differential equationsdx dy + 2 =-ans: y=e-x+ce-2x 2. 2 sin 3 x x y dx dy x + = ans: x3y=c-cosx 3....

Mathematics: Lecture 1 مدرس مساعد ازهار مالك Differential Equations:

-1-

Differential Equations

A differential equation is an equation that involves one or more derivatives, or differentials. Differential equations are classified by:

1. Type: Ordinary or partial. 2. Order: The order of differential equation is the highest order derivative that occurs in

the equation. 3. Degree: The exponent of the highest power of the highest order derivative.

A differential equation is an ordinary D.Eqs. if the unknown function depends on only

one independent variable. If the unknown function depends on two or more independent variable, the D.Eqs. is a partial D.Eqs..

2

22

2

2

xya

xy

¶¶

=¶¶ is a partial D.Eqs..

Ex1:

35 += xdxdy 1st order-1st degree

Ex2: 522

3

3

÷÷ø

öççè

æ+÷÷

ø

öççè

ædx

yddx

yd 3rd order-2nd degree

Ex3:

05sin4 2

2

3

3

=++ xydx

ydxdx

yd 3rd order-1st degree

Exercise: Find the order and degree of these differential equations.

1. 0cos =+ xdxdy ans:1st order-1st degree

2. 043 2 =+ dyydx ans:1st order-1st degree

3. 22

2

yydx

yd=+

4. 22 2)( xyy =¢+¢¢ 5. xyyy =¢¢+¢¢¢ 2)(2

Solution

The solution of the differential equation in the unknown function y and the independent variable x is a function y(x) that satisfies the differential equation. Ex: Show that y=c1 sin 2x+c2 cos 2x is a solution of y''+4y=0 sol:

y=c1 sin 2x+c2 cos 2x y'= 2c1 cos 2x. - 2c2 sin 2x. y''=-4 c1 sin 2x-4 c2 cos 2x -4 c1 sin 2x-4 c2 cos 2x+4(c1 sin 2x+c2 cos 2x)=0 \ y is a solution


-2-

Note:

The solution in example above is called general solution since it's contain an arbitrary constant c1 and c2, i.e. the general solution of differential equation is the set of all solutions, and the particular solution is any one of these solutions. Exercise:

1. Show that y=3e2x-e-2x is a solution to y''-4y=0 2. Determine whether y(x)= 2e-x+xe-x is a solution of y''+2y'+y=0 3. Determine whether y= x2-1 is a solution of (y')4+y2=-1

Ordinary Differential Equations:

Ordinary Differential Equations are equations involve derivatives. A. First Order D.Eqs.

1- Variable Separable. 2- Homogeneous. 3- Linear. 4- Exact.

1- Variable Separable:

A first order D.Eq. can be solved by integration if it is possible to collect all y terms with dy and all x terms with dx, that is, if it is possible to write the D.Eq. in the form

0)()( =+ dyygdxxf

then the general solution is:

cdyygdxxf =+ òò )()(

where c is an arbitrary constant. Ex.1:

Solve yxedxdy +=

Sol.: yx ee

dxdy

×=

dxeedy x

y =

dxedye xy òò ×=-

ceedxedye xyxy +=Þ=-×- -- òò - )( Ex.2:

Solve )1()1( 2 +=+ yxdxdyx

Sol.:

dxx

xy

dyò ò +

=+ 1)1( 2


-3-

cxxy

dxx

dxy

++-=+

-=

-

- ò ò1lntan

11tan

1

1

Ex.3: Solve (1) )( 2 Lxydxdy

-=

Sol.: 1ddy 1

ddy ,u +=Þ=-=-

dxdu

xdxdu

xxyput ….. (2)

òò =-

Þ=+ dxudxdu

1udu 1 2

2

[ ]

cxeuu

cx

cx

dxduuu

+=+-

+=+

+=+

=úûù

êëé

+-

+-

\ òò

2

11

1u1-u ln

21

1)(uln -1)-(u ln21

12/1

12/1

Exercise: Separate the variables and solve.

1. x(2y-3)dx+(x2+1)dy=0 ans: (x2+1)(2y-3)=c

2. dy=ex-y dx ans: ey=ex+c

3. sin xdxdy +cosh 2y=0 ans: sinh 2y-2cosx=c

4. xeydy+ 012

=+ dxy

x ans: ey(y-1)+ 2

2x +ln |x|=c

5. 12 =dxdyxy ans:

cxy += 2

123

32

2- Homogeneous:

Some times a D.Eq. which variables can't be separated can be transformed by a change of variables into an equation which variables can be separated. This is the case with any equation that can be put into form:

)(

xyf

dxdy

= …(1)

Such an equation is called homogenous.

Put uxuxy

=Þ= y , dxduxu

dxdy

×+= and (1) becomes

)( ufu

dxdux =+×


-4-

Ex.1:

Solve xy

yxdxdy 22 +

=

Sol.:

homo. 1 2

2

Þ+

=

xyxy

dxdy Put u

dxdux

dxdyu

xy

+×=Þ=

uuu

dxdux

21+=+× Þ

uuu

dxdux

221 -+=×

udx

dux 1=× , òò =×

xdxduu

cxcxu+=Þ+= ln

2xy ln

2 2

22

Ex.2: Solve the homogenous D.Eq 02 =+ ydxxdy

Sol.: xy

dxdyydxxdy 22 =Þ= put u

dxdux

dxdyu

xy

+×=Þ=

uudxdux 2=+× cux =- ||ln||ln c

yxc

ux

=Þ=Þ2

Exercise: Show that the following differential equations are homogenous and solve.

1. (x2+y2)dx+xy dy=0 ans: x2(x2+2y2)=c

2. x2dy+(y2-xy)dx=0 ans: cx

xy-

=ln

3. 0)( =-+ xdydxyxe xy

ans:

cex xy

=+-

||ln 3 - Linear

The equation of the form Qypdxdy

=×+ where P and Q are functions of only x or

constant is called linear in y and .dxdy

Find integrating factor ò=Pdx

efI .).( , then the general solution is

ò=× dxQfIfIy . .).(.).(


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Ex.1: Solve xexxy

dxdy

×=-

xexxQx

xP .)(,1)( =-=

xeefI xdx

x 1.).( ln1

==ò= --

Solution is

ò ××=× dxxexx

y x11

cexy x +=

Ex.2:

Solve xyxdxdy

=+ .

P=x, Q=x

2

2

.).(x

xdxeefI =ò=

Solution is

solution theis 1 222

22

222

22

xxx

xx

ceyceey

dxxeey-

+=Þ+=×

××=× ò

Exercise:

1. xeydxdy -=+ 2 ans: y=e-x+ce-2x

2. 2sin3

xxy

dxdyx =+ ans: x3y=c-cosx

3. ydyydxxdy =+ ans: y

cyx +=2

4- Exact

The equation 0),(),( =+ dyyxNdxyxM is said to be exact if xN

yM

¶¶

=¶

¶

General Solution is

)dy containsnot do ( xNintermsMdxc ò ò+=

Ex.1:

Show that the following D.Eq. are exact D.Eq. a) 0)2()23( 232 =++++ dyyxxdxxyyx


-6-

xN

yM

xxxNxx

yM

¶¶

=¶

¶

+=¶¶

+=¶

¶ 23 , 23 22

\ The D.Eq. is exact. b) 0)cos(()]sin()cos([ =+++++ dyyxxdxyxyxx

)cos()sin(

)cos()sin(

yxyxxxN

yxyxxy

M

+++-=¶¶

+++-=¶

¶

\ the D.Eq. is exact.

Ex.2: Is the D.Eq.

2

)( 22

xyyx

dxdy +

-= exact or not?

Sol. dxyxxydy )(2 22 +-=

exact is theD.Eq.,

2 , 2

\¶¶

=¶

¶

=¶¶

=¶

¶

xN

yM

yxNy

yM

Q

Ex.3:

Solve the exact D.Eqs. in Ex.1(a) above 0)2()23( 232 =++++ dyyxxdxxyyx Sol.

c

y

ydxxyyxc

=++

×+×+×=

++= ò ò

23

223

2

yyxy xissolution the3y 2

2 x2

3 x3y

dy2)23(

2

Ex.4:

Solve 0)()( 2 =+++ dyyxdxyx Sol.

exact is theD.Eq.

1 , 1

\

=¶¶

=¶

¶xN

yM


-7-

x)dycontainsnot do (ò ò+= NintermsMdxc

c

xy

ydxyx

=++

++=

++= ò

3yxy

2x issolution the

3y

2 x

dy)(

32

32

2

Exercise:

1. (2+yexy)dx+(xexy-2y)dy=0 ans: c=2x+exy-y2 2. (tanx+tany)dy+(ysec2x+secx tanx)dx=0 ans: c=y tanx-lncosy+secx 3. (2xy+y2)dx+(x2+2xy-y)dy=0 ans: x2y+y2x-y2/2=c

Problems: Solve the following differential equations: 1- 0)1(ln 2 =++ dyxydxy 2- 022 =- -+ dxedye xyyx 3- 0)2()2( =-++ dyyxdxyx

4- dxxyxydyx ))(cos ( 2+=

5- dxxyydyxyx )lnln1()ln(ln -+=- 6- 0)12( 2 =--+ dxxydyx 7- 0)cos-siny ( cos 2 =+ dyyxdxy 8- 0)12()1( 22 =++++ dyyxydxy

9- 0)()ln( =+

++ dyy

yxdxyex

10- 0)(21)1( 22 =+++ dyeyxdxex yy

References:

1- Calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت) 4- Modern Introduction Differential Equations, Schaum's Outline Series.


-8-


-9-

B. Second Order Differential Equations: The second order linear differential equations with constant coefficient has the genral

form is:

)(xFcyybya =+′+′′ …(1),

where a, b and c are constants.

If 0)( =xF then (1) is called homogenous.

If 0)( ≠xF then (1) is called non homogenous.

Ex:

1) y''-x2y'+sinx y=0 is linear, 2nd order, homo.

2) y''-(y')2+ y=sinx is non linear, 2nd order, non homo.

3) y''+2yy'=lnx

a) Homogeneous.

b) Nonhomogeneous.

- Undeterminant coefficients.

- Variation of parameters.

a) The Second order linear homogenous D.Eq. with constant coefficient:

The general form is

0=+′+′′ cyybya …(2)


The general solution

Put y'=Dy and y''=D2y in eq. (2) (D is an operator)

⇒ a D2y+bDy+cy=0

⇒ 0y)cbDaD( 2 =++ (using D-operator)

now substitute D by r and leave y then

02 =++ cbrar

is called characteristic equation of the differential equation and the solution of this equation

(the roots r) give the solution of the differential equation where


-10-

a

acbbr2

42 −−=

m

There are two values of r :

1- real (equal and not equal). 2- complex.

Case 1: If 042 facb − then r1 and r2 are distinct (r1≠ r2) and real roots, and the general

solution is xrxr ececy 2121 +=

Case 2: If 042 =− acb then r 21 == rr , and the general solution is:

rxexccy )( 21 +=

Case 3: If 042 pacb − then the roots are two complex conjugate roots βα ir ±= , 1−=i ,

and the general solution is:

)sincos( 21 xcxcey x ββα +=

Ex.1: Solve 032 =−′−′′ yyy

Solution:

032 =−′−′′ yyy

3 031 01

0)3)(1(y , y , 1y , 032 22

=⇒=−−=⇒=+

=−+=′′=′==−−

rrrr

rrrrrr

the general solution is xx ececy 3

21 += −

Ex.2: Solve 096 =+′−′′ yyy

Solution:

096 =+′−′′ yyy

3 0)3( 096

212

2

==⇒=−

=+−

rrrrr

xexccy 321 )( +=∴


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Ex.3: Solve 0=+′+′′ yyy

Solution:

0=+′+′′ yyy

231-

23-1-

1.21.1.41

1c 1,b 1,a 012

i

br

rr

±=

±=

−±−=

====++

23 ,

21-

23

21

==±−

= βαir

)23sin

23cos( 21

21

xcxceyx

+=∴−

Exercise: solve

1. 4y''-12y'+5y=0 ans:y=c1e(1/2)x+ c2e(5/2)x

2. 3y''-14y'-5y=0 ans:y=c1e5x+ c2e(-1/3)x

3. 4y''+y=0 ans:y=c1cos(x/2)+ c2sin(x/2)

4. y''-8y'+16y=0 ans:y=c1e4x+ c2xe4x

5. y''+9y=0 ans:y=c1cos3x+ c2sin3x


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b) The Second order linear non homogenous D.Eq. with constant coefficient:

The general form is: )(xFcyybya =+′+′′ …(3)



If yh is the solution of the homo. D.Eq. 0=+′+′′ cyybya , then the general solution of

eq. (3) is:

integral)r (porticula

function)tary (complemen

p

hph

yyyyy +=

hy )i is y homo.

py )ii (use the table)

Methods of finding py :

There are two methods:

1) Undetermined coefficients:

In this method py depends on the roots r1, and r2 of characteristic equation and on the

form of )(xF in eq. (3) as follows:

)(xF Choice of py M.R. nkx

nth degree polynomial 0

22

11 kxkxkxk n

nn

nn

n ++++ −−

−− L 0

pxke pxce p

xkorxk

ββ

cossin

xcxc ββ sincos 21 + βim

Note: For repeated term (root), multiply by x .

Ex.1: Use the table to write py

1) 2n , 3k , 3)( 2 === xxF

012

2 kxkxky p ++=


-13-

2) c 21-k , e

21)( 3x- ⇒=

−=xF

xp cey 3−=

3) 3 , 2k , 3x cos 2)( === βxF

3xsin c 3x cos 21 += cyp

4) 2c , 3k , 2e - 53x-3)( 3x2 −=−=+= xxF

xp cekxkxky 3

012

2 +++=

5) sin 21 cos2)( xxxF −=

sin x c x cos 21 += cyp

6) 2 cos sin)( xxxF −=

2xsin B 2x cosA sin x c x cos 21 +++= cyp

Ex.2: Solve 242 xyyy =−′−′′ …. (1)

Solution:

y'' –y'-2y=0

the char. Eq. r2-r-2=0

2r ,1r 0)2r)(1r(

21 =−==−+

xxh ececy 2

21 += −

f(x)=4x2 is polynomial of second degree then

212

012

2

2 , 2

(2) ...

kykxkykxkxky

pp

p

=′′+=′⇒

++=

Substitution gives

4) (2)2(2 201

22122 xkxkxkkxkk =++−+−

-3k 022:2k 022:.2 42:.

0012

112

222

=⇒=−−=⇒=−−−=⇒=−

kkkconstkkxofcoeff

kkxofcoeff

3-2x2 2 +−= xy p

322)(yy 2221hg −+−+=+= − xxececy xx

p


-14-

Ex.3: ey2yy x3=−′−′′

Solution:

(1) .... ey2yy x3=−′−′′

0y2yy =−′−′′

1,2 0)1)(2( 02

21

2

−==⇒=+−=−−

rrrrrr

)( 22

1xx

h ececy −+= , Put

3x3

3

9ce , 3

(2) ....

=′′=′

=

px

p

xp

ycey

cey

Substitute In (1)

9ce3x-3ce3x-2ce3x=e3x

9c-3c-2c=1⇒41c 14 =⇒=c

In (2)⇒ 41 3x

p ey =

xxxp eececy 3

22

1hg 41yy ++=+= −

Modification rule قاعدة التعديل

nkxxF) اذا كان 1 .x السابق في py يضرب ← 0 وكان احد جذري المعادلة القياسية = )(=2 ( - a اذا كان pxkexF .x السابق في py يضرب ← p وكان احد جذري المعادلة القياسية = )(=

b اذا كان - pxkexF .2x السابق في py يضرب ← p وكان جذري المعادلة القياسية = )(=

) اذا كان 3

=xsin k x cos

)(ββk

xF0 وكان , == αβir m .x السابق في py يضرب ←

Ex.4:Solve y''+y= sinx

Solution:

y''+y=0

r2+1=0, r2=-1 ⇒ r = ± i, α=0, β=1

yh=c1cosx+c2sinx

yp=x(c3cosx+c4sinx), y'p=x(-c3sinx+c4cosx)+(c3cosx+c4sinx)


-15-

y''p=x(-c3cosx-c4sinx)+(-c3sinx+c4cosx)+(-c3sinx+c4cosx)

Substitution gives

-2c3sinx+2c4cosx=sinx

-2c3=1⇒ c3=-1/2, 2c4=0⇒c4=0

xxxcxcyg cos2

sincos 21 −+=

Exercise: Find the general solution

1) y''=9x2+2x-1

2) y''-y'-2y=sin2x

3) y''-5=3ex-2x+1

4) y''+2y'+y=3e-x

5) y''-y'-2y=x2-x


-16-

2- Variation of parameters.

Let yh=c1u1+c2u2 be the homogenous solution of )x(Fcyybya =+′+′′ and the particular

solution has the form 2211p vuvuy += where v1 and v2 are unknown functions of x which

must be determined, first solve the following linear equations for v'1 and v'2:

v'1u1+ v'2u2=0

v'1u'1+ v'2u'2=F(x)

which can be solved with respect to v'1 and v'2 by Grammar rule as follows

)x(Fu0u

D,u)x(Fu0

D,uuuu

D1

12

2

21

21

21

′=

′=

′′=

and D

Dv,DDv 2

21

1 =′=′

by integration of v'1 and v'2 with respect to x we can find v1 and v2.

Ex.1:

Solve x3ey2yy =−′−′′ ……. (1)

xxh ececy 2

21 += − , hence

xx

xx

eueu

eueu2

22

2

11

2

=′⇒=

−=′⇒= −−

yp= v1u1+v2u2

)(

0

2211

2211

xFuvuv

uvuv

=′′+′′=′+′

x3x22

x1

x22

x1

e)e2(v)e(v

0)e(v)e(v

=′+−′=′+′

−

−

Solving this system by Cramer rule gives

x2x3x

x

2x5

x2x3

x2

1x

x2x

x2x

eee0e

D,ee2e

e0D,e3

e2eee

D =−

=−===−

=−

−

−

−

xx2

xx

x2

2

x4x41

x4x

x5

1

e31e

31ve

31

e3e'v

,e121e

31ve

31

e3e'v

∫

∫

==⇒==

−=−

=⇒−

=−

=


-17-

x3x22

x1

x3x2xxx4p

e41ececy:issolutiongeneralthe

e41ee

31ee

21y

++=

=+−=∴

−

−

Ex.2: solve

y''+y=secx

Solution:

y''+y=0

r2+1=0 ⇒r2=-1 ⇒ r = ± i α=0, β=1

yh=c1cosx + c2sinx, u1=cosx, u2=sinx, f(x)=secx

yp= v1u1+v2u2

= v1cosx +v2sinx then

xsec)x(cosv)xsin(v

0)x(sinv)x(cosv

21

21

=′+−′=′+′

1xsecxcosxsecxsin

0xcosD

,xtanxcos

1xsinxsecxsinxcosxsecxsin0

D

,1xsinxcosxcosxsinxsinxcos

D

2

1

22

==−

=

−=−=−==

=+=−

=

∫∫

==⇒=

=−

=⇒−=−

=′

xdxv1'v

|xcos|lndxxcosxsinvxtan

1xtanv

22

11

yp = ln |cosx| cosx + x sinx

yg = c1cosx + c2sinx + ln |cosx| cosx + x sinx

Exercise

1. y''-2y'+y = ex lnx

2. y''-2y'+y = 5

x

xe

3. y''+4y=sin22x


-18-

Problems: Find the general solution

1- xeyyy x ln42 −=+′+′′

2- -1(0)y , 0y(0) cos15sin23204 =′=−=+′+′′ ttyyy

3- 3(0)y , 1-y(0) 434 3 =′==+′−′′ xeyyy

4- xxyy 2 +=+′′

5- 3

x

xe12yy2y =+′−′′

6- xyy 2 sec 44 =+′′

References:

1- calculus & Analytic Geometry (Thomas).

2- Calculus (Haward Anton).

3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)

4- Modern Introduction Differential Equations, Schaum's Outline Series.


1

C. Higher order Differential Equations: How to find roots of an equation:

Let 022

11 =++++ −−

nnnn axaxax K be an eq. of degree n, we denote this eq. by 0)( =xf then:

1) r is a root of the eq. 0)( =xf if 0)( =rf .

2) r is repeated root of the eq. 0)( =xf if 0)( =′ rf .

3) If r is a root of the eq. 0)( =xf , then r must be a factor of na .

4) If r is a root of the eq. 0)( =xf , then )( rx − divides )(xf .

Ex.: Find all roots of 01834 23 =−−+ xxx

Solution: 18 9, 6, 3, 2, 1, : 18 mmmmmm=na

0186168)2( =−−+=f , 2 is a root of the eq. There are two methods to factorize f(x): long division & fast division.

First method: Fast division 1 4 -3 -18 2 ↓ 2 12 18 1 6 9 0

096 2 =++⇒ xx

3 ,3 ,2 0)3()3)(2( 0)96)(2(

321

2

−=−===++−=++−⇒

xxxxxx

xxx

The roots are 2, -3, -3

Second method: long division

01834 23 =−−+ xxx

0)3()3)(2( 0)96)(2( 2

=++−=++−⇒

xxxxxx

9x6x

18x9 18x9 x12x6

x3x6 x2x

18x3x4x)2x(

2

2

2

23

23++

±−

±

−

±

−−+−

0m

m

m


2

Higher order linear Differential Equations: The general form with constant coefficient is:

(1) ...... )(1)1(

1)( xFyayayay nn

nn =++++ −− K

If 0)( =xF then (1) is called homogenous, otherwise (1) is called nonhomogenous.


The methods of solving second order homogenous D.Eqs. with constant coefficients can be

extended to solve higher order homogenous and nonhomogenous D.Eq. with constant coefficients.

a) Homogenous: the characteristic equation of nth order homogenous D. Eq.: 01

)1(1

)( =++++ −− yayayay nn

nn K is: 01

11 =++++ −

−nn

nn ararar K Let r1, nrrr , ... , , 21 be the roots of characteristic equation then: 1) If nrrr , ... , , 21 are all distinct then the solution is:

xrn

xrxr necececy +++= K2121h

2) If 1r repeated m times, then hy will contain the terms:

... 111 121

xrmm

xrxr excxecec −+++ 3) If some of roots are complex ( βα ir m= ) then yh will contain

xexcxc αββ ) sin cos( 21 +

Ex.1: solve y''' -3y''+2y'=0

Solution:

1 , 2 , 0 01)-2)(-( 0)23( 023

321

223

===⇒==+−⇒=+−

rrrrrrrrrrrr

are all distinct

xx

xrxrxr

ececcyecececy

32

21h

321h

321

++=

++=⇒


3

Ex.2:

y(4)-3y'''+3y''-y'=0

3m 1 , 0 01)-(

0)133( 033

4321

3

23234

=⇒====⇒=

=−+−⇒=−+−

rrrrrr

rrrrrrrr

xxx

rxmmmxr

excxececcyexcxcxcecy

24321h

14

23

321h

)(1

+++=

+++=⇒ −−−

Ex.3:

012223)4( =+′+′′−′′′− yyyyy

factor a is 2)-( root a is 2 012223 234

rrrrrr⇒=

=++−−

1 -3 -2 2 12 2 ↓ 2 -2 -8 -12 1 -1 -4 -6 0

064 23 =−−−⇒ rrr

factor a is 3)-( root 3 , 064)(2( 23 rrrrrr ⇒==−−−−⇒ )

1 -1 -4 -6 3 ↓ 3 6 6 1 2 2 0

022 2 =++⇒ rr

1 , 11 ,3 ,2 0)22( )3)(2(

21

2

=−=−====++−−

βαirrrrrrr

m

xxx excxcececy −+++=⇒ ) sin cos( 433

22

1h

b) Nonhomogenous: the general form of nth order nonhomogenous differential equation is:

(1) ...... )(1)1(

1)( xFyayayay nn

nn =++++ −− K

The general solution is py+= hg yy


4

Methods of finding yp:

1) Undetermined coefficients

We can extend the methods of solving second order non homogenous D.Eqs. with constant

coefficients to solve higher order nonhomogenous D.Eq. with constant coefficients.

Ex.1: y(4)-8y''+16y=-18sinx

Solution:

py+= hg yy

y(4)-8y''+16y=0

r4-8r2+16=0 ⇒ (r2-4)2=0 ⇒ r2=4 ⇒ r=±2

yh=c1e2x+c2xe2x+c3e-2x+c4xe-2x

let yp=Acosx+Bsinx, y'p=-Asinx+Bcosx, y''p= -Acosx-Bsinx

y'''p= Asinx-Bcosx, y(4)p= Acosx+Bsinx

Acosx+Bsinx+8Acosx+8Bsinx+16Acosx+16Bsinx=-18sinx

25Acosx+25Bsinx=-18sinx

25A=0⇒A=0

25B=-18⇒B=-18/25

xsin2518-xececxececy 2x-

42x-

32x

22x

1g +++=

2) Variation of parameters

In this method, the particular solution yp has the form yp=v1u1+v2u2+… +vnun

Where u1, u2, …, un are taken from yh=c1u1+c2u2+… +cnun.

To find v1, v2, …, vn, we must solve the following linear eqs. For v'1, v'2, …, v'n:

)(uv uvuv0uv uvuv

0uv uvuv0uv uvuv

)1(nn

)1(22

)1(11

)2(nn

)2(22

2)-(n11

nn2211

nn2211

xfnnn

nn

=′+…+′+′=′+…+′+′

=′′+…+′′+′′=′+…+′+′

−−−

−−

M


5

Ex2: solve y'''+y'=secx

Solution:

Let y'''+y'=0

r3+r=0 ⇒ r(r2+1)=0 ⇒ r=0, r2=-1 ⇒ r1=0, r=±i

yh=c1+c2cosx+c3sinx

u1=1, u2=cosx, u3=sinx, f(x)=secx

xxxxx

inxosx

sec)(sinv)cos(v)0(v0)(cosv)sin(v)0(v

0svcvv

321

321

321

=′−−′+′=+−′+′

=′+′+′

1xcosxsinxsinxcos0

xcosxsin0xsinxcos1

D 22 =+=−−

−=

xsec)xcosx(sinxsecxsinxcosxsec

xcosxsin0xsinxcos0

D 221 =+=

−−−=

1xsecxcosxsinxsec

xcos0

xsinxsec0xcos00xsin01

D2 −=−=−

=−

=

xtanxsecxsinxsecxcos

0xsin

xsecxcos00xsin00xcos1

D3 −=−=−−

=−−=

∫ +==⇒==′ )xtanxln(secxdxsecvxsecDDv 1

11

∫ −=−=⇒−==′ xdx1v1D

Dv 22

2

∫ =−=⇒−==′ xdxxvxDD

v coslntantan 33

3

yp= ln (secx+tanx)-x cosx-ln cosx sinx

yg= c1+c2cosx+c3sinx+ ln (secx+tanx)-x cosx - ln cosx sinx


6

Exercise: Solve 1) y'''-6y''+12y'-8y=0

2) y'''-y=0

3) y(5)-2y(4)+y'''=0

4) y'''-6y''+2y'+36y=0

5) y(4)+8y'''+24y''+32y'+16y=0

6) y(4)-4y''+4y=0

Problems: Find the general solution y'''-6y''+12y'-8y=0

1) y(4)+8y''+16y=0

2) y(4)+y=x+1

3) y'''-3y'+2y=ex

4) y(4)-16y=0

5) y'''-y'=4x3+6x2

References:

1- Calculus & Analytic Geometry (Thomas).

2- Calculus (Haward Anton).

3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)

4- Modern Introduction Differential Equations, Schaum's Outline Series.

Mathematics: Lecture 4 مدرس مساعد ازهار مالك Matrices:

1

Matrices: When a system of equations has more than two equations, it is difficult to discuss

them without using matrices and vectors. The size of the matrix is described by the number of its row and columns. A

matrix of n rows and m columns is said to be mn × matrix.

[ ] m.1,2,...,j ,n 1,2,...,i ,

21

22221

11211

===

=

×

ij

mnnmnn

m

m

a

aaa

aaaaaa

A

L

M

L

L

Types of matrices: Square matrix: It is a matrix whose number of rows are equal to the number of columns ( mn = ). For example:

224251

×

=A ,

33081123031

×

=B

Diagonal matrix: It is a square matrix which all its elements are zero except the elements on the main diagonal. For example:

=

100090004

A

Identity matrix: It is a diagonal matrix whose elements on the main diagonal are equal to 1, and it is denoted by In. For example:

=

=

1001

,100010001

23 II

Transpose matrix: Transpose of A is denoted by )( TA , means that write the rows of A as columns in At. For example:

23

T

32 3-1172-3

A , 312

173

××

=

−−

=A


2

Matrix addition and multiplication If ][ and ][ ijij bBaA == and both A & B are mn × matrices, then ][][][ ijijijij babaBA +=+=+ Ex.1:

=

+

−7423

5432

2011

For any scalar (number) c , we can multiply A by c as follows:

]ca[]c[acA ijij ==

Ex.2:

−=

−6033

2011

3

A matrix with only one column, 1×n in size, is called a column vector, and one

of only one row, m×1 in size, is called a row vector.

Matrices multiplication Let A be an n× k matrix and B be a k×m matrix then C=AB is an n×m matrix,

where the element in the ith row and jth column of AB is the sum

.,...,2,1,...,2,1,...1

2211 pjandmibabababacn

kkjiknjinjijiij ===+++= ∑

=

Ex.3

Suppose

2332312

173

××

=

−−

=1-1

302-5

B , A then

2210131416

×

−

=AB

Determinants With each square matrix A we associate a number det(A) or |aij| called the

determinant of A, calculated from the entries of A as follows: For n=1, det(a)=a,


3

For n =2, 122122112221

1211det aaaaaaaa

−=

Minors To each element of a 33× matrix there corresponds a 22× matrix that is obtained by deleting the row and column of that element. The determinant of the 22× matrix is called the minor of that element.

For a matrix of dimension 3×3, we define

3231

222113

3331

232112

3332

232211

333231

232221

131211

333231

232221

131211

detaaaa

aaaaa

aaaaa

aaaaaaaaaa

aaaaaaaaa

⋅+⋅−⋅==

.

,,

3231

2221

3331

2321

3332

2322

13

1211

a of minor the is

a of minor the is a of minor the is where

aaaa

and

aaaa

aaaa

Ex.4:

Find the determinant of each matrix

a)

− 52

31

115231

=−

b)

12642

012642

=

Ex.5: Find the determinant of A where:

−−

−=

970642531

A

Sol.: By choosing the first column we get

62

6453

09753

)2(9764

1970642531

)det(

=

−⋅+

−−

⋅−−−

⋅=−

−−

=A


4

Ex.6: Evaluate the determinant of A if:

−−

−=

970642531

A

Solution: By choosing the second row we get

62

7031

69051

49753

)2(970642531

)det(

=

−⋅−

−+

−−

−−=−

−−

=A

Note that 62 is the same value that was obtained for this determinant in Example above. Note:

If a matrix A is triangular (either upper or lower), its determinant is just the product of the diagonal elements: Linearly Dependent and Linearly Independent Definition: the vectors v1, v2, …, vm are linearly dependent if | v1 v2 … vm|=0, and if | v1 v2 … vm| ≠ 0 then v1, v2, …, vm are linearly independent. Ex1: Let v1=(3, 6, -1); v2=(8, 2, -4); v3=(1, -1, 1), determine whether v1, v2, v3 are linearly dependent or not. Sol: Since

068)224()16(8)42(341

261116

81412

3141126

183≠−=+−+−−−=

−−+

−−

−−

−=

−−−

then v1, v2, v3 are linearly independent Ex2: Let v1=(2, 4, 6); v2=(1, 3, 3); v3=(1, 2, 3), determine whether v1, v2, v3 are linearly dependent or not. Sol: Since

0606)1812()1212()69(23634

3624

3323

2336234112

=−−=−+−−−=+−=

then v1, v2, v3 are linearly dependent


5

Exercise: 1) Determine whether the given vectors are linearly dependent or linearly independent.

a) (3,2);(1,-1) b) (4,-3,1);(10,-3,0);(2,-6,3)

2) Find determinant of the following matrices

a)

−

3001020000101014

b)

−−−

−

1231101

123

References: 1- Calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)


1

Solving a system of linear equations Let A be a matrix, X a column vector, B a column vector then the system of linear

equations is denoted by AX=B. The augmented matrix

The solution to a system of linear equations such as

6352

=+−=−

yxyx

Depends on the coefficients of x and y and the constants on the right-hand side of

the equation. The matrix of coefficients for this system is the 22× matrix

−1321

If we insert the constants from the right-hand side of the system into the matrix of

coefficients, we get the 32× matrix

−−65

1321

We use a vertical line between the coefficients and the constants to represent the

equal signs. This matrix is the augmented matrix of the system also it can be written as:

−=

−65

1321

yx

Note: Two systems of linear equations are equivalent if they have the same solution set.

Two augmented matrices are equivalent if the systems they represent are equivalent. Ex.1: Write the augmented matrix for each system of equations.

a) 042

3 25

=+−=+

=−+

zyxzx

zyx

−

−

035

412102111

b) 0

6 1

==+=+

zzyyx

− 561

100110011


2

We'll take two methods to solve the system AX=B 1) Cramer's rule

The solution to the system

222

111

cybxacybxa

=+=+

Is given by where and DD

yDDx yx ==

and ,22

11

22

11

22

11

caca

Dbcbc

Dbaba

D yx ===

Provided that 0≠D Notes:

1. Cramer's rule works on systems that have exactly one solution. 2. Cramer's rule gives us a precise formula for finding the solution to an

independent system. 3. Note that D is the determinant made up of the original coefficients of y and x .

D is used in the denominator for both y and x . xD is obtained by replacing the first (or x ) column of D by the constants 21 c and c . yD is found by replacing the second (or y ) column of D by the constants 21 c and c .

Ex.1: Use Cramer's rule to solve the system:

32423

−=+=−

yxyx

Sol.: First find the determinants yD and ,, xDD :

-178--932

43 , -26-4

1324

7)4(31223

==−

===−

−=

=−−=−

=

yx DD

D

By Cramer's rule, we have

717y and

72

−==−==DD

DDx yx

Check in the original equations. The solution set is

−− )

717,

72( .


3

Ex.2: Solve the system:

532932

=−−=+

yxyx

Sol.: Cramer's rule does not work because

0)6(632

32 =−−−=

−−=D

Because Cramer's rule fails to solve the system, we apply the addition method:

140 532

932

==−−

=+yx

yx

Because this last statement is false, the solution set is empty. The original equations are inconsistent. Ex.3: Solve the system:

14106 753

=−=−

yxyx

Sol.: Cramer's rule does not apply because

0)30(3010653

=−−−=−−

=D

Multiply Eq.(1) by -2 and add it to Eq.(2)

00 14106

14106

==−

−=+−yxyx

Because the last statement is an identity, the equations are dependent. The solution set is{ }753),( =− yxyx . Ex.4: Use Cramer's rule to solve the system:

532

3)1(32 −=

−=+−xyyx

Sol.: First write the equations in standard form, CByAx =+

523

032 −=+−

=−yxyx

Find yD and ,, xDD :

-100--10

5302

, -1515-02530

5942332

==−−

===−

−=

−=−=−

−=

yx DD

D

Using Cramer's rule, we get

25

10 and 35

15=

−−

===−−

==DD

yDDx yx

Because (3,2) satisfies both of the original equations, the solution se is )}2,3{( .


4

2) The Gaussian Elimination method When we solve a single equation, we write simpler and simpler equivalent

equations to get an equation whose solution is obvious. In the Gaussian elimination method we write simpler and simpler equivalent augmented matrices until we get an augmented matrix in which the solution to the corresponding system is obvious.

Because each row of an augmented matrix represents an equation, we can perform the row operations on the augmented matrix. Elementary Row Operation: 1. Construct the augmented matrix (A:B). 2. Interchange two rows (Ri ↔ Rj). 3. Multiply any row by a constant different from zero (Ri ↔ kRi) 4. Add a constant multiply of any row to another row (Ri ↔ Ri + kRj) Ex.1:

Use Gaussian elimination method to solve the system (two equations in two variables):

12113

=+=−

yxyx

Sol.: Start with the augmented matrix:

−111

1231

222111

7031

RR +=′

−

−1-2R

22311

1031

RR7

1 =′

−

−

1213

21001

RRR +=′

−

3

This augmented matrix represents the system 2=x and 3−=y . So the solution set to the system is ( ){ }3,2 − . Ex.2: Use Gaussian elimination method to solve the system (three equations in three variables):

43

632

=−−=−+

−=+−

zyxzyx

zyx

Sol.:

−

−−−

−

463

113111

112


5

1

R

−

−−−

−↔

43

6

113112111

2R

33

22

RR

RR

+=′

+=′

1-3R

1-2R

14156

24033011

−−

−−

−1

2R3

1- 2R

−−−−

=′

1456

240110111

33

11

RR

RR

+=′

+=′

24R

2-R

651

20010

01

−−1

1

3R2

1-3R

−−=′

351

100110

001 3R2R

−+=′

321

100010001

2R

This augmented matrix represents the system 1=x , 2=y and 3−=z . So the solution set to the system is ( ){ }3,2,1 − .

Ex.3: Solve the system

433

1=+−

=−yx

yx

Sol.:

−

−41

3311

13R

−+=′→

71

0011

22 RR

R2 corresponds to the equation 0 = 7. So the equations are inconsistent, and there is no solution to the system. Ex.4: Solve the system

226

13=+

=+yx

yx

Sol.:

21

2613

2R12R-

→ +=′

01

0013

2R

In the R2 of the augmented matrix we have the equation 0 = 0. So the equations are dependent. For ordered pair that satisfies the first equation satisfies both equations. The solution set is { }13),( =+ yxyx

Exercises: Solve the following systems:

1) 13

3−=+−

=+yx

yx

2) 363

12=+

=+yx

yx

3) 22

142

=+−=−+

=++

zyxzyx

zyx


6

Matrix Inverse The matrix A has an inverse denoted by A-1 if |A|≠0 where A.A-1=I. We'll take two

methods to find A-1 where A is an n×n matrix. 1) By Gauss elimination method(Using row operations):

1. Construct the augment matrix (A:I) 2. Use row operation until we have (I:A-1)

Ex1: Use Row operation to find A-1 if A=

4112

1001

4112

→ 11 21 RR = →

10

021

41211

−→−=′1

21

021

270211

122 RRR

−=′

42

71

021

10211

72

22 RR

−

−

=−=′

72

71

71

74

1001

21

211 RRR

A-1 =

−

−

72

71

71

74

Ex2: Find A-1 if A =

−

−

204201

312

−

−

100010001

204201

312

11 21 RR =→

−

−

100010

0021

204201

23

211

→R2= R2-R1

→R3=R3-4R1

−

−

−

−

−

102

0121

0021

42027

210

23

211

→R2=2R2

−−

−−

−

102021

0021

420710

23

211


7

→R1= 1221 RR + →R3=-2R2+R3 33 10

1

140021010

1000710

2401

RR =→

−−−

−

−−−

−

101

520

021010

100710201

→ R1=2R3+R1→ R2=R2+7R3

−

−−

101

520

107

541

51

510

100010001

−

−−=∴ −

101

520

107

541

51

510

1A

2) By Cofactor Method (Using determinant of the matrix)

The cofactor of the element aij of the matrix A = (aij) is defined by cij = (-1)i+j Aij

where Aij is the determinant of the matrix that remains when the row i and the column j are deleted.

To find the inverse of a matrix whose determinant is not zero 1- construct the matrix of cofactors of A, cof (A) = cij 2- Construct the transposed matrix of cofactors called the adjoin of A = adj (A) =

cof (A)T

3- then A-1 = )(det

1A

adj A

4- to check your answer A.A-1 = I or A-1.A = I

Ex.: Use determinant to find A-1 where A =

4112

A-1 = )(1 AadjA

7184112

=−==A

Cof(A) =

−

−2114

C11 = (-1)1+1 |4| = 4 C12 = (-1)1+2 |1| = -1 C21 = (-1)2+1 |1| = -1 C22 = (-1)2+2 |2| = 2

Adj(A) =

−

−=

−

−2114

2114 T


8

−

−

=

−

−=∴ −

72

71

71

74

2114

711A

Ex2: Find A-1 if A =

−

−

204201

312

Solution:

−−

−=

=+=−−

=

172482

0100)(

1082204201

312

Acof

A

1,7)7)(1(,2,4,8,2)2(,0,1010)1(,0

333231

232221

131211

==−−==−=−==−−==−=−==

cccccc

ccc

−−−==

1407810220

tcof Adj(A)

−

−−=∴ −

101

520

107

541

51

510

1A


9

Problems: 1) write the augment matrix to the following systems then find the solution:

a) 3333

22221

−=−+−=+−

=+−

zyxzyx

zyx

b) 8333

122

=−+=+−

=−+

zyxzyx

zyx

c)

5102

2212

41

41

32

2

=+−=−

=−=

xxxxxx

x

2) Find the inverse of each following matrix

a)

−

−

425253112

b)

−−

−

121112311

References: 1- Calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)

Mathematics: Lecture 6 مدرس مساعد ازهار مالك Vectors:

1

Vector: A vector is a matrix that has only one row – then we call the matrix a row vector

– or only one column – then we call it a column vector.

A row vector is of the form: [ ]n21 a ... a aa =

A column vector is of the form:

=

mb

bb

bM

2

1

A quantity such as force, displacement, or velocity is called a vector and is represented by a directed line segment

A vector in the plane is directed line segment. The directed line segment AB

has initial point A and terminal point B; its length is denoted by AB . Two vectors

are equal if they have the same length and direction.

Component form

If v is a two dimensional vector in the plane equal to the vector with initial

point at the origin and terminal point ),( 21 vv ,then the Component form of v is:

),( 21 vvv =

If v is a three dimensional vector in the plane equal to the vector with initial

point at the origin and terminal point ),,( 321 vvv , then the Component form of v is:

),,( 321 vvvv =


2

The numbers 321 and , vvv are called the components of v .

Given the points ),,( 111 zyxP and ),,( 222 zyxQ , the standard position vector

),,( 321 vvvv = equal to PQ is

) , , ( 121212 zzyyxxv −−−=

The magnitude or length of the vector PQv = is the nonnegative number

212

212

212

23

22

21 )( )( )( zzyyxxvvvv −+−+−=++=

The only vector with length 0 is the zero vector )0,0(0 = or )0,0,0(0 = . This

vector is also the only vector with no specific direction.

Ex.: Find a) component form and b) length of the vector with initial point )1, 4, 3(−P

and terminal point )2, 2, 5(−Q

Solution:

a) )1-2 , 4-2 , 35( +−=v

The component form of PQ is 1) , 2- , (-2=v

b) The length or magnitude of PQv = is 3 9 )1( 2)( )2( 222 ==+−+−=v Vector Addition and Multiplication of a vector by a scalar

Let ),,( 321 uuuu = and ),,( 321 vvvv = be vectors with k a scalar.

Addition:

) , , ( 332211 vuvuvuvu +++=+


3

Scalar multiplication: ),,( 321 kukukuku =

If the length of ku is the absolute value of the scalar k times the length of u .

The vector uu −=− )1( has the same length as u but points in the opposite direction.

If ),,( 321 uuuu = and ),,( 321 vvvv = , ) , , ( 332211 vuvuvuvu −−−=−

Note that uvvu =+− )( and the difference vu − as the sum )( vu −+

Ex.:

Let )1,3,1(−=u and )0,7,4(=v , find

a) vu 32 + b) vu − c) u21

Solution:

a) 2) 27, (10, 0) 21, (12,2) 6, ,2(32 =+−=+ vu

b) 1) 4,- ,5(−=− vu

c) 1121

21 ,

23 ,

21

21

=

−

=u

Properties of vector operations:

Let wand v, u be vectors and b and a be scalars.

1) uvvu +=+ 2) )()( wvuwvu ++=++

3) uu =+ 0 4) 0)( =−+ uu

5) 00 =u 6) uu =1


4

7) uabbua )()( = 8) avauvua +=+ )(

9) buauuba +=+ )(

Unit vectors

A vector v of length 1 is called unit vector. The standard unit vectors are:

)1,0,0( , )0,1,0( , )0,0,1( === kji

kvjvivvvv

vvvvvvv

321

321

321321

)1,0,0()0,1,0()0,0,1(

),0,0()0,,0()0,0,(),,(

++=++=

++==

We call the scalar (or number) 1v the i-component of the vector v , 2v the

j-component of the vector v , and 3v the k-component. In component form,

),,( 1111 zyxP and ),,( 2222 zyxP is

kzzyyxxPP )( )j( )i( 12121221 −+−+−=

If 0≠v , then

vvu = is a unit vector in the direction of v , called the direction of the

nonzero vector v .

Ex.:

Find a unit vector u in the direction of the vector )1,0,1(1P and )0,2,3(2P .

Solution k-2j2i )1-0( 0)j-2( )i13(21 +=++−= kPP

3 9 )1(- (2) )2( 22221 ==++=PP

kjikjiPPPPu

31

32

32

322

21

21 −+=−+

==


5

The unit vector u is the direction of 21PP . Midpoint of a line segment

The Midpoint M of a line segment joining points ),,( 1111 zyxP and ),,( 2222 zyxP is the point

+++

2)( ,

2)( ,

2 )( 212121 zzyyxx

Ex.:

The midpoint of the segment joining )0,2,3(1 −P and )4,4,7(2P is

)2,1,5(2

40 , 2

42 ,2

73=

++−+

Product of vectors u & v are vectors,

There are two kinds of multiplication of two vectors:

1- The scalar product (dot product) u.v. The result is a scalar.

2- The vector product (cross product) u×v. The result is a vector.

1) The dot product In this section, we show how to calculate easily the angle between two vectors

directly from their components. The dot product is also called inner or scalar

products because the product results in scalar, not a vector.

Def.: The dot product vu ⋅ of vectors ),,( 321 uuuu = and ),,( 321 vvvv = is:

332211 vuvuvuvu ++=⋅


6

Note:

Ex.: a)

7)2()53(

scalar 7)2(5)1(3)2,1()5,3(=+−⋅+

=+−=−⋅jiji

b)

6)25()43(

scalar 68151)2,5,1()4,3,1(−=++⋅+−−=+−=⋅−

kjikji

Angle between two vectors

The angle θ between two nonzero vectors ),,( 321 uuuu = and ),,( 321 vvvv = is given by

θcos⋅⋅=⋅ vuvu

⋅

⋅= −

vuvu cos 1θ where )(0 πθθ ≤≤

Ex.: Find the angle between two vectors in space

98cos

98cos

441414422cos

1-=⇒=

++⋅++++

=⋅

⋅=

θθ

θvuvu

0 , 11.1 =

⋅⋅⋅

==

⋅⋅⋅

jkkjji

kkjj

ii

kjikjiurrrr

22v , 22 +−=+−=


7

Ex.: Find the angle θ in the triangle ACB determined by the vertices

⋅

=

=+−=

=−+−=

=+−−=⋅

=−−=

−

13294cos

13 )3()3(CB

29 )2()5(

4 (-2)(3))2)(5( CB

(-2,3)CB and )2,5(

1

22

22

θ

CA

CA

CA

Orthogonal vectors Vectors ),,( 321 uuuu = and ),,( 321 vvvv = are orthogonal (or perpendicular)

if and only if 0=⋅ vu

Ex.: a) )6,4( vand )2,3( =−=u are orthogonal because 0=⋅ vu b) kjkji 42 vand 23 u +=+−= are orthogonal because 0=⋅ vu

c) 0 is orthogonal to every vector u since

0

),,()0,0,0(0 321

=⋅=⋅ uuuu

Properties of the Dot product

If wand , vu are any vectors and c is a scalar, then

1) uvvu ⋅=⋅

2) )()()( vuccvuvcu ⋅=⋅=⋅

3) wuvuwvu ⋅+⋅=+⋅ )(

4) 2uuu =⋅

5) 00 =⋅u

C(5,2) and B(3,5) , )0,0(=A B(3,5)

C(5,2)

A


8

Vector projection Vector projection of u onto v

vv

vuuprojv

⋅= 2 …… (1)

uprojv ("The vector projection of u onto v")

Ex.:

Find the vector projection of kjiu 236 ++= onto kjiv 22 −−= and the

scalar component of u in the direction of v.

Solution:

We find uprojv from eq.(1):

kjikjikjivvvvuv

vvuuprojv 9

898

94- )22(

94- )22(

441466 2 ++=−−=−−

++−−

=⋅⋅

=

⋅=

We find the scalar component of u in the direction of v from eq.(2):

Problems: 1) Let )5,2( vand )2,3( −=−=u . Find the a) component form and b) magnitude

(length) of the vector.

1. vu 52 +−

2. vu54

53

+

2) Find the component form of the vector:

a. The vector PQ where Q(2,-1) and )3,1(=P .

b. The vector OP where O is the origin and P is the midpoint of

segment RS , where )3,4(S and (2,-1) −==R .

c. The vector from the point )3,2(=A to the origin.

d. The sum of AB and CD , where

, )0,2(B , (1,-1) ==A and (-1,3)=C )2,2(D −=


9

3) Let wand , uv as in the figure: find a) vu + , b) wvu ++ , c) vu − and

d) wu −

4) Find the vectors whose lengths and directions are given. Try to do the calculation

without writing:

Length Direction

a. i 2

b. k- 3

c. kj54

53

21

+

d. kji73

72

76 7 +−

5) Find a) the direction of 21PP and b) the midpoint of line segment 21PP .

a. )5,1,1(1 −P and )0,5,2(2P

b. )0,0,0(1P and )2,2,2(2 −−P

6) Find uv ⋅ , uv , , the cosine of the angle between uandv , the scalar

component of u in the direction of v and the vector uprojv .

a) kjiv 542 +−= , kjiu 542 −+−=

b) kiv )54()

53( += , jiu 125 +=

c) jiv +−= , kjiu 232 ++=

d) jiv += 5 , jiu 172 +=

e)

=3

1 , 2

1v ,

−=

31 ,

21u

v

u

w


10

7) Find the angles between the vectors:

a) kjiu +−= 22 , kiv 43 +=

b) jiu 73 −= , kjiv 23 −+=

c) kjiu 22 −+= , kjiv ++−=

8) Find the measures of the angles between the diagonals of the rectangle whose

vertices are D(4,1) and C(3,4) , B(0,3) , )0,1(=A

References: 1- Advanced Engineering Mathematics (Erwin Kreyszic)- 8th Edition. 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)


11

2) The Cross product

The cross product is also called vector product because the product results a

vector.

Def.:The cross product n )sin u( θvvu =× , n unit vector (normal) perpendicular to the plane. Note: The vector vu × is orthogonal to both and vu

Parallel vectors

Nonzero vectors and vu are parallel if and only if 0 =× vu .

Properties of the cross product

If wand , vu are any vectors and r , s are scalars, then

1) ))(()()( uvrssvru ×=×

2) wuvuwvu ×+×=+× )(

3) uwuvuwv ×+×=×+ )(

4) ( )vuuv ×−=×

5) 00 =× u

Notes:

jkiikijkkj

kijji

=×−=×=×−=×

=×−=×

)()(

)(

0=

×××

kkjj

ii

n

vu ×

-n

uv ×


12

Calculating Cross product using determinants

If k321 ujuiuu ++= and k321 vjvivv ++= , then

321

321

vvvuuukji

vu =×

Ex.:

Find vu × and uv × if k2 ++= jiu and k34 ++−= jiv

Solution

10k6j2i )-( 10k6j--2i

34-12

14-12

- 1311

134112

−+=×=×+=

+=−

=×

vuuv

kjikji

vu

Ex.: Find a vector perpendicular to the plane of )0,1,1( −P , )1,1,2( −Q and )2,1,1(−R .

Solution

The vector PRPQ × is perpendicular to the plane because it is perpendicular

to both vectors.

kjikjiPQ −+=−−+++−= 2 )01()11()12(

kjikjiPR 222- )02()11()11( ++=−+++−−=

6k6i

22-21

22-1-1

- 221-2

222121

+=

+=−

−=× kjikji

PRPQ

Ex.: Find a unit vector perpendicular to the plane of )0,1,1( −P , )1,1,2( −Q and

)2,1,1(−R .

Solution

Since PRPQ × is perpendicular to the plane, its direction n is a unit vector

perpendicular to the plane


13

2

1 2

1 266k6i ki

PRPQPRPQn +=

+=

×

×=

Calculating the Triple scalar product (volume): also called Box product

321

321

321

)(wwwvvvuuu

wvu =⋅×

Ex.:

Find the volume of the box determined by k2 −+= jiu , k32 +−= iv and

k47 −= jw .

Solution

23- 470

302121

)( =−

−−

=⋅× wvu

The volume is 23 )( =⋅× wvu units cubed. Lines and Planes in Space

In the plane, a line is determined by a point and a number giving the slope of

the line. In space a line is determined by a point and a vector giving the direction of

the line.

Equations for a line

Suppose that L is a line in space passing through a point ),,( 0000 zyxP

parallel to a vector k321 vjvivv ++= . Then L is the set of all points ),,( zyxP for

which PP0 is parallel to v .

L

v


14

The standard equation of the line through ),,( 0000 zyxP parallel to

k321 vivivv ++= is:

∞∞+=+=+= pp t- , , , 000 tvzztvyytvxx

and ) , , (),,( 000 tvztvytvxzyx +++=

Ex.:

Find the equations for the line through )4,0,2(− parallel to k242 −+= jiv .

Solution

With ),,( 0000 zyxP equal to )4,0,2(− and k321 vjvivv ++= equal to

k242 −+= jiv

24 , 4 , 22 tztytx −==+−=

Ex.: Find the equations for the line through )3,2,3( −−P and )4,1,1( −Q .

Solution

The vector k734 +−= jiPQ is parallel to the line and equation with

)3,2,3(),,( 000 −−=zyx give

73 , 32 , 43 tztytx +−=−=+−=

We could have choose )4,1,1( −Q

74 , 31 , 41 tztytx +=−−=+=

An equation for a Plane in space

Suppose that plane M passes through a point ),,( 0000 zyxP and is normal to

the nonzero vector kCBjAin ++= . Then M is the set of all points ),,( zyxP for

which PP0 is orthogonal to n.

Thus, the plane through ),,( 0000 zyxP normal to kCBjAin ++= has equation:

0)()()( 000 =−+−+− zzCyyBxxA

or 000D e wher, CzByAxDCzByAx ++==++

n


15

Ex.:

Find an equation for the plane through )7,0,3(0 −P perpendicular to

k25 −+= jin .

Solution

2225072155

0)7)(1()0(2))3((50)()()( 000

−=−+=+−++

=−−+−+−−=−+−+−

zyxzyx

zyxzzCyyBxxA

Notice in this example how the components of k25 −+= jin become the

coefficients of zandyx , in equation 2225 −=−+ zyx . The vector

kCBjAin ++= is normal to the plane DCzByAx =++ .

Ex.:

Find an equation for the plane through )0,3,0( and )0,0,2( , )1,0,0( CBA .

Solution

We find a vector normal to the plane and use it with one of the point to write

an equation for the plane.

The cross product:

6 2 3 130102 kji

kjiACAB ++=

−−=× is normal to the plane.

6623

0)1(6)0(2)0(3=++

=−+−+−zyx

zyx

Lines of intersection

- Two lines are parallel if and only if they have the same direction.

- Two planes are parallel if and only if their normals are parallel.

- The planes that are not parallel intersect in a line.

Ex.:

Find a vector parallel to the line of intersection of the planes 15263 =−− zyx

and 522 =−+ zyx .


16

Solution

The line of intersection of two planes is perpendicular to both planes' normal

vectors 21 and nn and therefore parallel to 21 nn × . i.e. 21 nn × is a vector parallel to

the planes' line of intersection.

51 2 41 21226321 kji

kjinn ++=

−−−=×

Ex.: Find the point where the line , , tztytx +=−=+= 12238 intersects the plane

6623 =++ zyx

Solution

The point

+−+ 1 , 2 , 2

38 ttt

( ) ( )

-1t -88t

66t64t-6t8

6 6 2

==

=+++

=++−+

+ ttt 122

383

The point of intersection is 0) , 2 , 32( ),,( 1 =−=tzyx

Angles between planes

The angle between two intersecting planes is defined to be the angle

determined by their normal vectors.

Ex.:

Find the angle between the planes 15263 =−− zyx and 522 =−+ zyx

Solution

The vectors k2631 −−= jin and k222 −+= jin

are normals to the planes. The angle between them is

=

⋅=

−

−

214cos

cos

1

21

211

nnnnθ


17

Problems:

1) Sketch the coordinate axes and then include the vectors vuvu and , × as

vectors starting at the origin

a. j , i == vu

b. k j ,k -i +== vu

c. 2ji , j- i2 +== vu

d. ji , j i −=+= vu

2) In the triangle that determined by the points and , RQP , find a unite vector

perpendicular to plane PQR .

a. )1,1,3( and )3,1,2( , )1,1,1( −RQP

b. )2,2,1( and )1,1,0( , )0,2,2( −−−− RQP

3) Let k5 +−= jiu , k5−= jv and k3315 −+−= jiw . Which vectors, if any,

are:

a. Perpendicular?

b. Parallel?

4) Find equations for the lines:

a. The line through the point )1 ,4 ,3( −−P parallel to the vector k++ ji .

b. The line through )1,0,1( and )1,2,1( −− QP .

c. The line through the origin parallel to the vector k2 +j .

d. The line through the point )1 ,2 ,3( − parallel to the line

3 , 2 , 21 tztytx =−=+=

e. The line through )1 ,1 ,1( parallel to the z-axis.

f. The line through )5 ,4 ,2( perpendicular to the plane 21573 =−+ zyx

g. The line through )0 ,7 ,0( − perpendicular to the plane 1322 =++ zyx

h. The line through )0 ,3 ,2( perpendicular to the vectors k32 ++= jiu

and k543 ++= jiv

i. The x - axis.


18

j. The z - axis.

5) Find equations for the planes:

a. The plane through )1 ,2 ,0(0 −P normal to k23 −−= jin

b. The plane through )3 ,1 ,1( − parallel to the plane 73 =++ zyx

c. The plane through )1,2,0( and )2,0,2( , )1,1,1( −−

d. The plane through )5 ,4 ,2(0P perpendicular to the line

4 , 31 , 5 tztytx =+=+=

e. The plane through )1 ,2 ,1( −A perpendicular to the vector from the

origin to A .

6) Find the plane determined by the intersecting lines:

∞∞−=+=−=

∞∞−=+=+−=pp

pp

s- 22 , 21 , 41 :2t- 1 , 2 , 1 :1

szsysxLtztytxL

7) Find a plane through )1 ,1 ,2(0 −P perpendicular to the line of intersection of the

planes 22 , 32 =++=−+ zyxzyx .

8) Find a plane through the points )1 ,2 ,3( , )3 ,2 ,1( 21 PP perpendicular to the

plane 724 =+− zyx .

9) Find the angles between the planes:

a. 222 , 1 =−+=+ zyxyx

b. 132 , 105 −=+−=−+ zyxzyx

10) Find the point in which the line meets the plane.

a. 632 , 1 , 3 , 1 =+−+==−= zyxtztytx

b. 12436 , 22 , 23 , 2 −=−+−−=+== zyxtztyx

References:

1- Calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)


19

Double Integral : Lecture 8 مدرس مساعد ازهار مالك

1

Double Integral Definition: let R be closed region in the (x, y )- plane. If f is a function of two variables that is define on the region R, then the double integrals on R is written by

dAff y)(x, A)y,(x LimR

n

1rrrr

0An

r

∫∫∑ =∆=→∆

∞→

dydxاذا كانت المنحنيات بهذه الصيغة يؤخذ المقطع شاقولي

dx y)(x, y)(x,2

1R

dyfdAfb

a

y

y∫ ∫∫∫ =

dxdyاما اذا كانت المنحنيات بالشكل التالي يؤخذ المقطع افقيا

dxdy y)(x, y)(x,2

1R∫ ∫∫∫ =d

c

x

x

fdAf

R

R

a b

y1

y2

R

d

c

x2 x1


2

Examples:

1) Evaluate )dydx8xy 1(3

0

2

1∫ ∫ +

i) sketch: since l verticadxdy ⇒ 2y , 1y == ii)

{ }

{ }

{ }

57=+=+=

+=

+=

++=

+−+=

+=+

∫

∫

∫

∫∫ ∫

54)(3(0)-6(9))(3

)2

12(

dx 12x1

dx 4x][1-16x][2

dx )]1(41[]4x(4)[2

)2y8x(y)dydx8xy 1(

3

0

2

3

0

3

0

3

0

3

0

2

1

23

0

2

1

xx

x

dx

2) Evaluate ∫∫ −

R

dAyx )2( 2 over the triangular R enclosed by

3y ,x 1y , 1 =+=−= xy

i) sketch:

(-1,0) & (0,1) , (1,0) & (0,1) 1 x 0y , 1 x 0y

1y 0 x , 1y 0 xx1y , 1

⇒⇒−=⇒==⇒=

=⇒==⇒=+=−=

ifififif

xy


3

∫ ∫∫∫−

−

−=−3

1

1

1

22 )2()2(y

yR

dxdyyxdAyx

[ ] [ ]{ }

{ }

2112

)1()1()1()1( )(

3

1

322232

3

1

22223

1

1

1

22

∫

∫∫

−+−+−+−+−=

−−−−−−−=−=−

−

dyyyyyyyyy

dyyyyyyydyxyxy

y

624418 −=

−+−=+−

−+−

=

+−=

+−= ∫

32

21

28118)

32

21(18

218

)3

24y2(

)22(

3

1

34

3

1

23

y

dyyy

3) Evaluate dy dx e 2

0

1

2

x 2

∫ ∫y

Reverse the order of integration Since dxdy horizontal

2x y 2y x =⇒=

1 x = 2 0 fromy for →

1-e e e

)02(e

edx dy e dy dx e

011

0

x

1

0

x

2

0

1

0

x1

0

2

0

x2

0

1

2

x

2

2

222

=−==

−=

==

∫

∫∫ ∫∫ ∫

e

dxx

dxyxx

y


4

4) Evaluate dy dx x

sin x 0∫ ∫π π

y

From left yx =

From right π=x

value of y , from x 0 ⇒

reverse the order

2 1)-(-1- cos - sin

sin sin

d dy sin dy d sin

00

000

0 00

====

⋅=⋅=

=⇒

∫

∫∫

∫ ∫∫ ∫

ππ

ππ

ππ π

xdxx

dxxx

xdxyx

x

xx

xxx

x

x

x

y

5)

[ ] [ ]

[ ] sin44 −=+=

+−=−=

=

∫∫

∫ ∫∫ ∫

20

22

2

0

22

00

2

0 0

22

0

22

ysin -

2y cos 2 xycos 2

dxdysin xy 2dydxsin xy 2

y

dyyydyy

yy

y

y

x

6) Write an equivalent double of integration reversed dx )dyy(x 22∫ ∫−

+

−

+0

1

2x

x

∫ ∫ ∫ ∫

∫ ∫

− −

−

+

−

+++=

+

1 0 2 0

2

2222

0

1

2

)()(0 1

22 dx )dyy(x

y y

x

x

dxdyyxdxdyyx

x=π

y=π

(-1,1)


5

7) Draw the region bounded by y=ex, y=sinx, x=π, x=-π and evaluate its area.

π−ππ−π

π

π−

π

π−

π

π−

π

π−

−=π−−π+−

+=

−=

=

=

∫

∫

∫ ∫

eeee

xe

dxxe

y

dydxA

x

x

e

x

e

x

x

x

)cos(cos

cos

)sin(

sin

sin

8) Find the area bounded by y=-x, y=-3x and x=y+4. Solution:

262

)6

42

(

)3

()344(

0

2

222

3

22

0

2

2

3

0

2 34

2

3

4

34

2

3

4

3

0

23

=+−

+++

+−+++=

+=

+=

−

−

−

−

−

−

−

−−

−

−

+−

−

−

+

− −

−

−

∫∫

∫∫

∫ ∫ ∫ ∫

yyyyy

dyyydyy

dyxdyx

dxdydxdyA

yy

y

y

y

y

(1,-3)

(2,-2)


6

Problems

1) dxdy 1)(xy

x 1

0

1

02∫ ∫ +

2) dxdy e xy 2 ln

0

1

0

y2

∫ ∫ x

3) dxdy xycos

x1

20

2

∫ ∫π

π

x

4) dydx e 8 ln

1

y ln

0

yx∫ ∫ +

5) dxdy yx

2

1

2

∫ ∫x

x

6) dydx xy cosy 1

0 0∫ ∫

π

7) dydx e 4

0

2x 3

∫ ∫y

8) θθ

πθ

ddr cosr 2

0

sin

0∫ ∫

9) Evaluate ∫∫R

dA , R: 1st quadrant bounded by x=2y & y=2x

10) Evaluate ∫∫R

xydA , R: the region bounded by x=2y & xy =

11) Evaluate ∫∫−

+R

dAyx 21

2 )1( , R: the region in the 1st quadrant enclosed by:

2y x= , 4=y , 0=x

12) Evaluate ∫∫R

dA )(y sin 3 , R: the region bounded by x=y , 2=y & 0=x

13) dydx e x1

0

1xy2∫ ∫

y

References 1- calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)

Mathematics: Lecture 9 مدرس مساعد ازهار مالك Polar Integral:

Polar Coordinates and Graphs

Polar Coordinate system

Each point P can be assigned polar Coordinates (r, θ) where:

1) r is the distance from the pole (origin) 0 to the point P. r is positive if

measured from the pole along the terminal side of θ and negative if

measured along the terminal side extended through the pole.

2) θ is the angle from the Initial ray to (op). The angle θ is positive if the

rotation is counterclockwise and negative if the rotation is clockwise.

Review in trigonometric functions:

functions odd

cot)cot(tan)tan(csc)csc(sin)sin(

−=−−=−−=−−=−

θθθθθθθθ

functionseven sec)sec(cos)cos(

=−=−

θθθθ

1cotcsc1tansec1cossin

22

22

22

=−

=−

=+

θθ

θθ

θθ

xxxx

yxyxyx

xxxxyxyxxxxxxxxyx

2

22

tan1 tan22 tan y if

tan tan1tantan)tan(

sin cos )cos(2 y if sinsin cos cosy) cos( cos sin 2)sin(2 y ify sin cosy cos sin)sin(

−=⇒=

±=

−=⇒=±=

=⇒=+=+

mm

m

22cos1sin

22cos1cos

2

2

xx

xx

−=

+=


Converting from polar to rectangular form and vice versa We have the following relationship between rectangular Coordinates

(Cartesian) ),( yx and polar Coordinates ),( θr : 222 ryx =+

xy

xy

sryry

rxrx

1tan or tan

in or sin

cos or cos

−==

==

==

θθ

θθ

θθ

θd d rrdxdydydx

dA ⇒

=

Cartesian Coordinates

)(xfy =

Polar Coordinates

)( θfr =

Graphing polar equations

Sketch

i) symmetric about x-axis if replacing θ by )( θ− does not change the function.

ii) Symmetric about y-axis if replacing θ by )( θπ − does not change the function.

iii) Symmetric about the origin if replacing r by )( r− does not change the function.

iv)

2

0

M

π

πθ =


Ex.1: Converting an equation from Cartesian form to polar form

0422 =−+ yyx Since 222 ryx =+ and θin sry =

θθ

θθθ

sin 4ronly keep and 0r discaredcan we 0,4sin-r ofgraph in the included is pole thebecause pole. theis 0r ofgraph the

sin4or 00)sin4( 0sin4 04

2

22

====

===−=−

=−+⇒

rrrr

rryyx

Ex 2: Converting an equation from polar form to Cartesian form

θcos 3 −=r

θcos 3 2 rr −= Multiply both sides by r

03 3

22

22

=++⇒

−=+⇒

xyxxyx

Ex 3: Converting an equation from polar form to Cartesian form

r cos(θ-π/3)=3 r(cosθ cos(π/3)+ sinθ sin(π/3))=3

63323

21

3sin23cos

21

=+⇒=+

=+

yxyx

rr θθ

Ex 4: Converting an equation from polar form to Cartesian form r=4cosθ

xyxrr 4cos4 222 =+⇒= θ


Some important curves

}

}

}

cos

rose Leafed 4 cos , sin


cordioid ) sin (1 , ) sin -(1 ) cos(1 , ) cos-(1

cos , sin ,

22 θ

θθ

θθ

θθθθ

θθ

2

22

33

ar

arar

arar

arararar

circleararar

=

==

==

+==+==

===

Standard Polar Graphs

1) Circle a) a=r

2=r

M

2 2

2 4

2 0

=⇒=

=⇒=

=⇒=

r

r

r

πθ

πθ

θ

b) θsin ar = i) replace θ by -θ θθ sin )(-sin arar −=⇒=∴ Not symmetric about x-axis ii ) replace θ by π-θ θθπ sin )-(sin arar =⇒=∴ symmetric about y-axis iii ) Not symmetric about the origin.

2a

6

a 2

0 0

π

π

θ r


c) θ cos ar = i) replace θ by -θ θθ cos )(- cos arar =⇒=∴ symmetric about x-axis

2a

3

0 2

a 0

π

π

θ r

2) Cardioids a) ) cos(1 θ+= ar Symmetric about x-axis

2a

32

23a

3

0

a 2

2a 0

π

ππ

π

θ r

Rapid polar sketching Ex: Sketch ) cos(1 4 θ+=r

θ varies from

Cos θ varies from

4 cos θ varies from

) cos(1 4 θ+=r varies from

0 to π/2 1 to 0 4 to 0 8 to 4 π/2 to π 0 to -1 0 to -4 4 to 0 π to 3π/2

-1 to 0 -4 to 0 0 to 4

3π/2 to 2π 0 to 1 0 to 4 4 to 8


b) ) sin (1 θ+= ar H.W

) sin -(1 ) cos-(1

θθ

arar

==

EX.: Find the area of the region enclosed by the cardioids

) cos-(1 θar =

2

3)2sin41

21sin2(

))2cos1(21cos21()coscos21(

)cos1(2

2

2

0

00

2

0

2

cos1

00

0

cos1

0

2r

πθθθθ

θθθθθθ

θθθ

θ

π

ππ

π

θ

π

π θ

=++−=

++−=+−=

−==

=

∫∫

∫∫

∫ ∫−

−

dd

dd

rdrdA


Problems

1) Converting equations from Cartesian form to polar form

a) 0622 =−+ xyx

b) 22 5 xyy −=

c) xy 42 =

d) 12 =xy

2) Converting an equation from polar form to Cartesian form

a) 0sin 2 =+ θr

b) 1sin 4cos3( −=− θθr

c) 4 =r

d) 4

πθ =

3) a) sketch ) sin (1 5 θ+=r

b) sketch 2 cos 8 θ=r

4) change the Cartesian integral into an equivalent polar integral. Then

evaluate the polar integral

a) ∫ ∫−

−1

1

1

0

2

x

dydx

b) ∫ ∫−

−

−−

1

1

1

1

2

2

x

x

dydx

c) ∫ ∫−

+1

0

1

0

22

2

)( y

dxdyyx

d) ∫ ∫6

0 0

y

dydxx

5) Use polar coordinate ∫ ∫−

++

a xa

0 02/322

22

)yx(1dydx


6) Find the area of the region R that lies inside the cardioid ) cos(1 θ+=r

and outside the circle 1=r .

7) Find the area of the region R that lies inside the cardioid

) cos(12 θ+=r and outside the circle 3=r .

8) Find the area of the region R that lies inside the circle ) (sin 4 θ=r and

outside the circle 2=r .

9) Find the area of the region R cut from the first quadrant by the cardioid

) sin (1 θ+=r .

10) Find the area of the region common to the ) cos(1 θ+=r and

) cos(1 θ−=r .

References: 1- calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)


1

Polar Coordinates and Graphs 3) Leaves rose

θn cos ar = , θnsin ar =

}

}

cos



22 θ

θθ

θθ

2

22

33

ar

arar

arar

=

==

==

1- Number of leaves:

θn cos ar = , θnsin ar =

a) If n even → No. of leaves =2n b) If n odd→ No. of leaves =n 2- The major axes for the first leaf :

a) 001 cos =⇒=⇒= θθθ nn b)

nn

221sin πθπθθ =⇒=⇒=n

3- Limit for the first leaf (begin and end)

a)

−=⇒

−=

=⇒=⇒=

nn

nn

22

220 cos πθπθ

πθπθθn

b)

=⇒=

=⇒=⇒=

nn

nπθπθ

θθθ

000sin n


2

4- Complete drawing the other leaves: (360/No. of leaves) from major

axes. Ex.1: θ2 cos =r

1- No. of leaves = 4 2- 00212cos =⇒=⇒= θθθ

3-

−=⇒

−=

=⇒=⇒=

422

422

02cos πθπθ

πθπθθ

4- 24

360 π= , this mean every

2π leaf repeated.

Ex.2: θ2sin =r

1- No. of leaves = 4 2-

42212sin πθπθθ =⇒=⇒=

3-

=⇒=

=⇒=⇒=

22

0020sin πθπθ

θθθ2

4- 24


2π leaf repeated.


3

Ex.3: θ3 cos =r


3-

−=⇒

−=

=⇒=⇒=

623

623

03cos πθπθ

πθπθθ

4- 3

23


32π leaf repeated.

Ex.4: θ3sin =r



3-

=⇒=

=⇒=⇒=

33

0030sin πθπθ

θθθ3

4- 3

23


32π leaf repeated.


4

Ex.5: θ4 cos =r


3-

−=⇒

−=

=⇒=⇒=

824

824

04cos πθπθ

πθπθθ

4- 48

28

360 ππ== , this mean every

4π leaf repeated.

Ex.6: θ4sin =r

Ex.7: θ5 cos =r


5

Ex.8: θ5sin =r


102515sin πθπθθ =⇒=⇒=

3-

=⇒=

=⇒=⇒=

55

0050sin πθπθ

θθθ5

4- 45

25

360 ππ== , this mean every

52π leaf repeated.

Ex.9: θ2 cos a 22 =r


3-

−=⇒

−=

=⇒=⇒=

422

422

02cos πθπθ

πθπθθ

4- 2

22

360 π= , this mean every π leaf repeated.


6

Ex.10: θ2sin a 22 =r



3-

=⇒=

=⇒=⇒=

22

0020sin πθπθ

θθθ2

4- ππ==

22

2360 , this mean every π leaf repeated.

Problems

11) Find the area of the region enclosed by the cardioid θ2 cos 42 =r .

12) Find the area of the region enclosed by the cardioid )3 (cos12 θ=r .

13) Find the area of the region in the first quadrant bounded by 1=r and

θ2 sin=r , 24πθπ

≤≤

References: 1- calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)


7

Graphs of some problems: 6) Find the area of the region R that lies inside the cardioid ) cos(1 θ+=r


7) Find the area of the region R that lies inside the cardioid

) cos(12 θ+=r and outside the circle 3=r .

8) Find the area of the region R that lies inside the circle ) (sin 4 θ=r


4

3

2

4

2


8

9) Find the area of the region R cut from the first quadrant by the cardioid

) sin (1 θ+=r .

10) Find the area of the region common to the ) cos(1 θ+=r and ) cos(1 θ−=r .

11) Find the area of the region enclosed by the cardioid θ2 cos 42 =r .

12) Find the area of the region enclosed by the )3 (cos12 θ=r .

13) Find the area of the region in the first quadrant bounded by 1=r and

θ2 sin=r , 24πθπ

≤≤


9

Mathematics: Lecture 11 مدرس مساعد ازهار مالك Fourier series:

1

Fourier Series: Are series of cosine and sine terms and arise in the important practical task of

representing general periodic functions.

Periodic functions: A function )(xf is called periodic if it is defined for all real x and if there is

some positive No. T such that

)()( xfTxf =+

The No. T is called a period of )(xf .

Fourier said If )()( xfTxf =+ , T: periodic No. Then

∑∞

=

++=

1

0 2sin 2cos 2

)(n

nn xTnbx

Tnaaxf ππ

Where nn baa & , 0 are Fourier coefficients and

dxxfT

aB

A

)( 20 ∫=

dxxTnxf

Ta

B

An 2cos )( 2 π

∫=

dxxTnxf

Tb

B

An 2sin )( 2 π

∫=

BxA pp Notes:

No.integer , ,....) 2 , 1 , 0 ( , 0 sin nnn ±±==π

==

= ... , 4 , 2 , 0 1

... , 5 , 3 , 1 1- os

nn

nc π

....) , 3 , 2 , 1 , 0 ( allfor 1 2 os ±±±== n ,nnc π

sin )(-sin cos )(- cos

odd xxeven xx

==


2

EX.: Write Fourier series for π2 0 , )( pp xxxf =

ππ 2 0 - 2 T ==⇒ First we find nn baa & , 0

[ ] πππππ

ππ

2 0421

21

22

)( 2

22

0

22

0

0

=−=⋅==

=

∫

∫

xdxx

dxxfT

aB

A

dby , cos 1

2

2cos 22

2cos )( 2

2

0

2

0

vudxnxx

dxxnx

dxxTnxf

Ta

B

An

∫

∫

∫

=

=

=

π

π

π

ππ

π

π

[ ] 0 )11(12 cos1 cos 11

sin 1sin 1 1

22

2

02

2

0

2

0

=−==⋅=

−⋅= ∫

ππ

ππ

ππ

ππ

n0 cos-n

nnx

n

dx nxn

nxn

x

0a =∴ n

xnxdxxnx

dxxnxb

B

A

B

An

sin dv , u , sin 1

2

2sin 22

===

=

∫

∫

π

ππ

π

n

nxnn

xnxn

nxn

x

2b

sin1)012(1 1

d cos1 ) cos1( 1

n

2

02

2

0

2

0

−=

+−⋅

−=

−−

−⋅= ∫

π

ππ

ππ

π


3

)xx x

)xx x

)xx x

a ,nxxf n

L

L

L

+++−=

+−

+−

+−

+=

++++=

==⇒

−=

−=

−=

∑∞

=

3sin 312sin

21sin ( 2

3sin 322sin

22sin

12(

3sin b2sin bsin (b

0sin b 22)(

32b ,

22b ,

12b

321

1nn

321

π

π

ππ

Fourier even & odd functions 1) If )(xf is even , then

∫

∫

⋅=

⋅=

=

B

B

dxxTnxf

T

dxxfT

0n

00

n

2cos)( 22a iii)

)( 22a ii)

0b i)

π

2) If )(xf is odd , then

2sin)( 22b ii)

0aa i)

0n

n0

∫⋅=

==B

dxxTnxf

Tπ

Def.: A function )(xf is even if )()( xfxf =− for all x . For example, 2)( xxf = . A function )(xf is odd if )()( xfxf −=− for all x . For example, 3)( xxf = . Notes: - If )(xf symmetric about y-axis ⇒ even.

txfxfxfxxfxxf constan==== )( , )()( , cos)( , )( 2 - If )(xf symmetric about origin ⇒ odd.

sin)( xxf = EX.:

Write Fourier series for −

= 2 0 , 1 0 2- , 1

)(pp

pp

xx

xf

4 (-2) - 2 T ==⇒


4

i) From sketch ⇒ symmetric about origin ⇒ odd.

0aa n0 ==⇒

1)- (cos2- 2

cos 2-

d 2

sin 1

d 4

2sin 1 242b

2

0

0

0n

ππ

ππ

π

π

nn

xnn

xxn

xxn

B

B

==

⋅=

⋅⋅=

∫

∫

) cos1(2b n ππ

nn

−=∴ …. (1)

To find 1b , put n = 1 in eq. (1)

ππ

πππ

πππ

54))11(

52b

0))11(52b ,

34))11(

32b

0))11(22b , 4))1(1(2b

5

43

21

=+=

=−==+=

=−==−−=∴

.....) x2

1 x2

1 x2

4

..... x2

4 0 x2

4 0 x2

4

x2

nxf

+++=

+++++=

=⇒ ∑∞

=

ππππ

ππ

ππ

ππ

π

5sin 5

3sin 3

(sin

5sin 5

3sin 3

sin

sin b )( 1n

n

Notes:

y)] cos(y) cos([21 cos cos

cos 2cosy) cos(y) cos(

add sinsin cos cosy) cos( sinsin cos cosy) cos(

−++=

=−++

+=−−=+

xxyx

yxxx

yxyxxyxyxx

We can obtain yx sinsin by subtraction.


5

)sin()sin(21ysin cos

ysin cos2)sin()sin(

nsubtractio y sin cosy cos sin)sin(y sin cosy cos sin)sin(

yxyxx

xyxyx

xxyxxxyx

−++=

=−++

−=−+=+

EX.:

Write Fourier series for

=

23

2 , 0

2

2- , os

)(ππ

ππ

pp

pp

x

xxcxf

πππ 2 2

2

3 T =+=⇒

)(xf is even in 0b 2

2

-n =⇒

ππpp x

This is true if and only if the other interval = 0

ππππ

ππ

2 ]01[2 sin 2 cos 222 )( 22a

0

2

0

2

00 =−==⋅=⋅= ∫∫

B

xdxxdxxfT

....(2) 0)22

(sin1

10sin)22

(sin1

11

)(sin 1

1)(sin 1

11

])( cos )[cos(21 2

....(1) n cos cos 222

2

2ncos )( 22a

2

0

2

0

2

0

2

0n

−−

−+

−+

+=

−

−++

+=

−++=

⋅=

⋅=

∫

∫

∫

πππππ

π

π

π

π

π

π

π

π

nn

nn

nxxn

nxxn

dxnxxnxx

dxxx

dxxxfT

To find 1a , put n = 1 in eq. (1)


6

21 ))00(0

2(1

2sin211 )2 cos(1

212 cos2a

2

0

2

0

2

01

=+−+=

+=+⋅=⋅=∴ ∫∫

ππ

πππ

πππ

xxdxxdxx

in eq.(2)

ππ

πππ

32 )1

31(1

}2

sin{}2

3sin{ 311a2

=+−

=

−

−=

0 }}0{21-0{

411a3 =

=

π

ππ

πππ

152 )

31

51(1

}23sin{

31}

25sin{

511a4

−=−=

−

−=

+−++=

+=⇒ ∑∞

=

......xxx1

xnxf

4 cos 15

22 cos 32 cos

21

cos a 2a )(

0nn

0

πππ

Problems: Write the Fourier series for the following functions:

1)

−=

2 , 0 ,

)(ππ

πpp

pp

xaxa

xf

2) ππ - , )( pp xxxf =

3)

−+

= 0 ,

0 - , )(

ππππ

pp

pp

xxxx

xf

4) ππ - , sin)( pp xxxf =


7

5) −

= 2 0 , 1

0 2 , 0)(

pp

pp

xx

xf

6) −−

= 1 0 , 20 1 , 1

)(pp

pp

xxx

xf

7)

−

=

23

2 , 0

2

2 ,

)(ππ

ππ

pp

pp

x

xkxf

8)

−

=

23

2 , -

2

2 ,

)(πππ

ππ

pp

pp

xx

xxxf

9) ππ pp xxxf - , )( = . 10) ππ pp xxxf - , )( 3= .

11) −

= 0 , 1

0 - , 1 )(

ππ

pp

pp

xx

xf

12)

−

=

23

2 , 0

2

2 ,

)(ππ

ππ

pp

pp

x

xxxf

13) Find the Fourier series of the function )(xf which is assumed to have the period π 2 . 14) Find the Fourier series of the function )(xf

π/2 π -π π/2

1

k

π -π


8

15) Find the Fourier series of the function )(xf References: 1- Advanced Engineering Mathematics (Erwin Kreyszic)- 8th Edition. 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)

k

-k π 2π

Mathematics: Lecture 12 مدرس مساعد ازهار مالك Partial Differentiations:

1

Partial Differentiations

0z)y,f(x,or ),( == yxfZ

==∂∂

==∂∂

yy

xx

fZyZ

fZxZ

1st partial derivatives

2

2

2

2

2

2

=∂∂

∂

=∂∂

∂

==∂∂

==∂∂

xy

yx

yyyy

xxxx

Zyx

Z

Zxy

Z

fZyZ

fZxZ

2nd partial derivatives

yxxy ZZ = Ex.1

If yxZ = , find , yZ

xZ

∂∂

∂∂

functionpower constant x , ln , constant y y y1-y ⇒⋅⋅=∂∂

=∂∂ dyxx

yZx

xZ

Ex.2

If xyZ 1tan−= , show that xyZ=yxZ

22 yxy

+−

=−

⋅+

=xy

xy

Zx

2

2

1

1

(1) )()(

2.)1)((222

22

222

22

Lyxxy

yxyyyxZ yx +

−=

++−+

=


2

22 yxx+

=⋅+

=x

xy

Z y1

1

1

2

2

(2) )()(

2.)1)((222

22

222

22

Lyxxy

yxxxyxZxy +

−=

+++

=

(1) & (2) are equal

Properties: 1) If y),g( , )( xf == υυω

rulechain

or

∂∂

⋅∂∂

=∂∂

∂∂

⋅∂∂

∂∂

⋅∂∂

=∂∂

yy

xf

xx

υυωω

υυ

υυωω

2) If s)h(r,y , s)g(r, , ),( === xyxfω \

rulechain

∂∂

⋅∂∂

+∂∂

⋅∂∂

=∂∂

∂∂

⋅∂∂

+∂∂

⋅∂∂

=∂∂

sy

ysx

xs

r

ry

yrx

xr

ωωω

ωωω

3) Total differential If ,.....),,( zyxf=ω

...

...

+++=

+++=

dzdydxdordzfdyfdxfd

zyx

zyx

ωωωω

ω

Ex.1

If 222),,( zyxxyzzyxf +++==ω , Find dxdω

By property (3)


3

dxdzzxy

dxdyyxz

dxdxxyz

dxd

dzzxydyyxzdxxyzddzdydxd zyx

)2()2()2(

)2()2()2(

+++++=

+++++=

++=

ωω

ωωωω

Ex.2

If (1) ..... )()( ctxgctxf −++=ω , Show that 2

22

2

2

xc

t ∂∂

=∂∂ ωω

There are two methods to solve this Ex. First method: Let , sctxrctx =−=+ Eq.(1) becomes (1) ..... )()( sgrf +=ω

)()()(

)()(

csgcrfts

ssg

tr

rrf

t−⋅′+⋅′=

∂∂

⋅∂

∂+

∂∂

⋅∂

∂=

∂∂ω

( ) ( )

( ) ( ) (2) ...... )()(

)()()(

)()(

22

2

2

2

sgrfct

csgccrfc

ts

ssgc

tr

rrfc

t

′′+′′=∂∂

∴

−⋅′′−⋅′′=

∂∂

⋅∂

∂−

∂∂

⋅∂

∂=

∂∂

ω

ω

1)(1)(

)()(

⋅′+⋅′=∂∂

⋅∂

∂+

∂∂

⋅∂

∂=

∂∂

sgrfxs

ssg

xr

rrf

xω

11)(11)(2

2

⋅⋅′′+⋅⋅′′=∂∂ rgrf

xω

In eq.(2)

2

22

2

2

xc

t ∂∂

=∂∂ ωω


4

Second method: t) مباشرة بالنسبة لـ 1نشتق معادلة (

[ ] (2) .....

)()()()(

)()()(

22

2

2

2

gfct

ccctxgccctxft

cctxgcctxft

′′+′′=∂∂

∴

−⋅−⋅−′′+⋅⋅+′′=∂∂

−⋅−′+⋅+′=∂∂

ω

ω

ω

[ ] (3) .....

11)(11)(

1)(1)(

2

2

2

2

gfx

ctxgctxfx

ctxgctxfx

′′+′′=∂∂

∴

⋅⋅−′′+⋅⋅+′′=∂∂

⋅−′+⋅+′=∂∂

ω

ω

ω

From (2) & (3)

2

22

2

2

xc

t ∂∂

=∂∂ ωω

Ex.3

If

=

xyfxz n , Show that nz

yzy

xzx =

∂∂

+∂∂

(1) ..... )( )(

)( )(

)( )()(

1

12

12

xyfxn

xyfyx

xzx

xyfxn

xyfyxxx

xzx

xyfxn

xy

xyfx

xz

nn

nn

nn

⋅+′⋅−=∂∂

⋅

⋅+′⋅⋅−⋅=

∂∂

⋅

⋅+−

⋅′⋅=∂∂

−

−−

−

(2) ..... )(

0)1()(

1

xyfyx

yzy

xxyfx

yz

n

n

′⋅=∂∂

⋅

+⋅′⋅=∂∂

−


5

From (1) & (2)

nzxyfnx

yzy

xzx n

=

=∂∂

⋅+∂∂

⋅

)(

Ex.4

Express r∂

∂ω and s∂

∂ω in terms of s & r if 22 zyx ++=ω ,

rrsrx 2z , s lny , 2 =+==

srzr

szr

s

rz

zry

yrx

xr18441222211 +=++=⋅+⋅+⋅=

∂∂

⋅∂∂

+∂∂

⋅∂∂

+∂∂

⋅∂∂

=∂∂ ωωωω

ssrz

ssr

sz

zsy

ysx

xs202121 22

−=⋅+⋅+

−⋅=

∂∂

⋅∂∂

+∂∂

⋅∂∂

+∂∂

⋅∂∂

=∂∂ ωωωω

Problems:

Find zf

yf

xf

∂∂

∂∂

∂∂ , ,

1) xyzzyxf 1-sin ),,( =

2) 22 yxcos2y)-x(2),,(+

=zyxf

3) Find υω

∂∂ when

2-2uy , 12 , if 0 , 0 2 υυωυ +=+−=+=== uxxyxu

4) If

+

= 22 yxxyfω , show that 0=

∂∂

+∂∂

yy

xx ωω

5) If ),( yxf=ω , and θθ siny , cos rrx == , show that

2222

2 )(1)( yx ffyrx

+=∂∂

+∂∂ ωω

6) If xzyxzyxf

dd find , z & 0),,( +==


6

7) Find the directional derivative of xyxyxf 1tan),( −= at (1,1) in the

direction of jiArrr

−= 2

8) In which direction is the directional derivative of 22

22

),(yxyxyxf

+−

=

9) The D.D. of ),( yxf at )2,1(0p in the direction towards )3,2(1p is

22 and the D.D. at )2,1(0p towards )0,1(2p is -3 , find D.D. at

0p towards the origin. References:

1- calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)


1

Partial Differentiations The Gradient & Directional Derivative

0z)y,(x, =f

rsunit vecto & , , fyf

xf kjik

zji

rrrrrrr

∂∂

+∂∂

+∂∂

=υ

The Directional Derivative of z)y,(x,f at ),,(p 0000 zyx in the direction of 321 kajaiaA

rrrr++=

gradient & AAu , uD.D. υυ

rr

rrrr

=⋅=⇒

Ex.1: Find D.D. of zxyf −−= 23x at (1, 1, 0) in the direction of

632 kjiArrrr

+−= Sol.: First we find υr

1z

)1)(1(2 2y

)1()1(33x

22

-

-2

2

=∂∂

=−=−=∂∂

=−=−=∂∂

f

xyf

yxf

22

yx

kji

kzfjfif

rrrr

rrrr

−−+=∴

∂∂

+∂∂

+∂∂

=∴

υ

υ

74

49 6-64

1 , 3694

632) 22(

AA

, uD.D.

=+

=

=⋅+++−+

⋅−−+=

⋅=

⋅=⇒

iikjikjirr

rrrrrr

r

rr

rr

υ

υ


2

Maxima, Minima & Saddle point y)(x,f will have M , m, S according to: 1) 0 , 0 == yx ff to find the suggested point (a,b).

2) 0)( 2yy fxyxx fff −⋅

Then (a,b) is M or m according to xxf negative or positive. 3) 0)( 2

yy pxyxx fff −⋅

Then (a,b) is a saddle point. 4) 0)( 2

yy =−⋅ xyxx fff Ex.1: Locate M,m & S (if any) 422yx 22 −+++−= yxxyf

...(2) 022...(1) 0 2y -2x

22 2y-2x

=++−=+

++−=+=

yx

yxff

y

x

multi (1) by 2 + (2)

(-2,-2) , -2y , -2 x 063x ==⇒=+⇒

1 , 2 , 2 yy −=== xyxx fff

0 31)2)(2()( 2yy f=−=−⋅ xyxx fff

Since m is (-2,-2) 0 ⇒fxxf Ex.2: Locate M,m & S (if any) axyf 3yx 33 −+=


3

...(2) 033...(1) 03ay -3x

33

3ay-3x

2

2

2

2

=−

=

−=

=

axy

axyff

y

x

From (1) a

2xy =⇒

In (2) 0x 2

4

=−⇒ axa

0) x(0

33

34

=−

=−⇒

axxax

ayyaxx

==∴==⇒

, 0 , 0

a)(a, & )0,0( ⇒

afyfxf xyxx 3 , 6 , 6 yy −===

1) afffat xyxx 3 , 0 , 0 (0,0) yy −===⇒

0 9)( 22yy pafff xyxx −=−⋅

point saddle a is (0,0)

2) afafafat xyxx 3 , 6 , 6 a)(a, yy −===⇒

0 a 27 a 9-a 36

9)6)(6()(222

22yy

f==

−=−⋅ aaafff xyxx

i) if m is a)(a, 0 0 ⇒⇒ ff xxfa ii) if M is a)(a, 0 0 ⇒⇒ pp xxfa Problems:

Find zf

yf

xf

∂∂

∂∂

∂∂ , ,

1) xyzzyxf 1-sin ),,( =

2) 22 yxcos2y)-x(2),,(+

=zyxf


4

3) Find υω

∂∂ when

2-2uy , 12 , if 0 , 0 2 υυωυ +=+−=+=== uxxyxu

4) If

+

= 22 yxxyfω , show that 0=

∂∂

+∂∂

yy

xx ωω

5) If ),( yxf=ω , and θθ siny , cos rrx == , show that

2222

2 )(1)( yx ffyrx

+=∂∂

+∂∂ ωω

6) If xzyxzyxf

dd find , z & 0),,( +==

7) Find the directional derivative of xyxyxf 1tan),( −= at (1,1) in the

direction of jiArrr

−= 2

8) In which direction is the directional derivative of 22

22

),(yxyxyxf

+−

=

9) The D.D. of ),( yxf at )2,1(0p in the direction towards )3,2(1p is

22 and the D.D. at )2,1(0p towards )0,1(2p is -3 , find D.D. at

0p towards the origin. References:

1- calculus & Analytic Geometry (Thomas). 2- Calculus (Haward Anton). 3- Advanced Mathematics for Engineering Studies (أ. رياض احمد عزت)

differential equationsdx dy + 2 =-ans: y=e-x+ce-2x 2. 2 sin 3 x x y dx dy x + = ans: x3y=c-cosx 3....

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