TOPI Diffrntial quations Mthods of thir intgration oncption of diffrntial quations An quation which spcifis a rlationship btwn a function, its argumnt and its drivativs of th first, scond, tc ordr is calld a diffrntial quation Thus a diffrntial quation could b of th form ( ) ( ) f (, ) d lassification of diffrntial quations: ) ordinar - partial A diffrntial quation is calld ordinar if th unknown function and its drivativs dpnd onl on on indpndnt variabl In a partial diffrntial quation th unknown function and its drivativs dpnd on at last two indpndnt variabls Eampl: Th tim-dpndnt Schrödingr quation, h i ( V ( ) E) ( ) 0 m t is a partial diffrntial quation as it contains two indpndnt variabls and t ) ordr Th ordr of a diffrntial quation is th highst n th drivativ prsnt in th diffrntial quation d A first ordr diffrntial quation is an quation that involvs,, and If d d quation it is calld a scond ordr diffrntial quation also appars in th d So th quation 4 is a first-ordr diffrntial quation, whras th quation 4 is a scondordr diffrntial quation d d Eampl: Th quation dscribing th harmonic oscillator, d 0 Asin( t) is a scond-ordr diffrntial quation as th highst drivativ is of scond ordr ) homognous - inhomognous A diffrntial quation is calld homognous if vr trm in it dpnds on th unknown function or its drivativs It is inhomognous if thr is at last on trm which dpnds onl on th indpndnt variabls or is a constant diffrnt than zro Eampl: Th diffrntial quation of a drivn harmonic oscillator, d 0 0 is inhomognous as th trm on th right onl dpnds on t 4) linar - nonlinar A linar diffrntial quation contains onl linar trms of th unknown function and its drivativs Eampl: Th diffrntial quation dscribing a oscillating pndulum, d g sin( ) 0 l is a common ampl of nonlinar quation Evr linar ordinar diffrntial quation is of dgr on, but th convrs is not tru An nonlinar ordinar diffrntial quation (ODE) can b approimatd b a linar ODE through a procss calld linarization ) dgr Th dgr of a diffrntial quation is th highst powr of th highst ordr drivativ Eampls 0 7 0 is an quation of dgr is an quation of dgr
First-ordr diffrntial quations Tp of quation Mthod of solution f () Dirct intgration d f ( ) g( ) Sparating th variabls d Homognous Substitut =v Linar quations Intgrating factor Th main problm of th thor of diffrntial quations is to find th unknown function which, substitutd into th diffrntial quation, turns it into an idntit Such a function is calld th solution or intgral of th diffrntial quation It is possibl that a diffrntial quation has no solution Mor usual is that th diffrntial quation has infinitl man solutions Evn th simplst kind of first ordr diffrntial quation has usuall an infinit manifold of solution Dirct intgration It is th appropriat mthod if ou hav a diffrntial quation lik: f () d whr f() is an function of (including th possibilitis that f()=constant or f()=0) Th function f is continuous on an opn intrval I With som simpl algbraic manipulations, th diffrntial quation solvd - intgrat both sids of th quation To solv a diffrntial quation mans to find a continuous function of th indpndnt variabl = F() that, along with its drivativs, satisfis th quation f () d Rmmbr that all th antidrivativs of f() form a famil of functions, which diffr from ach othr b a constant This famil is just th indfinit intgral of f, f ( ) d F( ) whr is an constant which is not dfind b th diffrntial quation This is th sourc of th non-uniqunss of th solution of diffrntial quation: w can giv th intgration constant an valu, and ach tim w spcif a diffrnt valu of w ar also slcting a diffrnt solution of th diffrntial quation Th famil of functions (solutions) is th gnral solution of th diffrntial quation, which dpnd on th constant of th initial condition Eampls d ( t) t ( t) t ( t) t; ( t) t d d d So ()= + sin d sin d sin d So ()=-cos()+ 4 d d d So ()= + cos ( cos ) d d ( cos ) d d d So ()=- - +0 +sin()+ cos d
Boundar conditions You ma rcall that w wr looking at th quation 4 and w solvd this to gt th solution ()=4+ d At th tim thr wr infinitl mor solutions to that quation Hr's on of thos othr solutions: ()=4+ Hr's anothr: ()=4+ And so on Ths ar all solutions bcaus if w diffrntiat ()=4+c, whr c is an constant, w find that 4, d sinc th constant disappars whn ou diffrntiat it This is probabl familiar to ou from intgration, whnvr w intgratd a function w alwas had to add on a "constant of intgration", usuall calld This mans that whnvr w solv a diffrntial quation (i w intgrat it to find th function that satisfis it) w can alwas add a constant on to th solution So if w want to nd up with a uniqu solution to a diffrntial quation, w nd anothr pic of information as wll as th quation it satisfis, in ordr to rmin th constant For ampl, if w wr askd for th solution of th quation 4 that satisfis (0)=0, thn thr would d b onl on possibl answr: ()=4 If w tr an of th othr solutions to th quation, ma b ()=4+ or ()=4+, sa, w find that (0) is not zro, but or rspctivl Such a condition, which is givn in conjunction with a diffrntial quation to fi th constant of intgration and so giv a uniqu solution, is calld a boundar condition Sparabl quations It is th appropriat mthod if ou hav a diffrntial quation lik: g( t) h( ) first ordr diffrntial quation with sparabl variabls In an quation with sparabl variabls, is a product of a function of t and a function of In particular, g(t) is a diffrntial quation with sparabl variabls in which h() is th constant Similarl, h() is a diffrntial quation with sparabl variabls in which g(t) is th constant An quation with sparabl variabls can b solvd b sparating th variabls and intgrating both sids of th quation In an quation of this form th variabls ar said to b sparabl W simpl sparat th and trms (dividing b h() and multipling b d) and intgrat, w gt ( ) ( ) f d h Man tchniqus to solv ordr diffrntial quations hav as thir guiding principl som stratg of rducing th quations to a sparabl form Eampl d Intgrating both sids givs ln ( t) t, whr is a constant of intgration Nt, w can solv for (t) b ponntiation both sids ln ( t) t, ( t) t Finall, w can gt rid of th absolut valu sign b rplacing which can b positiv or ngativ So our solution can b writtn as t ( t) Initial conditions t, which is alwas positiv, b a constant,
t d Th solution ( t) satisfis quation for an valu of Plugging t=0 into our solution tlls us that ( 0) That is, th valu of is qual to th valu of (0) For ampl, if w wantd th solution of d quation that satisfid (0)=, th answr would b t ( t) An quation of th form (0)=K, whr K is a constant is calld an initial condition for a diffrntial quation For sparabl diffrntial quations, it should b clar that th solution procss alwas introducs a constant of intgration An initial condition is ndd to rmin th valu of this constant Eampl ( )( ) d To solv this quation, th first stp as bfor is to mov th -trm awa from th right-hand sid on to th lft: ( ) d Th scond stp is to intgrat both sids: ( ) d Th intgrals ar of standard form and w can do thm, to obtain th rquird rlation btwn and Don't forgt th constant of intgration ln( ) ln( ) Eampl d Somtims w nd to rarrang th right-hand sid to gt it into th form of a function of tims a function of, as in this cas Hr w can tak a outsid a brackt on th top, so that th two variabls can b sparatd: ( ) d Thn to solv this quation, w again mov th -trm awa from th right-hand sid on to th lft: ( ) d As bfor, th nt stp is to intgrat both sids: ( ) d W nd to split up th intgral on th lft-hand sid bfor w can do it: and thn it simplifis into a form w can do: So w obtain th rquird rlation btwn and Don't forgt th constant of intgration ln Eampl 4
Eponntial growth or dca Lt a b a constant Th ponntial growth or dca quation dscribs a situation in which a variabl grows or shrinks at a rat proportional to th amount prsnt a d Sparat: a, ad d Intgrat and solv for : a ln a,, 0 a a W v rplacd with 0 If a>0, thn incrass as incrass: ponntial growth If a<0, thn dcrass as dcrass: ponntial dca Eampl d Sparat: intgrat: 0 d d 0; ; d ; d ; d ; d ; d ; d ; ln ; ln Obsrv that thr is on intgration stp, hnc onl on constant In th last lin is rplacd with o It would not b wrong to writ, but this is natr You can alwas rnam constant quantitis to mak th rsult look nicr Finall, th problm did not includ an initial condition; hnc, it can stoppd at, rathr than taking squar roots Without initial condition, it can t tll which squar root to tak 0 Substitut =v It is th appropriat mthod if ou hav a homognous diffrntial quations Eampl Find th gnral solution to th following: Solution Fist substituting v, so From substitution v v v v d v v d v v v
v d v v v d thn sparating variabls, w gt d v d v ln c, v substituting back, w gt Eampl Solution ln c Substitution is v, so v From substitution v d v d v v v v v v d d sparating variabls, w gt d d v ln =ln + Eampl Solution Fist substituting From substitution v d v v v d v d v, so d v v v v v v v v d v arctan v ln v tan ln =tan(ln +) 4 Intgrating factor
It is th appropriat mthod if ou hav linar diffrntial quations A diffrntial quation is calld linar if it contains th unknown function () and its drivativ linarl, i if it is of th form p q f () whr p(), q() and f () ar known functions of In th particular cas, whn f ( ) 0, th diffrntial quation is calld homognous; if f ( ) 0 diffrntial quation is calld non-homognous onsidr first th homognous linar diffrntial quation p ( ) q( ) 0 Multipling b d, dividing b p() and rarranging w gt q( ) d p( ) and hnc, intgrating, w gt q( ) ln d p( ) whr w hav dnotd th intgration constant b If w now ponntiat and st gnral solution in th following form: q( ) d p( ) ln, th linar, w gt th This rsult shows that on can alwas find th gnral solution of a homognous diffrntial quation b quadratur: all that rmains to do, givn th functions p() and q(),is to tak th intgral of thir ratio For th non-homognous cas, rplac th constant b a function, and substitut back into th quation to gt Eampls Find th gnral solutions of th following linar homognous diffrntial quations: 0 Solution d d d ln 0 Solution d d d ln ln ln sin cos 0 ln ln
Solution sin d cos cos( ) d sin() cos( ) d sin() ln lnsin lnsin Find th gnral solutions of th following linar non-homognous diffrntial quation у' + х у = х Solution To non-homognous quation corrspond homognous у' + х у = 0 its solving is standard d Sparat: d d ln ln ln ln ln Solution of homognous quation is Now rplac th constant b a function d d d d d d Insrt and in initial quation d d d d d d d t d td d d t t
Insrt valu of in homognous quation solution So, gnral solution is Erciss Indpndnt work in th class Find gnral and partial solution of th givn diffrntial quations: if 0, 0; if, ; if 0, ; 4 tan if, ; Find th solution of th givn diffrntial quations: sin ; 6 ( ) ( ) d 0; ; 7 ( a) d ; 4 ; 8 d ( ) 0; 4 ; 9 d ; d cost; ln Find th solution of th givn linar homognous diffrntial quations: 4 ; 4 4 0 4 Find th solution of th givn linar non-homognous diffrntial quations: 6 ; cos cos sin ; ; 4 ln ; Homwork Find th solution of th givn diffrntial quations: cos ; 4 ; d ( ) ( ) d 0; d ( ) 0; ; 6 sin ; d d 4 t sin( t ); 7 sint; ; 8 ; 6 ; 9 ; d d 7 t; 0 cos t ; t 8 ( ) ; t ; 9 tant; ; t 0 ; t( ); ; d 4 ; d
; ; d d sin sin sin cos ; 6 /d = - 4 Find th solution of th givn linar homognous diffrntial quations: d ; ( ) d ( ) 0; ; 4 tg ( ) Find th solution of th givn linar non-homognous diffrntial quations: у / - ó õ ; õ ; cos sin ; 4 ( ) 4 SUPLEMENTARY Indfinit intgrals, its proprtis Basic mthods of intgration oncption of indfinit intgrals Intgration is th invrs opration of diffrntiation onsidr for instanc th function f() = A function which has f() as drivativ is F() = W call F() th antidrivativ of a function f () whn th drivativ of F() is f(): F ( ) f ( ) Th function F() = is not th onl antidrivativ of f () = Also th functions G()= +0 and H()= ar antidrivativs This is so bcaus additiv constants vanish whn diffrntiating Howvr, two antidrivativ can diffr onl b an additiv constant Whras an antidrivativ is rmind onl up to a constant, th following qustion has a uniqu answr: F( ) F( ) f ( ) Eampl Find th antidrivativ F() of f()= with F()= (initial condition) Solution: An antidrivativ of f() must b of th form F()= + with a suitabl constant To hav F() = w must hav F() = + = + = = = Hnc F() = solvs th problm Notation and trminolog: If F() is an antidrivativ of f() ( F ( ) f ( ) ) on writs f() d F( ) g d and calls f() d th indfinit intgral bcaus th rsult is dfinit onl up to a constant Th function f() is calld th intgrand and is calld th constant of intgration Th prssion abov is rad as "th indfinit intgral of f() with rspct to " Ruls of intgration Tabl of intgrals Ths ruls ar just invrtd ruls of diffrntiation Th quation f() d F( ) can b writtn in th form d du u or g( ) d g( ) or f ( ) d f ( ) d Th diffrntial smbol d and th indfinit intgral smbol bhav as invrss to ach othr
onstant rul Rul of sums cf ( ) d c f ( ) d (( f ( ) g( )) d f ( ) d g( ) d (Th constant of intgration is lft out) Tabl of intgrals r r d, r - 8 d cot r sin d d ln 9 d arcsin arccos d 0 d arctan arccot 4 a a d, 0 a tan d lncos lna sin d cos cot d lnsin 6 cos d sin 7 d tan cos 4 d arcsin a a d arctan a a a Intgration mthods Th distinguish thr simplst intgration mthods: - dirct intgration; - variabl substitution intgration; - intgration b parts Dirct intgration Dirct intgration is intgration whn w us algbraic transformations and indfinit intgral proprtis and rduc subintgral prssions to basic intgration formulas (tabl of intgrals) Eampl ( ) d ( ln whr Eampl ( 7sin whr ) d d ln, d 7cos ) d d d 7cos d 7sin, Intgration b chang of variabls (Intgration b substitution) In this sction w shall show how to mak us of th hain rul for diffrntiation in problms of intgration Th hain rul will lad to th important mthod of intgration b chang of variabls Th basic ida is to tr to simplif th function to b intgratd b changing from on indpndnt variabl to anothr d With th abov ruls w cannot t intgrat simpl functions lik asil b gussd using th chain rul or Howvr, th solution can
Eampl d lt u du ud d du ( ) d du d du d Eampl d х lt u du ( ) d d du d х du d Eampl a d lt u du d d du d du d Eampl b d lt du d du d d du d u u du u du du u du u du ln u ln u u du u du u Eampl 4 d lt u du d d du ln u ln( ) du d u du d Eampl tan( ) d tan d Eampl 6 d lt u cos sin du (cos ) d d cos du sin d sin d du du u ln u ln cos
lt u du d du d d u u du d du Eampl 7 d lt u du d d udu u ( ) du d du d Eampl 8 d Th trick hr is to rwrit th intgrat as a prfct squar: d lt u d du ( ) du d ( ) u u d du Eampl 9 d Guidd b our princ with th last ampl, w will complt th squar This mans w will rwrit: d ( ) ( lt u d du d du d ) ( ) du tan ( u) tan ( ) u Eampl 0 d This ampl is prtt similar to th prvious on so w'll just show th ida: d d du ( ) 4 4 u Eampl cos d cos cos 4 du u sin tan u sin cos cos sin cos cos d ( cos ) cos d d d sin 4 tan
Aftr rading th ampls abov, if ou fl that this mthod rquirs a crtain amount of gussing and luck, thn ou'v got th ida Howvr, with princ, ou will quickl s how to mak ducatd gusss and it will go much mor smoothl Don't b afraid to mak som mistaks as ou go along, providd ou think about what wnt wrong whn ou do mak a mistak Rmmbr that ou can alwas chck our answr b diffrntiating Intgration b parts Intgration b parts is a usful stratg for simplifing som intgrals Intgration b parts is th invrs to th Product rul ( uv) uv uv for function u(), v() W can rwrit this as uv ( uv) uv or sinc uv is th antidrivativ of (uv) u vd uv vu d u uv Rmark W can s importanc of corrct appling of rul in intgration b parts Eampl Th intgral d can b simplifid b th idntification vrsion vrsion ) u ) u du d du d d ) v d d ) v d d v d v d vrsion d d This dos not simplif mattrs W gt mor difficult intgral vrsion Th valu of th intgral is thn d d ( ) (Answr) Eampl ln d This sms a bit mor confusing sinc it onl looks lik thr is on "part" to th intgration Howvr, th on part w do s would crtainl b asir to dal with if diffrntiatd Hr is th trick to us: vrsion vrsion ) u ln ) u du 0 du d ln d ) v ) v d ln d d d v ln d v d (initial intgral) vdu Thn it follows that ln d ln d ln To solv intgrals without mstics stu attntiv this rul! LIATE rul A rul of thumb lopd in 98 for choosing which of two functions is to b u is th LIATE rul According to th rul, whichvr function coms first in this list is to b u: L: logarithmic functions: ln, log (), tc I: invrs trigonomtric functions: arctan, arcsc, tc A: algbraic functions:, 0, tc T: trigonomtric functions: sin, tan, tc
E: ponntial functions:,, tc Th function which is to b b whichvr function coms last in th list You can rmmbr th list b th mnmonic LIATE Th rason for this is that functions lowr on th list hav asir antidrivativs than th functions abov thm Eampl d d d According LIATE rul u ; du d d; d This rsults in d d; v d lt d d d t t t Eampl 4 arctan d According LIATE rul u arctan arctan d d arctan ln du v d arctan d Eampl An intrsting ampl is: cos d (b analog sin d ) whr, strangl nough, th actual intgral dos not nd to b valuatd This ampl uss intgration b parts twic First lt: u cos du sin d cos d cos sin d d v d Now, to valuat th rmaining intgral, w us intgration b parts again, u sin du cos d Thn: sin d sin cos d d v d Putting ths togthr, w gt cos d cos sin cos d Notic that th sam intgral shows up on both sids of this quation So w can simpl add th intgral to both sids to gt: cos d cos sin c cos sin cos d whr, c (and = c/) is an arbitrar constant of intgration
Particular solutions W hav sn that an intgral producs a whol famil of solutions paramtrizd b In most applications, w ar givn an initial or othr condition and hnc find th valu of Th antidrivativ with known is calld a particular solution Eampl: Find a solution to ( 4 ) d givn that f() = Solution: W first find an antidrivativ: (4 ) d Now plug in for and for F to gt: = () - () + = - + So that = Th particular solution is F() = - + Erciss Indpndnt work in th class Find th intgral of th following functions: ( )( ) d; d; sin d; 6 sin d; d d ; 7 ; 4 4 d; cos 8 d Find th intgral of th following functions (intgration b substitution): cos d ; 7 d; d ; 8 cos d; sin cos d; 9 ( ) d; 4 ln d ; sin cos d; 6 d ; sin 0 ln d; tan d ; d 4 Find th intgral of th following functions (intgration b th part): ln d; cos d; d; 6 sin d; sin d; 7 ln d 4 arcsin( ) d; 8 d Homwork Find th intgral of th following functions: sin d d ; cos ; sin 6 d ; ln 6 ( ) d ; cos d ; 7 sin d ;
4 cos( ) d; 8 4 d ; cos sin d; 9 sin d; 6 sin d ; 0 sin( a b) d ; 7 sin d d ; cos ; cos 8 6 cos d ; ( ) d ; sin 9 sin d ; ( ) d ; 0 sin( ) d; 4 d ; sin cos d; cos d; cos d ; d 6 sin ; arctan d ; 7 d; d 4 ; 8 ; d 4 d; 9 d; d 6 d; 40 4 ; d cos 7 ; 4 d ; ln sin ( ln ) 8 ln d ; 4 d ; cos 9 d ; d 4 sin ; 4 0 ( ) d ; 44 sin( ) d ; cos( 8 ) d ; 4 d; d d ; 46 ; 4 d ; ( ln ) 47 ; ln d 4 cos( a b) d ; 48 cot d ;