Honours Introductory Maths Course 2011 Integration, Differential and Difference Equations

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2 Honours Inroducory Mhs Course 0 Inegrion, Differenil nd Difference Equions Reding: Ching Chper 4 Noe: These noes do no fully cover he meril in Ching, u re men o supplemen your reding in Ching. Thus fr he opimision you hve covered hs een sic in nure, h is opimising he vlue of funcion wihou ny reference o ime. In sic opimision nd comprive sic nlysis we mke he ssumpion h he process of economic djusmen leds o n equilirium, nd we hen exmine he effec of chnges of he exogenous vriles on he equilirium vlues of he endogenous vriles. Wih dynmic nlysis, ime is explicily considered in he nlysis. While we re no covering dynmic nlysis his poin, cerin mhemic ools re required for dynmic nlysis, such s inegrion nd differenil equions. Wihou hese ools, i ecomes impossile o consider prolems which re no sic in nure. We will e covering oh of hese opics in minly mhemicl wy, leving economic prolems for ler de. Inegrion is he reverse process of differeniion. If funcion F(x) hs firs derivive f (x) hen he inegrl of f (x) will yield F (x). The noion o denoe inegrion is s follows: f ( x ), where he inegrl sign is n elonged S. f (x) is referred o s he inegrnd, nd he sign reminds us h we re inegring wih respec o he vrile x. We go hrough he following explnion o deermine where his noion comes from. Suppose we re given n rirry funcion f (x) nd sked o find he re of he curve eween poins, for exmple he re under he curve f(x) eween 5 nd Figure. Wih liner funcion, his eques o finding he re of ringle nd recngle s follows:

3 Figure. However wih non-liner funcion he prolem ecomes slighly more complex. Wh we cn do, however, is emp o find he re under he curve using numer of pproximing recngles s follows: Figure. We le ech of he recngles hve equl widh nd we cll his widh x. Ech recngle hs heigh equl o he funcion vlue, for exmple he heigh of he ls recngle where x=0 is equl o f(0)=0. Thus he re of he ls recngle is equl o 0( x), s re of recngle equls lengh imes redh, nd here redh is x nd lengh is f(0)=0. The re of ny of he recngles is equl o lengh imes redh, which equls f(x) imes x, s ll recngles hve equl redh equl o x. To find he re under he curve we dd up he re of ech of he recngles. This gives us he expression: A n = i= f ( x ) x Where n = numer of recngles x i = he vlue of x ech poin = he sum of ll he res, sring from he firs one (i = ) nd ending he nh one (i = ). Oviously his sum will no e very ccure represenion of he re. Bu perhps if we mke our x smller, hen his expression will ecome more ccure represenion of he re under he curve, s here will e less overshooing y ech recngle. If iniilly we hd en recngles, he re given y he sum of hese recngles res would oviously e more of n over-esime (or mye underesime) hn if we douled he numer of recngles, nd hen summed heir re. The more recngles we use in his pproximing process he eer our esime for he re under he curve. For exmple, imgine we wish o find he re under he curve f(x) = x eween 0 nd. i

4 Figure.4 We could ke four recngles, ech wih redh x equl o 0.5, nd ke he heighs from he righ hnd side of ech recngle. Hence he heigh for ech recngle will e: (0.5) (0.5) (0.75) Therefore he enire re equls o 0.5(0.5) + 0.5(0.5) +0.5(0.75) +0.5() = 5/= If we doule he moun of recngles from four o eigh, we will use x of 0.5, nd he following righ hnd heighs (rememer he heigh of he recngle is given y he funcion vlue f(x)). (0.5) (0.5) (0.75) (0.5) (0.65) (0.75) (0.875) The corresponding ol re is given y he sum of ech of he res which is x muliplied y ech funcion vlue: The finl vlue we ge is As cn e seen in figure., using righ end poins for he recngles for n incresing funcion will give n over-esime, while using righ end poins for decresing funcion will yield n over-esime. Thus douling he numer of recngles while rying o esime he re under he grph f(x)=x will egin o ring our esime down o is rue vlue. I ppers h s he numer of recngles increses, our esimions ecome eer nd eer pproximions of he re. If we le he numer of recngles end o infiniy, we will oin perfecly ccure esime for he re under our grph. Our expression for he re under he curve now ecomes: n A = lim f ( xi ) x n i= This gives us he expression for he definie inegrl, which gives us wy of finding he re under he coninuous funcion f(x) eween x= nd x=: An explnion of he erminology: n n i = f ( x) = lim f ( xi ) x

5 The inegrion sign is n elonged S, nd ws so chosen ecuse n inegrl is limi of sums., re he limis of inegrion, is he lower limi of inegrion is he upper limi of inegrion. f (x) is known s he inegrnd. hs no mening y iself, u merely reminds us h we re inegring wih respec o he vrile x. Forunely when we wn o find he re under curve, we do no hve o go ino he long process of finding n expression for he sum of he re of n recngles: numer of heorems mke he process esier. Before we se ou he properies of he definie inegrl, some rules of inegrion re s follows: (see pge 49 nd onwrds in Ching for exmples).. The Power Rule n n+ x = x + c ( n ) n +. The Exponenil Rule x x e = e + c. The Logrihmic Rule = ln x + c ( x > 0) x Properies of he definie inegrl:. c = c( ) = f ( x) +. [ f ( x) + g( x)] g( x). cf ( x) = c f ( x) = f ( x) 4. [ f ( x) g( x)] g( x) c c 5. f ( x) = f ( x) + f ( x) Propery ses h he inegrl of consn funcion y=c is he consn imes he lengh of he inervl, s seen in figure.5 4

6 Figure.5 Propery sys h he inegrl of sum is he sum of he inegrls. The re under f+g is he re under f plus he re under g. This propery follows from he propery of limis nd sums. Propery ells us h consn (u only consn) cn e ken in fron of n inegrl sign. This lso follows from he properies of limis nd sums. Propery 4 follows from propery nd, using c=-. Propery 5 ells us we cn find he re under he grph eween nd c, y spliing i up ino wo res, eween nd, nd eween nd c. We now find our rule for evluing he definie inegrl: f ( x) = F( ) F( ) where he derivive of F(x) is f(x), i.e. F is ny ni-derivive of f. For exmple, if we differenie x F ( x) = we oin x Thus x = F() F(0) = = 0 =. 0 0 f ( x) = x, so F(x) is n ni-derivive of f(x). Therefore he re under he curve f(x) = x eween 0 nd, is equl o hird, or 0. recurring. Incidenlly his nswers our previous quesion which we emped using he sum of he res of n recngles. The fundmenl heorem of clculus moives his use of he evluion heorem. In shor, i ses h differeniion nd inegrion re opposie processes. Thus, if we sr wih funcion F(x), nd differenie i o oin f(x), [ i.e. F ( x) = f ( x) ], if we hen inegre he funcion f(x), he resul will e he iniil funcion F(x). Similrly if we inegre f(x) o oin F(x), [ i.e. f ( x) = F( x) + C where C is n rirry consn ]. Thus o find he inegrl of funcion f(x), we mus find he funcion which when differenied yields f(x). This heorem is very useful o us, s oherwise whenever we wish o find he vlue of he re h lies underneh curve, we hve o go hrough he enire process of finding he limi of he sum of he res of n pproximing recngles, which is ime consuming process! Prior o he discovery of he fundmenl heorem, finding res, volumes nd oher similr ypes of prolems were nigh on impossile. For compleeness, he fundmenl heorem is presened elow: More ou he rirry consn lile ler 5

7 The fundmenl heorem of clculus: Suppose f(x) is coninuous funcion on he closed inervl [,] x x d. If g ( x) = f ( ) d hen g ( x) = f ( x) i.e. g ( x) = f ( ) d = f ( x). f ( x) = F( ) F( ) where he derivive of F(x) is f(x), i.e. F is ny ni-derivive of f. Wh i sys, roughly speking, is h if you inegre funcion nd hen differenie he resul, you rerieve he originl funcion. We now need o discuss wo differen ypes of inegrls definie nd indefinie. A definie inegrl involves finding he inegrl of funcion eween wo numer limis i.e. f ( x). The nswer o definie inegrl is numer, s we know ccording o he evluion rule he nswer o his is jus he niderivive F(x) evlued eween nd, i.e. F()-F(). An indefinie inegrl yields funcion of x s is nswer (if we re inegring wih respec o x). An indefinie inegrl is n inegrl of he form f ( x) (i.e wihou upper nd lower limis) nd he soluion is f ( x) = F( x) + C where C is n rirry consn which cn ke on ny vlue. The reson we include he rirry consn is illusred in he following exmple. Given he prolem: x poenil soluion is x s his is n niderivive of he cuic funcion (If we differenie x we oin x. However x + 4 is lso soluion o his prolem, s is x 00. This is ecuse when differeniing hese expressions, he consn differenied moves o zero. So i would pper h he mos generl form o give he nswer o his prolem would e s follows: x = x + C, where C is n rirry consn. Jus smll noe on rirry consns when we dd wo ogeher, we oin hird one which hs ggreged he firs wo, when muliplying, dividing dding or surcing numer y/o/from n rirry consn, he resul is jus he rirry consn. However he rirry consn when muliplied y funcion of x, will sy s jus h: f ( x) + g( x) = F( x) + C + G( x) + C = F( x) + G( x) + C where C nd C re wo rirry consns, nd F(x) nd G(x) re wo ni-derives of f(x) nd g(x) respecively. f ( x) = [ F( x) + C] = F( x) + C = F( x) + C However: 6

8 x f ( x) = x[ F( x) + C] = xf( x) + Cx We now urn o some rules of inegrion (definie nd indefinie) nd hen some exmples.. cf ( x) = c f ( x). ( f ( x) + g( x)) = f ( x) + g( x) n+ n x. x = + C n + 4. = ln x + C x x x 5. e = e + C x x 6. = + C ln 7. sin x = cos x + C 8. cos x = sin x + C (n cnno equl ) Rememer o check he nswer o ny inegrion sum jus differenie i nd you should rrive ck he originl funcion. Some exmples:. = x + C. = x + C = C 5. x x = x x x 4 x 5 4 0x 7 = = = ( ) = [0 x + sin x] = cos x + C = x cos x + C x x x 8 6. ( x 6x) = 6 = x = (9) = d = 7. = 8+ (4) ln 9 5 = ( ln 0 ) d = 9 ( + ) d = ln 9 7

9 8. 4 ( 6) d = = A few more useful properies:.. f ( x) = f ( x) = 0 f ( x) We re going o e looking wo very useful echniques used in inegrion: use of susiuion, nd inegrion y prs. There re whole hos of oher echniques which cn e useful, however i is hese wo which re mos useful o us in economics. Inegrion using Susiuion We use susiuion, when he inegrnd conins funcion nd is own derivive. i.e: f ( x) f ( x) For exmple: ( x + 5x 0)(x 0x) + If his is he cse, we cn mke use of he following susiuion: We le u equl o he funcion whose derivive we cn spo (or cree, using consn: more ou his ler). + x Le u = x 5 0 Then we know h du (x 0x) = + When we hen susiue he vlues of u nd du ino he inegrl, we oin he following inegrl: udu which hs he nswer u udu = + C x + 5x u: u = 0 Therefore our finl nswer is ( x + 5x 0) ( x + 5x 0)(x + 0x) = + C Thus he generl rule soluion for he prolem is s follows: [ f ( x)] f f ( x) f ( x) = + C or more simply f = (. ) f + C 8

10 However he susiuion rule cn e used for more compliced exmples, when our funcion f(x) whose derivive we cn spo occurs inside noher funcion: For exmple: f g( f ) or (6x + 8) (x + 8x ) In hese cses he procedure does no chnge ll we sill mke he susiuion s follows: Le u = f(x) Therefore du=f`(x) And proceed s usul: Some exmples:. (6x + 8) (x + 8x ) u = x + x Le ( 8 ) Therefore du = ( 6x + 8) Therefore our rnsformed inegrl is s follows: u (x + 8x ) u du =u du = + C = + C. Someimes we cn find funcion whose derivive we cn cree s follows: However only if we inroduce consn funcion. For exmple: e (4 x+ 5 ) If we le u = (4x+5), we know du = 4. However while we hve he, we do no hve 4. This is esily solved however hrough he following mnipulion: 4 4 (4 e x + 5) This inegrl now conins funcion nd is derivive, hus susiuion cn e used: Therefore le u = (4x+5), nd du = 4. The inegrl ecomes: 4 u e du = e 4 u + C = e 4 (4x+ 5) + C Rememer, he susiuion of u ino his funcion is device h we employ. Therefore our finl nswer ough no o conin u, s he originl prolem does no conin i. Alwys rememer o susiue ck for u. Noe: Unlike differeniion, here exiss no generl formul giving he inegrl of produc of wo funcion i..o. he sepre inegrls of hose funcions. There is lso no generl formul giving he inegrl of quoien of funcions in erms of heir sepre inegrls. As resul, inegrion is rickier hn differeniion, on he whole. 9

11 Inegrion y Prs We use his echnique when we hve o inegre produc: E.g. f(x).g(x) When we re given his ype of exmple, we mke use of he following formul: f(x).g (x) = f(x).g(x) f (x).g(x) For exmple: xe x We pick f s he funcion which is esy o dierenie, nd g s he funcion which is esy o inegre. Ofen picking squred or cuic erm for your f is good ide, s f will hve power h is hen one lower, nd hence simpler. I is very good ide o mke yourself mini le wih f, g, f nd g, o keep hings srigh. Also, noe h when nding g, we do no oher wih he rirry consn. xe x Therefore: f = x f = g = e x g = e x xe x = xe x ()e x = xe x e x + C Thus we hve mnged o use he formul o inegre our originl quesion. (Cn lso use he lernive noion used in Ching: where vdu = uv udv) Anoher exmple: xsinx Therefore: f = x f = g = sin x g = cos x xe x = x cos x + cos x = x cos x + sin x + C

12 An exmple wih rick: ln x Oviously we do no know he inegrl of ln x, h is why we re using his mehod. So we mke ln x equl o f, nd we cn hen nd is derivive. Bu hen wh will our g e? Simple, mke i. f = ln x g = f = x g = x ( x x = x ln x x + C ln x = x ln x ) This is hndy rick which cn lso e used o nd he inegrls of some of he rigonomeric funcions. Some more exmples: xe x Therefore: f = x f = g = e x g = ex xe x = xex = xex ex ex 4 + C Which cn hen e re-wrien if we wn o: xe x = ex ( x ) + C Anoher exmple: x ln x f = ln x f = x g = x g = x

13 x ln x = x ln x = x ln x = x ( ) x x x ln x x 4 + C Anoher exmple: (ln x) f = (ln x) g = f = (ln x) x g = x (ln x) = x (ln x) = x (ln x) x. ln x ln x ( ) x = x (ln x) (x ln x x) + C The ls line uses resul h we proved few exmples go. Now le's ry denie inegrl using inegrion y prs: 0 e d f = f = g = e g = e We rs clcule he indenie inegrl, hen go ck nd susiue in he limis. e d = e + e d = e e + C Therefore: 0 e d = [ e e ] 0 = [ e e ] [ e 0] = e = e

14 You should now mke sure you cn do he inegrion prcice quesions. An exmple of n economic pplicion of inegrls: One simple pplicion is o nd `ol' quniy from `mrginl' quniy. Suppose rm hs mrginl cos C (x) = + e x where x denoes oupu. Then ol cos is: C(x) = ( + e x ) = x + 6e x + B, where B is he consn of inegrion. Dierenil Equions A DE is n equion conining funcion y = f(x) nd one or more of is derivives, i.e. y, y ec. I shows he relionship eween he funcion nd is derivives. If only he rs derivive dy d is presen, he dierenil equion is sid o e of he rs order. Some exmples include: y = y + y dy + xy = x y y = x As you cn see, hese equions conin dependen vrile y, some or more of is derivives y, y nd some occurrences of he independen vrile, in hese cses x. Our im wih dierenil equions is o solve for our funcion y, in erms of he independen vrile x, wih none of he derivives sill presen. How we do his depends lrgely on he ype of dierenil equion presen. Wh form DE kes will deermine how we solve i. Your min sk in his secion is o idenify he ype of DE. Once his is done i is usully firly simple o proceed from here. We will e considering 4 ypes of dierenil equions:. Direcly inegrle DEs. Seprle DEs. s Order Liner DEs 4. nd Order DEs

15 Direcly Inegrle Dierenil Equions: This ype of DE conins only one of he funcion's derivives, nd funcion of he independen vrile. For exmple:. y = x +. d y + x = e x. y = 6 Solving hese DE's jus requires h we inegre s mny imes s is necessry, including he necessry consns of inegrion. Eg:. y = x + y = x + x + C. d y d y + x = e x = e x x dy = ex x + C y = e x x 6 + Cx + D. y = 6 y = 6x + C y = x + Cx + D y = x + Cx + Dx + E This procedure gives us he generl soluion for hese DE's. To oin he specic soluion, i.e. wihou he presence of rirry consns, we require some iniil condiions. Eg: y(0) = or y (0) = To solve for he specic soluion o one of hese DE's, we use he iniil condiions s follows: y = 6 y(0) = y (0) = We know h he generl soluion o his prolem is: y = 6x + C y = x + Cx + D Therefore we susiue in he iniil condiions, oin some simple simulneous equions, nd solve for he wo rirry consns. y = x + Cx + D y(0) = = D y (0) = = C y = x + x + Direcly inegrle DE's re he simples ype of DE, nd sdly we won' encouner hem very ofen.

16 Seprle DE's The generl form for seprle DE is: dy = f(x).g(y) i.e. he RHS cn e sepred ino funcion of x imes funcion of y. Someimes you my hve o do some mnipulion o chieve his. E.g. ecomes which is seprle. ( + x)dy y = 0 ( + x)dy = y dy = y + x = y. + x E.g. ecomes xy 4 + ( y + ) e x dy = 0 dy = ex. xy4 y + which is seprle. To solve seprle DE, we rewrie i s follows, nd hen inegre oh sides wih respec o x nd y. dy = f(x).g(y) dy = f(x) g(y) g(y) dy = f(x) Since he LHS involves only y nd he RHS involves only x, he soluion mehod is known s sepring he vriles. If possile, one hen solves for y in erms of x, u if no, one simply leves he equion in simples form, mking sure he derivives re no longer presen. For exmple:. dy = y +x y dy = +x y dy = +x ln y = ln( + x) + C e ln y = e ln(+x)+c

17 . y = A( + x), where A = e C dy = x y ydy = x ydy = x y = x + C I is ofen hrd or lmos impossile o solve explicily for y in erms of x in seprle DE. We hen usully re conen o jus solve he DE, i.e. elimine he presence of derivives. You will wse lrge porions of ime in your exm if you do no relise his. E.g. dy d = + y 6 + ( + ) (y 6 + ) dy = y y = C I is impossile here o solve for y in erms of, so we jus leve i s is. An iniil vlue prolem: dy = y 4 y(7) = 5 y 4 dy = ln(y 4) = x + C Su in y(7) = 5 ln(5 4) = 7 + C ln = 7 + C C = 7 ln(y 4) = x 7 y 4 = e x 7 y = 4 + e x 7 Some more exmples:. + e x dy = 0 dy = e x dy = e x y = e x + C

18 . (x + ) dy = x + 6 dy = x+6 x+ dy = x++5 x+ dy = + 5 x+ y = x + 5 ln(x + ) + C The nex exmple uses inegrion y prs: e x dy = x dy = xe x For xe x : f = x f = g = e x g = e x xe x = xe x e x + C y = xe x e x + C Solve he dierenil equion dy d + y = 0, where is non-zero consn. Sepring he vrile gives y dy = ( )d. Inegring we oin : ln y = + C, where C is n rirry consn. Hence y = e C = e C e Seing A = e C, we my wrie he generl soluion s y = Ae, where A is consn. The nex ype of dierenil equion we encouner: Firs Order Liner Dierenil Equions: The generl form of rs order liner DE is: dy + P (x)y = Q(x) NB: If your equion is NOT in his form, you mus REWRITE i efore you cn work wih i. The generl mehod for liner rs order DE is s follows:. Pu i ino sndrd form. dy + P (x)y = Q(x). Idenify P (x) nd nd he inegring fcor I, which is equl o: I = e P (x)

19 . Muliply LHS nd RHS y I. e P (x) dy + e P (x) P (x)y = e P (x) Q(x) 4. The LHS ecomes he derivive of he produc of I nd y, so we rewrie i s such. We re essenilly doing he produc rule for diereniion here, u in REVERSE. Insed of going from (f.g) = f g + fg, we re moving from he RHS o he LHS. Thus he LHS of he equion in sep hree ecomes: d (I.y) = Q(x).I 5. We hen inegre oh sides. For he LHS, his is simple, s he inegrl nd derivive signs cncel ech oher ou, nd we simply oin I.y on h side. d (I.y) = I.y = [Q(x).I] [Q(x).I] 6. The ls sep is simply o solve for y. Like he seprle equions, i is someimes no possile o solve explicily for y in erms of x, nd we jus work owrds elimining derivives. Exmple: Given: x. dy 4y = x6 e x. Rewrie in sndrd form.. Find he inegring fcor. dy 4 x y = x5 e x P (x) = 4 x I = e 4 x 4 ln x = e ln x 4 = e = x 4. Muliply LHS nd RHS y i. 4 dy x 4 x 4 x y = x 4 x 5 e x x 4 dy 4x 5 y = xe x 4. Rewrie he LHS s d (I.y). d ( x 4 y ) = xe x

20 5. Inegre oh sides. d ( x 4 y ) = xe x x 4 y = xe x e x + C 6. Solve for y (which is possile in his cse). y = x 5 e x x 4 e x + Cx 4 Anoher exmple: (Noe: lhough i is seprle DE u we solve i using he mehod for rs-order liner DE's). (x + 9) dy + xy = 0 dy + x (x + 9) y = 0 I = e = e x (x +9) x (x +9) = e ln(x +9) = e ln (x +9) = x + 9 Now muliply he LHS nd RHS y he inegring consn: x + 9 dy + x x + 9 (x + 9) y = 0 d ( x + 9.y) = 0 d ( x + 9.y) = x + 9.y = C 0 y = C x + 9 Anoher exmple, using n iniil vlue: Solve x dy + y = x, where y() = 0. Soluion: dy + x y =

21 Exmple : dy + y = 9 dy + 4y = I = e 4 = e 4x 4x dy e + 4e4x y = e 4x d [ e 4x.y ] = e 4x d [ e 4x.y ] = e 4x e 4x y = e4x 4 + C y = 4 + Ce 4x Homogeneous nd Non-Homogeneous Cses of Dierenil Equions: The equion dy + y = 0, where is some consn, is sid o e homogenoeus on ccoun of he zero consn erm. This equion cn e re-wrien s The soluions o his DE re s follows: dy y d = y() = Ae [generl soluion] y() = y(0)e [denie soluion] Non-Homogenous Cse: When consn kes he plce of he zero in he ove equions, we hve non-homogeneous liner dierenil equion: dy d + y = The soluion o his equion consiss of he sum of wo erms, one of which is clled he complemenry funcion (denoed y y c ) nd he oher known s he priculr inegrl, denoed y y p. In his cse:

22 y c = Ae [generl soluion] y p =, 0 [denie soluion] Generl soluion of he complee equion is he sum of y c nd y p : y() = y c + y p = Ae + Try he following exercise nd see if you ge he sme nswer: xy + y = e x y() = The soluion is: x.y = e x (x ) + Our ls ype of prolem involves: 4 nd Order Dierenil Equions Generl Form: Where, nd re consns. y + y + y = nd order DE's cn involve cses where, nd re no consn u for our purposes we don' need o worry ou hese. A generl soluion for he DE ove is: y = y c + y p where y c denoes he complemenry funcion nd y p denoes he priculr inegrl. We nd y c nd y p using he following se of rules: In solving for he priculr inegrl, we cn disinguish eween dieren cses: Cse : If is non-zero y p =

23 Cse : If = zero y p = Cse : Boh nd re zero y p = And o nd y c, we nd he soluion o he homogeneous version of he DE, i.e. y c is he soluion o: Which is he sme s sying h = 0. y + y + y = 0 The complemenry soluion o he second-order dierenil equion kes he form: y() = A e r + A e r where A, A re wo rirry consns nd r nd r re he roos from he chrcerisic equion: r + r + = 0 given y r, r = ( ± ) 4 In oher words: To solve his, we form he corresponding chrcerisic equion: r + r + = 0 And solve for r nd r, which re he roos of he chrcerisic equion. These hen yield he following generl soluions for y c : In he cse r nd r, re unequl: r r y c = Ae r + Be r In he cse where r nd r, re equl: r = r y c = Ae r + Be r

24 There is lso he cse where r nd r, re complex roos, which we ignore for our purposes. Hopefully you ll hve no forgoen he formul o solve qudric equion, u in cse you hve: So our nl soluion is: x + x + c = 0 x = ± 4c y = y c + y p Where we hve found y c nd y p using he relevn rules. Some exmples: y + y y = 0 Therefore: = = = 0 We nd he pproprie cse for y p, which is he rs one, i.e. is non-zero. y p = = 0 = 5 We hen nd he soluion for y c which is he soluion o: This is he cse of unequl roos: y + y y = 0 r + r = 0 (r + )(r ) = 0 r = r = r r y c = Ae r + Be r = Ae + Be y = y c + y p y p = 5 y = 5 + Ae + Be Given iniil condiions, we could lso solve for A nd B. For exmple, given h y(0) =, y (0) =

25 We cn now solve for he specic soluion: y = 5 + Ae + Be = 5 + A + B 7 = A + B y = Ae + Be = A + B 7 = A + B = A + B 9 = A A = B = 4 y = 5 + e + 4e Exmple Two: (This is cse of repeed roos) y + 6y + 9y = 7 y(0) = 5 y (0) = 5 = = 6 = 7 y p = 7 9 = y + 6y + 9y = 0 r + 6r + 9 = 0 (r + )(r + ) = 0 r = r = r y c = Ae r + Be r = Ae + Be y p = y() = y c + y p = Ae + Be + y () = Ae + Be Be y(0) = 5

26 5 = + A A = y (0) = 5 5 = A + B B = y = + e + e

27 DIFFERENCE EQUATIONS Wh re difference equions? Wheres differenil equions del wih prolems in coninuous ime, difference equions re concerned wih prolems in discree ime. Here he vrile is llowed o ke ineger vlues only, mking he concep of derivive or differenil no longer pproprie. Insed he pern of chnge of he vrile y mus e descried y differences, rher hn y derivives or differenils of y(). When we re deling wih discree ime, he vlue of vrile y will chnge only when he vrile chnges from one ineger vlue o he nex, such s from = o =. Menwhile nohing is supposed o hppen o y. I is herefore more convenien o inerpre he vlues of s referring o periods rher hn poins of ime, wih = denoing period, = denoing period nd so forh. Then we cn regrd y s hving one unique vlue in ech ime period. In view of his inerpreion, he discree-ime version of economic dynmics is ofen referred o s period nlysis. y In difference equions, he pern of chnge is represened y he difference quoien. cn only ke ineger vlues, so if we compre he vlues of y in wo consecuive periods, we mus hve =. For his y reson he difference quoien cn e simplified o he expression y ; his is clled he firsdifference of y. This refers o he re of chnge of y in period, which is equl o y + y. We cn herefore define he forwrd difference operor y y = y + y. Applicion of he operor my e regrded s he discree-ime counerpr of differeniion wih respec o ime. Difference equions re herefore like differenil equions, excep h derivives re replced y differences. This he discree-ime nlogue of he differenil equion dy + y =, where nd re d consns, is he firs-order difference equion y + y =. Reclling he definiion of he operor, we my wrie his equion s: y + cy =, where c=-. + Since his equion simply reles he vlue of y in period o is vlue in he previous period, i cn e wrien in he equivlen form: y + cy = 7

28 Solving firs-order difference equions. In his secion we re concerned wih difference equions which conin y nd y, u no y furher erms in he sequence. + + or Before looking generl mehod o solve difference equions, we will consider n ierive mehod, which will ddiionlly provide some insigh ino he nure of he soluion o difference equion. Priculrly in he cse of firs order difference equions, simple ierion of he differencing or recursive rule plys ou he recursive rule over numer of ime periods, in order o see wheher we cn depic generl chrcerision of he ime ph in y(). For exmple: y = y + +, y = 0 5. From his equion we cn deduce sep-y-sep h: y = y + 0 y = y + = ( y + ) + = y + () 0 0 y = y + = [ y + ()] + = y + () And in generl, for ny period y = y + () = This equion indices he y vlue of ny ime period nd herefore consiues he soluion of he difference equion. This mehod is crude nd quickly runs ino limiions nd essenilly corresponds o he process of direc inegrion, which is fesile for cerin ypes of differenil equions. For his reson we hve o use more generl soluion mehods. Generl Soluions To find he generl soluion of he difference equion mens finding formul giving ll sequences { y } which sisfy y + cy =. As in he differenil equion cse his will conin n rirry consn which will e ied down if we specify he vlue of y for some priculr. We sr wih he simple cse where = 0. Then y + = cy for ll nd hence: y = cy 0 y = cy = ( c) y 0 y = ( c) y 0 nd so on. I follows h: y = ( c) y for = 0,,

29 To summrise, if = 0, (i.e. liner, homogenous difference equion) he generl soluion o y + cy = + is ( ) y = A c, where A is n rirry consn which my e inerpreed s y 0. In he cse where 0, ( liner, non-homogenous DE) we use mehod similr o he one we used o solve he nlogous differenil equion. The generl soluion will consis of he sum of wo componens: priculr soluion y p, which is ny of he complee non-homogenous equion y + cy =, nd complemenry funcion y c, which is he generl soluion of he reduced equion y + cy = 0. Firs we consider he complemenry funcion: Our experience wih he exmple ove suggess h we migh ry soluion of he form y = A( c). Now we mus look for he priculr soluion, which reles o he complee equion y + cy = +. We noe h for y p we cn choose ny soluion of y + + cy =. Therefore, if ril soluion of he simples form = k ( consn) cn work ou, no rel difficuly will e encounered. y Now, if y = k, hen y will minin he sme consn vlue over ime nd we mus hve y = + k. If we susiue hese vlues ino y + cy =, we ge: k + ck = nd k =. + c + Since his priculr vlue of k sisfies he equion, he priculr soluion cn e wrien s: yp ( = k) = ( c ). + c Since his is consn, sionry equilirium is indiced in his cse. Wh if c =? If c =, he priculr soluion y p = is no defined nd some oher soluion of he nonhomogenous equion y + + c + cy = mus e found. In his cse we use he rick of rying soluion of he form y = k. This implies h y + = k( + ). Susiuing hese ino y + cy = we find: k( + ) + ck = nd k = = [ecsue c = ] + + c y ( ) p = k = + 9

30 This form of he priculr soluion is non-consn funcion of, i herefore represens moving equilirium. Adding yc nd y p ogeher, we my now wrie he generl soluion in one of he wo following forms: y = A( c) + [generl soluion, cse of c ] + c y = A( c) + = A + [generl soluion, cse where c = ]. To elimine he rirry consn., we hve o use he iniil condiion h y when 0 = y0 =. Leing = 0, we hve: For c = : y = A + nd A = y0 + c + c y = ( y0 )( c) + + c + c nd for c y = A, so he definie version of his equion is: 0 y = y + c 0 Exmple: Find he soluion of he difference equion y + y =, which sisfies he oundry condiion y 0 =. As priculr soluion y we ry This sisfies he equion provided : k k =. k = 4 We cn find I.e. y c p y = k ( consn). y y rying soluion y = A( c) = A( ), where A is n rirry consn. The generl soluion o he difference equion is herefore: y = 4 + A( ) = 4 + A I remins o use he oundry condiion y 0 = o find A. Seing =0 in he generl soluion we hve 0

31 = 4 + A A = nd y = 4 Second-Order Difference Equions A second-order difference equion is one h involves n expression of y, u conins no differences of order higher hn. The symol second-difference s follows: y = y = y + y ( ) ( ) = ( y y ) ( y y ) = y y + y + + A simple vriey of second-order difference equion kes he form: y + y + y = c () + + y, clled he second-difference is n insrucion o ke he This equion is liner, non-homogenous nd wih consn coefficiens, nd consn erm c. As efore, he soluion hs wo componens: priculr soluion y p, nd complemenry funcion y c, The priculr soluion defined s ny soluion of he complee equion cn e found simply y rying soluion of he form y = k. Susiuing his vlue ino () ove, gives: k + k + k = c c k = + + Thus, so long s ( + + ) 0, he priculr soluion is: c y ( = k) =, where + p + + Exmple: Find he priculr soluion of y y + 4y = Here =, = 4, c = 6 6 Since +, yp = = + 4 Cse where + = :

32 Here he ril soluion y = k reks down nd i is necessry o ry y Susiuing his ino () ove nd keeping in mind h we now hve: y y + + = k( + ) = k( + ) k( + ) + k( + ) + k = c nd c c k = = [since + = ] ( + + ) = k insed. c Thus we cn wrie he priculr soluion s y p ( = k) =, where + =,. + Exmple: Find he priculr soluion of y + y y =. The Complemenry Soluion: + + To find he complemenry funcion, we mus concenre on he reduced equion: () y + y + y = The soluion procedure involves solving he for he roos of he chrcerisic equion. In his cse he chrcerisic equion is + + = nd i possesses wo chrcerisic roos: 0, ± 4 =. Three possile siuions my e encounered wih regrd o he chrcerisic roos:. Disinc rel roos, i.e.. If he chrcerisic equion hs go wo disinc rel roos, he soluion is: y = A + A, where A nd A re consns c. Repeed rel roos, i.e. = If he chrcerisic equion hs co-inciden roos, follows: y = A + A = ( A + A ) = A c = = =, hen he soluion is s BUT since he wo componens hve collpsed ino single erm, we re shor of consn. We herefore hve o supply he missing componen. We use he rick of muliplying y he vrile. The complemenry funcion for he repeed-roo cse is herefore:

33 y = A + A c 4. Complex roos. If < 4, we hve complex conjuge roos. We won del wih his possiiliy in his course. SYSTEMS OF DIFFERENTIAL & DIFFERENCE EQUATIONS Refer o Ching, Chper 8 So fr we hve solved only snd-lone differenil nd difference equions. However, we my e confroned wih simulneous differenil nd difference equions. When? These rise from se of inercing chnges, e.g. in muli-secor mrkes when chnges in one mrke ffecs condiions in noher. To del wih sysems of equions, we hve o rnsform higher-order equions ino more mngele form (i.e. ll of he firs-order). Suppose we hve: y + y + y = c + + Then we cn le: x = y + x = y + + This gives us: x + x + y = c x + y + = 0 Similrly we cn rnsform n h order differenil equion ino sysem of n firs-order equions. Given he differenil equion y ''( ) + y '( ) + y( ) = 0 we inroduce new vrile x() defined y x( ) = y '( ) [implying x '( ) = y ''( )] Then we cn re-wrie he differenil equion s he following sysem of wo firs-order equions: '( ) ( ) ( ) 0 x + x + y = y '( ) x( ) = 0 We herefore hve o rnsform ny higher order difference/ differenil equion ino sysem of firs-order difference or differenil equions.

34 Solving Sysems of Dynmic Equions: The mehods for solving simulneous differenil equions nd simulneous difference equions re quie similr. We ll only consider liner equions wih consn coefficiens.. Simulneous Difference Equions Suppose we hve x + 6x + 9y = 4 + y + x = 0 Since priculr soluions represen ineremporl equilirium vlues, le us denoe hem y x nd y. As efore we firs ry consn soluions, nmely x + = x = x nd y+ = y = y. This works in he presen cse, ecuse when we susiue hese ino he sysem of equions we ge: 7x + 9y = 4 x + y = 0 x = y = 4 (If hese consn soluions don work hen you hve o ry soluions of he form x = k, y = k ec. For he complemenry funcions, we should using our previous experience dop ril soluions of he form: x = m nd y = n, where n nd m re rirry consns nd he se represens he chrcerisic roo. I is uomiclly implied h x = m nd y = n To simplify mers we re using he sme se for oh vriles, lhough heir coefficiens re llowed o differ. I is our im o find he vlues of, m nd n h cn mke he ril soluions ove sisfy he x + + 6x + 9y = 4 reduced homogenous version of. y x = 0 + Upon susiuing he ril soluions ino he reduced versions of cncelling he common fcor 0, we oin wo equions: x + + 6x + 9y = 4 nd y x = 0 + 4

35 ( + 6) m + 9n = 0 m + n = 0 This cn e considered s liner homogenous-equion sysem in wo vriles m nd n (considering s prmeer for he ime eing). Rule ou he rivil soluion of m = n = 0, y requiring he coefficien mrix of he sysem o e singulr. Th is, we require he deerminn of he mrix o e equl o zero: = + + = ( = = ) = de The ove equion is clled he chrcerisic equion nd is roos he chrcerisic roos, of he given simulneous difference-equion sysem. Given he vlue of, we cn ge he vlues of m nd n from ( + 6) m + 9 n = 0. m + n = 0 Since he sysem is homogenous, n infinie numer of soluions for (m, n) will emerge, expressile in he form of he equion m = kn, where k is consn. In fc, for ech roo i, here will in generl e disinc equion m i =k i n i,. Even wih repeed roos, we should sill use wo such equions m =k n, nd m =k n in he complemenry funcions. Moreover, wih repeed roos, we recll h we cn wrie he complemenry funcions s: x = m ( ) + m ( ) y = n ( ) + n ( ) The fcors of proporionliy eween m i nd n i mus of course sisfy he given equion sysem which mndes h = +, i.e. y n ( ) + n ( + )( ) m ( ) + m ( ) + + Dividing hrough y ( ) : n n ( + ) = m + m or x ( n + n ) n = m + m Equing erms wih on wo sides of he equls sign nd similrly for he erms wihou, we find h 5

36 m = ( n + n ) nd m = n If we now wrie n = A nd n = A hen i follows h: m = ( A + A ) m 4 = A 4 4, Thus he complemenry funcions cn e wrien s: x = ( A + A )( ) A ( ) c 4 4 = A ( ) A ( + )( ) nd 4 y = A ( ) + A ( ) c 4. Simulneous Differenil Equions The soluion o firs-order liner differenil equion is similr. The only mjor modificion is o chnge he ril soluions o: x( ) = me nd y( ) = ne r r r r This implies h x '( ) = rme nd y '( ) = rne. 6