Itroductio to Differetial Equatios Defiitios ad Termiolog Differetial Equatio: A equatio cotaiig the derivatives of oe or more depedet variables, with respect to oe or more idepedet variables, is said to be a differetial equatio (DE). Classificatio b Tpe: If a equatio cotais ol ordiar derivatives of oe or more depedet variables with the respect to a sigle idepedet variable is said to be a ordiar differetial equatio (ODE). EX: d + 5 = e, + 6=, ad + = + dt dt A equatio ivolvig partial derivatives of oe or more depedet variables of two or more idepedet variables is called a partial differetial equatio (PDE). EX: u u u u u u u v + =, = +, ad = t t Most tet will write ordiar derivatives usig either Leibiz otatio 3 d d,,... 3 or prime otatio,,... Usig prime otatio the previous ODE s ca be writte + 5= e ad + 6= Remember the otatio for 4 th, 5 th, etc. derivatives Other otatios: d s Newto s dot otatio: = 3 becomesɺɺ s= 3 dt Subscript otatio: the previous secod PDE becomes u = u u tt t Classificatio b Order: The order of a differetial equatio (either ODE or PDE) is the order of the highest derivative i the equatio. EX: d + 5 4 e = 3 is a secod order
We ca epress a th ordiar differetial equatio i oe depedet variable b geeral form F = ( ) (,,,,... ) where F is a real valued fuctio of + variables. We ca solve this equatio for the highest derivative b d ( 1) = f(,,,... ) where f is a real valued cotiuous fuctio ad is referred to as the ormal form of a differetial equatio Classificatio b Liearit A th order ordiar differetial equatio is said to be liear if F is liear i,,, (). This meas that a th order ODE is liear whe 1 d d a( ) + a 1( )... a 1 1( ) a( ) g( ) + + + = The two importat cases are a1( ) a( ) g( ) + = d ad a( ) + a 1( ) + a( ) = g( ) Characteristic Two Properties of a Liear ODE The depedet variable ad all its derivatives,, are of the first degree, that is, the power of each term ivolvig is 1 The coefficiets a, a 1,., a, of,, deped at most o the idepedet variable 3 d EX: ( ) + 4=, + =, ad + 5+ e 3 A oliear ODE is oe that is ot liear. Solutios of a ODE: a fuctio φ, defied o a iterval I ad possessig at least derivatives that are cotiuous o I, which whe substituted ito a th order ODE reduces the equatio to a idetit, is said to be a solutio of the equatio o the iterval. Occasioall it will be coveiet to deote the solutio b a alterate smbol ()
Iterval of Defiitio The iterval I i the previous defiitio is called the iterval of defiitio, the iterval of eistece, the iterval of validit, or the domai of the solutio ad ca be ope closed or ifiite. EX: Verificatio of a solutio Verif that the idicated fuctio is a solutio of a give differetial equatio o (-, ). a. / = ½ ; = (1/16) 4 b. + = ; = e A solutio i which the depedet variable is epressed solel i term of the idepedet variable ad costats is said to be a eplicit solutio. The previous solutios are eplicit Implicit solutio of a ODE: A relatio G(,)= is said to be a implicit solutio of a ODE o a iterval I, provided there eists at least oe fuctio φ that satifies the relatio as well as the differetial equatio o I. EX: Verificatio of a Implicit Solutio The relatio + = 5 is a implicit solutio of the differetial equatio = O the iterval -5 < < 5. B implicit differetiatio we obtai d + d = d 5 or + = Families of Solutios: Whe solvig 1 st order DE s F(,, ) =, we usuall obtai a solutio cotaiig a sigle arbitrar costat or parameter c. A solutio cotaiig a arbitrar costat represets a set G(,,c) = of solutios called a oe parameter famil of solutios. Whe solvig a th order DE we seek a th parameter famil of solutios. A solutio of a differetial equatio that is free of arbitrar parameters is called a particular solutio. E: = c 1 cos is a solutio to = (1 parameter) E: = c 1 e + c e is a solutio to + = ( parameter) Geometricall, the geeral solutio of a first order differetial equatio represets a famil of curves kow as solutio curves, oe for each assiged value of the arbitrar costat.
If C = is the geeral solutio to + = the the solutio curves would look like the followig Solvig differetial equatios aalticall ca be difficult or eve impossible. However, there is a graphical approach ou ca use to lear a lot about the solutio of a differetial equatio. Cosider a differetial equatio of the form = F(, ) where F(,) is some epressio i ad. At each poit (,) i the - plae where F is defied, the differetial equatio determies the slope = F(, ) of the solutio at that poit. If ou draw short lie segmets with slope F(,) at selected poits (,) i the domai of F, the these lie segmets for a slope field. A slope field shows the geeral shape of all the solutios ad ca be helpful i gettig a visual perspective of te solutios to a differetial equatio. Iitial Value Problems We are ofte iterested i problems i which we seek a solutio () of a DE so that () satisfies prescribed coditios. This is called a iitial valued problem (IVP). The values of () at a sigle poit (, ) are called iitial coditios. First ad Secod Order IVP s Solve: f(, ) = Subject to: ( ) =
Solve: d = f(,, ) ( ) =, ( ) = Subject to: 1 E: If = ce is oe parameter famil of solutios of the simple 1 st order DE =. All solutios are defied o the iterval (-, ). If we impose a iitial coditio () = 3, The = ad = 3 we ca fid a solutio to the IVP. If we place the iitial coditio (1) = - we ca fid aother solutio. major questios: Does a solutio to the problem eist? If the solutio eists, is it uique? Picard s eistece of a uique solutio Let R be a rectagular regio i the - plae defied b a b, c d that cotais the poit (, ) i its iterior. If f(,) ad f are cotiuous o R, the there eists a iterval I cetered at ad a uique fuctio () defied o I satisfig the IVP. E: If = c 1 cos4t + c si4t is a two parameter famil of solutios of + 16 =. Fid a solutio of the IVP π π + 16=, ( ) = -, ( ) = 1 Separable Variables: Review of Itegratio: k = k+ C for costat k + 1 = + C + 1 du l u c u = + udv= uv- vdu for all 1 Solutio b Itegratio: Cosider the first order differetial equatio f(, ) = Whe f does ot deped o the variable, that is f(,) = g(), the differetial equatio g( ) = ca be solved b itegratio. If g() is a cotiuous fuctio, the itegratig both sides gives us
where G() is a atiderivative of g(). Solve: a. 1 e = + b. =si = g( ) = G( ) + c Separable Equatio: A first order differetial equatio of the form g = ( ) h( ) is said to be separable or to have separable variables. 3+ 4 Separable: = e No-separable: si = + Separable equatios ca be writte g( ) = or h( ) = g( ) h( ) the h( ) = g( ) + c If we follow this procedure to solve separable variables a oe parameter famil of solutios is obtaied. EX: a. Solve (1 + ) = b. Solve the IVP =, ( 4) = 3 c. Solve e si = d. Solve 4 3 ( ) e + + = Uless coveiet leave i implicit form, however the iterval over which the solutio eists i ot eas to see. Make sure whe separatig the variables the divisors are ot zero E: Losig a Solutio Solve -4 =, () = E: A IVP Solve ( e - )cos e si =, () =
Homogeeous Equatios: Defiitio: If a fuctio has the propert that f t t (, ) = t f(, ) For some real umber, the f is said to be a homogeeous fuctio of degree. E: a. f(, ) = 3+ 5 b. 3 f(, ) = + 3 3 c. f(, ) = + + 1 c. f(, ) = + 4 If f (,) is a homogeeous fuctio of degree, otice that we ca write f(, ) = f 1, ad f = f 1 (, ), where f (1,/) ad f (/,1) are both homogeeous of degree. 3 E: a. f(, ) = 6 b. f(, ) = c. f(, ) = + 3+ Defiitio: A differetial equatio of the form M(, ) + N(, ) = Is said to be homogeeous if both coefficiets M ad N are homogeeous fuctios of the same degree. Method of Solutio: A homogeeous D.E. ca be solved b usig either substitutio = u or = v, where u ad v are ew depedet variables to reduce the homogeeous D.E. ito a separable first order D.E. Sice homogeeous Which gives E: a. Solve = u = u+ du M(, ) + N(, ) = M(, u) + N(, u)( u+ du) = M(, 1 u) + N(, 1 u)( u+ du) = [ M(, 1 u) + un(, 1 u)] + N(, 1 u) du= N(, 1 u) du + = M(, 1 u) + un(, 1 u) ( + ) + ( ) = b. Solve ( ) =
3 4 4 c. Solve + ( + ) = d. Solve e 1 1 = +, () = Eact Equatios: Differetial of a Fuctio of Two Variables: If z = f(,) is a fuctio of two variables with cotiuous first partial derivatives i a regio R of the plae, the its differetial is f f dz= + (1) I the special case whe f(,) = c, where c is a costat, the (1) implies f f + = I other words, give the oe parameter famil of fuctio f (,) = c, we ca geerate a first order DE b computig the differetial of both sides of the equalit. For eample, if 5 + 3 = c, the () gives the first order DE ( 5) + (-5 + 3 ) = or 5 = (3) 5 + 3 Eact Equatio: A differetial epressio M(,) + N(,) is a eact differetial i a regio R of the plae if it correspods to the differetial of some fuctio f(,) defied i R. A first order DE of the form M(,) + N(,) = is said to be a eact equatio if the epressio o the left had side is a eact differetial. E: 3 + 3 = is a eact equatio, because its left-had side is a eact differetial: d(1/3 3 3 ) = 3 + 3 The Criterio for a Eact Differetial: Let M(,) ad N(,) be cotiuous ad have cotiuous first partial derivatives i a rectagular regio R defied b a << b, c << d. The a ecessar ad sufficiet coditio that M(,) + N(,) be a eact differetial is M N = (4) ()
Method of Solutio: Give a equatio i the differetial form M(,) + N(,) =, determie whether the equalit i (4) holds. If it does, the there eists a fuctio f for which f = M(, ) We ca fid f b itegratig M(,) with respect to while holdig costat: f(, ) = M(, ) + g( ) (5) where the arbitrar fuctio g() is the costat of itegratio. Now differetiate (5) with respect to ad assume f/ = N(,): f = M(, ) + g ( ) = N(, ) This gives g ( ) = N(, )- M(, ) ) (6) Fiall itegrate (6) with respect to ad substitute the result i (5). The implicit solutio of the equatio is f(,) = c It is importat to realize the epressio N(, )- M(, ) ) is idepedet of, because N N M N(, ) M(, ) = M(, ) = = E: a. Solve + ( 1) = b. Solve (e cos) + (e cos + ) = cos si c. Solve =, () = (1 ) Itegratig Factors: It is sometimes possible to covert a o eact differetial equatio ito a eact equatio b multiplig b some fuctio µ(, ) called a itegratig factor. However the resultig eact equatio µ M(, ) + µ N(, ) = Ma ot be equivalet to the origial i the sese that a solutio of oe is also the solutio of the other. It is possible for a solutio to be lost or gaied as a result of the multiplicatio. E: Solve ( + ) + l = usig 1 µ (, ) =, o(, )
The hard part is fidig u, oe wa is to use partial DE, we cat do that et. Cosider this, b the Criterio M(,) + N(,) = will be eact iff (um) = (un), subscripts deote partial derivatives. Usig the product rule we get u N u M = (M N )u M, N, M, N, are kow fuctios of ad. So aother wa is to assume u is a fuctio of oe variable. I this case we get the two scearios: M(,) + N(,) = If (M N )/N is a fuctio of aloe, the a itegratig factor is M N N u( ) = If (N M )/M is a fuctio of aloe, the a itegratig factor is N M M u( ) = E: + ( +3 ) = (this is ot eact) Liear Equatios: A first order DE of the form e e a1( ) a( ) g( ) + = (1) is said to be a liear equatio i the depedet variable. Stadard Form: B dividig both sides of the previous equatio b the lead coefficiet we obtai the stadard form: + P( ) = f( ) () We seek a solutio, o I for which both coefficiet fuctios P ad f are cotiuous. Itegratig factor: with differetials we ca write the stadard form as + = With liear equatio ou ca have the propert that ou ca alwas fid a factor such that + = (3) is a eact differetial equatio. If it is eact the
= or = Lookig a little closer this is a separable equatio that we ca use to fid = = = We do t use a costat of itegratio because (3) is uaffected b a costat multiple. If we use this itegratig factor i (3) we get + = rewrite as = the itegrate or = + = + Solvig a Liear First Order Equatio: 1. Put a liear equatio ito the stadard form. From the stadard form idetif P() ad the fid the itegratig factor 3. Multipl the stadard form of the equatio b the itegratig factor, the left had side of the resultig equatio is automaticall the derivative of the itegratig factor ad. d P( ) P( ) e = e f 4. Itegrate both sides of this last equatio. E: a. Solve -3 = 6 b. Solve -4 = e c. Solve ( 9) + = d. Solve, () 3 + = = e. Solve, (1) + = = ( )
1 f. =, ( ) = + Equatios of Beroulli ad Ricatti This sectio does ot look at a particular tpe of DE but istead 3 specific equatios that ca be tured ito DE we alrea kow how to solve. Beroulli s Equatio: P( ) f( ) + = where is a real umber, otice that for = ad =1 the equatio is liear. (We alrea kow how to solve that ad we would t use the proceedig method) If we ca write the equatio as 1 + P( ) = f( ) 1 Let w=,, 1 the dw = (1 ) (use implicit diff where is a fuctio of ) substitute this ito the altered Beroulli equatio to produce a liear equatio of the form: dw 1 + P( ) w = f( ) (1 ) dw (1 ) P( ) w (1 ) f( ) + = 1 solve this for w the use w= ou get a solutio to the Beroulli equatio. 1 E: Solve + = E: Solve e = E: Solve (1 ) + = Ricatti s Equatio: the oliear differetial equatio P( ) Q( ) R( ) = + + if 1 is a kow particular solutio of the Ricatti Equatio the the substitutios = 1+ u 1 du ad = + ito the Ricatti Equatio leads to
1 du + = P( ) + Q( )( 1+ u) + R( )( 1+ u) 1 the sice P( ) Q( ) 1 R( ) 1 = du ( Q 1R) u Ru + = this is actuall ow a Beroulli equatio with =, it ca the be reduced to the liear equatio dw ( Q ) 1 R w R + + = b substitutig w= u 1. E: Solve = +