Let s look again at the first order linear differential equation we are attempting to solve, in its standard form:

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1 Th Ingraing Facor Mhod In h prvious xampls of simpl firs ordr ODEs, w found h soluions by algbraically spara h dpndn variabl- and h indpndn variabl- rms, and wri h wo sids of a givn quaion as drivaivs, ach wih rspc o on of h wo variabls. Thn jus ingra boh sids and simplify o find h soluion y. Howvr, his procss was fasibl only bcaus h quaions in qusion wr a spcial yp, namly ha hy wr boh sparabl, in addiion o bing firs ordr linar quaions. Thy do, howvr, illusrad h main goal of solving a firs ordr ODE, namly o us ingraion o rmovd h y -rm. Mos firs ordr linar ordinary diffrnial quaions ar, howvr, no sparabl. So h prvious mhod will no work bcaus w will b unabl o rwri h quaion o qua wo drivaivs. In such insancs, a mor labora chniqu mus b applid. How do w, hn, ingra boh sids? L s look again a h firs ordr linar diffrnial quaion w ar amping o solv, in is sandard form: y + p() y g(). Wha w will do is o muliply h quaion hrough by a suiably chosn funcion µ(), such ha h rsuling quaion µ() y + µ()p() y µ()g() (*) would hav ingra-abl xprssions on boh sids. Such a funcion µ() is calld an ingraing facor. 008, 01 Zachary S Tsng A-1-15

2 Commn: Th ida of ingraing facor is no rally nw. Rcall how you hav ingrad sc(x) in Mah 141. Th ingral as givn could no b ingrad. Howvr, afr h ingrand has bn muliplid by a suiabl from of 1, in his cas (an(x) + sc(x))/ (an(x) + sc(x)), h ingraion could hn procd qui asily. an x+ sc x an x+ sc x sc x an x+ sc sc x+ an x sc x dx sc x dx dx x du u ln u + C ln sc x+ an x + C Now back o h quaion µ() y + µ()p() y µ()g() (*) On h righ sid hr is xplicily a funcion of. So i could always, in hory a las, b ingrad wih rspc o. Th lf hand sid is h mor inrsing par. Tak anohr look of h lf sid of (*) and compar i wih his following xprssion lisd sid-by-sid: µ() y + µ()p() y µ() y + µ () y Th scond xprssion is, by h produc rul of diffrniaion, nohing mor han (µ() y). Noic h similariy bwn h wo xprssions. Suppos h simpl diffrnial quaion µ()p() µ () could b saisfid, w would hn hav µ() y + µ()p() y µ() y + µ () y (µ() y) Trivially, hn, h lf sid of (*) could b ingrad wih rspc o. (µ() y + µ()p() y) d (µ() y) d µ() y 008, 01 Zachary S Tsng A-1-16

3 Hnc, o solv (*) w ingra boh sids: (µ() y + µ()p() y) d µ()g() d µ() y µ()g() d (**) Thrfor, h gnral soluion is found afr w divid h las quaion hrough by h ingraing facor µ(). Bu bfor w can solv for h gnral soluion, w mus ak a sp back and find his (almos magical!) ingraing facor µ(). W hav sn on h las pag ha i mus saisfis h quaion µ()p() µ (). This is a simplr quaion ha can b solvd by our firs mhod of spara h variabls hn ingra: p( ) µ ( ) µ ( ) p() d ln µ() + C p ( ) d ln µ ( ) C p( ) d Cµ ( ) 1 008, 01 Zachary S Tsng A-1-17

4 This is h gnral soluion, of cours. W jus nd on insanc of i. Sinc any nonzro funcion of h abov form can b usd as h ingraing facor, w will jus choos h simpls on, ha of C 1 1. As a rsul p( ) d µ ( ). Onc i is found, w can immdialy divid boh sids of h quaion (**) by µ() o find y(), using h formula y( ) µ ( ) g( ) d + µ ( ) ( C) No: In ordr o us his ingraing facor mhod, h quaion mus b pu ino h sandard form firs (i.. y -rm mus hav cofficin 1). Els our formulas won work. Commn: As i urns ou, wha w hav jus discovrd is a vry powrful ool. As long as w ar abl o ingra h wo rquird ingrals, his ingraing facor mhod can b usd o solv any firs ordr linar ordinary diffrnial quaion. 008, 01 Zachary S Tsng A-1-18

5 Exampl: W will us our nw found gnral purpos mhod o again solv h quaion y r y k, r 0. Th quaion is alrady in is sandard form, wih p() r and g() k. Th ingraing facor is r d r µ ( ). Th gnral soluion is r r k r k r ( k d) + C C 1 y + r r r Tha is i! (I looks slighly diffrn, bu his is indd h sam soluion w found a lil arlir using a diffrn mhod.) 008, 01 Zachary S Tsng A-1-19

6 Exampl: W hav prviously sn h dircion fild showing h approximad graph of h soluions of y y. Now l us apply h ingraing facor mhod o solv i. Th quaion has as is sandard form, y + y. Whr p() 1 and g(). Th ingraing facor is d µ ( ). Th gnral soluion is, hrfor, y 1 ( d) ( d) ( C) C. 008, 01 Zachary S Tsng A-1-0

7 Summary: Solving a firs ordr linar diffrnial quaion y + p() y g() 0. Mak sur h quaion is in h sandard form abov. If h lading cofficin is no 1, divid h quaion hrough by h cofficin of y -rm firs. (Rmmbr o divid h righ-hand sid as wll!) 1. Find h ingraing facor: µ ( ) p( ) d. Find h soluion: y( ) µ ( ) g( ) d + µ ( ) ( C) This is h gnral soluion of h givn quaion. Always rmmbr o includ h consan of ingraion, which is includd in h formula abov as (+ C) a h nd. Lik an indfini ingral (which givs us h soluion in h firs plac), h gnral soluion of a diffrnial quaion is a s of infinily many funcions conaining on or mor arbirary consan(s). 008, 01 Zachary S Tsng A-1-1

8 Iniial Valu Problms (I.V.P.) Evry im w solv a diffrnial quaion, w g a gnral soluion ha is rally a s of infinily many funcions ha ar all soluions of h givn quaion. In pracic, howvr, w ar usually mor inrsd in finding som spcific funcion ha saisfis a givn quaion and also saisfis som addiional bhavioral rquirmn(s), rahr han jus finding an arbirary funcion ha is a soluion. Th bhavioral rquirmns ar usually givn in h form of iniial condiions ha say h spcific soluion (and is drivaivs) mus ak on crain givn valus (h iniial valus) a som prscribd iniial im 0. For a firs ordr quaion, h iniial condiion coms simply as an addiional samn in h form y( 0 ) y 0. Tha is o say, onc w hav found h gnral soluion, w will hn procd o subsiu 0 ino y() and find h consan C in h gnral soluion such ha y( 0 ) y 0. Th rsul, if i could b found, is a spcific funcion (or funcions) ha saisfis boh h givn diffrnial quaion, and h condiion ha h poin ( 0, y 0 ) is conaind on is graph. Such a problm whr boh an quaion and on or mor iniial valus ar givn is calld an iniial valu problm (abbrviad as I.V.P. in h xbook). Th spcific soluion husly found is ofn calld a paricular soluion of h diffrnial quaion. Graphically, h gnral soluion of a firs ordr ordinary diffrnial quaion is rprsnd by h collcion of all ingral curvs in a dircion fild, whil ach paricular soluion is rprsnd individually by on of h ingral curvs. To summariz, an iniial valu problm consiss of wo pars: 1. A diffrnial quaion, and. A s of iniial condiion(s). W firs solv h quaion o find h gnral soluion (which conains on or mor arbirary consans or cofficins). Thn w us h iniial condiion(s) o drmin h xac valu(s) of hos consan(s). Th rsul is a paricular soluion of h quaion. 008, 01 Zachary S Tsng A-1 -

9 008, 01 Zachary S Tsng A-1-3 Exampl: Solv h iniial valu problm y y 3 4, y(1). Firs divid boh sids by. y y 4 p ) (, and g 4 ) (. Th ingraing facor is ln ln ) ( d µ. Th gnral soluion is ( ) ( ) C d d y C + + Apply h iniial condiion y(1) C C 0 + C C Thrfor, y +.

10 Exampl: Solv h iniial valu problm cos() y sin() y 3 cos(), y(π) 0. Divid hrough by cos(): y an() y 3 p() an() and g() 3 Th ingraing facor is ( ) an( ) d µ. (Wha is his funcion?) Us h u-subsiuion, l u cos() hn du sin()d: sin( ) d du an( ) d ln u + C ln cos( + C cos( ) u ) Nar 0 π, cos() is posiiv, so w could drop h absolu valu. an( ) d ln(cos( )) Hnc, µ ( ) cos( ). ( sin( ) sin( ) ) 1 3 y( ) 3 cos( ) d d cos( ) cos( ) 3 cos( ) ( sin( ) + cos( ) + C) 3 an( ) + 3+ C sc( ) y(π) 0 6π an(π) C sc(π) C 3 + C C 3 Thrfor, y() 3 an() + 3 3sc(). 008, 01 Zachary S Tsng A-1-4

11 Th Exisnc and Uniqunss Thorm (of h soluion a firs ordr linar quaion iniial valu problm) Dos an iniial valu problm always a soluion? How many soluions ar hr? Th following horm sas a prcis condiion undr which xacly on soluion would always xis for a givn iniial valu problm. Thorm: If h funcions p and g ar coninuous on h inrval I: α < < β conaining h poin 0, hn hr xiss a uniqu funcion y φ() ha saisfis h diffrnial quaion y + p() y g() for ach in I, and ha also saisfis h iniial condiion y( 0 ) y 0, whr y 0 is an arbirary prscribd iniial valu. Tha is, h horm guarans ha h givn iniial valu problm will always hav (xisnc of) xacly on (uniqunss) soluion, on any inrval conaining 0 as long as boh p() and g() ar coninuous on h sam inrval. Th largs of such inrvals is calld h inrval of validiy of h givn iniial valu problm. In ohr words, h inrval of validiy is h largs inrval such ha (1) i conains 0, and () i dos no conain any disconinuiy of p() nor g(). Convrsly, nihr xisnc nor uniqunss of a soluion is guarand a a disconinuiy of ihr p() or g(). No ha, unlss 0 is acually a disconinuiy of ihr p() or g(), hr always xiss a non-mpy inrval of validiy. If, howvr, 0 is indd a disconinuiy of ihr p() or g(), hn h inrval of validiy will b mpy. Clarly, in such a cas h condiions ha h inrval mus conain 0 and ha i mus no conain a disconinuiy of p() or g() will b conradicing. 008, 01 Zachary S Tsng A-1-5

12 If so, such an iniial valu problm is no guarand o hav a uniqu soluion a all. Exampl: Considr h iniial valu problm solvd arlir cos() y sin() y 3 cos(), y(π) 0. Th sandard form of h quaion is y an() y 3 wih p() an() and g() 3. Whil g() is always coninuous, p() has disconinuiis a ±π/, ±3π/, ±5π/, ±7π/, According o h Exisnc and Uniqunss Thorm, hrfor, a coninuous and diffrniabl soluion of his iniial valu problm is guarand o xis uniquly on any inrval conaining 0 π bu no conaining any of h disconinuiis. Th largs such inrvals is (3π/, 5π/). I is h inrval of validiy of his problm. Indd, h acual soluion y() 3 an() + 3 3sc() is dfind vrywhr wihin his inrval, bu no a ihr of is ndpoins. 008, 01 Zachary S Tsng A-1-6

13 How o find h inrval of validiy For an iniial valu problm of a firs ordr linar quaion, h inrval of validiy, if xiss, can b found using his following simpl procdur. Givn: y + p() y g(), y( 0 ) y Draw h numbr lin (which is h -axis).. Find all h disconinuiis of p(), and h disconinuiis of g(). Mark hm off on h numbr lin. 3. Loca on h numbr lin h iniial im 0. Look for h longs inrval ha conains 0, bu conains no disconinuiis. Sp 1: Draw h -axis. Sp : Mark off h disconinuiis. Sp 3: Loca 0 and drmin h inrval of validiy. 008, 01 Zachary S Tsng A-1-7

14 Exampl: Considr h iniial valu problms (a) ( 81) y y sin(), y(1) 10π (b) ( 81) y y sin(), y(10π) 1 Th quaion is firs ordr linar, so h horm applis. Th sandard form of h quaion is 3 5 sin( ) y + y sin( ) wih p( ) and g ( ). Boh hav disconinuiis a ± Hnc, any inrval such ha a soluion is guarand o xis uniquly mus conain h iniial im 0 bu no conain ihr of h poins 9 and 9. In (a), 0 1, so h inrval conains 1 bu no ± 9. Th largs such inrval is (9, 9). In (b), 0 10π, so h inrval conains 10π bu nihr of ± 9. Th largs such inrval is (9, ). Rmmbr ha h valu of y 0 dos no mar a all, 0 alon drmins h inrval. Suppos h iniial condiion is y(100) 5 insad. Thn h largs inrval on which h iniial valu problm s soluion is guarand o xis uniquly will b (, 9). Lasly, suppos h iniial condiion is y(9) 88. Thn w would no b assurd of a uniqu soluion a all. Sinc 9 is boh 0 and a disconinuiy of p() and g(). Th inrval of validiy would b, hrfor, mpy. 008, 01 Zachary S Tsng A-1-8

15 Dpnding on h problm, h inrval of validiy, if xiss, could b as larg as h nir ral lin, or arbirarily small in lngh. Th following xampl is an iniial valu problm ha has a vry shor inrval of validiy for is uniqu soluion. Exampl: Considr h iniial valu problms ( ) y + y 0, y(0) α. Wih h sandard form y ' y , 10 h disconinuiis (of p()) ar ± Th iniial im is 0 0. Thrfor, h inrval of validiy for is soluion is h inrval ( , ), an inrval of lngh unis! Howvr, h imporan hing is ha somwhr on h -axis a uniqu soluion o his iniial valu problm xiss. Diffrn iniial valu α will giv diffrn paricular soluion. Bu h soluion will ach uniquly xis, a a minimum, on h inrval ( , ). Again, according o h horm, h only im ha a uniqu soluion is no guarand o xis anywhr is whnvr h iniial im 0 jus happns o b a disconinuiy of ihr p() or g(). Now suppos h iniial condiion is y(0) 0. I should b fairly asy o s ha h consan zro funcion y() 0 is a soluion of h iniial valu problm. I is of cours h uniqu soluion of his iniial valu problm. Noic ha his soluion xiss for all valus of, no jus insid h inrval ( , ). I xiss vn a disconinuiis of p(). This illusras ha, whil ousid of h inrval of validiy hr is no guaran ha a soluion would xis or b uniqu, h horm nvrhlss dos no prvn a soluion o xis, vn uniquly, whr h condiion rquird by h horm is no m. 008, 01 Zachary S Tsng A-1-9

16 Nonlinar Equaions: Exisnc and Uniqunss of Soluions A horm analogous o h prvious xiss for gnral firs ordr ODEs. Thorm: L h funcion f and f y b coninuous in som rcangl α < < β, γ < y < δ, conaining h poin ( 0, y 0 ). Thn, in som inrval 0 h < < 0 + h conaind in α < < β, hr is a uniqu soluion y φ() of h iniial valu problm y f (,y), y( 0 ) y 0. This is a mor gnral horm han h prvious ha applis o all firs ordr ODEs. I is also lss prcis. I dos no spcify a prcis rgion ha a givn iniial valu problm would hav a soluion or ha a soluion, whn i xiss, is uniqu. Rahr, i sas a rgion ha somwhr wihin hr has o b par of i in which a uniqu soluion of h iniial valu problm will xis. (I dos no prclud ha a scond soluion xiss ousid of i.) Th boom lin is ha a nonlinar quaion migh hav mulipl soluions corrsponding o h sam iniial condiion. On h ohr hand i is also possibl ha i migh no hav a soluion dfind on pars of h rgion whr f and f y ar boh coninuous. Exampl: Considr h (nonlinar) iniial valu problm y y 1/, y(0) 0. Whn 0, f y is no coninuous. Thrfor, i would no ncssarily 6 hav a uniqu soluion. Indd, boh y and y 0 ar funcions ha 36 saisfy h problm. (Vrify his fac!) 008, 01 Zachary S Tsng A-1-30

17 Exrciss A-1.: 1 4 Find h gnral soluion of ach quaion blow. 1. y y 4. y + 10 y y' y 0 4. y y 5 16 Solv ach iniial valu problm. Wha is h largs inrval in which a uniqu soluion is guarand o xis? 5. y + y, y(0) 6. y 11y 4 6, y(0) 9 7. y y +, y(1) 5 8. ( + 1) y y 3 +, y(0) 4 9. y + ( 6 ) y 0, y(0) y + 4y, y() ( 49) y + 4 y 4, y(0) 1 / 7 1. y y +, y(0) y + y, y(0) y + 4y 4, y() an() y sc() an () y 0, y(0) π 16. ( + 1) y + y 0, y(3) 1 008, 01 Zachary S Tsng A-1-31

18 17 0 Wihou solving h iniial valu problm, wha is h largs inrval in which a uniqu soluion is guarand o xis for ach iniial condiion? (a) y(π) 7, (b) y(1) 9, (c) y(4). 17. ( + 5) y y + ( 8)( 1) 3 y y sc( / 3) + 3 ( 6)( + 1) 19. ( + 4 5) y + an() y (4 ) y + ln(6 ) y 1. Find h gnral soluion of y + y. Thn show ha boh h iniial condiions y(1) 1 and y(1) 3 rsul in an idnical paricular soluion. Dos his fac viola h Exisnc and Uniqunss Thorm? 008, 01 Zachary S Tsng A-1-3

19 Answrs A-1.: 1. y 4+ C 3 / y + + C y C xp 3 4. y + C 5. y + 3, (, ) y, (, ) y + ln + 4, (0, ) 8. y ( + 1)(ln( + 1) 4), (, ) 9. y 8 xp( 3 ), (, ) y 4, (, 0) y, (7, 7) ( 49) 1. y 6 3 3, (, ) y + cosh( ), (, ) 14. y , (, 0). π sc( ) sc( ) y π, (π/, π/) y, (, ) (a) (3, 6); (b) (1, 3); (c) (5, 1). 18. (a) (0, 3π/); (b) (0, 3π/); (c) (3π/, 3). 19. (a) (3π/4, 5π/4); (b) no such inrval xiss; (c) (5, 5π/4). 0. (a) (, 6); (b) (, ); (c) (, ). + C 1 1. y ; hy boh hav y as h soluion; no, diffrn iniial condiions could nvrhlss giv h sam uniqu soluion. 008, 01 Zachary S Tsng A-1-33

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