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# Ordinary Differential Equations

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المعادلات التفاضلية العادية اعداد احمد حيدر كلية العلوم قسم الفيزياءجامعة المنيا
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م ي ح ر ل ا ن م ح ر ل ه ا ل ل م ا س In the Name of Allah Most Gracious Most Merciful Ordinary Differential Equations ه ي عاد ل ا ه ي ل ض ا ف ت ل ا دلات عا م ل اPrepared by Ahmed Hyder Ahmed Faculty of science
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بسم الله الرحمن الرحيمIn the Name of Allah Most Gracious Most Merciful

Ordinary Differential Equationsالعادية التفاضلية المعادالت

Prepared byAhmed Hyder Ahmed

Faculty of scienceDep. of physics

INTRODUCTIONDifferential equations introduce in different fields of science and its importance appear 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 method of solution of ordinary differential equations as introduction to study it and to be as base to study theoretical physics and understand the physical meaning of relations.

LET’S UNDERSTANDAhmed Hyder Ahmed - faculty of science - Dep. of physics

TO my mother , my brothers

and my best friendAbd El-Razek

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 twoor 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 Equations

• 2- Homogeneous Equation

• 3- Exact Equations

• 4- Linear Equations

• 5- 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

1 (x) / 1 (x) dx + 2 (y) / 2 (y) dy = 0

by integrating we find the solution of this equation.

Ex) find general solution for xy dx - x dy = 0

1/x dx = 1/y dy by integration

ln y = lnx + c

and this the solution of equation

2

2 -Homogeneous Equationthe 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 degree

so 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

the required condition of equation to be exact equation is

/x = / yand its general solution is

dx + dy = cnote : - take the repeating factor one time only

If /x = / y the equation will be not exact to

convert it to be exact multiply it by integral factor

as following .

integrating factor is

(x) = exp [ 1/ ( / y - / x) dx ]

(y) = exp [ 1/ ( / x - / y) dy ]

i . (2x + 3cosy) dx + (2y - 3x siny) dy = 0

ii . (1 - xy) dx + (xy – x ) dy = 0

solution

i . it is exact so

(2x + 3cosy) dx = x + 3x cosy

(2y - 3x siny) dy = y + 3x cosy

sol is x + 3x cosy + y = c

2

Examples- :

2

2

2 2

ii. ( 1-xy) / x = -x = (xy – x ) / y = y - 2x

so it is not exact

since (x) = exp[ 1/ ( / y - / x) dx ]

then (x) = exp[ 1/(xy-y) (-x-y+2x) dx]

= exp( -1/x ) = exp (-lnx) = exp ln x

= 1/ x by multiplying this value on equation ii

(1/x – y) dx + (y - x) dy = 0 this equation is exact

(1/x – y) dx + (y - x) dy = c

lnx – xy + y /2 = c

note :- we took the repeating factor one time only

-1

2

4 -Linear Equations Linear Equation form is

dy/dx + P(x) y =Q(x)

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

= exp p dx by multiplying integrating factor on Linear Equation form dy/dx + P(x) y = Q(x) this equation is exact

so the general solution is :-

y = Q dx + c

Ex) y + y sec x = cos x note y = dy / dx

solution ;-dy/dx + P(x) y =Q(x) , p (x) = sec x

= exp sec x dx = exp (ln sec x + tan x)

= sec x + tan x

but general solution is y = Q dx + c

(sec x + tan x ) y = (sec x + tan x) . cos x dx

= (cos x + sin x cos x) dx

= sin x + ½ sin x + c

2

2

5 -Bernoulli Equations Bernoulli Equation form is

dy /dx + P(x) y = Q(x) ynote :- if n = 0 the Bernoulli Equation will be linear equation

if n =1 Bernoulli Equation will be separable equation

to solve Bernoulli Equation

a) divide Bernoulli Equation over y

1/ y dy /dx + P(x) y = Q(x)

b) put z = y then dz / dx = (1 - n) y dy/dx

1/(1-n) dz/dx + P(x) z = Q(x)

dz /dx + (1 - n) P(x) z = (1 - n) Q(x)

this is linear equation and its solution as in pervious slide

n

n

n - 1

(1 – n)

Ex : - y dy/dx – y / x = 3sin xsolutiondy/dx + y / x = 3 sin x . yput z = y and then dz/dx = 2y dy/dxdz/dx + 2z / x = 6 sin x this equation is linear = exp p dx = exp 2 / x dx = exp 2lnx = exp lnx = xthe general solution will be x y = 6 x sin x dx + c

x y = 6 (-x cosx + sin x + cosx) + c

2 2

2

2

2

2

2

-1

2

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 pif we can analysis the equation then the equation will be

acceptable solution on p

ex) x p + 3x py + 2y =0

sol

(xp + y) (xp + 2y) = 0

then xp + y = 0 or xp + 2y = 0

x dy/dx + y = 0 or x dy/dx + 2y = 0

dy/y = - dx/x dy/y = -2 dx/xlny + lnx = lnC1 lny + lnx = lnC2

xy – C 1 = 0 x y – C 2 = 0 (xy – C 1)(x y – C 2) = 0 and this the solution of the equation

2 2 2

2

2

2

2 -Acceptable solution on y

If we can not analysis the equation then the equation

will be acceptable solution on y or xfirstly , To solve the equation that acceptable solution on y

their are three steps :-

1- Let y be in term alone .

2- By diifferentiation 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 .

ex) 3y = 2px - 2p / xsoly = 2/3 p x - 2/3 p / x , dy/dx = pby differentiation with respect to xdy/dx = p = 2/3 x dp/dx - 2/3 1/x 2p dp/dx + 2/3 p / x1/3 p - 2/3 p / x = (2/3 x - 4/3 p / x) dp/dx by multiplying X 3p - 2 p / x = 2(x - 2p / x) dp/dx by multiplying X xp x - 2p = 2(x - 2px) dp/dxp (x - 2p) = 2x(x - 2p) dp/dx

(x - 2p)(p - 2x dp/dx) = 0x - 2p = 0 or p - 2x dp/dx = 0

2 dy/dx = x p / 2x dp/dx 2dy = x dx dp/p = dx / 2x 2y = x / 3 + c lnp = ½ lnx i.e. p = xto delete p from two equation substituting about p on origin equation

y = 1/6 x

2

2 2

2 2

2 2

2

2 2

3

2

2

22

2

2

3

3

2

3 -Acceptable solution on x

secondly, To solve the equation that acceptable solution on x

their are three steps :-

1- Let x be in term alone .

2- By diifferentiation 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 .

Ex) x = p + pby differentiation with respect to y

dx/dy = dp/dy + 3p dp/dybut dx/dy = 1 / p 1/ p = (1 + 3p ) dp/dydp/dy = 1/ p (1+3p )dy = ( p + 3p ) dp

y = ½ p + 3/4 p x = p + p the origin equation

we can not delete p from the last tow equations so this the parametric solution.

3

2

2

2

3

2

3

4

Lagrange’s Equation form

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

ex ) y = 2xp + p

dy/dx = 2p + 2x dp/dx + 2p dp/dx

p = 2p (2x + 2p) dp/dx

- p = (2x + 2p) dp/dx

1 = (- 2x / p - 2) dp/dx

dx/dp = -2x / p - 2

dx/dp + 2x / p = -2 linear differential equation

= exp 2/p dp integrating factor

exp 2lnp = p

p x = -2 p dp

p x = -2p / 3 + C

4 -Lagrange’s Equation

2

2

2

22

3

Note the method of solution

in the example

Clairaut’s Equation is special case of Lagrange’s Equation

Clairaut’s Equation form :-

y = x p + f (p)

ex) y = p x + a/p

dy/dx = p + x dp/dx - a/p dp/dx

p = p + (x - a/p ) dp/dx

(x - a/p) dp/dx =0

dp/dx = 0 or (x - a/p ) = 0

p = c & p = a/x

y = x c + a/c

y = p x a / p + 2px a/p

y = p x + a / p + 2ax , y = a + a x + 2ax

y = 4a x single solution (parabola)

5 -Clairaut’s Equation

2 2 2 2 2

2

2

2

2 2 2 2 2

Note the method of solution

in the example

2

2

6 - Linear homogeneous Differential Equations with Constant Coefficients

(ao D + a1 D + a2 D + ….. + an) y = f (x)D = d/dx , ao,a1, a2,…..,an constants

L(D) y = f (x) non-homogeneousbut L(D) y = 0 homogeneous

then L() = 0 assistant equationRoots of this equation are 1 , 2 , 3 ,……,n

This roots take different forms :-1- if roots are real and different each other then the complement solution is

y c = C1 exp 1 x + C2 exp 2 x +……...+ Cn exp n x2- if roots are real and equal each other then complement solution is

y c = exp x (C1 + C2 x + ……..+ Cn x )3- if roots are imaginary then complement solution is

y c = exp x (C1 cos x + C2 sin x)

n n-1 n-2

examples-:

1) y - y = 0

( D - D) y = 0

( - ) = 0 ( - 1) = 0

( - 1)( + 1) = 0

1 = 0 , 2 = 1 , 3 = -1

y c = C1 + C2 exp -x + C3 exp x

2) y + 3y + 2y = 0

(D + 3D +2) y = 0

+ 3 + 2 = 0

( + 1) ( + 2) = 0

1 = -1 , 2 = -2

y c = C1 exp -x + C2 exp -2x

3) (y + a )y = 0

(D + a ) y = 0

+ a = 0

= ± a i

y c = C1 cos ax + C2 sin ax

4) (D + 2D + 2 )y = 0

( - 1) ( +1) = 0

1 = 1 , 2 = ± i

y c = C1 exp x + C2 cos x + C3 sin x

3

3 2

2

2 2

2 2

2

2

L(D) y = f (x) non-homogeneous

the general solution of non-homogeneous is

y = y c + y p

y c complement solution [solution of homogeneous equation L(D) y = 0 look last slide]

y p particular solution is y = f (x)

We knew how to get the complement

solution last slide

Now to get the particular solution

It depend on the type of function

we will know the different types

and example to every one

7 -Linear non-homogeneous Differential Equations with Constant Coefficients

L(D)

1

NoteL(D) is differential effective

1 / L(D) is integral effective

y = f (x)i) if f (x) is exponential functiony = exp ax = ; L (a) = 0 ; D a

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

iii) if exponential function multiplied f(x)y = f (x) exp ax = exp ax f (x)

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

L(D)1

L(D)1

L (a)

exp ax

L(D)1

L (D + a)

exp ax

L(D)1

L (D + a)1

L(D )1

L (- a )

122

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

multiplied exponential function

y = exp ax sin x or cos x = exp ax sin x or cos x

vi ) If f (x) is polynomial

L(D) = f (x)

y = f (x)

and then use partial fractions and use the following series:-

(1 - x) = 1 + x + x + x +…..

(1 + x) = 1 - x + x - x +……

L(D )1

L (D + a)

1

L(D )1

-1 2 3

-1 2 3

solved example :-y + y - 6y = 8 exp 3x(D + D - 6) y = 0 + - 6 = 0( - 2)( + 3) = 01 = 2 , 2 = -3

yc = C1 exp 2x + C2 exp -3xyp = 8 exp 3x y = 8 exp3x D a

yp = 8/6 exp 3x = 4/3 exp 3xgeneral solution y = yc + yp

y = C1 exp 2x + C2 exp - 3x + 4/3 exp 3x

D + D - 6

1

9 + 3 - 61

2

2

2

general problems1) (xy - y) dx +(x y - x) dy = 0

2) (2xy + tany) dx + (x + sec y )dy = 0

3) (exp x +4y )dx + (4x - siny) dy = 0

4) (x + y ) dx + xy dy = 0

5) x dx/dy +y =y ln x

6) 3 dy/dx - 2y = y exp 3x

7) y = x p + p

8) y = x /p +p

9) y - y - 4y + 4y = 0

10) y + 2y - 8y =0

2 2

2 2

2

4

2

Find

gen

eral

sol

utio

n of

the

follo

win

g eq

uation

AHM

ED H

YDER

AHM

EDFA

CULT

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SCIE

NCE

2 2

11) (D - 6D +9) y = exp 3x

12) (D + 2D - 2D - 1)y = 4 exp -x

13) (D - 5D + 6)y= x exp 2x

14) (D + D - 12)y = x exp x

15) (D - 7D + 12)y = (x - 5x ) exp 2x

16) (D - 7D + 12 )y = 8 exp (5x) sin 2x

17) (D - 4D + 8)y = exp (2x) cosx

18) (D + 4D)y = x - 7

19) (D - D + 1)y = x + 6

20) (D - 10D + 25)y = 30 x - 3

TRY TO SOLVE PROBLEMS ALONE GOODLUCK

2

34

32

2 2

2

2

232

2

2

2

2

3

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 advance and perfecting of mathematics are closely related by prosperity of the nation”

Ahmed Hyder Ahmed - Faculty of science - Dep. of Physics

[email protected]

Ahmed Hyder Ahmed

Nuclear Physics Lab

Faculty of Science Minya universityMinya CityEgypt

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