Prfac Hr ar m onlin nots for m diffrntial quations cours that I tach hr at Lamar Univrsit. Dspit th fact that ths ar m class nots, th should b accssibl to anon wanting to larn how to solv diffrntial quations or nding a rfrshr on diffrntial quations. I v trid to mak ths nots as slf containd as possibl and so all th information ndd to rad through thm is ithr from a Calculus or Algbra class or containd in othr sctions of th nots. A coupl of warnings to m studnts who ma b hr to gt a cop of what happnd on a da that ou missd. 1. Bcaus I wantd to mak this a fairl complt st of nots for anon wanting to larn diffrntial quations I hav includd som matrial that I do not usuall hav tim to covr in class and bcaus this changs from smstr to smstr it is not notd hr. ou will nd to find on of our fllow class mats to s if thr is somthing in ths nots that wasn t covrd in class.. In gnral I tr to work problms in class that ar diffrnt from m nots. Howvr, with Diffrntial Equation man of th problms ar difficult to mak up on th spur of th momnt and so in this class m class work will follow ths nots fairl clos as far as workd problms go. With that bing said I will, on occasion, work problms off th top of m had whn I can to provid mor ampls than just thos in m nots. Also, I oftn don t hav tim in class to work all of th problms in th nots and so ou will find that som sctions contain problms that wrn t workd in class du to tim rstrictions. 3. Somtims qustions in class will lad down paths that ar not covrd hr. I tr to anticipat as man of th qustions as possibl in writing ths up, but th ralit is that I can t anticipat all th qustions. Somtims a vr good qustion gts askd in class that lads to insights that I v not includd hr. ou should alwas talk to somon who was in class on th da ou missd and compar ths nots to thir nots and s what th diffrncs ar. 4. This is somwhat rlatd to th prvious thr itms, but is important nough to mrit its own itm. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using ths nots as a substitut for class is liabl to gt ou in troubl. As alrad notd not vrthing in ths nots is covrd in class and oftn matrial or insights not in ths nots is covrd in class. 7 Paul Dawkins 1 http://tutorial.math.lamar.du/trms.asp
Eact Diffrntial Equations Th nt tp of first ordr diffrntial quations that w ll b looking at is act diffrntial quations. Bfor w gt into th full dtails bhind solving act diffrntial quations it s probabl bst to work an ampl that will hlp to show us just what an act diffrntial quation is. It will also show som of th bhind th scns dtails that w usuall don t bothr with in th solution procss. Th vast majorit of th following ampl will not b don in an of th rmaining ampls and th work that w will put into th rmaining ampls will not b shown in this ampl. Th whol point bhind this ampl is to show ou just what an act diffrntial quation is, how w us this fact to arriv at a solution and wh th procss works as it dos. Th majorit of th actual solution dtails will b shown in a latr ampl. Eampl 1 Solv th following diffrntial quation. - 9 + ( + + 1) d d Solution Lt s start off b supposing that somwhr out thr in th world is a function (,) that w can find. For this ampl th function that w nd is, + + 1-3 3 Do not worr at this point about whr this function cam from and how w found it. Finding th function, (,), that is ndd for an particular diffrntial quation is whr th vast majorit of th work for ths problms lis. As statd arlir howvr, th point of this ampl is to show ou wh th solution procss works rathr than showing ou th actual solution procss. W will s how to find this function in th nt ampl, so at this point do not worr about how to find it, simpl accpt that it can b found and that w v don that for this particular diffrntial quation. Now, tak som partial drivativs of th function. -9 + + 1 Now, compar ths partial drivativs to th diffrntial quation and ou ll notic that with ths w can now writ th diffrntial quation as. d + (1) d Now, rcall from our multi-variabl calculus class (probabl Calculus III) that (1) is nothing mor than th following drivativ (ou ll nd th multi-variabl chain rul for this ). d ( (, ( ) )) d So, th diffrntial quation can now b writtn as 7 Paul Dawkins ii http://tutorial.math.lamar.du/trms.asp
d ( (, ( ) )) d Now, if th ordinar (not partial ) drivativ of somthing is zro, that somthing must hav bn a, c. Or, constant to start with. In othr words, w v got to hav 3 + + - c 1 3 This thn is an implicit solution for our diffrntial quation! If w had an initial condition w could solv for c. W could also find an plicit solution if w wantd to, but w ll hold off on that until th nt ampl. Oka, so what did w larn from th last ampl? Lt s look at things a littl mor gnrall. Suppos that w hav th following diffrntial quation. d M (, ) + N(, ) () d Not that it s important that it b in this form! Thr must b an on on sid and th sign sparating th two trms must b a +. Now, if thr is a function somwhr out thr in th world, (,), so that, M, and N, thn w call th diffrntial quation act. In ths cass w can writ th diffrntial quation as d + (3) d Thn using th chain rul from Calculus III w can furthr rduc th diffrntial quation to th following drivativ, d ( (, ( ) )) d Th (implicit) solution to an act diffrntial quation is thn, c (4) Wll, it s th solution providd w can find (,) anwa. Thrfor, onc w hav th function w can alwas just jump straight to (4) to gt an implicit solution to our diffrntial quation. Finding th function (,) is clarl th cntral task in dtrmining if a diffrntial quation is act and in finding it's solution. As w will s, finding (,) can b a somwhat lngth procss in which thr is th chanc of mistaks. Thrfor, it would b nic if thr was som simpl tst that w could us bfor vn starting to s if a diffrntial quation is act or not. This will b spciall usful if it turns out that th diffrntial quation is not act, sinc in this cas (,) will not ist. It would b a wast of tim to tr and find a nonistnt function! So, lt's s if w can find a tst for act diffrntial quations. Lt's start with () and assum that th diffrntial quation is in fact act. Sinc its act w know that somwhr out thr is a function (,) that satisfis 7 Paul Dawkins 3 http://tutorial.math.lamar.du/trms.asp
M N Now, providd (,) is continuous and it s first ordr drivativs ar also continuous w know that Howvr, w also hav th following. M M ( ) ( N) N Thrfor, if a diffrntial quation is act and (,) mts all of its continuit conditions w must hav. M N (5) Likwis if (5) is not tru thr is no wa for th diffrntial quation to b act. Thrfor, w will us (5) as a tst for act diffrntial quations. If (5) is tru w will assum that th diffrntial quation is act and that (,) mts all of its continuit conditions and procd with finding it. Not that for all th ampls hr th continuit conditions will b mt and so this won t b an issu. Oka, lt s go back and rwork th first ampl. This tim w will us th ampl to show how to find (,). W ll also add in an initial condition to th problm. Eampl Solv th following IVP and find th intrval of validit for th solution. - 9 + ( + + 1) d, ( ) - 3 d Solution First idntif M and N and chck that th diffrntial quation is act. M - 9 M N + + N 1 So, th diffrntial quation is act according to th tst. Howvr, w alrad knw that as w hav givn ou (,). It s not a bad thing to vrif it howvr and to run through th tst at last onc howvr. Now, how do w actuall find (,)? Wll rcall that M N W can us ithr of ths to gt a start on finding (,) b intgrating as follows. Md OR Nd Ú Ú 7 Paul Dawkins 4 http://tutorial.math.lamar.du/trms.asp
Howvr, w will nd to b carful as this won t giv us th act function that w nd. Oftn it dosn t mattr which on ou choos to work with whil in othr problms on will b significantl asir than th othr. In this cas it dosn t mattr which on w us as ithr will b just as as. So, I ll us th first on. (, ) 9 d 3 3 h( ) Ú - - + Not that in this cas th constant of intgration is not rall a constant at all, but instad it will b a function of th rmaining variabl(s), in this cas. Rcall that in intgration w ar asking what function w diffrntiatd to gt th function w ar intgrating. Sinc w ar working with two variabls hr and talking about partial diffrntiation with rspct to, this mans that an trm that containd onl constants or s would hav diffrntiatd awa to zro, thrfor w nd to acknowldg that fact b adding on a function of instad of th standard c. Oka, w v got most of (,) w just nd to dtrmin h() and w ll b don. This is actuall as to do. W usd M to find most of (,) so w ll us N to find h(). Diffrntiat our (,) with rspct to and st this qual to N (sinc th must b qual aftr all). Don t forgt to diffrntiat h()! Doing this givs, + h ( ) + + 1 N From this w can s that h ( ) + 1 Not that at this stag h() must b onl a function of and so if thr ar an s in th quation at this stag w hav mad a mistak somwhr and it s tim to go look for it. W can now find h() b intgrating. h + 1d + + k Ú ou ll not that w includd th constant of intgration, k, hr. It will turn out howvr that this will nd up gtting absorbd into anothr constant so w can drop it in gnral. So, w can now writ down (,). 3 3, - 3 + + + k + + 1-3 + k With th cption of th k this is idntical to th function that w usd in th first ampl. W can now go straight to th implicit solution using (4). 3 + + 1-3 + k c W ll now tak car of th k. Sinc both k and c ar unknown constants all w nd to do is subtract on from both sids and combin and w still hav an unknown constant. 3 + + 1-3 c-k 3 + + 1-3 c 7 Paul Dawkins 5 http://tutorial.math.lamar.du/trms.asp
Thrfor, w ll not includ th k in anmor problms. This is whr w lft off in th first ampl. Lt s now appl th initial condition to find c. 3-3 + + 1-3 - 3 c fi c 6 Th implicit solution is thn. 3 + + 1-3 - 6 Now, as w saw in th sparabl diffrntial quation sction, this is quadratic in and so w can solv for () b using th quadratic formula. ( 1) ( 1) 41 ()( 3 3 6) - + ± + - - - 1 () 1 1 5 4 3 - + ± + + + Now, rappl th initial condition to figur out which of th two signs in th ± that w nd. - 1± 5-1± 5-3 ( ) - 3, So, it looks lik th - is th on that w nd. Th plicit solution is thn. - + - 4 + 3 + + 1 1 5 Now, for th intrval of validit. It looks lik w might wll hav problms with squar roots of ngativ numbrs. So, w nd to solv 4 3 + 1 + + 5 Upon solving this quation is zro at 11.8155764 and 1.396911133. Not that ou ll nd to us som form of computational aid in solving this quation. Hr is a graph of th polnomial undr th radical. 7 Paul Dawkins 6 http://tutorial.math.lamar.du/trms.asp
So, it looks lik thr ar two intrvals whr th polnomial will b positiv. - < -11.8155764-1.396911133 < Howvr, rcall that intrvals of validit nd to b continuous intrvals and contain th valu of that is usd in th initial condition. Thrfor th intrval of validit must b. - 1.396911133 < Hr is a quick graph of th solution. That was a long ampl, but mostl bcaus of th initial planation of how to find (,). Th rmaining ampls will not b as long. Eampl 3 Find th solution and intrval of validit for th following IVP. + 4 3- - 1 8 Solution Hr, w first nd to put th diffrntial quation into propr form bfor procding. Rcall that it nds to b and th sign sparating th two trms must b a plus! + 4-3- So w hav th following + 4+ - 3 7 Paul Dawkins 7 http://tutorial.math.lamar.du/trms.asp
M + M 4 4 N - N 6 4 and so th diffrntial quation is act. W can ithr intgrat M with rspct to or intgrat N with rspct to. In this cas ithr would b just as as so w ll intgrat N this tim so w can sa that w v got an ampl of both down hr. (, ) 6d 6 h( ) Ú - - + This tim, as opposd to th prvious ampl, our constant of intgration must b a function of sinc w intgratd with rspct to. Now diffrntiat with rspct to and compar this to M. + h ( ) + 4 M So, it looks lik h 4 fi h 4 Again, w ll drop th constant of intgration that tchnicall should b prsnt in h() sinc it will just gt absorbd into th constant w pick up in th nt stp. Also not that, h() should onl involv s at this point. If thr ar an s lft at this point a mistak has bn mad so go back and look for it. Writing vrthing down givs us th following for (,)., - 6+ 4 So, th implicit solution to th diffrntial quation is - 6+ 4 c Appling th initial condition givs, 64-48- 4 c c 1 Th solution is thn - 6 + 4-1 Using th quadratic formula givs us ( ) 6± 36-4 4-1 6± 36+ 48-16 3 6± 9+ 1-4 3 3± 9+ 1-4 3 Rappling th initial condition shows that this tim w nd th + (w ll lav thos dtails to ou to chck). Thrfor, th plicit solution is 7 Paul Dawkins 8 http://tutorial.math.lamar.du/trms.asp
3+ 9+ 1-4 3 Now lt s find th intrval of validit. W ll nd to avoid so w don t gt division b zro. W ll also hav to watch out for squar roots of ngativ numbrs so solv th following quation. 3-4 + 1 + 9 Th onl ral solution hr is 3.17361577. Blow is a graph of th polnomial. So, it looks lik th polnomial will b positiv, and hnc oka undr th squar root on - < < 3.17361577 Now, this intrval can t b th intrval of validit bcaus it contains and w nd to avoid that point. Thrfor, this intrval actuall braks up into two diffrnt possibl intrvals of validit. - < < < < 3.17361577 Th first on contains -1, th valu from th initial condition. Thrfor, th intrval of validit for this problm is - < <. Hr is a graph of th solution. 7 Paul Dawkins 9 http://tutorial.math.lamar.du/trms.asp
Eampl 4 Find th solution and intrval of validit to th following IVP. t -t-( - ln( t + 1 )) ( 5) t + 1 Solution So, first dal with that minus sign sparating th two trms. t - t+ ( ln( t + 1 ) - ) t + 1 Now, find M and N and chck that it s act. t t M - t M t + 1 t + 1 t N ln( t + 1) - Nt t + 1 So, it s act. W ll intgrat th first on in this cas. t t, Û Ù - tdt ln t + 1 - t + h ı t + 1 Diffrntiat with rspct to and compar to N. ln t + 1 + h ln t + 1 - N So, it looks lik w v got. This givs us Th implicit solution is thn, h - fi h - + - - t, ln t 1 t ln t + 1 -t - c 7 Paul Dawkins 1 http://tutorial.math.lamar.du/trms.asp
Appling th initial condition givs, - 5 c Th implicit solution is now, ( ) t t ln + 1 - - - 5 This solution is much asir to solv than th prvious ons. No quadratic formula is ndd this tim, all w nd to do is solv for. Hr s what w gt for an plicit solution. t -5 () t ln t + 1 - Alright, lt s gt th intrval of validit. Th trm in th logarithm is alwas positiv so w don t nd to worr about ngativ numbrs in that. W do nd to worr about division b zro howvr. W will nd to avoid th following point(s). ( t ) ( t ) ln + 1 - ln + 1 W now hav thr possibl intrvals of validit. t + 1 t ± - - < <- - - - < < - t 1 t 1-1 < t < 1 1 Th last on contains t 5 and so is th intrval of validit for this problm is graph of th solution. - 1 < t <. Hr s a Eampl 5 Find th solution and intrval of validit for th following IVP. 7 Paul Dawkins 11 http://tutorial.math.lamar.du/trms.asp
3 3 3 3 3-1+ + 3 1 Solution Lt s idntif M and N and chck that it s act. M 3-1 M 9 + 9 3 3 3 3 3 N + 3 N 9 + 9 3 3 3 3 3 So, it s act. With th propr simplification intgrating th scond on isn t too bad. Howvr, th first is alrad st up for as intgration so lt s do that on. 3 3 3 (, ) 3 Ú - 1d - + h( ) Diffrntiat with rspct to and compar to N. 3 3 3 3 + 3 + h ( ) + 3 N So, it looks lik w v got h fi h Rcall that actuall h() k, but w drop th k bcaus it will gt absorbd in th nt stp. That givs us h(). Thrfor, w gt., - 3 Th implicit solution is thn Appl th initial condition. Th implicit solution is thn 3 - c 3 1 c - 1 This is as far as w can go. Thr is no wa to solv this for and gt an plicit solution. 7 Paul Dawkins 1 http://tutorial.math.lamar.du/trms.asp