2 nd class Advance Mathematics and numerical analysis الریاضیات المتقدمة والتحلیل العددي الماده:م.د.عبدالمحسن جابرعبدالحسین

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1 Save from: d cla Advace Mathematic ad umerical aali استاذ الریاضیات المتقدمة والتحلیل العددي الماده:م.د.عبدالمحسن جابرعبدالحسین

2 CHAPTER ONE Defiitio -Partial Derivative If f i a fuctio of the variable, ad i the regio plae the Partial Derivative of f with repect to (w. r. to, at poit (, i lim f (, f (, f/ = Ad (w. r. to at poit (, i f/ = lim f (, f (, To fid f/ i impl regard a cotat i f (, ad Differetial (w. r to i writte i form f/ = z/ = F/ or D X f= Z =F Uig ame wa to fid f/ i impl regard a cotat i f (, ad Differetial (w. r. to i writte i form f/ = z/ = F/ or D f = Z =F Sice a partial derivative of fuctio twice variable to obtai ecod partial derivative a - f/ = f - f/ = f - / ( f/ = f/ = f 4- / ( f/ = f/ = f 5- / ( f/ = f/ = f 6- / ( f/ = f/ = f Note I It i ea to eted the partial derivative of fuctio of three variable or more / ( f/ = f/ = f Theorem If f (, ad it' partial derivative f, f, f,ad f are defie i regio cotaiig a poit (a, b ad are all cotiuou at (a, b, the f = f.

3 Eample Let f(, = The fid f, f, f,ad f Solutio f = + f = - + f =. Problem -Let f(, = e - i + e co +8 The fid f, f -Fid f ad f at poit (,/ if f = 4 -If f(, = e - i(/ +. The fid f, f, f, F ad f 4-If U = +arc ta(z, the fid U, U ad U z. 5-If V = + + z + Log(z, the fid V, V, V z,v ad V zz. 6-If f =, the fid f, f. 7-Prove that U = U If a-u = i + co b-u = L -Chai Rule - Fuctio of oe variable If = f (, ad = (t, = (t the = t t - Fuctio of two or three variable i a- If Z = f(,, = (t, = (t the Z = t Z t Z + t b- If Z = f(,, w, = (t, = (t, w = w (t the Z = t Z t Z + Z + t w w t Eample Let f(, = e, = rcoθ, = r iθ Fid fr ad f, i term r ad θ. Solutio

4 f = f + f r r r f = e f = e = coθ r = iθ r f = e coθ+ e iθ r r co co = r i co e = r co co r i e. f = f = -r iθ = rcoθ + f f = e (-r iθ+ e (rcoθ r co co r co co = r i e + r co e r co co = r e (co i r co co = r co e. Problem Fid f t, i the followig fuctio - f(, = -, = e t, = t-6 - f = -, = co t, = e -t - f = /, = l t, = cot t 4- f = e l( -, = t, = t- t 5- f = 6- f =, = i - t, = it, = coh t, = ih t 7- f = i(+ - z, = i t, = t e 8- f = ta (, = e t co t, = e t i t, z = l t 4

5 9- f =, = e t, = t-6 Fid U, U ad U z, i the followig fuctio tah r - U = (, r = i z, = co z -U = l( r+ + tif, r =, = z, t= z - The Total Differetial The total differetial of fuctio W = f(,, z,..( I defied to be f dw = d + f d + f z dz Or dw = f d + f d + f z dz I geeral the total differetial of fuctio W = f(,, z, u,..,v i defied b dw = f d + f d + f z dz+ f u du+.+ f v dv where,, z, u.ad v are idepedet variable. But if, ad z are ot idepedet variabl but are them ca elve give b = (t, = (t, z = z (t, the we have d = t dt, d = t dt, dz= z t dt. Or i the form:- = (r,, = (r,, z = z(r,. The we ha d = r d = r dz = r z dr+ dr+ z dr+ d d..( d The ( become i cae W = f(,, z = f((r,, (r,, z(r, = f(r,. The from ( dw = w d + dw =[ r ad ( we obtai:- w d + w z dz dr+ d ] w + [ r dr+ d ] w +[ z z r dr+ d] w z 5

6 w =[ w + r w + r z z r w ] dr+ [ w + Eample Fid the total differetial of fuctio W = + + z if = r co, = r i ad z = r Solutio dw = w d + w d + w z dz = d + d + z dz d = r dr+ d, or w + z z ]d.(4 d = r dr + d =codr - r i d d = r dr + d = i dr + rcod, dz = z r dr + z d = dr. Now dw = [codr - r i d] + [i dr + rcod] + r [dr.] = rco [codr - r i d] + r i [i dr + rcod] + r [dr.] dw = [rco ]dr - r i d] + [i dr + rcod] + r [dr.] = [rco + r i +r] dr + [-r i co+ r i co ] d + r dr.], dw = 4r dr]. Problem If U = f(, Fid du, i the followig:- - U = L + L if, = e -t, = e t. - U = ta - + = if, = t, = t-. - U = i(+ +co,, = π +t, = π-4t. 4- U = + + 6, = t-, = 4t-. 5- U =, = ech t, = coth t. tah r 6- U = (, r = i z, = co z 7-U = l( r+ + tif, r =, = z, t= z. 8- If f(, = co+ e. Prove that f = f. 9- If f(, = ta (. Prove that f + f. = - If f(, = e - co. Prove that f + f. = 6

7 - If W= i(+ ct. Prove that W tt = c W. - If W= co(+ ct. Prove that W tt = c W. - If W= L(+ ct+ co(+ ct. Prove that W tt = c W. 4- If W= ta(- ct. Prove that W tt = c W. 7

8 CHAPTER TWO Differetial Equatio (d.e Itroductio:- Defiitio. Differetial Equatio (d.e If i a fuctio of, where i called the depedet variable ad i called the idepedet variable. A differetial equatio i a relatio betwee ad which iclude at leat oe derivative of with repect to (w.r.to. Which ha two tpe:- - Ordiar d.e. If (d.e ivolve ol a igle idepedet variable thi derivative are called ordiar derivative, ad the equatio i called ordiar (d.e. - Partial d.e. If there are two or more idepedet variable derivative are called partial derivative, ad the equatio i called partial (d.e. For eample (a d d d e d d d (b u/ + u/ = df (c + = i d (d - + = The Order of (d.e I that the derivative of highet order i the equatio for eample (a order (b order (c order (d order? Solutio of Differetial Equatio A relatio betwee the variable that occur i (d.e that atifie the equatio i called a olutio or whe ad it' derivative are replace through out b f( ad it' derivative for eample Show that = aco +bi, of derivative a olutio of (d.e + 4 =.. (, Where a ad b are arbitrar cotat. Solutio Sice = aco +bi, =-ai + bco -4aco -4bi, put ad i ( -4aco -4bi + 4(aco +bi = 8

9 =, the thi olutio called the geeral olutio. Eercie Show that each equatio i a olutio of the idicated (d.e ( = where = c + c + c e ( + = where = c l+ c ( +9 = 4co where = co (4 - = e where = e (5 = ec where = ta. Firt Order Differetial Equatio The firt order differetial equatio take i the form:- M(, d+n(, d = ( Where M ad N are fuctio of ad or both. To olve thi tpe of (d.e, we coider the followig method:- -Variable Separable A (d.e ca be put i the form:- f(d +g(d =, or ad derivative of i term ad derivative of i aother term. Thi equatio called Variable Separable, thi equatio ca be olve b take the itegral of two ide of thi equatio f(d + g(d =c, where c i arbitrar cotat. Eample Solve d=d d-d= (d-d=, d d d, b itegral of two ide d c, L-l=c, l c, c e c,. c Problem Solve the followig differetial equatio:- - (-d +( +d= - ( +d + d 9

10 d - = e - d d 4- = d 5- e ecd+cod=. -Homogeeou Differetial Equatio (H.d.e The differetial equatio a form M(, d+n(, d =, Where M ad N are fuctio of ad i called (H.d.e if atif the coditio M (k, k =k M(, N (k, k =k N(, Where k i cotat. For eample - ( - d + d= M = -, N = M(k,k =(k - (k = k k = k ( k (M N (k,k = (k = k ( k (N. The equatio i (H.d.e. - Solve (-d +d = M = -, N = M(k,k =(k- (k= k( k(m N (k,k = k ( k (N. The equatio i ot (H.d.e. If the equatio i homogeeou we ca olve b the followig method:- Put (H.d.e i the form d = f(/ ( d Let v= /. (4 d Put (4 i ( = f(v..(5 d From (4 =v, ad d = dv+vd, divided b d

11 d d dv d v d, ice = f (v from (5 d dv dv f(v v f (v-v d d dv d (f(v-vd =dv f ( v v d dv. Or v f ( v d dv (6 f ( v v After olvig replace v b /. Eample Solve ( + d + d= Solutio Sice thi equatio (H.d.e. Now d= - ( + d, d =, put =v d d v v = = d ( v v v f(v = ' v d dv, v f ( v d dv, v v v d vdv v, b itegral both ide l + /l(+v = c, l + /l(+ / = c l c, c c, where c c e Problem Solve the followig differetial equatio:- ( d -(+ d= ( ( + +d +(-d =

12 ( ( -4 d +( -6 d = (4 ( e / -e / d+ e / d= (5 (+ e / -e / d+ e / d= -Eact Differetial Equatio The differetial equatio a form M(, d+n(, d =...(7 There i fuctio f(,=c....(8, Which a olutio of (7. df(,=...(9. From total partial differetial equatio df(,= f/ d + f/ d (. From 7,8,9 ad, f/ =M... ( f/ = N Now f/ = M/, f/ = N/, Sice f/ = f/ M/ = N/,. (. Which coditio of eact? To olve equatio (7 we mut fid f which the olutio of equatio (7. f/ =M from (, f =M, f= M + A(.(, Where A( i fuctio of. We mut fid A( f/ = N = / [ M ]+ A (, ice f/ = N, from (, A ( = N- / [ M ] A ( = {N- / [ M ]}+c..(4 Now put (4 i ( which complete olutio. Eample Solve the followig differetial equatio d = d Solutio (- d ( -d =, N= -, M =, M/ = N/ = -, f/ = M, f= M + A( = ( + A(

13 f= ( ( / + A(.(* (we mut fid A(, f/ = - + A ( = N= - A ( =, A( = +c put i (* F= ( ( / + +c. Problem Solve the followig differetial equatio:- ( ( + d +( + d= ( ( - d - (-d = ( (co+d +( + d = (4 ( e +d+ ( + e d= (5 ta d+ ec d=. Itegratig Factor If the equatio M d+n d =. I ot eact, the there i uch that M d+ N d =.(* I eact the ( M/ = ( N/. To fid ( i called itegratig factor. Theorem I M N (i If f ((fuctio of, or cotat. N The = e f(d. N M (ii If g( (fuctio of, or cotat. M The = e g(d. (iii If i fuctio of ad, the there i o geeral method to fid (itegratig factor. Eample Solve the followig differetial equatio d +(+ -d= Solutio M=, N= +. M = N =, ot eact

14 N M, i fuctio of M The = e g(d d l = e e (d +(+ -d= d +( + - d= M=, N= + -. M = =N = eact f/ = M, f= M + A( = ( + A( f= + A(. (* We mut fid A(, f/ = + A ( = N= + - A ( = -, A( = - 4 /4 +c put i (* f= /4 +c. Eample Solve the followig differetial equatio d =- d Solutio d= (-d (-d-d= M=-, N=. M = - N =, ot eact M N, i fuctio of N The = e f(d d = e e ((- d-d= e e (-d- e d= M= e - e, N= -e. M = -e =N =- e eact f/ = M, f= M + A( = (e - e + A( f= e -e - e + A(. (* We mut fid A(, f/ = -e + A ( = N= -e A ( =, A( = c put i (* f= e -e - e +c. 4

15 4- Firt Order Liear Differetial Equatio If the equatio a form:- d +P( = Q( (5 d Where P ad Q are fuctio of. To olve equatio (5, we mut fid (I where I= e pd { I i itegratig factor}. Now multiple both ide of (5 b I d + P d = Q d (5 e pd { d + P d = Q d } e pd d + e pd P d = e pd Q d } d [e pd ]=Q e pd d (6 b itegrate (6 e pd = Q e pd d +c, Which the olutio of (5, or the olutio i I = IQ d +c Eample Solve the followig differetial equatio d + = d Sol Sice P =/, Q = I= e pd = I= e d/ = e l =. The olutio I = IQ d +c = d +c, = +c = + c/ Problem Solve the followig differetial equatio:- ( + = e ( + = ( + cot = co (4 + = - + (5 - ta =. 4.The Beroulli Equatio The equatio d +P( = Q( (*, if. d I imilar to L-equatio i called Beroulli Equatio. We hall how how traform thi equatio to liear equatio. I fact we mut reduce thi equatio to liear, product (* b ( - or 5

16 d [ +P( = Q( ] - d d - +P - = Q. (**. d Let w = - dw = (- d or dw = d put i (** dw +P - = Q, ( d or Which i L-equatio dw + (-P w = (- Q d Eample Solve the followig differetial equatio d + = d Sol d [ + = ] - d d - + =. (#, d Let w = - dw =- - d -dw = - d, put i (# dw w - + = d dw w - = -. d P = -, Q = -, I= e pd = I= e - d/ = e -l =. The olutio Iw = IQ d +c, w = - d +c, 6

17 w = - l + c,. Sice w = - = = - l + c, =. ( c l Problem Solve the followig differetial equatio:- ( - / = 4 ( +/ = ( + = e i (4 + / = 5 / (5 + =. Secod Order Differetial Equatio Special Tpe Certai tpe of ecod order differetial equatio uch that d d F(,,, (7 d d Ca be reduced to firt order equatio b a uitable of variable:- Tpe I Equatio with depedet variable whe equatio a form d d F(,,.. (8 d d It ca be reduced to firt order equatio b uppoe that:- d d dp p =, = d d d The equatio (8 take the form dp F(, p, =, d Which i of the firt order i p, if thi ca be olved for p a fuctio of a? p = q(,c. The ca be foud from oe additioal itegratio d = ( d +c= p d +c = q(,c d +c. d Tpe II Equatio with idepedet variable whe equatio (7 doe ot cotai eplicit but ha the form 7

18 d d F(,, =.. (9 d d The ubtitutio to ue are :- d d dp dp d dp p =, = d =. = p d d d d d The equatio (9 become dp F(, p, p =, d Which i of the firt order i p. Which olutio give p i term of, ad the further itegratio give the olutio of equatio (9. Eample Solve the followig differetial equatio d d d + =... (* d d d dp Let p =, = d p, put i (* d d dp p +p =, p d dp + =, d dp +d =o. b itegratio p+ = c d + = c d d = c - d d = d -l(c =+c c c l(c = -+c c = e = ce - = c - ce - Eample Solve the followig differetial equatio X d d d d Let + d + = d d d = /. (** d d p = ad d d = dp, put i (** d 8

19 dp + p = /, which liear i p, d I= e pd = I= e d/ =. Ip = IQ d +c Ip = (/ d +c = l+c p =l +c P = (l/ + c/ d Let = (l/ + c/ d d= [(l/]d +( c/d, = (l / + c l +c Problem Solve the followig differetial equatio:- ( + = ( + = ( + = (4 - = (5 + w =, where w cotat. Homogeeou-Secod Order (D. E With Cotat Coefficiet Coider liear equatio with cotat coefficiet which i the form:- +a +b= ( where a, b are cotat. How to olve thi equatio we hall ow fid how to determie m uch that = e m i a olutio of ( the = me m ad =m e m, put i ( m e m +a me m +be m = e m (m +a m+b =, ice e m, the m +a m+b =. (. Which called characteritic equatio. The we aw that e m i a olutio of ( m i root of (. Note The geeral olutio of (, there i three cae:- Cae i If m =m i equatio (, the olutio of ( (homogeeou equatio i h = (c + c e m Cae ii If m m i equatio (, the olutio of ( (homogeeou equatio i m m h = c e c e 9

20 Cae iii If m ad m root ( m= + i where i= i equatio (, the olutio of ( (homogeeou equatio i h = e c co c i ( E i Solve +4 +4=..(* Sol let = e m, = me m ad =m e m, put i (* m e m +4 me m +4e m = e m (m +4 m+4 =, ice e m, the m +4 m+4 = Which called characteritic equatio? (m+ = m =m = -, the olutio of (* i h = (c + c e - E ii Solve + -6=..(** Sol let = e m, = me m ad =m e m, put i (* m e m + me m -6e m = e m (m + m-6 =, ice e m, the m +m-6 = Thi called characteritic equatio (m+(m- = either m = - or m =, the olutio of (** i h = c e - + c e E iii Solve -4 +5=..(** Sol let = e m, = me m ad =m e m, put i (* m e m -4me m +5e m = e m (m -4 m+5 =, ice e m, the m -4m+5 = Thi called characteritic equatio m =, m = i + i, =, =, h = e (c co + c i.

21 No-Homogeeou-Secod Order (D. E With Cotat Coefficiet Coider the equatio which i the form:- +a +b= f( ( where a, b are cotat. To fid the geeral olutio of (. We fid olutio of homogeeou part +a +b= (, let h be olutio of (. The the olutio of ( take b added the olutio h to a aother pecial olutio p of (uch that the geeral olutio of ( become ( = h + p Method of Udetermied Coefficiet The coditio of thi ma that the form f(, ma be gueed for eample f( ma be a igle power of a polomial a epoetial fuctio a i, coi or um fuctio. The geeral olutio of (o- H. D.E become. ( = h + p. We tudet to fid h.we ca elect p from the followig table. Table k f ( (,,... p ke k i k co k k k,..., k, k p are co ta t ce, c co ta t mco i... k k m ad co ta t p mod p i How to ue the table to fid p:- (a If f( fuctio of firt colum of table, the we take p from ecod colum which correpodig it. (b If f( i um of two fuctio of -t colum the we elected p um of fuctio of -th colum which correpodig. (c If the umber of f( i root of h we mut modif the olutio of p,

22 Modificatio Rule If the umber lited i the table the lat colum root of h (H-part of equatio (. The the fuctio i ecod colum of table mut be multiplied b m where m i the multiplicit of the root i that equatio [hece for a ecod order equatio m ma be equal or ]. Eample B ue table write p where (a f(= (b f(= e (c f(= 4i (d f(= co +i (e f(= e + Sol (a p = k + k + k +k (b p = ce (c p = m co +i (d p = m co +i (e p = ce +k +k Eample Fid the geeral olutio of the followig (d.e -4= 8.(* Sol -4=.(** let = e m, = me m ad =m e m, put i (** m e m - 4e m = e m (m - 4 =, ice e m, the m - 4 = Thi called characteritic equatio m =4 m =, ( or m =, m = -; h = c e - + c e -, p = k + k +k, we mut fid k, k ad k p = k + k +k, p = k + k ad p = k put i (* k -4(k + k +k = 8-4k =8, k -4k = k =-, k =, k =- p = - - ( = h + p ( = c e - + c e

23 Variatio of Parameter Coider the equatio which i the form:- +a +b= f( (4 Where a, b are cotat, f( be a fuctio of. To olve (4 (a Fid h (olutio of (H-part, h = c u + c u (5 Where c ad c are arbitrar cotat, ad u ad u are two fuctio a form:- let e m or e m e co or e i, which olutio of (H-part. (b We replace c ad c b fuctio of a v ad v the (# become h = c v + c v..(6, Which olutio of (4, h = v u v u + v u + v u h = (v u + v u + (v u + + v u. (7, from thi h = v u + v u, ad v u + + v u =..(8 Now h = v u + v u + v u + v u (c Now put h, h ad h i (4 v u + v u + v u + v u + a[v u + v u ]+ b[c v + c v ]=f( v [u +a u +b u ] + v [u +a u +b u ]+ v u + v u =f( Sice +a +b=, u +a u +b u =o, u +a u +b u =o, ad v u + v u =f(. {the value i bracket vaih becaue b hpothei both u ad u are olutio of homogeeou equatio correpodig to (4. The the equatio (4 atif b equatio v u + v u =... (9 v u + v u =f(. (. (d B olve (9 ad ( we fid two ukow v, v ad we fid v,v b itegral. (e The geeral olutio of (o- H. D.E (4 i Y( = v u + v u Eample Fid the geeral olutio of the followig (d.e - -= e -. (*

24 Sol ( Fid the geeral olutio of (H.d.e - -=, or m m- =, (m-(m+ = h = c e + c e -, u = e, u = e -, u = e, u = -e - v u + v u = # v u + v u =f(. ## v e + v e = # v e - v e - = e -. ## v e = e - v =/e - v = -/9e - +c, From (#v e = - v e -, or v =- v e v = -/ v = -/+c Sice Y( = v u + v u Y( =(-/9e - +c e + (-/+c e. Problem Solve the followig differetial equatio:- ( +4 = ( - 4 = 8 ( - - =co (4-4 + = e (5 + = ec. Problem Solve the followig differetial equatio:- -(-d +( + d= - ( + d + d = d - Si +coh = d 4 - d = d d 5- L = d 6- ( e d+ 7- d= d + d = 4

25 d 8- = (+ cc d d 9- = e - d - e ec d + co d = (H. d. e - ( + d +d = - d + ( d = e / +d d = 4 ( + d +( d = d 5 - = + co ( d 6- d-d= 7- d +( - d = (Liear d. e d 8- + = e - d i 9- + = - - =e / - d+ d = i d - d +d = d - (- + 4 (- =+ 4- Coh d+ ( ih+ e d= 5- e d +( e d = 6 (- d + d = 7 ( + d+ ( + d = (Eact d. e Ue the give itegratig factor to make (d. e eact the olve the equatio 8 - (+ d d =, (I = 9 d + d =, (I = or (I = Solve (eact d. e - ( + d + (+ d = ( e +e d + ( + e d = ( + d + ( + d = ( 5

26 - (+ d (- 4- d+ d+ d = 5- d- d = d 6- ( +- d +d = d = 7- (e +l + d +( + l +i d = 8- ( - d + ( ta - + ihd = i 9- d + d = (Secod- Order 4- + = = = = = = i = e = e = i = = e 5- += ec 5- += ta 54- += cot 6

27 CHAPTER THREE Laplace Traformatio (L. T Defiitio. Let f (t be fuctio of variable t which defie o all value of t uch that (t >. The Laplace traformatio of f (t which writte a L {f (t} i F( = L {f (t} = e t f ( t dt... ( Note The Laplace traformatio i defie i ( i coverge to value of, ad o defie if the itegral i ( ha o value of. Laplace Traformatio of Some Fuctio:- Uig the defiitio ( to obtai the followig traform:- - If f(t = Solutio Sice L {f (t} = e t t I f ( t dt = e =+ e = L ( =. at -If f (t = e Solutio Sice L {f (t} = L {f (t} = t at e e dt = = a l { e at } = e t f ( t dt = L {f (t} = ( a t e dt = a. ( a t e dt ( a t I e a Note Let f(t be fuctio ad c cotat the (i L {cf (t} = cl {f (t} (ii L {f (t f (t} = L {f (t} L {f (t} -If f(t = co(wt 4-If f(t = i(wt. 7

28 Solutio From Euler formula iwt e = co (wt + ii (wt. iwt L{ e } = L{ co (wt} + i L{i (wt} (*. But { iwt l e } =, from ( iw iw = iw iw = iw iw w = w +i w w l { e iwt } = w +i w w, from (* L{ co (wt} + i L{i (wt}= l e } = From thi - L {co (wt} = 4- L {i (wt} = 5-If f(t = ih (wt. 6-If f(t = coh (wt. Solutio w w w { iwt w +i w w Sice ih = [ e e - ], coh = [ e + e - ]. Now ih (wt = [ e wt e -wt ], L {ih (wt} = {L ( e wt L( e -wt }, = { w - w }. w w w L {ih (wt} = w = = w w, ad 8

29 L {coh (wt} =. w Eample Fid L{8-6e t + e -4t +5it+7coht} Solutio L(8= 8L(= 8 = 8. L (6 e t = 6L ( e t = L ( e -4t = 4 6 L {5i (t} = 5L {i (t} = 5 = 9 7 L {7coh (t} = 7L {coh (t} = 9 Laplace Traformatio of Differetial 5 9 Theorem_:- If f(ti cotiuou fuctio of epoetial o [o, whoe derivative i alo epoetial the the (L.T of f (t i give b formula L {f (t} = Proof e t f t dt udv = uv- vdu. t Let u= e t du =- e dt, dv = f (t dt v =f(t, e t f t dt = =o- e f( + t e f e t f ( t dt ( I t + t e f ( t dt = -f( + L{f(t}. Where L {f (t} = L {f (t} f (. e t f ( t dt = L {f (t}. Corollar If both f(t ad f (t are cotiuou fuctio of epoetial order o [o,, ad if f (t i alo epoetial the :- 9

30 L {f (t} = L {f (t} f ( - f ( Proof L{ f (t} = L {f (t} = L {f (t} - f ( = [ L {f (t} f (] - f ( = L {f (t} f( - f (. Now i geeral L { f (t} = L {f (t} - f( f - (. Problem Prove that L {t! } =, where =,,,.., ad!= (-(-..(-. Ad! =. Propertie of L. T ( Shiftig If L {f (t} = f( = L{ e at Eample Fid L{ e -4t cot} Solutio f(t} f(-a f(t =cot, a=-4, the f( = L {f (t}= L{ cot} = 9 L{ e -4t ( 4 cot} = ( ( 4 9 ( L. T of Itegral: = 4 ( 4 9 t L { f (u du} = f ( Eample t Fid L { Solutio ihtdt}

31 F (u = L{iht} = 4 t, L { ihtdt}= 4 =. ( 4 ( Multiplicatio b t If L {f (t} = f(, the L {t f (t} = (- Eample Evaluate L {t e t } Solutio f(t= e t d f ( d L ( f(t = L ( e t = =f( f (= f/ = ( f ( = f/ = ( L {t e t } = (- = ( ( (4 Diviio b t If L {f (t} = f(, ad L { f ( t t }= f(udu Eample Evaluate i t L { } t Solutio lim i t t =, eit. t f(t = it L{it} = t lim f ( t eit t =f(

32 f(u= L { u f ( t }= L { t = Π/ - ta -. Eample Let f(t =cot. Fid L {f (t } Solutio i t } = t u du = ta - u I = ta - - ta - L {f (t} = L {f (t} f ( - f ( f(t =cot, f(=co(=, f (t = -6it, f ( = -6i(=, L {f (t}= L{ cot} =, 9 L {f (t}= L{ cot} = L{ cot} = =f(. 9 L {f (t} = [ ] [] - 9 = L {f (t} =. 9 Uit Step Fuctio u a (t Defiitio. The uit tep fuctio i defied b:- whe t < a u a (t = whe t > a If a=, the u (t = If a=, the U (t = Defiitio.. whe t < Whe t >. whe t < whe t >

33 To fid Laplace Traformatio of uit tep fuctio (L { u a (t} i defied a :- Sice whe t < a u a (t = whe t > a L { u a (t} = e t f ( t dt = e t u a ( t dt a t t = e ( dt e ( dt = e a = - [ e -a ] = a e a L { u a (t} =. Defiitio.. To fid the term of the uit tep fuctio i defied b:- whe <t < f(t = whe <t < - whe <t <. f(t = [u u ] + [u u ] - [u u ] = u u + u u - u +u t I f(t = u + u u +u Problem Fid Laplace Traformatio of f (t which defie i (Defiitio... Solutio Sice f(t = u + u u +u L{f(t} = L{u (t} + L{u (t} L{u (t} +L{u (t} e ( = + e - e + e

34 e e e = L. T of Periodic Fuctio If f (t i Periodic fuctio of period T> atif uch that f(+t = f(, the L {f (t} = T e t e f ( t dt T Gamma Fuctio Defiitio.4 If ( >, the the gamma ( become:- (= t e t dt.( Importat Propertie of gamma fuctio (i (+ = ( ii (+ =! (iii ( = (Π Table I Some elemetar fuctio f (t their Laplace Traform L {f (t} = f(. 4

35 5 ( ( 4 ( ( ( ( ( ( ( ( ( ( } ( { ( ( ( ( } ( { ( ih 9 coh 8 i 7 co 6 5!,,,..., 4! ( } ( { ( t at poitive t f t f t f du u f f t L t f t L t a a at a at w w wt w wt a e t t t f t f L t f

36 Problem Fid Laplace traform (f( of the followig fuctio :- - f(t = i t - f(t = t 4 e t - f(t = e -t coh t ih t 4- f(t = t 5- f(t = t e t 6- f(t = t+4 7- f(t = t +at +b 8- f(t = ( a+bt 9- f(t = t e -t - f(t = (e t -4 - f(t = t e at iat - f(t = coh at co at whe <t < -f (t = 4 whe <t. 4 -Prove that 5- Prove that t e t (a = L {a+bt} = i t dt (b = L {t co at} = a b 5 a ( a, Ivere Laplace Traformatio If L {f (t} = f(. The we call f(t i the ivere of (L. T of fuctio f( ad which writte a: f(t = L - {f(}, for eample L ( e t = =f(. L - {f(} = L - { } = e t. Some Propertie of Ivere L. T 6

37 We ee the L. T of firt (9 i table I, we ca ivere there Laplace to fid ivere of thi for eample L( = f(=, L - {f(} = L - { } =, Eample Fid f (t, if f( = Solutio f(t = L - { 5 5 } = 5L - { Eample Fid f (t, if f( = Solutio f( = +, f (t = L - { } + L - { }, = cot + it. Partial Fractio } = 5 e -t If we wat to fid the ivere traform of a ratioal fuctio a f (, where f ad g g( are polomial which the degree of f le tha degree of g the. We ca take advatage of partial traform i eail foud a ee i eample:- Eample Fid the ivere Laplace traform (f (t, if f( = ( Solutio f( =, ( A B C D = + (, = A +B + A + B + C +D, B =, B+D =, A+ C =, D = - A= C=, = -, ( 7

38 f (t = L - { ( = t it. } = L - { } - L - { }, Eample Fid the ivere Laplace traform (f (t, if f( = Solutio 6 f( = = 6 ( ( S+ =A - A + B + B, -A +B =, A+B =, A B =. -A +B = A + B = + 5B= B =/5, A = /5, f (t = L - { F(} = /5L - { } - /5L - { }, = /5 e -t +/5 e t Problem Fid f(t { the ivere Laplace traform} of the followig:- ( f( =, ( f( = (, ( f( = (4 f(= 6 4, ( 4, (5 f( = , 4 (6 f( =, 8

39 (7 f( =, 9 (8 f( = 9, (9 f( =. 6 Applicatio of Laplace Traformatio Liear (D. E With Cotat Coefficiet To olve L- o homogeeou (d. e of order with cotat coefficiet. We ue ame wa a ecod- order (d. e a form :- a + a + a = f( (* Where a, a ad a are cotat, which atif iitial coditio: (= A ad ( =B (** Where A ad B are choice cotat. Eample Fid the olutio of the followig (d.e b (L. T + + = (* Which atifie iitial coditio? (= ad ( =. Solutio L { (t} = { (} ( - ( L { (t} = { (} (. Put i (* L { (t} = ( Sice L {} = ( { (} ( - ( +[ { (} (] + ( = B ue ( = ( =, ( + + ( = (+ ( + (, ( + + ( = ++ = +4, 4 4 ( = = ( ( S+4 = A+ B + A + A, A + B = A +B = 4 - -B =- B = A =-, A B =. 9

40 ( =, (t = L - { (} = L - { (t = e -t - e -t } - L - { }, Eample Fid the olutio of the followig (d.e b (L. T = (i Which atifie iitial coditio? (= ad ( =. Solutio L { (t} = { (} ( - ( L { (t} = { (} (. L { (t} = ( Sice L {} = ( Put i (i { (} +4[ (] + 4 ( = [ ] ( = + =, ( = ( ( = ( = A(+ + B, 4 4 (, ( = B =-½ ad A =½, ( =, A B,

41 (t = L - { (} = /L - { } - /L - { (t = ½ - ½e -t. }, Problem Fid the olutio of the followig (d.e b (L. T, which atifie the give iitial coditio:- ( =, at (= ad ( =, ( =, at (= ad ( =, ( - =, at (= ad ( =, ( =, at (= ad ( =, (5-9= it, at (= ad ( =, (6-9 = e t, at (= ad ( =, (7 + 4 = it, at (= ad ( =, ( =4 cot, at (= ad ( =5,

42 CHAPTER FOUR Fourier erie Periodic Fuctio Defiitio. The fuctio f( atif the coditio f(+t = f( For all value of where T i real umber the f( i called Periodic fuctio, ad if T leat poitive umber atifie (, the T i called periodic umber of fuctio. We ca fid that:- F( = f(+t = f(+t = (+T =..= (+T. Ad F( = f(-t = f(-t = (-T =..= (-T. Thi mea that F( = ( T, where iteger. Some Propertie of Serie -f(+t = f( Periodic fuctio - =No of term poitive iteger. - Co = if eve (, 4, 6 - if odd (,, Co =, 5- Si = i =, 6- Co = Co (-. Some Importat Itegral:- - i d = co d =, where iteger. - i m i d = ½ [co(m- co(m+]d =.

43 = i d =½ Co i d = = ½ co d= [ co]d =, where iteger. i d =. [co co d =. Fourier erie Suppoe that f( i periodic fuctio to, ad i periodic umber of it. Ad the fuctio f( i defied o the iterval (< <. The we ca write f( i the form:- f(= a + a co + a co +.+ a co + b i + b i+..+ b i ( Thi mea that f(= a + (a co + b i. ( = (a co + b i.. (, Such that a = a = b = f( d f( co d, =,,,. f( i d. The erie ( i called Fourier erie of the fuctio f(. If the fuctio f( defied o iterval - < <, the a = f( d a = f( co d, =,,,. b = f( i d.

44 Eample Fid Fourier erie of the fuctio f( =, from = to = or (< <. Solutio Ue the rule to fid a, a ad b, a = f( d = d = f( d a = ] =. 4 a = f( co d = co d, i co = [ ] ] = i co co [ ] [ ] i co = [ ] =. f( i d = b = b = = co i [ ] =. The equatio ( become: f(= - i ( ] i d f(= - ( i + i + i +.. Eve ad Odd Fuctio If f( = f(-, i called eve fuctio. If f(- = -f(, i called odd fuctio. Fourier erie of Eve ad Odd Fuctio If f( i eve whe {, 4, 6 co, i, f(. 4

45 If f( i odd whe {,, 5,, i. (i If f( i eve the b = (ii If f( i odd the a = a =. Eample Fid Fourier erie of the fuctio f( =, for (- < <. Solutio Sice f(- = - = -f(, the fuctio i odd. a = a =. b = f( i d = b = co i = [ ] ] co = [. ] = co i d = i d = (- The the erie become: f( = f(= - b i (- i ( f(= ( i - i + i -.. Eample Fid Fourier erie of the fuctio a = f( = Solutio if < <. if < <. f( d = f(d + f( d 5

46 = i = d + d = ] + ] d a = /. a = f( co d = f( co d + = co d + co d, = a =. b = i ] + co d, ] i co [ ] ] =. i d + i d f( co d co = co = [ ( b = co ] + ( ] =,, ], a = /, a =. b =, b = b =, i i 5 f(= / - [ i ]. Half-Rage Serie If we wat fid Fourier erie o iterval (< <, doe ot o all iterval (- < <, the we ca fid the Fourier erie b :- - Fourier Coie erie or f( a eve fuctio a:- 6

47 f(= a + a co + a co +.+ a co. - Fourier Sie erie or f( a odd fuctio a:- f(= b i + b i+..+ b i Such that a = a = b = f( d f( co d, =,,,. f( i d. Eample Fid coie Half-rage erie for the fuctio defied a f( =, for < <. Solutio Ue the rule to fid a ad a a = f( d = d = f( d a = a = = = ] =. f( co d = i co [ ] (co co d, ] a = 4 if eve. if odd 7

48 4 co f(= - [ co + co ]. Eample 4 Fid ie Half-rage erie for the fuctio defied a f( =, for < <. Solutio Ue the rule to fid b b = f( i d = b = co i = [ ] ] co = [. ] = co i d = i d = (- The the erie become: f( = f(= - b i (- i ( f(= ( i - i + i i

49 CHAPTER FOUR Partial Differetial Equatio Partial Differetial Equatio (P. D.E Partial Differetial Equatio are Differetial Equatio i which the ukow fuctio of more tha oe idepedet variable. Tpe of (P. D.E The followig ome tpe of (P. D.E:- -Order of (P. D.E The order of (P. D.E i the highet derivative of equatio for eample:- U = U Firt-order (p. d. e. u u 4 Secod -order (p. d. e. t -The Number of Variable For eample:- U = U tt (two variable ad t. U = U rr + U (Three variable t, r ad. U r r + r -Liearit The (P. D.E i liear or o-liear, i liear (P. D.E if u ad whoe derivative appear i liear form (o- liear if product two depedet variable or power of thi variable greater tha oe. For eample {the geeral ecod L. P. D.E i two variable} Au + Bu + Cu + Du + Eu + Fu + G = (* Where A, B, C, D, E, F ad G are cotat or fuctio of ad for eample u tt + e -t u =it (Liear u = u (Liear u u + u = (No-Liear u + u + u = (No-Liear. 4-Homogeeit If each term of (P. D.E cotai the ukow fuctio ad which derivative i called (H. P. D.E otherwie i called (o-h. P. D.E, i pecial cae i (* i homogeeou if [ G =]. Otherwie ohomogeeou. Au + Bu + Cu + Du + Eu + Fu = (H. P. D.E Where A, B, C, D, E ad F are cotat or fuctio of ad. Eample 9

50 Determie which (L. P. D.E i, order ad depedet or idepedet variable i followig:- u u 4 t Liear ecod degree u, depedet variable, ad t are idepedet variable. r r Liear - degree( r, depedet variable, ad are idepedet variable. w w rt No-Liear - degree( w, depedet variable, r, ad t are idepedet variable. Q Q Q 4 z Liear - degree( Q, depedet variable,, ad z are idepedet variable, homogeeou. u u 5 ( ( t No-Liear - degree( u, depedet variable, t ad are idepedet variable, homogeeou. Solutio of (P. D.E A olutio of (P. D.E mea that the value of depedet variable which atified the (P. D.E at all poit i give regio R. For Phical Problem, we mut be give other coditio at boudar, thee are called boudar if thee coditio are give at t= we called them a iitial coditio it order. For a liear homogeeou equatio if u, u u are olutio the the geeral olutio ca be writte a (-th order p. d. e u= c u + c u + + c u. Note i We ca fid the olutio of (P. D.E b equece of itegral a ee i the followig eample:- Eample Fid the olutio of the followig (P. D.E z Solutio

51 ( z z B itegrate (w. r. to give c( z Where c( i arbitrar parametric of. Alo b itegrate (w. r. to give ( ( c c z Where c( i arbitrar parametric of. Eample Fid the olutio of the followig (P. D.E z Solutio B itegrate (w. r. to give ( c z B itegrate (w. r. to give ( ( 6 c c z ( ( 6 c F z. Eample4 Fid the olutio of the followig (P. D.E 6 u With boudar coditio, u(,= -, u(,= 5-5 Solutio B itegrate (w. r. to give ( c u B itegrate (w. r. to give ( ( 4 g c u ( ( 4, ( g h u g h u ( ( 4 (, ( 5 4 ( g h ( ( 5 4 4, ( g g u 5 5 ( ( 4 6, ( g g u ( 6 7 ( g g

52 u(, Formatio of (P. D.E A (P. D.E ma formed b a elimiatig arbitrar cotat or arbitrar fuctio from a give relatio ad other relatio obtaied b differetiatig partiall the give relatio. Note ii Suppoe the followig relatio:- z z z z 4 z z z z p q r t z 5 z Eample 5 Form a Partial Differetial Equatio from the followig equatio:- Z= ( -a +(-b.( Solutio z z z z ( -a ( -b Eq( become Z ( z ( z Z ( z ( z 4 4Z ( p ( q Eample 6 Form a Partial Differetial Equatio from the followig equatio:- Z= f( +.( Solutio Z =f ( + Z =f ( + Eq( become z, z - Z + Z = p -q =

53 Eample 7 Form a Partial Differetial Equatio from the followig equatio:- Z= a+b+a +b. (. Solutio Z =a Z =b Eq( become Z= Z + Z +( Z +( Z Z= p + q +( p +( q Eample 8 Form a Partial Differetial Equatio from the followig equatio:- v= f( -ct +g(+ct Solutio v = f ( -ct+g (+ct v t = -cf ( -ct+cg (+ct v = f ( -ct+g (+ct v tt = c f ( -ct+c g (+ct v tt = c [f ( -ct+g (+ct] v tt = c v, or v v c Oe dimeioal Wave equatio t Solutio of Firt Order Liear (P. D. E Let the Partial Differetial Equatio a form:- Pp+ Qq =R... (4 Where P, Q ad R are fuctio of, ad z. So the olutio of thi equatio i the ame a the olutio of imultaeou d d dz...(5 P Q R Eq (5 are calle LaGrage Auiliar Equatio or (characteritic equatio. A oluatio of Eq(5, ca be writte a U(,, z = c, V(,, z = c The geeral olutio writte a F (U, V =, or F (c, c =. Note iii To olve Eq(5, we ote that:-

54 (i If P or Q or R equal to zero the d or d or dz equal to zero repectivl, For eample If R= dz = Qd =Pd from Eq(5, which ca eail to olve it. (iii cae parable the variable i problem, the we ca write characteritic Eq(5, i the followig form d d dz d d dz P Q R P Q R We lected the value of λ, µ ad β uch that give λp +µq + βr =, λd +µd + βdz =. Which helpe to fid of Solutio of (P. D.E. Eample 9 Solve the followig Partial Differetial Equatio zp+zq = Solutio Suppoe the followig relatio:- Where z z z p, ad z q P= z, Q= z, ad R= d d d d z z L=l =l c c = V. (6 d dz d dz d= zdz z z zdz = c d z c +c z =c z - =c = c =V The geeral olutio F (c, c =, or F (, z - =. Eample 4

55 Solve the followig Partial Differetial Equatio (+zp (+zq =-. (7 Solutio P= +z, Q= -(+z, ad R= - d d dz d d dz z ( z ( z ( z ( d d dz Where λ =, µ =, β =. d +d + dz =. + + z = c = U. For λ =, µ =, β =-z d d zdz d +d - zdz =. + - z =c = c =V The geeral olutio F (c, c =, or F ( + + z, + - z =. Eample Solve the followig :- z Z + z Z +( + = Solutio z Z + z Z = -( + d d - z z d d d d l- l=lc L lc - d z c. dz ( From ( = c 5

56 d z d z dz ( c dz ( c -( + c d=zdz ( + c d+zdz = ( c z c + c +z = c, + +z = c, where c = c. The geeral olutio F (c, c =, or F (, + +z = Problem Fid the olutio of the followig Partial Differetial Equatio:- - p+q = - p-q =z - zp- zq = 4- (+zp +(+zq =+ 5- ap +bq+cz = 6- ( +z - p - q +z = Theorem If u u.. are olutio of equatio F(,..u=, The U= c u + c u + i olutio alo, where u= c, c.are cotat. Method of Variable Sparable Let the Partial Differetial Equatio a f f F(,..u=. Let the geeral olutio of above equatio i Let u (, t = XT, or u (, t = X(T(t Be olutio of (P. D.E where X if fuctio of ol, ad Y fuctio of ol. A ee i the followig problem:- Eamp Solve the followig Partial Differetial Equatio with boudar coditio u u With boudar coditio. u (, = 4e - -e -6. (8 Solutio To olve Eq(8 uppoe 6

57 u (, t = XT. Be olutio of (8 where X if fuctio of ol, ad Y fuctio of ol. u u YX, XY dx dy X Y d d Put i eq (8 YX + XY = X Y X Y Now let X Y = c X Y X = c X Y = c Y X -CX =, Y -CY =, X= a e c, Y= a e -c u (, t = XT= a a e c-c = Be c(-, where B= a a, are cotat. Now let c ( c ( u = b e, ad u = b e olutio of (8 (theorem c ( c ( u= u + u = b e + b e,from boudar codio c c u (, = b e + b e = 4e - -e -6 b =4, b =-, c =, c =6 ( 6( u (, = 4e -e Eample Fid the olutio of followig [Heat equatio] b uig partial differetial equatio:- u u.... (9 t With boudar coditio. ( u (, t =, ( u (, t =, for all t, ( u(, = 5i +i - i4 Solutio Let u (, t = XT. Be olutio of (9 7

58 u XT t u TX Put i( XT = TX.... ( We ca write (i the form:- T X T X Let T X =c T X Where c be cotat T -ct=, X - cx= there three cae OF C ( C=,C> ad c< CaeI. If c= T =, T= c ad X =,X= c + c U= TX=c (c +c U=A+B Where A=c c, B= c c U(,t= B= U(, t=a U(,t= A= A= U Which trivial olutio c CaeII. If C> Te -c =c T =c e ct c c X= c e c e u (, t = XT, = c e ct c c c e c e ( u = e ct c c ( Ae B e A= c, c, ad B= c, c U(,t= e ct ( A + B= e ct A + B= A= -B 8

59 U(,t=B e -ct c c ( e e U(,t=B e Ct ( c e e c = If B= A= U= Which trivial olutio B c c e e = e c e c e c c c c e e e which impoible ice e There i o olutio if C >. CaeIII. If c<, let c=- k k k T +k T=, X + k k t X= T = c e, X= c cok + c ik. k t U(,t= c e ( c cok + c ik k t U(,t= e ( A cok + B ik. Where A=c c, B= c c k t U(,t= e ( A = k t A, becaue e k t U(,t= B e ( ik. k t U(,t= B e ( ik. = k t Sice B, e ik = k=, where =... k U(,t= B U(,t= b U(,t= b U(,t= b t e t 5 e t 5 e t 5 e ( i ( i ( i ( i = B t 5 e c c e, ( i = U(,t= b t 5 e ( i t 5 + b e ( i t 5 + b e ( i 9

60 U(,= b i + b i + b i 5i +i - i4 b =5, b =, + b = -, =5, =, =4 U(,t= 5 9 t e i t + e 8 t i - e i 4 = Eamp 4 Fid the olutio of followig [Wave equatio] b uig partial differetial equatio:- u u ( 4 With boudar coditio. t ( u (, t =, ( u (L, t =, for all t, L, (4 u(, = f(. (5 u t = g(, at t=. Solutio Let u (, t = XT. Be olutio of ( where X if fuctio of ol, ad Y fuctio of ol. u XT t u TX Put i( XT =4 TX T X 4T X Let T X k 4T X Where k be cotat T -4k T=, X - k X=(there three cae CaeI. If k T =, T=at+b X =,X= c +d

61 U= TX=(at+b(c+d U(,t= (at+b( d= at b b U(,t=(at+b c U(L,t=(at+b cl= cl= L c c +d= U(,t= CaeII. If k T -4k T=, X - k X= T= a e kt + b e -kt, X= c e k + d e -k U(,t=(a e kt + b e -kt ( c e k + d e -k U(,t=(a e kt + b e -kt ( c + d= c + d= d= -c U(,t=c(a e kt + b e -kt ( e k - e -k U(L,t=c(a e kt + b e -kt ( e kl - e -kl = If c= X= U= e kl - e -kl = e kl = e -kl e kl =, which impoible ice L, k There i o olutio if k CaeIII. If k k T +4k T=, X + k X= T= A cokt + Bikt, X= C cok + Dik. U(,t=( A cokt + Bikt( C cok + Dik U(,t=( A cokt + Bikt( C = c, becaue A cokt + Bikt U(,t=( A cokt + Bikt Dik U(L,t=DikL ( A cokt + Bikt= Sice A cokt + Bikt DikL= If D= U= DikL= kl=, where =... k L U(,t=Di U(,t= (A co ( A co L Where A =AD, B =BD L t + Bi t L t + B i t i. L

62 U(, t U (, t U (, t (A co U(, = f(. L t + B i t i. L f( = A i (, g(, U t g(= L. B ( i. Problem Fid the olutio of the followig Partial Differetial Equatio:- u u ( t With boudar coditio. u (, t =, u (, t =, for all t, u(, = 5i +i - i4. u u ( u (, = e, u u ( t With boudar coditio. u (, t =, u (п, t =, for all t, u(, = i - 5i4. With boudar coditio. u u (4 4 With boudar coditio. t (i u (, t =, (ii u (L, t =, for all t, L > (iii u(, = f(. (iv u t = g(, at t=. u u (5 4 With boudar coditio. (i u (, =, (ii u (, =, for all t, u (iii (, =, at t=.

63 (iv u(, = i - 4 i. CHAPTER FIFE Numerical Aali Solutio of No-Liear Equatio -Newto-Rapho Method for Approimatig Iterpolatio -Lagrage Approimatio Numerical Differetiatio ad Itegratio Approimate Itegratio Itegratio Equal Space -The Trapezoidal Rule 4-Simpo' Rule 5-Simpo' (/8 Rule Solutio of Ordiar Differetial Equatio Numerical Differetiatio 6-Euler Method The Step b Step Method 7-Modified Euler Method (Euler Trapezoidal Method 8-Ruge Kutta Method 9-Ruge- Kutta-Mero Method Stem of Liear Equatio -Cramer' Rule -Solutio of Liear Equatio b uig Ivere Matrice -Gau Elimiatio Method

64 -Gau Siedle Method 4-Leat Square Approimatio Numerical Aali Solutio of No-Liear Equatio -Newto-Rapho Method for Approimatig We ue taget to approimate the graph of = f(, ear the poit P (,, where = f(, i mall. Let + be the value of where that taget lie croe the -ai. Let taget = The lope betwee (, ad (,, i f `( =. ( Sice the taget lie croe the -ai, =, ad = f(, put i Eq ( which become f ( f `( =, - = = - f ( f ( f ( f (,. (. Put = + i Eq ( give f ( + = -. ( f ( Eq ( called Newto-Rapho Method, ca uig thi method b the followig -Give firt approimatig to root of equatio f( =. A graph of = f(. -Ue firt approimatig to get a ecod. The ecod to get a third, ad o o. To go from th approimatio to the et approimatio +, b uig Eq (, where f `( the derivative of f at. Eample 4

65 Solve the followig uig Newto-Rapho Method + =, tart with = -.5, error % =.5 % Where e % = Sol f( = f `( = - f( = f `( = - = - +, = -.5, += -,.5 (.5 f ( f ( = B ue e % = e % =.75 % = -4, from Eq (.75 (.5 = % % a e % = % B ue ame of ew of i Eq ( a f ( = - f (, = -.97, i ame we ca fid ad 4 which ue i the followig table f( f `( + e % % % % % To check the awer a:- += -+=. 5

66 Iterpolatio -Lagrage Approimatio Iterpolatio mea to etimate amaig fuctio value b takig a weighted average of kow fuctio value of eighborig poit. Liear Iterpolatio Liear Iterpolatio ue a lie egmet pae through two ditict poite (, ad (, i the ame a approimatig a fuctio f for which f( =,ad f( = b mea of firt-degree polomial iterpolatio. The lope betwee (, ad (, i Slope =m= (, P(!!!!!!!! (, The poit- lope formula for the lie = m( - + 6

67 =P( = m( - + = = +( - ( - + P ( = + (4 Each term of the right ide of (4 ivolve a liear factor hece the um i a polomial of degree. L, ( =, ad L, ( = (5 Whe =, L, ( = ad L, ( =. Whe =, L, ( = ad L, ( =. I term L, ( ad L, ( i Eq (5 called Lagrage coefficiet of polomial hazed o the ode ad, P ( = =f(,ad P ( = = f(. Uig thi otatio i Eq (4, ca be write i ummatio P ( = L, (+ L, ( P ( = k L k (. k Suppoe that the ordiate k = f( k. If P ( i ue to approimate f( over itervalle [ ]. Eample Coider the graph = f( =co( o ( =., ad =., to fid the liear iterpolatio polomial. Sol Now =f( = f(. = co (.=., ad =f( = f(. = co (.=.64, L, ( = L, ( = P ( =. =... =.. k L k (. k P ( = L, (+ L, (. =, ad. =... P ( = -(. + (.64.. P ( = -.8( Quadratic Lagrage Iterpolatio 7

68 Iterpolatio of give poite (,, (, ad (, b a ecod degree polomial P (, which b Lagrage ummatio a P ( = L, (+ L, ( + L, (. P ( = k k L k ( = k f ( L. k k ( ( ( L, ( =, ( ( ( ( L, ( = ( ( ( ( L, (= ( ( approimatig a fuctio f for which f( =,ad f( = b mea of ecod -degree polomial iterpolatio. Eample Uig the ode ( =, =.5 ad =4, to fid the ecod iterpolatio polomial for f( =. Sol We mut fid (.5( 4 L, ( = = (-6.5+, (.5( 4 ( ( 4 ( 4 4 L, ( = = (.5 (.5 4 ( (.5 ( L, (= =. (4 (4.5 Now f( = f( =.5, f( = f(.5 =.4, ad f( = f(4 =.5, ad P ( = k k L k ( = k f ( L. k k ( P ( = L, (+ L, ( + L, ( ( 4 4 ( =.5[-6.5+]+.4[ ] +.5 [ ]; P ( =[ ] +.5 f(= P ( =.5. f(= P ( =.5. Cubic Lagrage Iterpolatio 8

69 Iterpolatio of give poite (,, (,, (, ad (, b a third degree polomial P (, which b Lagrage ummatio a P ( = L, (+ L, ( + L, ( + L, (, P ( = k k L k ( = f ( k L k (. k ( ( ( L, ( =, ( ( ( ( ( ( L, ( = ( ( ( ( ( ( L, (=, ( ( ( ( ( ( L, (= ( ( ( Approimatig a fuctio f for which f( =,ad f( = b mea of third -degree polomial iterpolatio. Eample 4 Coider the graph = f( =co( o ( =., =.4, =.8 ad =., to fid the cubic iterpolatio polomial. Sol Now =f( = f(. = co (.=., =f( = f(.4 = co (.4=.9, =f( = f(.8 = co (.8=.6967, ad =f( = f(. = co (.=.64, ( ( ( (.4(.8(. L, ( = =, ( ( ( (..4(..8(.. L, ( = -.64( -.4( -.8( -., L, (= 7.958( -.( -.8( -., L, (= -5.44( -.( -.4( -. L, (=.946( -.( -.4( -.8. P ( = L, (+ L, ( + L, ( + L, (, P ( = k k L k ( = f ( k L k (. k P ( = L, (+ L, ( + L, ( + L, (, P ( = -.64( -.4( -.8 ( ( -.( -.8( ( -.( -.4( ( -.( -.4( -.8., I geeral cae we cotruct, for each k=,, we ca write 9

70 = if k = i L,k ( i = if k i Where ( (...( k ( k...( L,k (= ( k ( k...( k k ( k k...( k or ( i L,k (=. ( i k i i k Problem -If ( =, ( =5, (5 =5, ad (6 =. Fid the four poit Lagrage iterpolatio polomial that take ome value of fuctio ( at the give poit ad etimate the value of (4 at give poit. -Fit a cubic through the firt four poit (. =., (.7 =7.8, (. =4., (. =.ad (5.6 =5.7, to fid the iterpolated value for =. fuctio ( at the give poit ad etimate the value of (4 at give poit. -If f(. = , f(. =.686, f(.6 =.4554, f(.9 =.8886 ad f(. =.6. Ue Lagrage polomial to approimatio to f(.5. Numerical Differetiatio ad Itegratio Approimate Itegratio Itegratio Equal Space We begi our developmet of umerical itegratio b givig wellkow umerical method. If the fuctio f( uch a ature that b a f ( d caot be evaluated b method of itegratio. I uch cae, we ue method to approimatio to value. A geometric iterpolatio of b a f ( d i the area of the regio bouded b the graph of = f(, = a = b, ad =. We ca obtai a etimate of the value of itegral b ketchig the boudarie of the regio ad etimatig the area of the ecloed regio. -The Trapezoidal Rule

71 b We hall obtai a approimatio to a f ( d b fidig the um of area of trapezoid. We begi b dividig [a, b] ito equal ubiterval ad cotructed a trapezoid. Let the legth of the ordiate draw at the poit of ubdiviio b f, f,, f -, ad f ad the width of each trapezoid b Δ = fid the um of the area of the trapezoid i:- b a, we A= [ f + f ] Or b a f ( d = + [ f + f ] + + [ f - + f ] [ f + f + f + + f - + f ] (6 Eq (6 called The Trapezoidal Rule. Eample 5 Fid d, for = 6 b Trapezoidal rule Sol f(=, =, 6 = 6 h = = = h 6 6 =, f = = + h = = = = 4 =, f = 6, f = 6, f = 6 4, f4 = 6 =.979 ( 6 =.9 ( 6 =.8 ( 6 =.69 4 ( 6

72 5 5 =, f5 = = ( 6 6 =, f 6 = = =.5 ( A = h [ f + ( f + f + f + f 4 + f 5 + f 6 ] A = [ + ( ] A = [ + ( ] A = Simpo' Rule We obtai aother approimatio to b f ( d. We dividig the iterval a from = a to = b ito a eve umber of equal ubiterval. We ca drive the formula of Simpo b coected a three o-colliear poit i the plae ca be fittig with parabola ad Simpo' Rule i baed o approimatig curve with parabola a how i the followig:- Let the equatio of parabola a f = A + B + C. The area uder it from = -h to = h a b h f ( d = ( A B C d = h [ A B C ] a h h h h = A Ch= [Ah 6C]. Sice the curve pae through the three poit (-h, f, (, f ad (h, f f = Ah - Bh + C f = C f = Ah +Bh + C. From above equatio ca ee that C = f Ah - Bh = f - f Ah +Bh = f - f Ah = f +f - f. b Now the area b a f a f ( d i term of ordiate f, f ad f, we have h h ( d= [Ah 6C].= [ f +f - f +6 f ], or

73 b a h f ( d = [f +4f + f ]. (7 Eq (7 called Simpo' Rule of two iterval [the with h]. Now i geeral to eve umber of equal ubiterval b pa a parabola through [f, f ad f ], aother through [f, f ad f 4 ] ad through [f -, f - ad f ]. We the fid the um of the area uder the parabola. b a b h h h f ( d = [fa +4f + f ]+ [f +4f + f 4 ] + + [f- +4f - + f b ] h f ( d = [fa +4f + f +4f + f f - +4f - + f b ]. a b a Where h =, ad = eve. Ad the trucatio error for Simpo' rule i:- 5 ( b a (4 ( b a 4 (4 e = f ( c = h f ( c Eample6 Ue Simpo' rule to evaluate d, for = 6. Sol f(=, =, 6 = 6 h = = = h 6 6 =, f = = = + h =, f = = ( 6 = + h = + =, f = = =.9 ( 6, f = =.8 6 ( 6

74 4 = 5 = 4, f4 = 6 5, f5 = 6 6 =, f 6 = =.69 4 ( 6 =.59 5 ( 6 = =.5 ( A = h [f +4f + f +4f + f 4 +4f 5 + f 6 ] A = [ +4(.979 +(.9 +4(.8 + ( ( ] A = Simpo' (/8 Rule If f( approimated b polomial of higher degree the a accurate approimatio i computig the area o if the iterval divided ito ubiterval that ( i odd umber divided b ad b calculatig the area of three trip b approimatig f( b a cubic polomial a i Simpo' Rule. Ad for the formula we obtai the three eight rule b h f ( d = [fa +f + f +f + f 4 +f 5 + f f - +f - + f b ]. 8 a b a Where h =, ad = odd Ad the trucatio error i:- 5 ( b a (4 e r = f ( c. 648 Eample7 Ue Simpo' rule to evaluate 4 8 d, for = 6. Sol f(= 4, =, 6 = b a 6 h = = = = h 6 =, f = ( 4 = = ( 4 = = + h =, f = ( 4 = = + h = + =,

75 f = ( 4 =.4 6 = 4 = 5 =, f = ( 6 4, f4 = ( 6 5, f5 = ( 6 4 = = = =, f 6 = ( b a 6 4 =. 6 h f ( d = [fa +f + f +f + f 4 +f 5 + f 6 ]. 8 h A = [fa +(f + f +f 4 + f 5 + f + f 6 ]. 8 A =.4. Problem - Approimate 4 d, b the trapezoidal rule ad b the Simpo' rule, with = 6. - Approimate each of the itegral i the followig problem with = 4, b (i The trapezoidal rule ad (ii The Simpo' rule. Compare our awer with (a The eact value i each cae. (b Ue the error i term i Trapezoidal rule. (c Ue the error i term i Simpo' rule. ( ( ( (4 4 4 (5 d d d d d 5

76 (6 i d. Solutio of Ordiar Differetial Equatio Numerical Differetiatio Let f(, be a real valued fuctio of two variable defied for (a b, ad all real value of. 6-Euler Method The Step b Step Method Thi tart from = (, ad compute a approimate value of the olutio at for `(= f(, ( at = +h, i ecod tep compute the value of olutio at = +h = +h, where h i fied icremet, i each tep the computatio are doe b the ame formula uch formula uggeted b Talor erie h h ( + h = ( +h`( + ``( + ```( +.. f `(= f(, (, ``(= f `(, ( + h h ( + h = ( +h`( + ``( + For mall h ad eglected term of h, h.. ( + h = ( +h f(, = +h f(,, = +h f(,,.. + = +h f(,. Which called Euler' method for firt order. Eample 8 Ue Euler' method to olve the D. E f + ` ```( +.. 6

77 d = +4 -, with, = = 4, for = to =., h =.5 d work to (4D. Sol d f(, = = +4 - d + = +h f(,. =, =, = 4 = +h f(,. = 4 +.o5 f(, 4. = 4 +.o5[ +4-4 ]. = 4 -. =.9 = +h = +.5 =.5 = +h f(,. =.9 +.o5 [( (.5- ]. =.8 =.5+.5=. =.5, =.7 4 =., 4 =.67 5 =.5, 5 =.7. 7-Modified Euler Method (Euler Trapezoidal Method The Modified Euler Method give from modified the value of ( + at poit ( + b give the ew value ( + b the followig method = +h ( = +h f(,. ( h = + [ f(,. + f(, ( ], ( h = + [ f(,. + f(, ( ] (r+ h = + [ f(,. + f(, (r ], we ca go to five iteratio. Eample 9 Ue Euler' Modified method to olve the D. E d + = +4, with, = 4, for = (.5., work to (D. d 7

78 Sol Step f(, = +4 - ( = +h f(,. =, =, = 4 = +h f(,. = 4 +.o5 f(, 4. ( = 4 +.5[ +4-4 ]. ( =.9 ( h = + [ f(,. + f(, ( ],. 5 = 4 + ( =.96. [- 4 +(-.5 +4(.5 -.9] =.96 ( h = + [ f(,. + f(, ( ]. 5 = 4 + [- 4 +(-.5 +4( ] =.96 ( =.96 Step = +h = =.5 +. ( = +h f(,. =, =.5, =.96 ( = +h f(,. =.96+.5[(.5 +4( ] =.9 ( =.9 ( h = + [ f(,. + f(, ( ],. 5 = ( =.868. [(.5 +4( (. +4(. -.9] = ( h = + [ f(,. + f(, ( ]. 5 = ( =.84. [(.5 +4( (. +4( ] = 8

79 ( h = + [ f(,. + f(, ( ]. 5 =.96+ [(.5 +4( (. +4(. -.84] =.85 ( =.85. Step = +h =.+.5 =.5 =, =., =.85 ( = +h f(,. =.85+.5[(. +4(. -.85] =.75 ( =.75 ( h = + [ f(,. + f(, ( ],. 5 =.85+ [(. +4( (.5 +4( ] =.756 ( =.756. I ame wa we fid ( =.756. Step 4 4 = +h =.5+.5 =. =, =.5, =.756 ( 4 = +h f(,. = [(.5 +4( ] =.69 ( 4 =.69 ( h 4 = + [ f(,. + f( 4, ( 4],. 5 =.756+ [(.5 +4( (. +4(. -.69] =.699 ( 4 =.699. ( h 4 = + [ f(,. + f( 4, ( 4], 9

80 . 5 =.756+ [(.5 +4( (. +4( ] =.699 ( 4 =.699. The followig table give the above reulted of ad Problem Appl Euler' method to the followig iitial value problem. Do 5 tep. Solve the problem eactl. Compute the error to ee that the method i too iaccurate for Practical purpoe ( + o. = with ( =, h =.. ( =, with ( =, h =.. ( = with ( =, h =.. (4 = (+ with ( =, h =.. Fid the eacted olutio ad the error (5 + = with ( =, h =.. (6 = (+, with ( =, h =.5. (7 Ue Euler' method to fid umerical olutio of the followig d. e. (8 = 4 + -, with ( = 4, h =.5, fid to -decimal. 8-Ruge Kutta Method Whe d = f(, d + = + 6 [k + k +k + k 4 ] Where k = h f(,. h k k = h f( +, +. h k K = h f( +, + K 4 = h f( +h, + k..

81 Where h ad (, are give. Eample Ue Ruge Kutta Method to olve the D. E d = +, with =, =, with h =. work to (4D. d Sol d f(, = = + d = + 6 [k + k +k + k 4 ] k = h f(,. =, =, = k =. f(, =.[ +]=. k =. h k k = h f( +, +.. =. f( + K =..., + h k K = h f( +, + =.[.5+.5]. =. f(.5, + =.[.5+.55] K =.5. K 4 = h f( +h, + k =.[.,+.5] =.[.,.5] K 4 =.5. = + 6 [ ], =.4 = + 6 [k + k +k + k 4 ] k = h f(,. =, = +h = +. =., =.4 k =. f(.,.4=.[. +.4]=. k =. h k k = h f( +, +.. =. f(. + K =.8. K =.68. K 4 = ,.+ =.8