First Order Linear Differential Equations

Transcription

1 Firs Ordr Linar Diffrnial Equaions A firs ordr ordinary diffrnial quaion is linar if i can b wrin in h form y p() y g() whr p and g ar arbirary funcions of. This is calld h sandard or canonical form of h firs ordr linar quaion. W ll sar by amping o solv a coupl of vry simpl quaions of such yp. 008, 01 Zachary S Tsng A-1-9

2 Exampl: Find h gnral soluion of h quaion y y 0. Firs l s rwri h quaion as dy d y. Thn, assuming y 0, divid boh sids by y: Muliply boh sids by d: 1 dy y d dy y d Now wha w hav hr ar wo drivaivs which ar qual. I implis (as a consqunc of h Man Valu Thorm) ha h anidrivaivs of h wo sids mus diffr only by a consan of ingraion. Ingra boh sids: ln y C or, Whr C 1 C y ( C ) C C 1 is an arbirary, bu always posiiv consan. To simplify on sp farhr, w can drop h absolu valu sign and rlax h rsricion on C 1. C 1 can now b any posiiv or ngaiv (bu no zro) consan. Hnc y() C 1, C 1 0. (1) 008, 01 Zachary S Tsng A-1-10

3 Lasly, wha happns if our arilr assumpion ha y 0 is fals? Wll, if y 0 (ha is, whn y is h consan funcion zro), hn y 0 and h quaion is rducd o which is an xprssion ha is always ru. Thrfor, h consan zro funcion is also a soluion of h givn quaion. No xacly by a coincidn, i corrsponds o h missing cas of C 1 0 in (1). As a rsul, h gnral soluion is in h form y() C, for any consan C. Tha is, any funcion of his form, rgardlss of h valu of C, will saisfy h quaion y y 0. Whil hr ar infinily many such funcions, no ohr yp of funcions could saisfy h quaion. Th similar chniqu could also b usd o solv his nx xampl. 008, 01 Zachary S Tsng A-1-11

4 Exampl: For arbirary consans r and k, r 0, solv h quaion y r y k. W will procd as bfor o rwri h quaion ino qualiy of wo drivaivs. Thn ingra boh sids. Assuming r y k 0: dy d ry k dy ry k d dy ry k d Thrfor, 1 ln r ry k C Simplifying: ln ry k r C 1 ry k r C 1 ry k C r 1 C1, whr is an arbirary posiiv consan. Dropping h absolu valu sign: r C1 k C, C ± is any nonzro consan. ry 1 r Tha is, y ( C k) r C r r k r. 008, 01 Zachary S Tsng A-1-1

5 Lasly, i can b asily chckd ha if r y k 0, implying ha y is k h consan funcion, h givn diffrnial quaion is again r saisfid. This consan soluion corrsponds o h abov gnral soluion for h cas C 0. Hnc, h gnral soluion now includs all possibl valus of h unknown arbirary consan: C r k y, C is any consan. r r 008, 01 Zachary S Tsng A-1-13

6 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-1-14

7 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-1-15

8 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-1-16

9 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-1-17

10 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-1-18

11 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) 1 C. 008, 01 Zachary S Tsng A-1-19

12 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-1-0

13 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 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-1-1

14 008, 01 Zachary S Tsng A-1 - 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) 1 1 C 1 C 0 C C Thrfor, y.

15 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) 3 C sc(π) 0 3 C 3 C C 3 Thrfor, y() 3 an() 3 3sc(). 008, 01 Zachary S Tsng A-1-3