)+5y(s)- (0)-2y(8)=

(s =
2 3 = 5
>6 DI OS

Now, ing par “a el
EN

y= ET" Fi + +1 tr

1
Das; Speo da

mane} les) st)
jeans Beet

¢) Step-1: PIE EEE 6).
Ly"-3y'#2y }=L(6e"} > LO"}- -3L}+2Lty}= 6Le").
are Ha EAU INS) AN +28) =F E

II a ape E
6 6 ie
>= Er) EA ~
Step-2: Solve for y(0) from (0) = LUS)
To rl ive we Parl Fron Deomposion.
6 3.
Here, ED Sn pos
‘Then, by using cover-up method, find the constants A, Band C.
Ats=2,wehave A=2 ,ats=1,8=-3 andat s=-1,C=1.

ae 6 Ÿ
te man I
O 2 ste

urn eal {23,2
ae (genen Zr 251 +t}

we oe bear ae

A 115

2
(62 +2545)y(s) = 222543. yyy st +543
a

A _
wre (42545425842)

By partial fraction,
542543 Ao Cs+D

E: AAA
AAA 42545 SP 42542 O

So, == 3 a =e" (sin2t +sint)

nl 3° (pH

g) Liy"+4y+5y) = 25)» Lp") +4Ly)+SL(y)=25L41")
IO IA 5910)

Dr. EEE ER

2 +4545) 5744545

But by partial fraction decomposition, we have

et By Ce, DIE

PE eases) rer

AA As

>4=10828.0=2,D=-2,E

nl l TE ja afer 743)

2 N 2 s 43 1

ae xis) 5 Se)
2 (1 2 .( 5+2-2 )_43 pf al 1 )

30 AC D ETES

e E 2 cnt 22 sin) Qe (How?)

DE >

fied Makes ye
Ay Ly’ yp = Lire! } > LU")

> (s -25+1)y(s)

_ 6
290° DED

‚Therefore, using s-shifting and power properties of Laplace transform, we
af 6. 1 6 1
hi =1 | — |= Al | pe re
ave y(t) en 6L (7) wot a0"
Lay Le) > Lo" #20) +20)= LUE)
= 8'166) -y0)-y0) + Ayo) Ol »6)= ZT

s+
Ñ 1 3 +75? 4165 +12
SAA HF ee NO rer

Continue with partial fraction and complete it.
2. Using Method of Laplace Transform, solve
a) y"-y=te, YO=Y 0)=1 by y'-3y-4y=?,xO=2YO=!
o)y"edy's3y =e 90) = YO) =1 dy"s9y=sin2,,y(0)=7,YO=3

eyy"ty-2y=1e%,y(0) = 0,9'0)=9 Ny2y3y=e”,y(0)=0,y'(0)=0

à) Taking Laplace transform on both sides, we get
1

=»

Loy Le) > yo) YO-YO) =

q
[Te

sel TRE
ten 5

But by Partial Fraction decomposition,

A Br Sn
aye Ce CE

r Bee E
BE mie

EEE LT AN
Nana ‘iin |

IIA AO) 966)-0]-4y()

2 (61-35-4)9(6)=2s~54. 2. yg) 25-5 a _
o . e 6D "0D
Butby Partial Fraction decomposition,
2-5 _A,B
—— ta D = Bas
(oe) Aa
2 C,D E,FG
76-6+ wir)

Ths (= Tet 42 le vale +

2
©) Taking Laplace storm both sides, we get
Liy4y43y}= Le) > L{y"}+ L(4y)+LBy) = Le")

6546 5746846
($ 44543) =, re
(44543916) 210 Pen
wie E dz

LL gel gal,
tz
Sayed

Tf 1) 1 fi eee
3 Gr Gr} elas AE

2,13
Ly+9y} = L{si A 4 2
4) Liy"49y} = L{sin2} > y(s) maga hs
2/5 215 13,25 15, Ts

S44 549 5149 e 549 549

SOS

Soe pinza Tear

guy re) —

— 44
EE) 1-2 (5-2

1 dE 5
Die en eng

Ths, yok Es a 542] 16

516, V4 U3 vs Solel dh

3. Using Method of Laplace Trausform, solve the following IVPs.

DNIIOJO1I020 where f= { ai
0,1>1
1,051<2
0,122
nosı<a
01>x

5) y'#y= 10,70) =0,7(0)=0 se =|
©) y+y=10,/0=0,/0)=0 where 10%
aya o

Solution:
a) Fist observe that using uni step function /()=1-u(t-1).
Lagi) =101-ae-D) = LOL}

sos

Then L17-)=LJOI2L)-L=LVO ==>

sy) -Y 0-10)

o 0
FENTE) =p

ADA Gl
aro je Ces} ds, 3)

a TY 123

lad Bob plied Mae by Baha Fr, a ll A

seo le

sé +et-

b) First observe that

LIO) Jens (dt = Jens (at + fe" fd = jera =
a o 2 o

Then, LUN = LU) > LLO = LF} =
5

ie
aja
28 +999) 2 Pr CNT ONE)

tl |
PRE (cs 56 3)" das an)" ir 5)

1 .4,/BC dal pu Go
Gans ET JET EU 3 c=0

AAA | af) (59 11
Lac 5)" (e 5 )- NE a) gg x

11
11 dos3r-[2-L0053(-2) jul -2)
Therefore, y= 5— 5083 Gr gast -2 pe -2)

Now, by partial fraction,

) First observe that using unti step function f(t) =1—tu((—x) and using the t-

shitingpropey that LOIS ALU) vetas
Lf} = L{t-m(t-#)}= L{)}- = ll -m)

Then, L {y"+y}

> (5° +Dy(s)

)

sc)
(+1) sea) 3s? +),

>= oats) re (5 # 5)
Fer) sean 7+),

Now, by partial fraction,
no ED 0,B=1,C=0,D=-1
PEN ss À

aA, B+, 4=5,B=-",C=0

Seas stl
af re FN CR i
-2(5)- Ga [ 6 Far u)
zizsint- (most Mult-m)-sinle-)
= (¢—sint,0<t<2
= ame leere a

=y0)=

sind ay
0 non 5 A

1. mm ms
03320 3-2

à Ly 2y=h

Equation that involves an integral and derivative of a function is de
Integral (integro) DE. Such equations AI, also solved easly by us bare

‘Transforms of derivatives and integrals at the same time. is
125

Examples: Solve the integral-DEs of rT ee
poro» M0=r+ [sort
arroja 0 fe densi
Solution: a) Using Laplace en of derivatives and integrals,
vf 00 Oz

42
FE)

12
»rols-I=5+1>10)=

Ten, oe 2 )

ED),

B
But by Partial Fraction aa 23/2222 Therefore,
Fe

=) si sel 5
#42
EIGEN)

er)
s+i $
+) Using the definition of convolution the given equation is transformed into

«deL
2 2
0 =1+sint* y(¢). Now, by taking the Laplace transform both sides, we get
1,1 Liss?

1
a IO

Finally, taking inverse Laplace transform gives us
a 1) al lat
= LH) = rfi

HO)

CALCULUS OF VECTOR VALUED FUNCTIONS

3.1 Definition and Examples of VVF

A function F:R->R° of the form F()=iOi+. Sit LOK ER is called
Vector Valued Function. Here each fff, are called component functions.
‘Why we need Vector Valued Functions (VVF)?

‘Think about an object traveling in space along a curve C , particularly, in R’.
‘Then, its x,y,zcoordinates are changing through time as it changes position,
So, if we denote its coordinates as a function of time to be

= fi(t),y= fal.2= /,(0, then its position will be given by
(,y,2)=(f(,f(0,4(0)at any time + as shown in the diagram below.

Position :P(x,y,2) = 0.40.40)
= fit AO hk

7

The position of an object moving in space at
any met.

In vector notation, the position vector of the object from the origin to its
position is written as r()= f(Ol+fOi+ £(0K: But this is really what we
mean by a vector valued function. This means that to describe the motion of an
‘object in space, to know its velocity, speed, acceleration, distance moved and
other motion related concepts we need the concepts of vector valued functions.

ad bof plied Maha i Fatt iad Spa
3.2 Limit and Continuity of Vector Valued Functions

3.2.1 Limit of Vector-Valued Functions

Let F:R > R° defined by Fl) = AOi+LOI+ LOKER.
Then, Fhas a limit given by lim F(t) = L = ai+bj-+ok at t,if and only if the
coordinate functions f,,f,, f,have limits att, .
That means
lim F() =ai+b/+ck Im /(0=a lim (=, im (De
ie im FD =[lim A+ Lim f(O1 Lim f (01

Examples: Evaluate lim F{r)where

—V5-Fk at ty =2

i+

Ber = 1
D Fl0)=Inp=: pl

jrek at to=1

a, Disthtss wives
(Q0-ni+ 8-H j+(P-Nk, 123

Solution:

23, lim 6-7 212 lim F()=6i+3j-k
E

4) lim Inf does not exist. Hence, lim F(t)does not exist
9 lim(2r+1)=7, lim (21—1) = 5, lim 8=8 = Im FO = 71+5)+ 8k

lim =6, lim
mig oi

lino) =7, lin) =5, lim? -1)=8= lim FO = 7145) + 8
Here, tim F(0) = lim Fo). Hence im F(D = 71+ 57 +8.

ee A e

Ei
CHAPTER-3 à

CALCULUS OF VECTOR VALUED FUNCTIONS
3.1 Definition and Examples of VVF

A function F:R->R° of the form F(t)= fi(Oi+ AOI+LOKER is calles
Vector Valued Function. Here each fi, ,,/, are called component functions,
Why we need Vector Valued Functions (VVF)?
“Think about an object traveling in space along a curve €, particularly, in R°
Then, its x,y,z coordinates are changing through time as it changes position
So, if we denote its coordinates as a function of time to be
x= fit), = /,0,2 = J,(0), then its position will be given by
(5,1,2)=(4,00,4,00, £(0)at any time £ as shown in the diagram below.

Postion :P(x,y,2) = (f(0./:(5, 00)
LOS 10138 10138 hOk

3
‘The position of an object moving in space at
any time t,

In vector notation, the position vector of the object from the origin to its
position is written as r(t)= i+ f+ GOK. But this is really what we
mean by a vector valued function. This means that to describe the motion of an

‘object in space, to know its velocity, speed, acceleration, distance moved and
‘other motion related concepts we need the concepts of vector valued functions.

a a EL eM Fo yon omens ad ring
Basic Properties of Limits of VVF:
Suppose Fand Gare vector valued fu

0)lin(F()+G(0]=lim FCO +i 100)

€) linfF(9.G(o}= im Fim GO)

©) lim(F()XG(0] = lim F()xtim co
Notice that since the functions are vector valued, the operation in F()G()is
the usual dot product of vectors and the operation in F(t) G(?)is the ul

mnctions whose limit exits. Then,

cross product of vectors.
Examples:

2
O + Gr? +1)
Evaluate
i) lim[F(+ GC] ii) lim[F@).G@)]

iii) lim[F(Qx GO) iv) lim3G())

Solution: Here, Im F()=1+ j+klimG()=i+2/+3k.

‘Then, by the above properties,
i) Tim[F ()+ G(0)]= lim FO +limG()=21+3/+4k

ii) Iim[F() GO] = lim F(0.lim G(e) = 6
3
1 1lsi-2j+&
1 23

ii) in(F()xG(O) =Hin Flin GD) =

iv) lim[SG(] = 31+6/+9k

A vector valued function F is said tobe continous at aif and only if
limF)=F@)=f@it L@i+ ak.

This means, a vector valued function is continuous if and only if all of its
component functions are continuous.
Examples:
1. Check the continuity of the following VVFS at the indicted point.
a) F(t) =(cost)i+ (Sint) + (tant)k at t= and t=2
DA+ V4, 1<2

D FO= =
de poo

Solution:
en li 3 x. Do x
in FOSSES It Vite FO). Hence, Fis continuous at t=,
4

3
Buta r=5, 50) = tant is not continuous and thus F is not continuous.
) lim 2#=4, lim Jf+7 =3 => lim F()=4i+3j

are a
lim (St—6)=4, lim (? +1) =5 = Im FQ) = 4145;
Here, lim F() # lim F (0). Hence, lim F() does not ist
Therefore, the function is not continuousat f= 2.

Lut FE ye AAA ent, find a and 5.
bei+(4r-1)j+( 1,123

Solution: To be continuous atz =3, we have lim F(¢) = lim F(9 = FO).
2
Therefore, lim (+1) = tim br? 29b=2>5=5,

ImQa’ +) = lim(4t 1) =9 2a +3=11 => 0742

DEENTRAR

3.3 Derivatives and Integrals of Vector Valued Functions

3.3.1 Derivatives of Vector Valued Functions

Let F(t)= f(i+ fyi+ f(Ok,t € Rbe any vector valued function.
Then, the derivative of F denoted by F'(1) or D,[F(0)], is defined by
Dre On DFO AS
Examples: Find F'(t) and compute F'(2) where
A) FO=GP+9i-Sj+ +k FO A+ PK
Solution: P
a) F()=9Pi-Sj+(Q1+Dk, F()=361-Sj+5k.
b) Since f,(1)=|-2is not differentiable at r=2, F itself is not differentiable,
Properties: Suppose Fand Gare vector valued functions whose derivative
exits and let f be differentiable scalar function. Then,
i) Dot-Product Rule: D,{F(0.GO1=D[FOLGO+ FODIGO]
if) Cross Product Rule : D,{F()xG()] = D,LFWI« GO) + Fx DIGO!
iii) Chain -Rule: D {FO = SOF UO)
Examples: If /()= F(0.G(0,H(O=F()xG(), find FC), H'().Given that
F(t) =(2t+ Dit f+ (6-4), FO) =31+5k, G(1) = es Pk
Solution:
Here, FO) = C++ j+ (6 - FO) =3i+ j-3k, F()=2i+5k and
GQ) =H +P j-Pk => G()=i+ j—k, 0) =i+2j-3k
SO=FOGO+FOGO=fO=FOGH)+FOGH
=> f= (i+ 5h). + Jj 1) + Bit J -3k).G+2j -3k) = 11
i jk

2 1 Joana

hoa -

’ ii k
piro-rosennct o ie
1 2 -

ed bk pe aha eg Y do

3,32 Integration of Vector Valued o

La FO=AOAOIROK where ff, fare continuous on [9,5]
definite integral: [Bs Spas page Isa.

Definite integral: [ros (rois Joël
Examples:
[Let FD =100 +29 GF? —40)k. Find {Fir and compute i FA.
Solution: :
y fFUndi= [naar | Keddie | Aare
= front ri +f 2taej + [GP anar
= ri j+l -2r)k
+ 2
a fFodr =([poati+ [a+] Hark)”
1
a frortati+ Far [ar ana)"
HQC 2 paf =621+3j+k
2e Fl) =2i+5j+3rk, F() =21+5)+3k. Find FÜ).
Solution;
FO=[FWa= (faro + sani+ 3" dik
=Pi+SH+PK+C, (C=aite,j+ck)
Dar) = 21 + 5j-+3k = 1+5j+k+C=2+5}+3k => C=i42k-
This, FQ) = 11459) Pk + CP + SHH +2k

Pa y

TAC UT Fos

3.4 Space Curves and Length of Space Curves

Curves: The graph of any continuous vector valued function r(e) on a yg

interval [a,b] is said to be a curve.
For instance, line segments, circles, rectangles, ellipses are curves.
From now onwards, we use the notation Cto denote a curve and (?)to dengg
a vector valued function whose graph is the curve C. In such cases, we sayi
r(t) is the parameterization of Con[a,2]- We write as C:r(1),a tsb,
i) Closed curve: A curve is said to be closed curve ¡Fit has a
parameterization r(() on the closed interval[o,5] such thatr(a) =r(0). nal
words, a curveC is said to be closed if ts initial and terminal points concides
Examples: 4
a) The curve parameterized by C:r() = cos(2ati+-sin(2nt)/, 01 SIS
because r(0)=r(l).
b) The curve C:r()=11+17j,-1815 1is not closed because rl sr.
ii) Smooth curve: A vector valued function r()is sid to be smooth on a give,
interval [if r'is continuous on / and r'(()+Ofor all points in J (with the
possible exception at the end points of J ).
Furthermore, r()is said to be piecewise smooth on the interval Jif
expressible as the union of finite number of sub intervals such that (VJ
smooth on each of the sub intervals and the one sided derivatives 1, (pr)
‘exists at each interior points of] .
A curveC is said to be smooth if its parameterization r(¢) is smooth and it is
said to be piecewise smooth if its parameterization r(¢) is piecewise smooth.
Examples: ental
a) Consider the curve parameterized by CO =P + PE. *
Here, P(0)=21+2/+3Pis continuous, Besides, r1() Ofor any sbecause
the j component of (9) ean never be zero. Hence, thes enrve CS

post ma Teis! A

rized by C:r(1) (el +o s+0k.

sider the curve paramet

#) Cor :
E. 0 =2n1+U=0")) +34 kis continuous but, Y) =0at 1=Oand thus
0) ve not smooth. Thus, the curve Cisef isnot smooth.

ify that the curve parameterized by
Dj ++ Din(r+1) = on 051 SI is not smooth.
0 and thus r(/)is

y ve 5
cani
re, PO = 6 + (2e =D) + In + DK but (= Oat f=
smooth, Hence, the curve Citself is not smooth.

ve Cr(1)=2(1-cost)i+2(¢-sine)j,/ =[27,277- Here,

not
4) Consider Ihe cu
„(9 =2sind-+201-0ost) is continuous yet r()is not smooth because
ram =r (0) =r (7)=0and thus Citselfis not smooth, But if we break the
interval as 1=[-22,227]=[-27,0]U0,22] we get ¥ (0) # 0 for any

amer <0, 0<1-< 21. This means r(()is smooth on [-27,0]and [0,2].
‘Therefore, Cis piecewise smooth.

So,r(t)is piecewise smooth on[-27,27].
) Consider the curve parameterized by C:r(()= Pirgrók.

Here, (9) =2, 14 +2 is not continuous at *=0 and thus r()is not

smooth. Thus, the curve C itself is not smooth.
Common Parameterizations: If you remember, we have seen in Applied 1,

how to give the parametric equations of a line in space. Likewise, it is also
posible 10 express different curves using parameter. Here, under let's see
parametric equations of important figures which we will use frequently.

i) Parametric equations of a line segment:

Suppose L is a line segment connecting the points A(x», Yo,20) and

B(x,,,,2,) Then, its parametric equation is given by

rat

L:X=4+ (AB => L:ly= y, +bt for0St51 whereai+ bj+ dk = AB

=2. +0

In shor, the parametrization is r(0) = A+1AB

‘Tien Bock of ppd Mathematics. by Begashow I. For your comments and sagestions u
Example: Find the parametric equations of a line segment connecting q
points AZ,-1,1) and B(L3,-2).

Solution: Here, ai+bj+ck= AB =B-A=-i+4j-3k.
x=2-1

Hence, using (o, Yo» 20) =(2,-1,), we have L:4y ==1+41 for 0151
2=1-3

úi) Parametric equations of a circle:
Suppose is a circle given byC: x? + y? =r*.Then, its parametric equation i
given by C:x=rcost, yarsiny,for 051527 .

Similarly, ifthe center is not the origin like C:(x- a)? +(y-5)?

c: fe for Osıs2r

“|y=bersint,
Example: Find the parametric equations of a circle given by
aC:r+y'=4 b) C:(x-2° +(y+3) = 16.
Solution:
a) Here, r=2. Hence, C: x=2cost, y= 2sint,for O<1<2r
b) Here, and the center is (2,-3).
Hence, C:x=2+doost, y=-3+4sinyfor 0S1<27 *
iil) Parametric equations of an Ellipse:

a =1.Then, its parametric equation

for0sts2n.

Suppose E is an ellipse given by E
d

(x=acost,

Example: Find the parametrization of the ellipse E:9x* +4y? =36,
3

Solution: Here, first write in standard form as E: and identify a

land Book of Applied Mathematis- by Brgushow M. Fo your comments and
jv) Parametric equations of functions:

SupposeC: y= g(x), a <x <b Then, its parametric equation is given by
x=1y=800,0S1S5>C:r(O=1i+g(0)j, asıch.

Example: Find the parametrization of C: y= x*~2xfrom (0,0) to (23).
Solution: Here, x=1,y=1? -21,0<1<2=>C:r(f) =1i+(P -21)j,0<152
Oriented curves and types of orientations:

Suppose Cis a curve parameterized by C:x= f{1),y= g(9), a <t <b as
shown in the diagram below.

Suggestions ur RE

(Positive orientation)

B
© Types of orientations:
A [— Positive oprientation
5

L_ Negative orientation
a

B

(Negative orientation)

If we move from point A to point B along the curve C (in the direction of the
arrow), the value of the parameter £ is increasing from a to b. Such type of
direction in which we move or trace out the curve in the increasing order of the
Parameter values is said to be positive orientation. On the other hand, if we
move from point B to point A, the value of the parameter + is decreasing from
5 to aas shown in the lower diagram. Such type of direction in which we
Move or trace out the curve in the decreasing order of the parameter values is
said to be negative orientation. A curve C with either types of orientations
(Positive or neontive ic exi to he nriented curve.

{nd ok of pied that y Beas Fr yu cna a
by gas To yur commen ad pin 2506866700
jy) Parametric equations of functions:

Suppose
x=)

y =8(%), 0 $ x < b Then, its parametric equation is given by
M,ası<b>Cirt)=i+g()), asızb

2xfrom (0,0) to (28).
-21,0515223C:r(t) =ti+(P -21)j, 05152

Example: Find the p

ametrization of C:

Solution: Her
Oriented curves and types of orientations
Suppose Cis a curve parameterized by C:x

FS0,y=800), astsb as
shown in the diagram below

(Positive orientation) _

>

Y Types of orientations:

c
A yA f Positive oprientation

L_ Negative orientation

If we move from point A to point B along the curveC (in the direction of the
arrow), the value of the parameter £ is increasing from a to b. Such type of
direction in which we move or trace out the curve in the increasing order of the
parameter values is said to be positive orientation. On the other hand, if we
move from point B to point A, the value of the parameter # is decreasing from
b to aas shown in the lower diagram. Such type of direction in which we
move or trace out the curve in the decreasing order of the parameter values is
said to be negative orientation. A curve C with either types of orientations
(Positive or negative) is said to be oriented curve.

Lengths of Space curv

Just think How to find the total Jength of the curve below?

‘Suppose a particle moves from 4 to B along a circular helix

C:x(0) =0osti +sing +i for OS < 2: as shown in the diagram below,
:

„a Cdi) = cont Dei

8
0,02)

0.00)

uh
y

‘What idea do you have to compute this length?
‘Are Length: Let C be a curve having a piecewise smooth parameterization

%
rt) on [a,b]. Then the length L of C is defined by z=[[rokt.
Particularly, if r() =D 9 + 20%, l= Vx? + yO? +20" ‚the
arc length is given by L= Iron- j POSTER
Example: Find the ac length Lofa curve Con the given interval

a) roma ts jrrk;-Isısi

5) r(9) =cos4f +sin4t 3,0 <<

Lan y

Ord Mea yelg-ısısi

Ar()=eite*j+VIk;0<st51

à Ce coinprco ss 22

2 10-01

AMAS

0 pene PTE. k
‘Then, L= {role j ee? + yn? +2) at

Mr ra [lia [arras
4 3 a
pyr) =-4sin4ti+ 400s 4 -3k

he L= ff t= [ Vico arsine Sct
4 o

=] ons sin an 3a = [Vd = [5d = Se
NA a

Aral
CHOSE) LR LE

ñ ı f
= ora la fae
Tee, I RO +y 0 a i Arr ¡fa 5
a ade safe tenet
> e o

» » : +
O CITAN =
o te] ostrgenty sea Beraten] [ =

Av =4i+ 224k
3
Then, Le [rot =f ESOO a [lira
> > ®

-f ar jene]. =21
d 0

wie

ATAN al EERE THAR

Zu save
3.5 Tangent ‘and Normal Vectors and Curvatures

et be smooth curve with parameterization (0) on an open interval I.

‘Then, we have , .
i) Unit Tangent Vector: For any point fin J bre (1) #0, the unit tangent

vector T(p) to the curve Cis defined TO Fe or

il) Unit Normal Vector: If r@is ‘smooth e zes: the unit normal

vector N(t) to the curve Cis defined by N() = e vr
. ii) Curvature and radius of Curvature:

‘The curvature x of Cis defined vo. Furthermore, the radius of

curvature, p ata given point is given by Ad
Examples: Find the unit tangent, unit normal, curvature and give the radius of
‘curvature at the point 1=f, forthe curves whose parameterization is given .
a(t) =cos3A-Aj+sin3ikst, =a b)r(0)=sine'i+cose'j~ Skit, o

9 nd =2äH elek =0 ayrt)=2n+j+inrk;t,
OIrt)=eire'j+/Zik;n=0 D eQ=C +4)i+ 2
Solution: \

à) r'=-3sin36-4j+300s34k:r (Of = VOcos” 31+ sine 3 +16 =5

PO din, 4 ur
H =
re sen,

Agia, TER m.

Soden" mean CPE
2%

Ai Ge SA ce
ns DRE NET

«>

—

RAR

3,5 Tangent and Normal Vectors and Curvatures

Let C bea smooth curve with parameterization r(() ‘onan open interval I.

Then, we have
i) Unit Tangent Vector For ny point tin J wherer'(Q #0, the unit tangent

vector T(#) to the curve Cis defined by7(r) = Fa A

ii) Unit Normal Vector: If r(is Smooth and 7°() #0, the unit normal

vector N() to the curve Cis defined by N() = ror

ol

. ill) Curvature and radius of Curvature:
‘The curvature x of Cis defined wat. Furthermore, the radius of

curvature, ata given point is given by a=
Examples: Find the unit tangent, unit normal, curvature and give the radius of
curvature atthe point =f, for the curves whose parameterization is given .
a)r()=cos3-4j-+sin3k:ty =a b)r(t)=sine'i+cose’j—Sk;f, =0

0 d)r()= 28+? j+intk;
DP) aC +4)i+ 2s =1

o OP EP kit

dr=eire"i+ ait
Solution: 7

a) ¥'(t)=~3sin3i- 45+ 3c0531k;f'(0] = V9cos? 31 +9sin*31+16=5

TG
NEE sin3ik;x(t) = A

TO) „-dsmei-esmei__snei-snei
SO €
role
x)= Fol
oro= meager; RT +1 = rr =2+0

4 42. 2
Here, T(1): pa
O ea Gry one

A a
al;

“QF

mare
+
ro TO Re, die, e

FO er ea ea
D T= tdi 1

ae Fat Pe E

Le =

KT
nd Vector Fields

3.6 The concepts 0 Scalar a

3.6.1 Divergence and Curl of Vector Fields

rector to each point Pin a region R (in a plane or in

‘A function that assigns a Y
denoted by F= M+Nj+ Pk.

space) is called a vector field and usually
a) FY) 2x1 + y’jisa vector field in a plane

D) Fxy,2)= 271+ + ksa vector field in space.

Note that a vector field F formed by the gradient of a scalar function fs sid
to be gradient vector field and denoted by FaV Safir Sit be
Definition: Let F=Mi+Nj+ PK be a vector field such that all the partial

2. OMAN oP
derivatives —» = exist. Then,
ey ae

1) Divergence of Vector Fields:
‘The divergence of F denoted by divF orv.F is the scalar function defined by

de +N, +P.
In a plane, where P=MI+NJ, dvF =VF=M,+N,-
in) The Curl of Vector Fields:
or vxF is a vector function defined

Da eur of F which is denoted by carl
¿A Di aM,
by orlF = Vi Fey ze ae ye

Ve can express, ar In compact formas ar ==

Ze o -
ZQ\o~
Por

in a plane, where F =MI+N) air =(N, -M,)k
0, then the field Fis sid to be divergence fre o solenoidal.

ii) If carl F=YxF=0, then the field F said to be irrotational.

141

Examples: Find the divergence and fhe following vector fields.
JESSE D) FR D=ri+(Gx+prt)j-22k
yF6yD=7i-3 + Yzk d) F(x,y,z)=e" cos pie" sin yj+ay'k
OFQuy2)=3rityj-y2k N F(x,y,2z)= 91+ y jeoyzk

Solution:
De Aa Then,

i) divF = a O y +10

D ari=(N, -M,k=(09* -39*)k= 79"
b)i)divF =M, +N, +P, =2x+2"-62" =2x-52

i air =}

O) dvF=M,+N, +P, =-3x+y*
ij
pp 2
lex ay
je? -3y y
d)i)divF=V.F=M, +N, +P, =e" cosy-e" cos y=0

i ik

: 2 2 Alora
DR il

loosy -esiny 29"

6)i)divF =M, +N, +B,=1x-y° ii) oulF =2 + ye

2.Find a if F(x,y,2)=¢" sin y+ (2ay +e" 006 y))—Grks divergence fee,
Solution:

CPR 2 =0=a=
dur = Ces) Say cy ee 02a=3

= 2yi+3z7j-3yk

k
i) arr =| a

2|

- ET IRIARTE AA
3.6.2 Conservative Vector Fields and Test of Conservativeness
Definition: A vector field Fis said to be conservative if there exists à
differentiable function f such that F = Vf. The function f with this propeny

is called potential function for F .
Example: The function F(%,y,2)=3x
there exists a potential function [(x,9,2) =’ +3’ +?

Test of Conservativeness:
2) =Mi+Nj+Pxk is conservative vector field if and

21+3y*j+3z"kis conservative because
? for F such that F=Wf.

A vectror field F(x,y,
ti fsck
> 0.0 2
E MA
only if curl F(x, y,z): y e
IM N P
“Test of Conservative la the plane:
ma
=,

In a plane, F(x, 3) = Mi+ Nj is conservative only if curlF =0 > 7,
Procedures for Finding potential Function:

If a vector field F(x, y,z) = Mi +N.j+P.kis conservative, we can find its
‘potential function f such that F = W = f,4+J,5+ fk =Mi+Nj+Pk.
That means f,i+ J,I+ 5,6 =Mi+Nj+PK= f, =M, f, =N,f, =P.
Step-1: Variable separation: That means express the function f as a sum of
three functions in the from f(x, ¥,z) = K(x y, 2) +80%,2) + Wz).

Then, f, = hye Sy = hy +8, Sz = hy +8, +).

Notice about the functions k,g,h in the decomposition!

‘The functions are separated in such a way that # is the function of all the three
variables x, y,z the function g is the function of two of variables y,z and thus

g,=0. Again, his a function of only variable z and thus 4, = 0,4, =0.

een rete eee
=Mwith respect to x '

Step-2: Integrate /, =
That is k,=M> [ka
Step-3: Integrate f, =
1,4, +8, END 8, NE, = fe,dy=[(N-k,)ay

= 802)=J(N-k dy
Step-4: Integrate f, =k, +g. +h'(z)=Pwith respect to z. That is
K, 48,48 @)=P>M(2)=P-k, -~g, = [wd = [(P-k, -g,)dz

Ma = (x,y,z) = [Mar

iy +8, =Nwith respect to y . That is

Finally: Write f(x, y,2)=k(x,y,2) + 80,2) + H(z) using the above results.

Examples:
1. Determine whether the following vector fields are conservative or not for
these which are conservative, find the potential function.

a) F(x, y,z) = 4xe"i — 6 yj +(2x7e" -9)k

b) F(z, y,z) = 2xcos yi +(6yz - x? sin y)j + By? -4z)k
©) F(x,y,2) = (2xy +2*)i+x*j+(2x2 - 6 0083z)k

d) F(x, y,2) =(2y +e")i+ (x? +2yz)j+ (7 +x0*)k
©) F(x, y) = e+ (xe + y)j

SD) Foyer pe + y

zm
8) F(x,y,2)=-i-Fj+—k
yom Y

Solution:
a)Here, M=4xe*, N=-6y, P=2x%e* -9
isa eos k
a lo à ala à 2
(st Conservativeness: ewlF=|— a y El
M N P| fare -6y 2re-

Therefore, the vector field
Determine Potential Function:
Variable separation: Suppose J(x,y,2) = k(x,7,2)+ 80,2)+ (2).

First: Evaluate k(x,y,2)= fi Ma.

That is k(x, ‚n2)= [Mar = [Are’dr= 2x".

Second: Evaluate g(y,2) = [(N-K,)4y

Thatis g0,2)=[(N-4,)dy = [(-6y-O)dy= [6rd y= 39.

Third: Evaluate A(z)= |(P-k, -g,)de.

Thatis h(2)= [(P-k, ~g,)de= | Qx%e' -9- re” Ja = [-9 =-92.
Finally: Express f(x, yz) = k(x, y,z)+ g(y,2) + h(z) using the results.
Therefore, f(x,y,2) = 2x’e' -3y? 92.

b) Test Conservativeness: Here, M = 2xc0s y,N = 6yz—x*sin y,P=3y* 42.

ij kl | a i k

o 6 alla a a

ir=|2 2 2/2 7
‘Then curl F 0.

MN PI Preosy 6y7-rsny 3y°-4
‚Therefore, the vector field is conservative.
Determine Potential function: -
Variable separation: Suppose {(x, y,2)=Kk(x, y, 2)+ 8U,2)+H2).
First: Evaluate &(x,,2)= [Mar
That is K(x, y,2)= [Mar= [2xoos ye = x? cos y.
Second: Evaluate g(),2)= [(N-k, Jay
Tits 80,9 [ONE y= [Corr sy 43°08) = [may dye,
Third: Evaluate H(z) = (Pk, -8,)&.

—4zde =—22

Finally: Express f(x,

Therefore, f(x,y,2)

¢) Test Conservativeness: curl F=| À

Paytz? x 260053
‘Therefore, the vector field is conservative,
Determine Potential Function:
Variable separation: Suppose f(x, y,2) = k(x,y,z)+ g(y,2)+h(2).
First: Evaluate k(x,,2)

That is k(x,y,2)= [Made = [Cy +2")dr=x"y 4.
Second: Evaluate g(y,z)= {(N~k, )dy
That is g(y,2)= [(N-£ dy = [xt -x")dy= [Ody =e.
Third: Evaluate h(2)= [(P-K, - g,)de.
Thatis W(z)= |(P-k,-8,)d= [ (2x2 600532 -232)de
= J-6cos3zde=—2sin3z
Finally: Express (x, y,2) = A(X, y,2)+ 80,2) + (2) using the results,
Therefore, f(x, y,2)=x"y +22? -2sindz.
4) Test Conservativeness: Here, M=2xy+e",N =a" +2y2,P=y* +e"
Then, 22 aN aP _ aM ON _aM

y &'& aa y
Potential function:
Variable separation: Suppose (2,32) =(%,J,2) + 80,2) +2).

Fit: Evaluate k(x,y,2) = [Mde.
Thtis (3,2) = [Mac [Oy tet ama.

EEE A Sn

'5

id af do Si
Second: Evaluate g(3,2)= |(N -K,)d

That is g(»,2)= [(N=k Jay = (+22 dy = fasse ee
Í g.)k

Third: Evaluate A(z) = |(P-k,

se Pé = fade =0
80,2)+h(2) using the results

Hence, its is conservative,

Yen 2g0)=y>20)

Mae = flx,y)=2e +80

Le.
2

D First: Evaluate k(x,y,2)= [Mdr

Hence, f(x,y) = xe” +

That is k(x,y,2)= [Mde= [2

Second: g0,2)= [ON--4, )dy= [(1y2* -3"y2")dy= [Ody=e.

Third: A()= [P-k,-g,)e= f(e'y"z-2"y"2)de= fade =0. ii

Therefore, f(x, y. prete. la

8) Similarly, we get Fis conservative and fx, y,2)= es y

y

2. Find a so that F(x, y,z)=2ayi +(x" -3092)j+ (y? —3a)k is conservative.

Solution: To be conservative, we have curlF =0. y
i j k

O CO Iavsanı

; a à Ej +0 :

Pay x-397 y -

> y(3a+2)=0=a=-3

o 2

GST Nai eh ns ri
3.7 Line Integrals of Scalar and Vector Fields

3.7.1 Line Integrals of Scalar Fields

Definition: If fis a continuous scala field on a smooth curve C parameterized

by r@on [0,5], then the integral f(x, y,z)dsis said to be the line integral of
4

Jon C. Itis defined as f(x, y,2).ds = 600.20)" or.

As we see from this formula, to evaluate line integrals;

First: Compute ds = [r'(r)ldt where Irol= 00 +20

Second: Use the formula: L Say) = [D Oder by

replacing x,y,z in the function f with their parametric expressions.

Examples:
1. Evaluate the following line integrals

2) [O9*-332-a"2)ds where Cir()= 24H 215152.
») [+ yds where C:r()=24 +1" 0151.
©) [ 3<*yeds where Cort =v je2Pkosts1.

9 (Bas where Ceri)= 2006 + Ising) +h, 0sts2n.
Pay

9 [G+ay)de where C:r(t)=4oosti + singj+3ik,0<t <a.
D [rra where C:r(t)=(€' +Di+e -DJ0S1SIn2
Solution:

9) First: Compute ds = (ét: Here, r(()=24 +1 +20

Then, (0) = 214+ j+ 2k fe (Of = 8174 =3 ES

DATS

RE a
Express the integrand f as a function of / and apply the formula,

Secon:
2H +1j+21k we have x=2,y=1,2=21,

From the parametrization r(()=
Then, f(a,y,2)=89?-32-72 > /0=16

A
Therefore, | f(x,y2)45 O 60) = 302-27 <a,
i 1
b) Here,r'()=21425=) O]=V4+4? awe?
f
Hence, Lande] serra rra
0 à

61-81 = ~6"

i 1, 13
-2a neue] 2204 )=5
Or O=1425+27k fr Ofavirae 4 = (+27 =1424
1 ñ
Hence, [Ju »,2) = [1600.20 olas = [24° (1+ 20?)
2 ,

ora
fer arzt
dr) =-2sind+2cosg+K>|r@]= Vasin? r+ 4005" 1 + #5

RE ea Gr,
He N ar aa -+ fe dia

€) x{0) = 4cost,y(1)=4sint,2(1) =31,1"(0) =Asini+4c051)+3k,
ropa lo as
Hence, je Ses, node f(sHécsrspsdr= 25e
Nrozetediapr fale +e eV = Vie
Lerna-fe mena feta ez =n

5
4 (I-e*)

2 au the following line integrals.
Ré where C:x= DAS Osts3,

D [tae where Crsin+sing sink Ose <=.

9 [09-22 tod where C:r(t)

Atftiisrs2,
[+ Is where Crens'h+sh/0sr< 2.

Solution:
noise edit
Fiat

Hence, [465 y)ds = [reo on ares un n

m
A Ti = Blo
Se cost20 010157 [ro] = poss = Boost

So, [ f(e,y,2)d5= ot t-+siné)costdt = Je - -£

OrO=i+j+k == Viti] = Band f()=78 2 +7 268.
3

Tenir, [ ras = | foe Of = sf oar =2 A =
1 1

0) |'(0] =3 feos" rsin® ncos? 1 + sin” 1) = 3Vc0s*tsin’ = 3postsing|

Suez cos/20,sinr20 0005157 # = Hosts Seostsint

2 P
Hence, f(x, , dé =3 [oos!rsin ad +3 [sin'*1costat
: d

A

3, Think About: Suppose [.(5+y)ds = 307 where Cis a circle centerg
a

the origin. Find the radius of this circle.
Solution ‚red at the origin, it is parameterized by

since C is a circle center
ES
[6+ná= [Gersinfdt=107
è

ri) =reosti-+rsing, 01S 2-Hence,

Then, using the given value we havel0ar =307>r=3wnits.

Remark: In each of the above examples, we are given the curve Cwith is
parametric equations. But in many cases, we may need to evaluate a line
integral in which the curveC is not given in parametric form. In such cases, we
use the basic parameterization techniques that we discussed in Section 3.4.
Examples: Evaluate the following line integrals.

a). [ (e+ y+2)ds where Cis the line segment from (0,L1)to (2,2,3).

+) [pos where Cis the circle way

9 [erde where Cis the upper half of the circle x?+y?=25 in CCW
direction:

d) IN where Cis the graph of y = x’ for 0Sx<1

©) [,@~5y2)db where Cis the line segment from (1,0,1) to (0,3,2).
Solution: In each ofthe problems, the curve Ci not given in parametric form.
So, our first task is to find the parametric representation of the curves.

2) Using the parametric equations for line segment, the line segment from
(04,10 (2,2,3)is parameterized by

r=2y=1+0,2=1+2, 0StSh ro) =2+j+2=>]r(0]=493=3

1 1
Hs, [ 1059.24 = | 140008 Of =farsa-Z
a >
b) The circle xi 4 y? = dis parameterized by x=20061, y = 2sin,Osı 527
and thenr'() 2sin + 20051 =] (0) =2

fr
=2 |4costsinndt =4 [sin id = 20062" =
penes, [ds Í cos 2" =0

¢) The upper half of the circle x? + y? = 25 in CCW direction means the semi-
cie from the postive x-axis to the negative x-axis as shown below,

The circle is parameterized by C:x=Scost,y=Ssint,0<t & x (observe that
we used the interval to be O<¢<r but not O<1<2m because C is the
semicircle not the full circle) and then r'(£) = -Ssins + 5c0s j >lr(=5

Hence, [(x+ y)ds = ficos: +Ssinnar =25(sint —cosr) = 50
>
d) Here, 4 ‘Suggests a parameterization of x=t,y =f for O<1<1.
Thus f(x,y) = +99 > [(9=V1+90 OSB
Pr

ts Sa ae De 8

+ D I
9) The line segment from (1,0,1)to (03,2)is Parameterized by
al-hy=3yz=141, 05151, P()=1+3j+ fr = VTT

1
Hence, _#(3,y,2)ds = Tis —1)-15101+0)jat= fair -9nar
> 4
'
sie? e), „a

2

|
|

fr
Here, [ nus =2[ seostsintd = Jan =- 202 =0

©) The upper half of the circle x* + y? =25 in CCW direction means the semi
circle from the positive x-axis to the negative x-axis as shown below,
. y

The circle is parameterized by C:x=Scost,y=Ssint,0StS x (observe that
we used the interval to be O<t< but not 0S1S2r because C is the
semicircle not the full circle) and then 1'(/) =—Ssindi +5cos =>] r'(0)] =5

Hence, [+ = 5] Sens + Sima = (int xs =50
2

d) Here, y= x? suggests a parameterization of x=t,y=f°for OSt<l.
Thus pe SEO
ñ s|
to [rm je Ef +
° ho
à Tene segment fom (10,1)t0 (03;2)is parameterized by
alot y=3 2=1+1,0S15L, O O Til
1
ees [fared isis = AT a? oa
+ D

er -2¢

e Integrals of Vector Fields

Mi + Nj+ Pk be continuous vector field defined on a
8.

Definition: Let F
smooth curve C parameterized by r(t) = x(Ni+ y()j+z()k,a Sí

Then, the line integral of Fon C denoted by [far, is defined as

[Far=[FO,20,20)1'Odt.
: *(0)i+ y'()j + z'(Ok and the dot product

More precisely, using r'(1) =
À
operator in Fr (dr, we get [Far = [(M «(+ 5 (0+P2 Mr.

À, we have

Using the notation, x'() = E we

* A
[fa = [F6 »0,20)r Oat = || ( ue + ve + ri

Evaluation Procedures: To evaluate [Fr ;
First: Express the given field F as F = F(0) in terms of the parameter ¢ and

evaluate the dot product Fr'dt using r'(¢)=x'(Oi+ y'(Oj+2'(Ok.

Second: Evaluate the integral using, [F.dr = | F()r' (dt.

Examples: Evaluate the line integral [Fae where

a) F(x,y,2) =8x°yzi+52j-Axyk where C:r(()=54+Pj+Pk0S1 Sl.

5) F(x,y,z) = 22 + xj+ y"k and C is boundary of x? + y? = 4in the xy plane
©) F(%,y,2) = 2i- y —xk and C:r(0) = Si -sint—cosk,0 <1< 7/4

in the xy plane.

me
d) F(x, y,2) = 3yi +3xj and C is the ellipse re E

©) F(x, y,2)=ye"i+xcos y'e”k and C:r()=ei+ej+3k,0<r<1.
Cite j+3k0S1s1

I Fx, y,2) = ye" it xe” j+ x" kandC: r(r
AAA 5

solution:
d_ à

=4yN=F 2) =P, do, d _
Here, 0) = 2, =F, 2(0) a Ya ur

Fromthe field F, M(x, y,2) =8x"yz, M(x, y,2) = 52, P(x y,2) =-49.
SM) =", NW) =50,P)=-4 Fu) = 8145? jar

Wo de dye)
wwe [par [ug en raf tain
b)Inthexy plane thecircle Cis parameterized by , r(s) = 2 cos + 2sinÿ,z=0
for0S1S 2 sothatr'(t) =-sinti + 2008 ÿ.
Besides, M(x, y,2) = 2z, N(x, y,2)= x, P(x, y,2)=y?,
Then, M(t) = 0, V(t) =2c051, P(t) = 4sin* rand F(0) = 2c0s + 4sin° dk

y xy 2
Hence, [Fa = [FW (Ode = [ 40s" ıdı = 2 [(1+ cos 29) =4r

2 o a
2 j*_102-21

lo 4

6) Here, [Far = [(-sintcost—Ssint)dr os
a

4) The ellipse is parameterized by r(t) = 3costi + 2sinÿ,0 <1 < 2x

Hence, [ Fr = frorwa = Teıssn’ 1+18c0s? )dt
= ‘ 2% e
=18/(cos*1—sin* )at =18 [cos 2d = 9sin 24, =0
a >

j m
o [Far= je 2 + costero” )0jar [ee di= de L = ke à

Dee ad 026,10 aE 5
Using the substitution, u=e” => du =2e”e” dt we have

j flee see he dae | sé e.
[Far=[rorod=flee +e | h

EAN 15

DS

Additivity of Line Integrals:
‘Suppose an oriented curve C is not smooth but piece wise smooth composed of

smooth curves C,,C,..,C, such that C=C, UC, U..UC, as shown

Then the line integral over Cis given by f Far =) [Far.
ë

ma

Particularly, for C=C, UC, UC,, [Far = [Fdr+ [Far + [Far
E a 4 à

C: Piecewise smooth curves

D C=GUCU..UC, M c=aUGUC,

Examples: Evaluate [Far where
a) F(x, y) =xi+ yj and Cis the triangle in a plane with vertices (0,0), B(,0)

and C(0,!) oriented counter clockwise (CCW).

D F(x,y,2)=x1—yj+(x+2)k and Cis the triangle in space with vertices
| A0,0,0),B(0,,0) and C(0,0,1) oriented counter clockwise (CCW).

Solution:

3) Con:

the diagram to determine the parametric equations of the curve.

Gix=ty=00<rs1
Gix=l-1y=10s1s1
Gix=0y=1-10<1<1

onc Jaja

f ı
on c:]par= [ancora] | (==

mec, [Far = [Fae + [Far+ [Far=3-

+) Consider the One to determine the parametric equations of the curve.

C=C,UC, UC, where

(c= (00,1)
Beil! 0<ts

e NA G:x=1-ty ts
G:x=0,y=1-12=10S151

Fang G:x=ny=0z=1-s0<tsl

Fairs [id=

On G [pan] fta-ne-n-nar “ju =-0nC,
onc, :[far= jean -Djdt= je =F

Therefore, [Fdr= = [pact [pars [Fae ie

3.7.3 Line Integral in Differential Form
A line integral of the form [Ma+ Ndy + Pdz is kown as differential form.
ë

The procedure of evaluation of this differential form is the same as the
procedure of a line integral forthe vector field F = M+ Nj+ Pk

f
1022227
Mara

That is [ Mae + Nay + Pas = [F
E

Examples:

1. Evaluate the following line integrals

0) [soe + (er) +2 de where Cir) =(+Di+((-Dj+Pk-1s182

0) [ (et + yd (x+y), where Cis ane segment from (0) o (23).
O) [rod +642 #74, where Cis ar ofthe parabola y=xtin the

planez=2 from 4(0,0,2)to B(L12).
d) | yde+ 2dy +a, where C consists of the line segment from (2,0,0)to

(6,4) followed by a line segment from (34,5) to (3,4,0).
©) [+20 p+ (x+.x7e" dh where Car Vii+ingj +i sts4
Y) [ -spete+ dy He where C:x= 0051, y=sin/,2=31,0<1<2x.
Solution: .

a) Here, x()=1+1, (0) =1-L,2(0)=
Besides M(1)=1?-1,N(Q=0 +1+1,P(0)=14

2
So, Sparer agree [Peret +20 ja
7 4

ph

1

= {er +2? soja =| 2,
fe ars) Grete

ey
2

sis

py alin pment inte plane fom (Qi)t0 (23)is parameterized by
142 OSI land de= 24h dy = 2c

yd

pesos LC + y)de+ er 2y)y= 2 (€" +8r+3)dr=e*+13

her,

the curve from 4(0,0,2)to B(1,12)is parameterized by
2 222, OS1<land dx=dl, dy =2dl, de =0.

Hence, [pd + er? (0 +20 +40)dt ==

2 [can where Cis consisting ofthe ine segment fom 2.0,0)t0
(345) flowed by aline segment from (345)t0 (340).
Her the line segment from (2,0,0) to (3,4,5)is parameterized by
Gu=t+2,y=4,2=5%, 0x1 Sland the line segment from (34,5)10 (340)
isparameterzed by C, :x=3,=
Evaluate the line integral on each segement separately:

086; yess [Ear [iras = es 10; À
: °
y

On yo e+ 2dy ade = [Fr = | -15dt==15

‘Thus, by additivity of line integral, [Fate = [Fr+ [Far =

loreto +0)dt=2In4+15
‘ade sin dy costa de = 3.

Nes, [otra = esa ernst a)a= four =1r.
: :

DR |
y

2, Evaluate the following line integrals
a) [ote + or and Cirli
b) [edad + neds and C:r()=fi+4j+2ik,-Isısi

"+ ejem sts)

DJ Saar + y dy +37 de where C: =sinf, y =0051,2= "osısz

a AE where Cr +
ey

e) [pde—ady where Cis the parabola y =x* for OSx$3
Solution:

ae (ira t-te [0-20 -T- =
4 è h

ı a 6
byHere,[ etd taney + gade= [(e + +4? yar =e ++

of ar. ya +32 = [era

e) Here, the curve is parameterized by x =

2
o fra

RE Tamez zz, PA
314 Fundamental Theorem of Line Integral

(For Conservative Vector Fields)

Let C be an oriented curve with intial point P(x, Ju,£o) and terminal point
QU, y,2)as shown in the diagram below.

Fundamental Theorem of Line integral:
Let F be a conservative vector field with arbitrary potential function f (that is

F=Vf). Then, a line integral from P to Q along a curve C is given by

[rara [yr d= 10)-10)= Ses )

?
This theorem tells us that if a vector field is conservative, then the line integral

depends only at the end points of the curve but no! he parameterization.
‘The basic advantage of this Theorem is that it helps us Lo evaluate the line
integral without finding the parametic equations ofthe curve. Even when itis
imposible to find the parameti equations of a eue, we can evaluate the ine
integral [F-dr provided that Fis ‘conservative.

Remarks:
i) If Fis conservative, then there is a potential function f such that F = Y.

ji) If Fis conservative, then the line integral [F>arisindependent of path.

2)- fo Yor%

ipa vector field F is conservative, then [F-dr =0 for every closed path.

160

474 Fundamental Theorem of Line Integral

(For Conservative Vector Fields)

Let C be an oriented curve with initial point P(x, y4,z9)and terminal point

04%), 2)as shown inthe diagram below.

95,32)

Fundamental Theorem of Line integral:
Let F be a conservative vector field with
F = Vf). Then, a line integral from P to Q along a curve C is given by

SCH)

arbitrary potential function f (that is

[ra jra=[ va 10-10=10
This theorem tells us that if a vector field is conservative, then the line integral

depends only atthe end points ofthe curve but not the parameterizations.
at it helps us to evaluate the line

The basic advantage of this Theorem is Uk
«gal without finding the parametric equations of the curve. Even when itis
impossible to find the parametric equations of a curve, we can evaluate the line

integral («dr provided that Fis conservative.

Remarks:
i) If F is conservative, then there is a potential function f such that F = Vf .

ji) If F is conservative, then the line integral [Faris independent of path.
iii) Ifa vector field F is conservative, then free =0 for every closed pathC.

ERATION 100

Allon tok pp Matematica Brash Fo yout to RSS
Examples
1. Verify that F is conservative and evaluate [F dr where
a) FU y2)= (2042 lija ae —arsinz2)k and Cis a curve from
(3,-1,1)10 (23.1 al
b) F(xy
©) For,
8) [pd + x'y'dywhere Cis acurve along x = 4,7 =3+15,0<1<1

ga certain path

and Cis the curve from (:

nite +y)j+ L4,0)10(1,2,4)

xzj+ayk and Cisa curve from (1,2,3)10(2,3,4)

9) |, y'de+39 "dy where Cis a curve along = from (LI)t0(2,4)

Solution:
a) Test Conservativeness:

=(22-22)j+(2x-2x)k=0

Plor

Pxy+z x 2uz-zsin.
Therefore, the vector field is conservative
Determine Potential Function:
Variable separation: Suppose f(x, y,2) = k(x, y,2)+8(),2)+h(2).
First: Evaluate k(x, y,2)= f Mar.
That is k(x,y,2)= [Mar = [Oy +2Mdr=x" yc.
Second: Evaluate 80.2)= [N-k,Jd
That is g0,2)= (0-40) = f(x) = [ody =e.
Third: Evaluate h(z)= |(P-k, - g, az.
Thatis Hz) =|(2= -sinz- 2x) = [-rsin mc = 005
Finally: Express f(x, ¥,2) =K(x,),2)+ g(y,z)+ A(z) using the results,
Therefore, f(x,y.2)=x7y +22" +0087.

| und Book of Spiel Mathematics by Basho Y. For por comments ad
ence, by Fundamental Theorem of Line integrals,

ons use OSSD

[peda] Par Sad JOH
+) Clearly Fis conservative. (Jusüfy!). Then,

3+7=20

J such that? =

ner some ups and downs we get _/(s,942
Hence, by Fundamental Theorem of Line int
(F-ar= [0-4 028-1140

©) Clearly F is conservative. (Justify!). Then, find f such that # = Vf
Here, we get (x, 352)

Hence, by Fundamental Theorem of Line Integrals,
[fae Ya 1039- f023=24-6=18.
4) In this case the line integral is given in differential form [Max + Nay.
For such situation, conservative test is applied by assuming as the force is of the
form F = Mi+ Nj. in our case, Fer yee yj.
Hence it is conservative. Hore, we get /(4,y) = qx". Besides, find the end

points of the curve. Observe that the end point of the curve corresponds to the
end points of the parameter. That is P=(0,3) at t= 04 14) at t=1
Hence, [Far = [¿U-dr=F0.9-J0D=8-0 ét.

N, So, Fis conservative,

e) Here, M, =3y",N, =3y M,

Besides, its potential function is f(x.
Therefore, [F-dr= 7024) SU)

2. Given F(x,y,2) =(2y +32)i +0 y + Gr
a) Show that F is conservative
») Find the potential function of F

nd ak Splash. Foro cannes
IB br Brgashae M For sand sega Tr
«by Fundamental Theorem of Line integrals, ds

[Far= [9 -dr=F030-SGzAD=15+
py ClearlyF is conservative, (Justfy!). Then, find Y such that F = Vf.

Henc
0.

‘After some ups and downs we get f(x.)

Hence, by Fundamental Theorem of Line integrals,
[F dr= [we dr= f(l,2,4)—f(-14,0) = 20-12 =8.
€) Clearly F is conservative, (Justify!). Then, find f such that F = Vf
Here, we get f(x.342)= 95 -
Hence, by Fundamental Theorem of Line Integrals,
[Far= [wear = (23,4) f(1,23) = 24-6 = 18.

4) In this case the line integral is given in differential form [ Mes + Nay.

For such situation, conservative test is applied by assuming asthe force is of the

form F = Mi+ Nj. In our case, F(x,y) =p itv.

Hence it is conservative. Here, we get /053)= ix y! Besides, find the end

tof the curve corresponds to the

points of the curve. Observe thatthe end pola
= (14) at t=1

end points of the parameter. Thatis P=(0.3) at t=
Hence, [F-dr= [oar fu9- 03) = 64-0 = 4.

N, So, F is conservative.

e) Here, M, = 3y2,N, =3y >M,=
Besides, its potential funetion is 0%) = x.

‘therefore, [Frdr 04 fü =128-1=127.
2.Given F&,y,2)= 229+ ia HERE.

a) Show that Fis conservative
by Find the potential function of F
o) Evaluate {Fr using FTLI where C is the curve from (0,0,0)10(1,3,)

e
en 162

Lei RTE mE erg
Solution:
i j k
ea? à à [eo
a) Test Conservativeness: CUT & x a

‘Therefore, the vector field F is conservative.
b) Determine Potential Function:
Variable separation: Suppose f(x, 942) = K(x,y,2)+ 802) +z).

First: Evaluate k(x,y,2) = [Mdr

That is A(xyy,2)= [Mér= fe +de y.

Second: Evaluate g0,3=/0: kb

Thatis g0,) = [0-4 = [0-4 jr.
Third: Evaluate h(2)= |(P-k,-8,)&-

Thatis Ha)=|P-k, ~g,)4 = [(Gx—6e* 30d: = [Bede =—3e*
Finally: Express f(x,y,2)=(x,y,2)+ 80,2) + A(z) using the results.
Therefore, f(x, py the Le,
©) By! u Feel Theorem of Line integral (FTLI), we have

els ad sesos se PAR

aso pe Me iaa
Solution:
ie, Mein y, N

MEN gr

Ga yn ae oF is conservative.

sos by part (i) ofthe above remark, the line integral is independent of path

Besides, the potential function is f(x, y) =e" In y. Hence, by Fundamental
on

Tieorem of Line integral, [F.dr = f(0,€°)- f(0)=2-0=2,

ay

+) Clearly Fis conservative. (Justify!) Here, we get /(x,,2)=3ye""

en 0m
Here, [Fdr= [Wdr=023)-0.02)=35-8=25.
wo aim

¢) Clearly F is conservative with potential function
Sx,y,2)= A In(1+x? + y? +2?) (Verify this!). Besides, from the parametric

OS1S1, we have
0<151>P=(00.0)2tt=0,0=(L1)at

equation r(t) = #i +07 j +1)
Hat =F, 20
Therefore, [F «dr = f(Q)- f(P) = 04) (000)=}in4= m2.

AAA MA 164

1, Foro art nd sen
ne Integrals

(and kako ppd Vater
3.7.4 Green’s Theorem for L
Let C be a simple closed piecewise smooth curve oriented counteclockWi

and let R be the region enclosed by Cas shown in the diagram below.

| Green's Theorem

Let M and N be functions of two variables having continuous partial
aN _ aM,

derivatives on R. Then, [ Af(x, y)ae+ M(x, yay = | I

lerivatives J (x, y)de+ M(x, dy Je > de

Altematively, if F(x,y)= M(x,yi+ Ms,pJiwhere Mand N have
E a en _ eM,
continuous partial derivatives on R, then [Fr = [[(E- 4.
e qa o

ions: The curve Cis simple means it does not intersect itself,

i) The con
closed means its initial and terminal points are the same (like a circle, a

triangle), piecewise smooth means the parametric equations of the curve are
differentiable except a finite number of points, oriented counter clockwise
(CCW) means the region R is always lying t the left of the path while moving
around C.Unless and otherwise stated assume the curve is oriented CCW.

ii) This Theorem states that the value of a line integral along a simple closed
and oriented curve Cis evaluated by a double integral over a simply connected
plane region À enclosed by the boundary of C.

iit) One of the advantage of Green's Theorem is that we can evaluate the line
integral [F «de without using the parameterization of the curve C. It is also
useful to evaluate difficult line integrals that cannot be evaluated using the

parameterization of the curve.
pun IRIS WT Mir OR UE emer ALAN ON LAY AAA 165

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