Ordinary differential equations

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Ordinary differential equations

  1. 1. In the Name of Allah Most Gracious MostMerciful Ordinary Differential Equations Prepared by Ahmed Haider Ahmed B.Sc. Physics - Dept. of Physics – Faculty of Science
  2. 2. TO my mother , my brothers and my best friend Abd El-Razek
  3. 3. 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
  4. 4. DefinitionsDifferential equation is an equation involving anunknown function and its derivatives.Ordinary Differential equation is differential equationinvolving one independent variable and its differentialsare ordinary.Partial Differential equation is differential equationinvolving two or more independent variables and itsdifferentials are partial.Order of Differential equation is the order of the highestderivative appearing in the equation.Degree of Differential equation is the power of highestderivative 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.
  5. 5. Solutions of First Order Differential Equations  1- Separable Equations  2- Homogeneous Equation  3- Exact Equations  4- Linear Equations  5- Bernoulli Equations
  6. 6. 1- Separable Equations (separation variable)General form of differential equation is (x ,y) dx + (x ,y) dy = 0By separation variableThen 1 (x) 2 (y) dx + 1(x) 2(y) dy = 0 1( x) 2 ( y) dx dy 0 1 ( x) 2 ( y)by integrating we find the solution of this equation.Ex) find general solution for xydx x 2 dy 0 1 1 dx dy by integration x y ln y ln x c
  7. 7. 2- Homogeneous EquationThe condition of homogeneous function is n f ( x , y) = f (x ,y)and n is Homogeneous degree (x ,y) dy + (x ,y) dx = 0and , is Homogeneous function and have the same degreeso the solution isput y = xz , dy = x dz + z dx and substituting in the last equationthe equation will be separable equation, so separate variables and then integrate to find the solution.
  8. 8. 3- Exact Equations (x ,y) dy + (x ,y) dx = 0The required condition of equation to be exact equation is x yand its general solution is dx dy cnote : take repeated factor one time only.
  9. 9. if x yThe equation will be not exact,to convertit to be exact multiply it by integral factoras following .integral factor is 1 ( x) exp ( y x ) dx 1 ( y) exp ( x y ) dy
  10. 10. Examplesi ) (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 = cii) (1 – xy) dx + (xy – x ) dy = 0
  11. 11. (1 xy) ( xy x 2 ) x y 2x x y So, its not exact 1Since ( x) exp ( y x ) dx 1Then ( x) exp ( xy y ) ( x y 2 x) dx 1 1 exp x dx exp ln x exp ln x 1 (by multiplyin g this value by equation ii) x 1 y dx ( y x)dy 0 thi equation is exact s x 1 x y dx ( y x)dy c yln x xy c 2 note :- we took the repeated factor one time only
  12. 12. 4- Linear EquationsLinear Equation form is dy P( x) y Q( x) dxthe integral factor that convert Linear Equations to exact equation is :- = exp p(x) dxby multiplying integral factor by Linear Equation form dy P( x) y Q( x ) his equation is exact t dxso the general solution is :- y= Q dx + c
  13. 13. dyEx) y y sec x cos2 x , y dxsolution dy p( x) y Q( x) p( x) sec x , Q( x) cos2 x dx exp sec xdx exp ln sec x tan x sec x tan xgeneral solution is y Q dx c (sec x tan x) y (sec x tan x).cos2 dx (cos x sin x cos x)dx 1 sin x sin x c 2
  14. 14. 5- Bernoulli Equation Bernoulli Equation form is dy n P( x) y Q( x) y dx note :-To solve Bernoulli Equation if n = 0 the Bernoulli Equationa) Divide Bernoulli Equation over y n will be linear1 dy equation. P( x) y(1 n) Q( x)y n dx dz dy if n = 1 Bernoullib) Put z y (1 n) then (1 n) y Equation will be dx dx separable 1 dz equation p( x) Q( x) (1 n) dxdz (1 n) p ( x) z (1 n)Q ( x)dxthis is linear equation and its solution as w e told before.
  15. 15. dy y2 Ex) y – 3 sin x dx x solution dy y – 3 sin x. y 1 dx x dz dy put z y andthen 2y dx dx dz z 2 6 sin x thisequation is linear dx x exp pdx exp 2 x dx exp 2 ln x exp ln x 2 x2the general solution will be x2 y 6 x 2 sin xdx c x2 y 6 ( x 2 cos x sin x cos x) c
  16. 16. 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.
  17. 17. 1- Acceptable solution on pif we can analysis the equation then the equation will be acceptable solution on pEx) x 2 p 2 3 xpy 2 y2 0Sol xp y xp 2y 0xp y 0 or xp 2y 0 dy dyx y 0 or x 2y 0 dx dx dy dx dy dx or 2 y x y xln y ln x ln c1 or ln y 2 ln x ln c2xy c1 0 or x2 y c2 0 ( xy c1 )( x 2 y c2 ) 0and this is the general solution of the equation.
  18. 18. 2- Acceptable solution on yIf 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 ythere 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 .
  19. 19. p2 dyEx) 3y 2 px 2 , p x dxSolution 2 2 p2y px by differentiation with respect tox 3 3 xdy 2 dp 2 1 dp 2 p2 p x 2pdx 3 dx 3 x dx 3 x21 2 p2 2 4 p dp p x , multiplyin g by 33 3 x2 3 3 x dx p2 p dpp 2 2 2 x 2 , multiplyin g by x 2 x x dx
  20. 20. 2 2 3 dppx 2 p 2( x 2 px) dx dpp( x 2 2 p) 2 x( x 2 2 p) dx 2 dp( x 2 p) p 2 x 0 dx dpx2 2 p 0 or p 2x 0 dx dp dp2 x2 p 2x dx dx 2 dp dx 2dy x dx p 2x x3 12y c ln p ln x p x 3 2to delete p from two equation substituting about p on origin equation 1 3 y x 6
  21. 21. 3- Acceptable solution on xsecondly, to solve the equation that acceptable solution on xthere 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 originequation and the equation that we got after second stepif we can not delete it the solution called the
  22. 22. dyEx) x p p3 , p dxby differentiation with respect toydx dp 2 dp 1 dx 3p , butdy dy dy p dy1 2 dp (1 3 p )p dydp 1dy p (1 3 p 2 ) dy ( p 3 p 3 )dp 1 2 3 4 y p p 2 4 x p p3 (the origin equation)we can not delete p from the last tow equations sothis the parametric solution.
  23. 23. 4- Lagrange’sEquationLagrange’s Equation form y = x g (p) + f (p)Ex) y 2 xp pdy dp dp 2 p 2x 2pdx dx dx Note dp dp the method of solutionp 2 p(2 x 2 p) p (2 x 2 p) in the example dx dx 2x dp dx 2x1 ( 2) 2 p dx dp pdx 2 x 2 linear differential equationdp p dp exp 2 integral factor pe 2 ln p p2 p2 x 2 p 2 dp 2 2 p3p x c 3
  24. 24. 5- Clairaut’s EquationClairaut’s Equation is special case of Lagrange’s EquationClairaut’s Equation form :- y = x p + f (p) a NoteEx) y px p the method of solution in the exampledy dp a dp p xdx dx p 2 dx a dpp p x p 2 dx a dp x 0 p dxdp a 0 o r (x 2 ) 0dx p
  25. 25. ap c & p x ay xc c 2 a2 ay p2 x2 2 2 px p p a2y2 p 2 x2 2ax , 2 y a ax 2ax p2y2 4ax single solution (parabola)
  26. 26. 6 - Linear homogeneous Differential Equations withConstant Coefficients(a0 D n a1 D n 1 a2 D n 2 ........ an ) y f ( x) dD , a0 , a1 , a2 , a3 ,..... n are constant a dx L(D) y = f (x) non-homogeneous but L(D) y = 0 homogeneous then L( ) = 0 assistant equationRoots of this equation are 1 , 2 , 3 ,……, nThis roots take different forms as following:-
  27. 27. 1- if roots are real and different each other then thecomplement solution is 1x 2x nx yc C1e C2 e ......... Cn e2- if roots are real and equal each other thencomplement solution is x r 1 yc e (C1 C2 x ......... Cn x )3- if roots are imaginary then complement solution is x yc e (C1 cos x C2 sin x )
  28. 28. examples :-1) y y 0 3)( y a ) y 0(D3 D) y 0 (D 2 a )y 2 0 3 2( ) 0 ( 1) 0 2 2 ( a ) 0 ( 1)( 1) 0 1 0, 2 1, 3 1 aiyc C1 C2 e x C3 e x yc C1 cos ax C2 sin ax2) y 3y 2y 0 4)(D 2 2 D 2) y 0( D 2 3 D 2) y 0 ( 1)( 2 1) 0( 3 2) 0 1 1, 2 i( 1)( 2) 0 x 1, 2 yc C1e C2 cos x C3 sin x 1 2 x 2xyc C1e C2 e
  29. 29. 7- Linear non-homogeneous Differential Equations withConstant CoefficientsL(D) y = f (x) non-homogeneousthe general solution of non-homogeneous is y = yc + ypyc complement solution[solution of homogeneous equation L(D) y = 0 look last slide]y p particular solution is Note L(D) is differential 1 effective y f ( x) 1 / L(D) is integral L( D) effectiveWe knew how to get the complementsolution 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 .
  30. 30. 1 Particular solution is y f ( x) L( D)i) if f (x) is exponential function 1 e ax y e ax ; L(a) 0 ; D a L( D) L(a )ii) special case in exponential function at L (a) = 0 1 e ax y e ax L( D) L( D a )iii) if exponential function multiplied f(x) 1 1 y f ( x)e ax e ax f ( x) L( D ) L( D a )iv) if f (x) is trigonometric function sin x or cos x 1 1 y 2 sin ax or cos ax 2 sin ax or cos ax L( D ) L( a )
  31. 31. v) if f (x) is trigonometric function sin x or cos x multiplied exponential function 1 1 y e ax sin x or cos x e ax sin x or cos x L( D) L( D a ) vi ) If f (x) is polynomial L( D) f ( x) 1 y f ( x) L( D) and then use partial fractions or use the following series : - 1 (1 - x) 1 x x2 x3 .. 1 (1 x) 1 x x2 x3
  32. 32. solved example :-y y 6 y 8e3 x(D2 D 6) y 0 2 6 0 2 3 0 1 2 , 2 3yc C1e 2 x C2 e 3x 1 1yp 2 8e3 x yp 8e3 x D D 6 9 3 6 D a 8 3x 4 3xyp e e 6 3general solution y yc yp 2x 3x 4 3xy C1 e C2 e e 3
  33. 33. general problems1) ( xy 2 y )dx ( x 2 y x)dy 02) (2 xy tan y )dx ( x 2 sec2 y )dy 03) (e x 4 y )dx (4 x sin y )dy 04) ( x 2 y 2 ) dx xy dy 0 dx5) y y 2 ln x dy dy6) 3 2 y y 4e3 x dx7) y x p p 2 x8) y p p9) y y 4 y 4 y 010) y 2y - 8y 0
  34. 34. 2 3x11) ( D 6 D 9) y e 4 3 x12) ( D 2D 2 D 1 )y 4e13) ( D 2 5 D 6) y x3 e 2 x 2 2 x14) ( D D 12) y xe15) ( D 2 7 D 12) y ( x 5 x )e 2 x 2 5x16) ( D 7 D 12) y 8e sin 2 x17) ( D 2 4 D 8) y e 2 x cos x 2 218) ( D 4 D) y x 719) ( D 2 D 1) y x3 6 220) ( D 10 D 25) y 30 x 3
  35. 35.  Finally , this course of ordinary differential equations is usefulto different student special students of physics. Theoreticalphysics required to be know the bases of mathematicsspecially differential equations such that quantum mechanicsdepend on Schrödinger equation and this equation isdifferential equation so this branch of physics depend upondifferential equations . I made slide of problems in differenttypes of differential equations to examine yourself .Finally don’t forget these words for Napoleon “the advancingand perfecting of mathematics are closely related byprosperity of the nation”
  36. 36. Ahmed Haider AhmedNuclear Physics LabFaculty of ScienceMinia universityMinia CityEgypt

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