First order differential equation Linear equation; Method of integrating factors

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1 First orr iffrntial quation Linar quation; Mtho of intgrating factors Exampl 1: Rwrit th lft han si as th rivativ of th prouct of y an som function by prouct rul irctly. Solving th first orr iffrntial quation From calculus for iffrntiating a prouct, w hav (4 + t ) y + ty = 4t. (0.1) (4 + t ) y + ty = ((4 + t )y); it follows that th quation (0.1) can b rwrit as ((4 + t )y) = 4t. (0.) Thus, vn though y is known, w can intgrat both sis of quation (0.) with rspct to t, thrby obtaining (4 + t )y = t + c, whr c is an arbitrary constant of intgration. Solving for y, w fin that y = t 4 + t + c 4 + t. This is th gnral solution of quation (0.1). Unfortunatly, most first orr linar iffrntial quation cannot b solv as illustrat in Exampl 1 bcaus thir lft han si ar not th rivativ of th prouct of y an som othr function. Howvr, Libniz iscovr that if th iffrntial quation is multipli by a crtain function µ(t), thn th quation is convrt into on that is immiatly intgrabl by using th prouct rul of rivativs, just as Exampl 1. Th function µ(t) is call an intgrating factor an our main task in this sction is to trmin how to fin it for a givn quation. Exampl : Multipli µ(t) an thn appli prouct rul. y + 1 y = 1 t 3. (0.3) Th first stp is to multiply quation (0.3) by a function µ(t), as yt untrmin; thus µ(t) y + 1 µ(t)y = 1 µ(t)t/3. (0.4) Th qustion now is whthr w can choos µ(t) such that th lft han si of abov quation is th rivativ of th prouct µ(t)y. For any iffrntial function µ(t) w hav (µ(t)y) = µ(t)y + µ(t) y. (0.5) Thus th lft han si of quation (0.4) an th right han si of quation (0.5) ar intical, provi w choos µ(t) to satisfy µ(t) = 1 µ(t). (0.6) 1

2 Our sarch for an intgrating factor will b succssful if w can fin a solution of quation (0.6). Rform quation (0.6) to µ µ = 1, intgrat on both si, w can gt that which givs log µ(t) = 1 t + c, µ(t) = c t/. (0.7) Th function µ(t) givn by (0.7) is an intgrating factor for quation (0.3). Sinc w o not n th most gnral intgrating factor, w will choos c to b 1 in quation (0.7) an us µ(t) = t/. Now w rturn to quation (0.3), multiply it by th intgrating factor t/, an obtain t/ y + 1 t/ y = 1 5t/6. (0.8) By th choic w hav ma of th intgrating factor, th lft han si of quation (0.8) is th rivativ of t/ y, so that quation (0.8) bcoms (t/ y) = 1 5t/6. By intgrating both sis of th abov quation, w obtain t/ y = 3 5 5t/6 + c, solving for y, w gt th gnral solution of quation (0.3), namly, y = 3 5 t/3 + c t/. Lt us xtn th mtho of intgrating factors to quations of th form y + ay = g(t). (0.9) Whr a is a givn constant an g(t) is a givn function. Procing as in Exampl, w fin that th intgrating factor µ(t) must satisfy µ = aµ. Thus th intgrating factor is µ(t) = at. Multiply th quation (0.9) by µ(t), w obtain at y + aat y = at g(t), or (at y) = at g(t). Intgrating on both sis of abov quation, w fin that at y = at g(t) + c,

3 whr c is an arbitrary constant. For many simpl function g(t), w can valuat th intgral in th abov quation an xprss th solution y in trms of lmntary functions, as in Exampl. Howvr, for mor complicat function g(t), it is ncssary to lav th solution in intgral form. In this cas, y = at t t 0 as g(s)s + c at, not that in quation abov w hav us s to not th intgration variabl to istinguish it from th inpnnt variabl t an w hav chosn convnint valu t 0 as th lowr limit if intgration. Th choic of t 0 trmins th spcific valu of th constant c but os not chang th solution. For xampl, plugging t = t 0 into th gnral solution formula shows that c = y(t 0 ) at0. Exampl 3: Fin th gnral solution of th iffrntial quation y y = 4 t (0.10) an plot th graphs of svral solutions. Discuss th bhavior of solutions as t. Equation (0.10) if of th form with a = ; thrfor th intgrating factor is µ(t) = t. Multiply th iffrntial quation (0.10) by µ(t), w obtian t y t y = 4 t t t, or ( t y) = 4 t t t. (0.11) Thn, by intgrating both sis of this quation, w hav t y = t + 1 t t t + c, whr w hav us intgration by parts on th last trm in quation. Thus th gnral solution of quation (0.10) is y = t + ct. (0.1) Th bhavior of th solution for larg valus of t is trmin by th trm c t. If c 0, thn th solution grows xponntially larg in magnitu with th sam sign as c itslf. Thus th solutions ivrg as t bcoms larg. Th bounary btwn solutions that ultimatly grow positivly an thos that grow ngativly occurs whn c = 0. If c = 0 an thn st t = 0, w gt that y = 7 4 is th sparation point on th y axis. Not that for this initial valu, th solution is y = t; it grows positivly, but linarly rathr xponntially. Th gnral first-orr linar iffrntial quation. y + p(t)y = g(t), whr p, g ar givn functions. To trmin an appropriat factor, w multiply quation by an as yt untrmin function µ(t), obtaining µ(t) y + p(t)µ(t)y = µ(t)g(t). (0.13) Follow th sam lin of vlopmnt as in Exampl, w s that h lft han si of quation (0.13) is th rivativ of th prouct µ(t)y, provi that µ(t) satisfy th quation µ(t) = p(t)µ(t). (0.14) 3

4 If w assum tmporarily that µ(t) is positiv, thn w hav 1 µ(t) = p(t) µ(t) an consquntly log µ(t) = p(t) + c. By choosing th arbitrary constant c to b zro, w obtain th simplst possibl function for µ, namly, µ(t) = xp( p(t)). Not that µ(t) is positiv for all t, as w assum. Rturning to quation (0.13), w hav Hnc (µ(t)y) = µ(t)y. (0.15) µ(t)y = µ(t)g(t) + c, (0.16) whr c is an arbitrary constant. Somtims th intgral in quation (0.16) can b valuat in trms of lmntary functions. Howvr, in gnral this is not possibl, so th gnral solution of quation is y = 1 t µ(t) ( µ(s)g(s)s + c), (0.17) t 0 whr again t 0 is som convnint lowr limit of intgration. Obsrv that quation (0.17) involvs two intgrations, on to obtain µ(t) an th othr to trmin y from quation (0.17). Exampl 4: Solv th initial valu problm { ty + y = 4t, y(1) =. (0.18) In orr to trmin p(t) an g(t) corrctly, w must first rwrit quation in th stanar form. Thus w hav y + y = 4t, (0.19) t so p(t) = /t an g(t) = 4t. To solv quation (0.19), w first comput th intgrating factor µ(t) µ(t) = xp( t ) = log t = t. On multiplying quation (0.19) by µ(t) = t, w obtain t y + ty = (t y) = 4t 3, an thrfor t y = 4t 3 = t 4 + c, whr c is an arbitrary constant. It follows that, for t > 0, is th gnral solution of quation from (0.18). y = t + c t (0.0) 4

5 To satisfy initial conition, st t = 1 an y = in quation (0.1), so c = 1; thus is th solution of th initial valu problm (0.18). It is important to not that whil th function y = t + 1 t, t > 0 (0.1) y = t + 1 t, t < 0 is part of th gnral solution of quation, it is not part of th solution of this initial valu problm. This is th first xampl in which th solution fails to xist for som valus of t. Again, this is u to th infinit iscontinuity in p(t) at t = 0, which rstrict th solution to th intrval 0 < t <. Draw th pictur for y = t + c t for iffrnt valu c. Exampl 5: Solv th initial valu problm y + ty =, (0.) y(0) = 1. (0.3) To convrt th iffrntial quation (0.) to th stanar form, w hav y + t y = 1. (0.4) Thus p(t) = t an g(t) = 1, th intgrating factor is µ(t) = xp(t /4). Thn multiply quation (0.4) by µ(t), so that t /4 y + t t /4 y = t /4. (0.5) hnc Intgrating on both sis, w obtain (t /4 y) = t /4. t /4 y = t /4 + c. (0.6) Th intgral on th right han si cannot b valuat by lmntary function, so w lav th intgral unvaluat. By choosing th lowr limit of th intgration as th initial point t = 0, w can rplac th quation (0.6) by t /4 y = t 0 s /4 s + c. (0.7) whr c is an arbitrary constant. It thn follows that th gnral solution of quation (0.) is givn by y = t /4 t By th initial conition, w can trmin that c = 1. 0 s /4 s + c t /4. (0.8) 5

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