Ordinary Differential Equations
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Transcript of Ordinary Differential Equations
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
Y OF
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
Ahmed Hyder Ahmed
Nuclear Physics Lab
Faculty of Science Minya universityMinya CityEgypt