DIFFERENTIAL EQUATIONS

Transcription

1 DIFFERENTIAL EQUATIONS PAGE # An equaion conaining independen variable, dependen variable & differenial coeffeciens of dependen variables wr independen variable is called differenial equaion If all he differenial coeffeciens are presen wr one independen variable hen he differenial equaion is called ordinar differenial equ If he are presen wr more han one independen variables hen i is called parial differenial equ d d d d 3 ODE z 0 PDE 3 Differenial equaion represens a famil of curve which share some common characerisics d eg : 5 d = 5 + c famil of lines wih slope 5 eg : d 0 d k famil of hperbola wih cenre as (0, 0) & pair of asmpoes as = 4 The famil of curves obained afer inergraing he given differenial equ, free from differenial coeffeciens & conaining arbirar consans is called general soluion or soluion or inegral curves or primiive If all he arbirar consans presen in general sol are subsiued wih some specific values hen he obained equ is called paricular sol 5 Order & Degree : The order of highes order differenial coeffecien presen in he equaion is called order of differenial equaion eg : 3 8 d d 3 d 0 d order = 3 The eponen of highes order differenial coefficien presen in he equaion is called he degree of differenial equaion provided all he differenial coefficiens are presen in polnomial form Order of a differenial equ is alwas defined while degree need no o be defined eg : d d d d order 3 degree 7 eg: 5 d 5 d 4 e d / d 3 order 5 degree undefined eg: 3 e 4 d d 4 4 order 4 degree 3 7 "" eg: 4 ' sin 0 order 4 degree 3

2 PAGE # / 3 eg: 5 "" ' order - 4 degree (iv) 3 sin " n order degree FORMATION OF DIFFERENTIAL EQUATION : NOTE : (A) If he equaion of famil of curves is given hen (i) coun he no of independen arbirar consans sa n The no of independen arbirar consans will be he order of differenial equ o be formed eg : = = 3 sin k k 3 sin k order k eg : = k3 ke k4 k = ke k 4 order 3 eg3 : = A sin Bcos order eg4 : A sin c cos D order eg5 : sin cos k3 k4 sin k k (ii) Differeniae he given equaion n imes Eliminae hese n arbirar consans from hese n + equaions, one given and n obained (iii) The eliminan of above will be req differenial equ (B) If he equaions are given in erms of language hen form he corresponding mahemaical equaion & hence differenial equaion hus obained is Reqd DE Q Form he differenial equ for follow famil of curves (i) k e k ' k e e ' k e ' e " 0 = (ii) = k e k e ' k e k e " k e k e = (iii) k e k e ' k e k e = k e k e

3 PAGE # 3 (iv) = = 3 k 5 3k = 3 (v) ab e c = a ke ' ak e a c a e ' ak a a ae ' e " 0 = a = a (vi) Famil of circles cenred a origin k ' 0 (vii) Famil of circles having cenre a line = & radius = k 4 k ' 0 k ' 4 ' (viii) Famil of parabolas ais // o -ais & he vere a (4, 5) 5 a 4 5 ' a 5 5 ' ' (i) Circle ouching -ais a origin r r r 0 r 0

4 PAGE # 4 ' ' 0 ' () c c ' c ' c ' ' ' ' SOLUTION OF DIFFERENTIAL EQUATION : A soluion of a differenial equ will saisf i idenicall eg : Find which one of he follow/s is/are sol of differenial equ " 3 0 (i) = a sin bcos (ii) 3 asin b cos (iii) 3 a sin bcos (iv) 4sin 3 Sol Checking opion (ii) " a sin b cos " 3 as bc as bc Sol Checking opion (iv) " 4sin 4s 4s For a general curve (a) Tangen d Y X d (b) Normal Y X d d (c) Inerceps made b angen on boh he aes are A :, 0 ' & B : 0, ' 3 (a) RP = Lengh of angen Inerceped porion of angen b/w p & -ais (b) PS = Inerceped porion of normal b/w p & -ais Lengh of Normal (c) RQ = subangen = Projecion of lengh of angen on -ais (d) QS = Projecion of lengh of normal on -ais Subnormal Subnormal = QS = ang =

5 PAGE # 5 ' Lengh of normal = PS = subangen = ' Lengh of angen = ' 4 d = d d Lengh of curve = AB = B d A B = d d ' d A METHODS OF SOLVING DIFFERENTIAL EQUATION : Variable separable : Take d on one side of equaion & d on oher side If all he erms conaining can be clubed wih d & he erms conaining can be clubed wih d hen i is variable separable form Separae he erms & hen inergrae (i) d sin cos 0 d sin d cos d cos sin c d cos an 3 0 d (ii) cos d 3co d sin 3n sin c (iii) sin 3 d d 0 sin 3 d d d d 0 sin 3 d sin3 d + = 0

6 PAGE # 6 3 sec d 3 3 an an + an c (iv) d d e 3 e sin d 3 e e e sin d 3 e d e sin d 3 e e cos c 3 (v) d d d n a b d e ab d a e e d b b a e d e d b a e e c b a (vi) Find c urve (which is no passing hrough origin) f or which lengh of normal a a poin is equal o radius vecor? Sol Lengh of normal = radius vecor ' ' = k (vii) Find curve for which segmen of angen conained b/w coordinae aes is biseced b he poin & he curve passes hrough (3, 4) A : ' A B p 0 ' '

7 PAGE # 7 d d n n c ( (3, 4) lies on he curve) = k = Q8 Find he curve pass ing hrough (, ) for which mid p of (lengh of ) angen is biseced a is p of wih -ais A P 0 0 ' d ' d n n c k (, ) k = 4 4 Q9 Find curve for which area bounded b curve, he coordinae aes & a variable ordinae is equal o he lengh of corresponding arc in s quad & given ha curve passes hrough (0, )? Ar OAPQ 0 PQ d = ' d 0 = f() f d 0 = ' ' ' = = d d n c

8 n PAGE # 8 e e e e e e & = (B) Reducial o Variable separable : d f a b c d le a + b + c = a + b = a bf a f d d d d sin 3 cos 3 4 d = + 3 = + 3 = d + 3(sin () + cos() + 4) = d d 3sin 3cos 4 d d 3 d d 4 d d 5 d an d

9 + = + = d PAGE # 9 + d d = + c = an 3 + = + = d d d d d = 4 + = ' d + d d d d 5 ' d an d

10 d = co PAGE # 0 c n sin 3 Variable separable using Trigonomeric subsiuion : (a) r sin & = r cos (b) r (c) an / (d) d d r dr (e) sec d d d r d d d T - : (a) r sec = r an (b) r (c) sin (d) d d r dr d d (e) cosd d d r secd d d d d = r dr r r sec d r dr r sec d n r r n sec an c d d d d 0 r cos r dr r sin r d 0 cos dr sind 0 dr an sec d 0

11 PAGE # r sec c 0 d d d d 3 4 d d d d 5 ' ' 3 r dr r d r r cos dr = sec r = an c d 4 r dr = r cosr d dr cos d r 5 c sin r ' ' r cos r sin' r r cos' r sin r + d d d r dr = d d d r d = r r dr r d d a b c d a b c ad bd cd ad bd cd bd cd ad cd ad bd 0 a b 0 If a d b d a d d

12 PAGE # eg: d d d 8d 9d 3d 4d 5d 8d 9d 3d 5d 4d d c eg : + 4 = eg3 : d 4 9 d 4 6 d 3 d 3 Inegrae erm b erm eg:4 : 3d 9 8 d 8 eg: 5: d 3 4 d 9 7 HOMOGENOUS DEs : The equaion of he form same degree d d P, Q, is called homogenous differenial equaion where P & Q are homogenous funcions of In his case RHS ca also be wrien as f [B cancelling n from numeraor & D r of RHS where n is he degree of P & Q] To solve pu / = = d d d f() = + d f() = d d f variable separable Q d 3 4 d 4 3 As i is homog diff equ using / =

13 PAGE # 3 d n d d 3 d d 4 sin d sin d 3 d d 5 d d d ln n c = / 3 d d d d 3 n = + c 4 d sin / d sin /

14 PAGE # 4 d sin sin n sin 5 d 3 d d = 3 n c 6 Find he curve for which subnormal is equals o summaion of coordinaes ' = d = 7 Find he curve for which raio of subnormal o he sum of coordinaes is equals o ha of ordinae & absicca? ' = 0 or = 8 If he area bounded b he curve b/w =, =, = 0 & he curve is = raio of cube of ordinae o absicca 3 d = 3 3 ' 3 3 ' 9 Find he curve for which angle formed b -ais a an p is wice he formed b polar radius of p of angenc wih -ais an = an '

15 PAGE # 5 ' / d 0 ' ' 0 = 4 4 = d / / e d e / d 0 / d e / d / e d e e d d 0 d d d Q3 d d REDUCIBLE TO HOMOGENOUS DIFFERENTIAL EQUATION :

16 PAGE # 6 T- d a b c d a b c = + h = + k d a b ah bk c d a b a h b k c If h & k are chosen such ha ah bk c 0 ah bk c 0 d a b d a b d a b a b E- d 5 d 3 4 = + h = + k d d 3 4 h + k 5 = 0 3h + 4k = 0 h =, k = d = 4 T-: If & is presen as f() & g() hen search f () wih d & g () wih d Q 3 d d = d d d d 4 5 8

17 PAGE # 7 3 d d c c d sec 3cos 4sin d cosec 4cos 5sin d d 0 d 4 e e e d 5 'sin cos sin cos d 3cos 4sin sin d 4cos 5sin d sin 3cos 4sin d cos 4cos 5sin = sin, = cos d 3 4 d = 3 c 3 d d 0 d d 0 d d 4 e d e d d e d e e e e e Now e & e d d Reducible separable mehod

18 5 'sin cos sin PAGE # 8 sin d sin cos sin d d d d = d d 7 d dy dx X Y X le X = X 7 Y = Y + 3 dx X = where Y = X Ulimael we will pu Y = Y + 3 & X = 7 LINEAR DE OF ST ORDER & FIRST DEGREE : General equaion : d p Q d Define inegraing facor IF = ep p d = e p d mulipl boh sides b IF p d e d p d e p Qe d d p d p d e Q e d p d pd p d e Q e d (i) NOTE : If an epression conains d d of ( Inegraing Facor) linearl added wih (where coefficiens could be funcions of ) hen i can be made perfec diff d d p Q is linear differenial equ where is dependen veriable and is independen variable d Q Q n n n d d d

19 PAGE # 9 d d d n Q3 Q4 d cos sin d Q5 d Q6 d 3 d Q7 d d 0 Q8 cos 3 Q9 d an d Q0 d Q d d 0 d / d 0 d Q sin cos sin d n d d IF = ep d = ep d = ep n = Using (i) = n d = n d = n c

20 PAGE # 0 IF = d ep n = ep nn = n From (i) n n d n n c d d 3 cos sin d d cos sin sin sin e sin e d 4 9 d n d d n d IF = = n d d an d e an an e an d 5 d d d IF = ep d ep n an c

21 PAGE # ep 6 IF = ep n d = ep = d d = d 7 d 3 d d d d IF = ep d d = an c = = an c d cos d 8 IF = ep d cos d sin c d d d 0 d 3 d 0 d 0 d 5 c d d

22 d d PAGE # IF = ep d e e e d d d co sin IF = ep co d ep nsin sin sin sin sin d DIFFERENTIAL EQUATION REDUCIBLE TO LINEAR DE : d f ' p f Q d f() = z (diff wr ) dz p z Q d Linear differenial equaion in z BERNOULI S EQUATION : d d n p Q d p Q n d n n z eg : ' ' / = d IF / d 3 '

23 PAGE # 3 3 ' n n 4 3 ' d 5 sin cos cos sin d 6 / ' e sin ' d d d d d IF = ep 4 = ep n 4 = / 4 / 4 / 4 d 3 d d n n / n d IF = / d 4 d d 3

24 d d PAGE # d 3 IF = ep 3d e 3 3 e 3 e d 5 d cos cos sin (cos ) d sin sin 6 d / e sin d / / e sin d = IF = ep d ep / / / e e e sin d e / e / cos c 7 Find he curve such ha he iniial ordinae of an angen is less han absicca of p of angenc b unis Y ' X Iniial ordinae = ' ' ', 0 pu = 0 8 Find he curve if he produc of iniial ordinae & absicca is consan? I O = ', 0 = ' c FORMING PERFECT DIFFERENTIALS : d d d

25 d d d PAGE # 5 d d 3 d d d d 4 5 d an d d Q Solve he followings : 3 d d d d = sec an d d sec an d d d d d sec an d sec an d = n sec an + n cos c 3 d d d d 3 cos d d 3 d d 4 d d 0 5 d d d 6 ' ' 7 d d sin cos d d sin d 0 cos cos

26 d d d d d / d PAGE # 6 = d d 3 cos d d cos d d sin c 4 d d d d 0 d d d 0 3 / c 3 5 d d d d d n c d d 6 d d d d d d d d

27 7 d d d 0 PAGE # 7 k 8 cos d d sin d sin d 0 an cos cos c d d d d 9 d d d d 3 / 3 / d d d an 0 c sin d 3 cos d 0 d 3 d cos d sin d 0 3 d d d cos 0 3 cos c d d d d d d d d d d ORTHOGONAL TRAJECTORY : (O T) : OT of a famil of curves is a famil of curves each member of which inersec wih each member of given famil a 90

28 PAGE # 8 To find equ of orhogonal rajecor s find differenial equ of given famil, hen replace d inegrae his DE o ge req OT Q Find orhogonal rajecor of follow : d d in his diff equ b d & hen (i) k ' 0 For OT = 0 ' ' 0 ' d d 0 d 0 = c (ii) r + = 0 For ORTHOGONAL TRAJECTORY 0 ' = 0 d d = 0 d d 0 n() n() = c n c = c ' (iii) / 3 / 3 / 3 a ' 0 / 3 / 3 For ORTHOGONAL TRAJECTORY 0 / 3 / 3 ' / 3 / 3 d d 0 '

29 4 / 3 4 / 3 k PAGE # 9 RATE CHANGE & FORMING CORRESPONDING DE : To analse he problem consider a ime inerval ie from o ( + ) In his ime inerval, a changing quani can be assumed as consan 3 Form DE for corresponding phsical phenomenon Inegrae he ob differenial equ The given informaion in he ques ma be used in limis of inegraion or for indefinie o evaluae arbirar consan Q Le a spherical ball losses is volume which is direcl proporional o is insananeous area Also a = 0, r = m & a = 3 monhs, r = m Find radius as funcion of ime? dv 4 r dv k 4 r d 4 r k 4 r dr 3r k 4r 3 dr k r = k + c = 0, r = c = = 3 r = 3k + = k = 3 r = 3 Q Le a cone is filled wih waer & waer evaporaes from i a he rae direcl proporional o suface area epossed o environmen proporionali consan being m/s Find he heigh of waer column as func of ime given a = 0, h = 00 m & he semiverical angle is 60 v r h han60 h 3 3 = h 3 dv r 6 h 3 d h 6h

30 dh 3h 6h PAGE # 30 dh h = + k h = 00 = 0 k = 00 h = + 00 Q3 Le a righ circular clinder of cross-secional area A is provided wih a circular opening a boom of area a, which is covered wih a diaphrogm z is opening a a consan rae & is compleel opened a he ime, = T The h of circular vessel is H & is comp filled column as a func of ime? (i) < T (ii) > T a a k a when = T k = a T a v a T reduced v ou Adh a v (v is he veloci of waer moving ou) Adh a gh dh a A gh T h A dh a g h T H 0 A h a h g H T 0 A a h H g T g a h H, < T 4TA > T A dh a gh Q4 Le a conainer conains v 0 L of fresh waer A ime = 0 a saled waer is run ino i a he rae v L/min, each lire of which conains a kg of sal A = 0 waer also sars flowing ou a he rae of v L/min hen find quani of sal presen in conainer as a funcion of ime, all he ime miure is kep uniform b sirring am = In ou

31 PAGE # 3 m av v = v0 v v v v v Volume of waer = 0 m dm av v v 0 v v dm v m av v v v ep 0 If = v v v v 0 v ep lnv v v v v = 0 = 0 v v v v v v v v v = m v v v 0 0 v vv av v v v v v v mv0 v v = av v v v 0 v vv v v v v v + c STQ: f () + p()f() Linear Diff Epression d Le for a given diff equaion f, d If g() is one of he soluion hen dg f, g d Q Le = f() be a soluion of differenial equ curve a a given poin? d 3 d T = + 4 = ( ) N 4 d 3 d 5 & i passes hrough (, 4) hen find equ of angen & normal o his

32 Q Le p() = 0 and d p d TPT p() > 0 p, PAGE # 3 dp p 0 d d p e 0 d f () > 0 f() ing hen if > f() > f() p() p() e p e e 0 p() > 0 Alernaive : d p e 0 d Inegraing boh sides wihin limis o p e p e 0 p() > 0 p e 0 Q3 Le f(,, c) = 0 be he inegral curves of he differenial equ d d g, k, where g & k are homogenous ep equ of same degree TPT angens drawn o all he inegral curves a heir p of inersecion wih a sraigh line passing hrough origin, will be parallel d d g, k, d d f d MT d, f m MT d d, f m Q4 Le wo curves are given = f() & = g() saisf he follow wo properies (i) angens drawn a he ps of equal absicca inersec on -ais (ii) normals drawn a he poin of equal absical inersec on -ais hen find he curves

33 df T Y f X d PAGE # 33 dg T Y g X d As boh passes hrough A f f ' g g' d N Y f X df d N Y g X dg As boh passes hrough B f f ' gg' (ii) f g c f g f g c (ii) f g f ' g' from (i) d f g f g d d f g d f g f g k (iv) from (iii) & (iv) k(f + g) = c f + g = c k f g = k c f = g = c k c k Q5 Le wo curves = f() & = g() are given where g() = f d & also angens drawn a ps wih equal absicca o boh curves inersec on -ais f ind he curves if f() passes hrough (0, ) & g() pas ses hrough 0, 4 As he inersec on -ais f g f ' g'

34 f ' g' f = kg() f g PAGE # 34 g() = f d g () = f() g () = kg() g' k g n g k c g() = f() = e kc kk e k k e k 0, 4 k = /4 (0, ) = k 4 k = 4 f() = 4 e g() = e 4 4 Q6 Le & are wo soluions of d p d = Q() & also z Q d TPT z = ae where a is arbirar consan d p d d p d z Q Q dz d z pz Q d d dz zq Q d dz Q z d dz Q d z Q nz d c z k e Q d

35 PAGE # 35 Q d z = k e