In the Name of Allah Most Gracious Most Merciful

Ordinary Differential Equations

Prepared byAhmed Haider AhmedB.Sc. Physics - Dept. of Physics – Faculty of Science

TO my mother , my brothers

and my best friendAbd El-Razek

PrefaceDifferential equations are introduce in different

fields and its importance appears not only in mathematics but also in Engineering , Natural science ,Chemical science , Medicine ,Ecology and Economy.

Due to its importance in different fields I collected the laws and methods of solution of ordinary differential equations as an introduction to study it and to be as base to study theoretical physics and understand the physical meaning of relations.

LET’S UNDERSTAND

Ahmed Haider Ahmed – B.Sc. Physics

Definitions Differential equation is an equation involving an unknown function and its derivatives.Ordinary Differential equation is differential equation involving one independent variable and its differentials are ordinary.Partial Differential equation is differential equation involving two or more independent variables and its differentials are partial.Order of Differential equation is the order of the highest derivative appearing in the equation.Degree of Differential equation is the power of highest derivative appearing in the equation.

particular solution of a differential equation is any one solution.

The general solution of a differential equation is the set of all solutions.

Solutions of First Order Differential Equations

1- Separable Equations2- Homogeneous Equation3- Exact Equations4- Linear Equations5- Bernoulli Equations

1 -Separable Equations (separation variable)

General form of differential equation is (x ,y) dx + (x ,y) dy = 0

By separation variable Then 1 (x) 2 (y) dx + 1(x) 2(y) dy = 0

by integrating we find the solution of this equation.

Ex) find general solution for

0)(

)(

)(

)(

2

2

1

1 dyy

ydx

x

x

cxy

dyy

dxx

dyxxydx

lnln

nintegratioby 11

02

2 -Homogeneous Equation

The condition of homogeneous function is

f (x , y) = f (x ,y)and n is Homogeneous degree

(x ,y) dy + (x ,y) dx = 0 and , is Homogeneous function and

have the same degreeso the solution is put y = xz , dy = x dz + z dx and

substituting in the last equation

the equation will be separable equation, so separate variables and then integrate to find the solution.

n

3 -Exact Equations

(x ,y) dy + (x ,y) dx = 0

only. timeonefactor repeated take: note

issolution general its and

isequation exact be oequation t ofcondition required The

cdydx

yx

dyy

dxx

yx

xy

)( exp)(

)( exp)(

1

1

. following as

factor integralby it multiply exact be it toconvert to

exact,not be illequation w The

if yx

integral factor is

Examples i ) (2x + 3cosy) dx + (2y – 3x siny)

dy = 0Solutionit is exact so, (2x + 3cosy) dx = x + 3x cosy (2y - 3x siny) dy = y + 3x cosyThe solution is x + 3x cosy + y = c

ii) (1 – xy) dx + (xy – x ) dy = 0

exactnot its So,

2)()1( 2

xyy

xxyx

x

xy

dxxyxxThen

dxx

yxy

xy

)2( exp)(

)( exp)( Since

)(1

1

cy

xyx

cdyxydxy

dyxydxyx

x

xxdx

x

x

2ln

)(

exact isequation s thi0)(1

ii)equation by value thisgmultiplyin(by 1

lnexplnexpexp

1

11

note :- we took the repeated factor one time only

4 -Linear Equations Linear Equation form is

the integral factor that convert Linear Equations to exact equation is :-

= exp p(x) dx by multiplying integral factor by Linear

Equation form

so the general solution is :-

y = Q dx + c

)()( xQ y x Pdx

dy

exact isequation his t)()( xQ y xP dx

dy

cxx

dxxxx

dxxxyxx

cdxQy

issolutiongeneral

xx

xxxdx

xxQxxpxQyxpdx

dy

solutiondx

dyyxxyyEx

sin2

1sin

)cossin(cos

cos).tan(sec)tan(sec

tansec

tanseclnexpsecexp

cos)( , sec)( )()(

, cossec )

2

2

2

5 -Bernoulli Equation Bernoulli Equation form is

nyxQ y x Pdx

dy)()(

before. told weassolution its andequation linear is this

)()1()()1(

)()()1(

1

)1( then Put b)

)()(1

yover Equation Bernoulli Divide a)

Equation Bernoulli solve To

)1(

)1(

n

xQnzxpndx

dz

xQxpdx

dz

n

dx

dyyn

dx

dzyz

xQ yx Pdx

dy

y

n

nn

note :- if n = 0 the Bernoulli Equation will be linear equation.

if n = 1 Bernoulli Equation will be separable equation

the general solution will be

2

22

1

2

lnexpln2expexpexp

linear isequation thissin62

2

.sin3

solution

sin3 )

x

xxdxpdx

x x

z

dx

dzdx

dyy

dx

dzthen y and put z

y xx

y –

dx

dy

xx

y –

dx

dyy Ex

x

c x x x x y x

cxdx x y x

)cossincos(6

sin622

22

Solution of 1st order and high degree differential equation-:

1- Acceptable solution on p.2- Acceptable solution on y.3- Acceptable solution on x.4- Lagrange’s Equation.5- Clairaut’s Equation.6- Linear homogeneous differential

Equations with Constant Coefficients.7- Linear non-homogeneous

differential Equations with Constant Coefficients.

1 -Acceptable solution on p

if we can analysis the equation then the equation will be acceptable solution on p

equation. theofsolution general theis thisand

0))((

0 0

lnln2ln lnlnln

2

02 0

02 0

02

023 )

22

1

22

1

21

222

cyxcxy

cyxorcxy

cxyorcxy

x

dx

y

dyor

x

dx

y

dy

ydx

dyxory

dx

dyx

yxporyxp

yxpyxp

Sol

yxpypxEx

2 -Acceptable solution on y

If we can not analysis the equation then the equation will be acceptable solution on y or x

firstly , to solve the equation that acceptable solution on y

there are three steps :-

1- Let y be in term alone . 2- By differentiation the equation with respect

to x and solve the differential equation .3- By deleting p from two equations (the origin

equation and the equation that we got after second step) if we can not delete it the solution called the parametric solution .

2

2

2

2

3

212

3

2

3

2

xrespect to ation withdifferentiby 3

2

3

2

, 223 )

x

p

dx

dp

xp

dx

dpxp

dx

dy

x

ppxy

Solutiondx

dy p

x

ppxyEx

22

2

2

2

by gmultiplyin , 222

3by gmultiplyin , 3

4

3

2

3

2

3

1

xdx

dp

x

px

x

pp

dx

dp

x

px

x

pp

3

2

2

2

2

22

322

6

1

equationorigin on pabout ngsubstitutiequation twofrom p delete to

ln2

1ln

3

32

2 2

2 2

02 02

02)2(

)2(2)2(

)2(22

xy

xpxpcx

y

x

dx

p

dpdxxdy

dx

dpxpx

dx

dpdx

dpxporpx

dx

dpxppx

dx

dppxxpxp

dx

dppxxppx

3 -Acceptable solution on xsecondly, to solve the equation that acceptable solution on

x

there are three steps :-1- Let x be in term alone . 2- By differentiation the equation with respect

to y and solve the differential equation .3- By deleting p from the two equations (the

origin

equation and the equation that we got after second step

if we can not delete it the solution called the parametric solution .

solution. parametric thethis

so equations last tow thefrom p deletenot can we

equation)origin (the

4

3

2

1

)3(

)31(

1

)31(1

1but , 3

y respect to ation withdifferentiby

, )

3

42

3

2

2

2

3

ppx

ppy

dpppdy

ppdy

dp

dy

dpp

p

dy

dx

pdy

dpp

dy

dp

dy

dx

dx

dypppxEx

4 -Lagrange’s EquationLagrange’s Equation form

y = x g (p) + f (p)

Note the method of solution

in the example

cp

xp

dppxppe

p

dp

p

x

dp

dx

p

x

dp

dx

dx

dp

p

x

dx

dppxp

dx

dppxp p

dx

dpp

dx

dpx p

dx

dy

pxp y

p

3

2

2

factor integral 2 exp

equation aldifferentilinear 22

22

)22

(1

)22()22(2

222

2 Ex)

32

222ln2

5 -Clairaut’s EquationClairaut’s Equation is special case of Lagrange’s

EquationClairaut’s Equation form :-

y = x p + f (p)Note

the method of solution in the example

0)(0

0

)

2

2

2

p

axr o

dx

dp

dx

dp

p

ax

dx

dp

p

axpp

dx

dp

p

a

dx

dpx p

dx

dy

p

a p x y Ex

(parabola)solution single 4

22

2

2

22

2222

2

2222

axy

ax a x a , yax p

a xp y

p

apx

p

a x p y

c

a x cy

x

ap & c p

6 - Linear homogeneous Differential Equations with Constant Coefficients

L(D) y = f (x) non-homogeneous but L(D) y = 0 homogeneous then L() = 0 assistant equation

Roots of this equation are 1 , 2 , 3 ,……,n

This roots take different forms as following:-

constant are ,.....,,, ,

)()........(

3210

22

110

n

nnnn

aaaaadx

dD

xfyaDaDaDa

1- if roots are real and different each other then the complement solution is

xn

xxc

neCeCeCy .........2121

2- if roots are real and equal each other then complement solution is

3- if roots are imaginary then complement solution is

).........( 121

rn

xc xCxCCey

)sincos( 21 xCxCey xc

examples-:

xxc eCeCCy

yDD

yy

321

321

23

3

1,1,0

0)1)(1(

0)1(0)(

0)(

0)1

xxc eCeCy

yDD

yyy

221

21

2

2,1

0)2)(1(

0)23(

0)23(

023)2

axCaxCy

ai

a

yaD

yay

c sincos

0)(

0)(

0))(3

21

22

22

xCxCeCy

i

yDD

xc sincos

, 1

0)1)(1(

0)22)(4

321

21

2

2

L(D) y = f (x) non-homogeneousthe general solution of non-homogeneous is

y = yc + yp

yc complement solution

[solution of homogeneous equation L(D) y = 0 look last slide]y p particular solution is

We knew how to get the complement

solution last slide. To get the particular solution it depends on the type of function we will know the different types and example to every one as following .

7 -Linear non-homogeneous Differential Equations with Constant Coefficients

NoteL(D) is differential

effective1 / L(D) is integral

effective

)()(

1xf

DLy

i) if f (x) is exponential function

)()(

1 issolution Particular xf

DLy

ii) special case in exponential function at L (a) = 0

iii) if exponential function multiplied f(x)

iv) if f (x) is trigonometric function sin x or cos x

aDaLaL

ee

DLy

axax ; 0)( ;

)()(

1

)()(

1

aDL

ee

DLy

axax

)()(

1)(

)(

1xf

aDLeexf

DLy axax

axaxaL

axaxDL

y cosor sin)(

1cosor sin

)(

122

v) if f (x) is trigonometric function sin x or cos x multiplied exponential function

vi ) If f (x) is polynomial

xorxaDL

exorxeDL

y axax cos sin)(

1cos sin

)(

1

321

321

1)1(

1)1(

-:series following theuseor fractions partial use then and

)()(

1

)()(

x x x x

.. x x x - x

xfDL

y

xfDL

solved example :-

xxx

pc

xxp

xp

aD

xp

xxc

x

ee C e Cy

y y y lution general so

e ey

e ye DD

y

eCeCy

yDD

eyyy

3321

33

332

32

21

21

2

2

3

3

4

3

4

6

8

8639

18

6

1

3 , 2

032

06

0)6(

86

2

general problems

Find

gen

eral

sol

utio

n of

the

follo

win

g eq

uation

AHM

ED H

AIDER

FACU

LTY

OF

SCIE

NCE

082)10

044)9

)8

)7

23)6

ln)5

0)()4

0)sin4()4()3

0)sec()tan2( )2

0)()( )1

2

34

2

22

22

22

y - yy

y yyy

pp

x y

p x py

eyydx

dy

xyydy

dx

xy dy dx yx

dyyxdxye

dyyxdxyxy

dyxyxdxyxy

x

x

330)2510()20

6)1()19

7)4()18

cos)84()17

2sin8)127()16

)5()127()15

)12( )14

)65()13

4122()12

)96( )11

2

32

22

22

52

22

22

232

34

32

xyDD

xyDD

xyDD

xeyDD

x eyDD

exxyDD

exyDD

e xyDD

e )y DDD

e yDD

x

x

x

x

x

x

x

الحمد لله الذي هدانا لهذا وما كنا لنهتدي لوال أن هدانا الله

Finally , this course of ordinary differential equations is useful

to different student special students of physics. Theoretical physics required to be know the bases of mathematics specially differential equations such that quantum

mechanics depend on Schrödinger equation and this equation is differential equation so this branch of physics depend upon differential equations . I made slide of problems in different types of differential equations to examine yourself .Finally don’t forget these words for Napoleon “the advancing and perfecting of mathematics are closely related by prosperity of the nation” Ahmed Haider Ahmed - Faculty of Science -

Dept. of Physics ahmedscience2020@yahoo.com

Ahmed Haider Ahmed

Nuclear Physics Lab

Faculty of Science Minia universityMinia CityEgypt