MATH 6101 Fall 2008 Newton and Differential Equations
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1 MATH 611 Fall 8 Newto ad Differetial Equatios
2 A Differetial Equatio What is a differetial equatio? A differetial equatio is a equatio relatig the quatities x, y ad y' ad possibly higher derivatives of y. Examples: y = x+ y y - y = (1 + x ) y + xy + 4x y= 1-Oct-8 MATH 611
3 Differetial Equatios How are differetial equatios used? Newto s Secod Law: Radioactive decay: dmv ( ) F = ma = m dt = dt dp dt =-kp ds 1-Oct-8 MATH 611
4 Differetial Equatios Newto s Law of Coolig: The wave equatio: dq h A ( T Tev ) dt = - P = c t The heat equatio: u k æ + + ö t ç x y z è ø u u u u - kç + + = 1-Oct-8 MATH 611 4
5 Differetial Equatios: Moder Solutio Cosider the followig simple differetial equatio: dp k P dt = Our moder process is Separatio of Variables itroduced by l Hospital i Oct-8 MATH 611 5
6 Separatio of Variables dp = k P dt dp = k dt P ó dp = k dt= k dt õ ò ò P l( P ) + C = kt+ C l( P) = kt+ C 1 + Pt () = e = Ae kt C kt 1-Oct-8 MATH 611 6
7 Newto s Method of Series dp dt = k P Now, assume that we ca express P as a fuctio of t by: P() t = å = a t The dp -1 = å a t ad our first equatio gives dt = 1 1-Oct-8 MATH 611 7
8 Newto s Method of Series dp k P dt = 1 at - = k at = kat = 1 = = å å å at + at+ at + 4at + = kat + kat+ kat + kat Settig correspodig powers of t equal, gives the followig 1-Oct-8 MATH 611 8
9 Newto s Method of Series a = ka 1 k k a = ka1 a = a1 = a k k a = ka a = a = a 4 k k 4 a = ka a = a a 4 = k k a = ka a = a = a! Oct-8 MATH 611 9
10 Newto s Method of Series k ( kt) Pt () = at = a t = a!! We will show later that Therefore, å å = = å = e x x = å =! ( kt) Pt () = aå = ae å =! kt 1-Oct-8 MATH 611 1
11 Newto s Method of Series Example : Fid a solutio to the followig equatio Let The y + y= y å = = åax -1 - ad ( 1) 1 y a x y a x = å = å - å å = = 1-Oct-8 MATH
12 Newto s Method of Series y + y= å ( - 1) ax - = -å ax = = y =-y a + ax+ 4 ax + 5 4ax + =-a -ax-ax -ax - This gives us that a =-a 4 a =-a 6 5a =-a 8 7a =-a a =-a 54 a =-a 76 a =-a 98 a =-a Oct-8 MATH 611 1
13 Newto s Method of Series 1 1 a =-a a =- a =- a! 1 1 a =- a a =- a =-! a a =-a a =- a = a = a 4 4 4! a =-a a =- a = a = a ! a =-a a =- a =- a =- a ! a 1 7 =- a 5 a 7 = a = a = - 7! a 1 1-Oct-8 MATH 611 1
14 Newto s Method of Series 4 y åax a a1x ax ax a4x = = = = a+ a1x- ax - a1x + ax + a1x - ax - a1x +!! 4! 5! 6! 7! = a æ 1- x + x - x + ö + a æ 1 x- x + x - x + ö! 4! 6! çè ø çè! 5! 7! ø Let cx ( ) 1! x 4! x 6! x ad 4 6 = sx ( ) x x x x! 5! 7! 5 7 = Oct-8 MATH
15 Newto s Method of Series This gives us that y= a c ( x ) + a s ( x ) Note that: [ ( ) ] [ ( ) ] 1 æ 1 ö æ ö cx ( ) + sx ( ) = 1- x + x - x + x x x x! 4! 6! çè ø èç! 5! 7! ø = 1- x + x + x - x - x + x - x + x + x +!!! 4!!4! 6!!!! 5! æ 1 ö (1 1) x æ ö = ç + - x !! 4!! x + çè ø èç!4! 6!!! 5! ø æ 1 1 1ö æ ö 4 6 = 1+ x x x çè ø çè ø æ 4 4 ö æ ö 6 = 1+ x + x x + èç 1 1 ø çè 6 6 ø = 1 1-Oct-8 MATH
16 Newto s Method of Series Also ote that: c () = 1 s () = Further ote that: d cos( x) =-cos( x) dx d si( x) =-si( x) dx We ca show that: tht c( x) = cos( x) s( x ) = si( x ) 1-Oct-8 MATH
17 Your Tur Solve by Newto s method (i.e., fid the first 5 o-zero terms): 1) 1 y = y+ 1 - x ) y = 1 + xy 1-Oct-8 MATH
18 Solutios 1 y = y+ 1 - x 1 1æ 1 ö 1æ1 ö 1æ 1 ö 4 5 yx ( ) = a (1 ) o+ + ao x+ ax o + ao x a + + o- x + + a o x + ç è ø 6ç4 6çè è ø ø y = 1 + xy yx ( ) = ao+ x+ ax o + x + ax o + x Oct-8 MATH
19 Solvig Algebraic Equatios How ca we solve a equatio of the followig form? y y x y x = For this techique we do the same: assume that y has a power series expasio i terms of x. y Where does this lead us? = åax = 1-Oct-8 MATH
20 Solvig Algebraic Equatios å å å = = = æ ö æ ö æ ö ax ax 4 x ax + x = çè ø èç ø çè ø How do we cube out a series? This seems to be the oly stickig poit to proceedig with this process. We will make a couple of defiitios ad assumptios. 1-Oct-8 MATH 611
21 Solvig Algebraic Equatios y = åax = Give the above series defie the tail series to be å k= t = åa x = a x + a x + a x + We the have that t = y ad t = a x + t Oct-8 MATH 611 1
22 Solvig Algebraic Equatios Give a algebraic equatio we will write it as F ( x, y ) = Ad we will begi our procedure by settig F ( x, y) = F( x, y) The kth iteratio of our process cosists of three steps 1) Extract the x k level equatio from F k (x,t k ) =. That is fid the coefficiets of x k ad set them equal to. ) Solve the x k level equatio for a k. ) Substitute t k = a k x k + t k+1 ito F k (x,t k ) = ad use the Biomial Theorem to obtai the ew equatio F k+1 (x,t k+1 ) 1-Oct-8 MATH 611
23 Solvig Algebraic Equatios Example: F ( x, y) = y + y-4 - x y+ x= F ( x, t ): t + t -4- x t + x= Step 1: Sice t =a +a 1 x+a x + ad the x level equatio igores all o-costat terms i F (x,t ): x level: ( )( ) a + a - 4= a + 4 a - 1 = a =- 4or a = 1 This will give us two solutios ad two equatios 1-Oct-8 MATH 611
24 Solvig Algebraic Equatios Solutio 1: a = 1 Now, replace t with 1+t 1 to get: ( ) ( ) ( ) 1+ t + 1+ t -4- x 1+ t + x= t + t + + t -4-x - t x + x= Igore all terms with degree > 1. 5t + t -x - t x + x= 5a 1 + = a = Oct-8 MATH 611 4
25 Solvig Algebraic Equatios F ( x, t ) = 5t + t -x - t x + x= Now, replace t 1 with - 5 x+ t 5t + t -x - t x + x= æ ö æ ö æ ö 5 - x+ t + - x+ t -x - - x+ t x + x= ç è 5 ø çè 5 ø çè 5 ø Igore all terms with degree > t+ t - xt-x t- x + x = a - 5 = 1 a = 15 1-Oct-8 MATH 611 5
26 Solvig Algebraic Equatios 4 1 F( x, t) = 5t+ t - xt-x t- x + x = Now, replace t with 15 x + t t+ t - xt- x t+ x - x = Igore all terms with degree > a + 65 = a = Oct-8 MATH 611 6
27 Solvig Algebraic Equatios So, we fid a series expasio for y of the form: y a x x x x = å = = Oct-8 MATH 611 7
28 Your Tur 1) Fid the solutio correspodig to a = -4 ) Fid the first four terms of the ifiite series expasio of y if y + xy = 1. 1-Oct-8 MATH 611 8
29 Your Tur 1) We ca solve the origial equatio for y ad the fid the series expasio for that. We should get the same aswer. y + y-4- x y+ x= y x y x + (- ) + ( - 4) = y = y = ( x -) (-x ) -4(x-4) x - x - x - x+ 4 ( ) Oct-8 MATH 611 9
30 Your Tur y y 1 4 x - + x -6x - 8x+ 5 = = 1- x + x - x + x x -- x -6x - 8x+ 5 = =- 4+ x+ x + x - x This is what we should have foud with our other process. 1-Oct-8 MATH 611
31 Your Tur ) Fid the first four terms of the ifiite series expasio of y if y + xy = 1. y + xy- 1= 1 x y= x x y= 1- x+ x + x Oct-8 MATH 611 1
32 Problem F x y y xy Stage : F ( x, t) = t + xt- 1 = a - 1= (, ) = + - 1= a = 1 Stage 1: I F (x,t ) replace t = 1 + t1 F x t t x t 1 (, 1) = (1 + 1) + (1 + 1) - 1 = 1 1 = = F ( x, t ) t t t xt x a + 1 = 1 a 1 1 =- 1-Oct-8 MATH 611
33 Problem Stage : I F 1 (x,t 1 ) replace t 1 1 =- x+ t F ( x, t ) = t + t + t + xt + x= æ 1 ö æ 1 ö æ 1 ö F( x, t) = ( - x+ t) + - x+ t t + = 1- + tx+ t + x - x+ t + x= ç è ø èç 1 ø èç ø 1 1 = t+ t + t - xt+ x t-xt - x = 7 a = a = 1-Oct-8 MATH 611
34 Stage : I F (x,t ) replace Problem t = x + t = t 1 1 F( x, t) = t+ t + t - xt+ x t-xt - x = F( x, t) = t+ t + t - xt+ x t-xt - x = 7 1 a - = 7 a 1 = 81 1-Oct-8 MATH 611 4
35 Stage 4: I F (x,t ) replace Problem 1 t = x + t F x t t t t xt x t xt x 7 (, ) = = æx ö æx ö æx ö æx ö F4( x, t4) = + t t t 4 - x + t 4 + ç è ø èç ø ç è ø ç è ø æ x ö æx ö ç 4 ç = x + t - x + t - x = ç81 81 è ø çè ø 7 = t + t 1 a4 - = 81 1 a 4 = 4 4 x t x t x t x t t4 xt4 xt xt4 x x x x x = = Oct-8 MATH 611 5
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