Second Order Linear Differential Equations

1 CHAPTER12 SecondOrderLinearDifferentialEquations a relationinvolvingvariablesx y y y . Asolutionis a functionf x suchthatthesubstitutiony f x y f x y f x givesanidentity. Thedifferentialequationissaidtobelineari f it is linearinthevariablesy y y . We have alreadyseen( )how tosolve firstorderlinearequations; ( )y ay by g x whereaandbareconstants,andg x is a differentiablefunctionofx. ,wesaw thata firstorderequationhasa one-parameterfamilyofsolutions,andthatth especificationofaninitialconditiony x0 y0uniquelydeterminesa , ,andnumbers y0 y 0, there isa uniquefunctionf x which solvesthedifferentialequation( )andsatisfiestheinitialconditionsf x0 y0 f x0 y tocompletelysolve equation( )whenthefunctionontherighthandsideis zero:( )y ay by 0 Thisis calledthehomogeneousequation.

2 Animportantfirststepis tonoticethatiff x andg x aretwo solutions,thensois thesum;infact,sois any linearcombinationA f x Bg x . Thus,onceweknow two solutions(they mustbeindependentin thesensethatoneisn t a constantmultipleoftheother)wecansolve y 0 y 0 4 y 0 1 Now, weknow thatcosxandsinxaresolutionsoftheequation ,sowetrya solutionoftheformy x Acosx Bsinx. Evaluatingatx 0, wefindthatA ,gettingy x Asinx Bcosx,andevaluatingatx 0, wefindB x 4 cosx thaty1 cosxis thesolutionwiththeparticularinitialvalue sy1 0 1 y 1 0 0 andy1 sinxis thesolutionwithy1 0 0 y 1 0 1 . Thenthesolutionwithinitialvaluesy 0 andy 0 is( )y x y 0 cosx y 0 y 0 withgiveninitialvaluesy 0 y 0 Nowexande xaresolutionsofthisdifferentialequation, sothegeneralsolutionis a t have aseasya timefindinga solutionlike ( ),sincethesefunctionsdonothave theinitialvalues1 0;0 1 respectively.

3 Howeverif weintroducethefunctions( )coshx 12 ex e x sinhx 12 ex e x thesedohave therightinitialvalues:( )cosh0 1 sinh0 0( )ddx coshx sinhx ddx sinhx coshxso cosh 0 0 sinh 0 ,thesolutiontoourproblemis( )y x y 0 coshx y 0 sinhx Thisparticulardifferentialequationcomesu psooftenthatit is importanttorememberthesefunctions,coshx sinhx, calledthehyperbolicfunctionsandtheirbasi cproperties:equation( )and( )cosh2x sinh2x 1 Becauseof( )thesefunctionsparametrizethestandardhyp erbola(andit is forthisreasonthattheyarecalledhyperbolic functions).We now bea rootoftheequation( )r2 ar b 0 Thenerxis a solutiontothehomogeneousequation:( )y ay by 0 Equation( )is calledtheauxiliaryequationof thedifferentialequation( ).

4 To verifythepropo-sition,lety erxsothaty rerx y r2erx. Substitutingintoequation( ):( )r2erx arerx berx erx r2 ar b 0if andonlyif r is a rootoftheauxiliaryequation. unfortunatelya quadraticequationdoesnotnecessarilyalway shave two realroots,sowehave realroots. If thediscriminanta2 4b 0,thentherearetwo realroots,andit is 6y 5y 0 y 0 4 y 0 1 Theauxiliaryequation,r2 6r 5 0 hastherootsr 1 5, soe xande ( )y Ae x Be 5xwithderivativey Ae x 5Be 5x Evaluatingatx 0, wehave 4 A B 1 A 5B. Solvingthispairofequations,wegetA 19 4andB 3 4, sooursolutionis( )y 194e x 34e functionx x t satisfiesthedifferentialequation( )x 2x 15x 0 Underwhatconditionsonthevaluesofxatt 0 willthisfunctiondecayto0 ast ?

5 Theauxiliaryequationr2 2r 15hastherootsr 3 t Ae 3t Be5t. ThiswilldecayatinfinityonlyifB , evaluatingxandx at0 givesustheequations( )x 0 A B x 0 3A 5B SettingB 0, theconditionbecomesx 0 3x 0 Ifthediscriminanta2 4b 0,thentherootsaretwo complex conjugatenumbers i i .Let s firstlookat thecasey y 1 0 are i, andwe d like to saythatthesolutionsarethefunctionseix e ix. Thisdoeswork,andallthealgebrainthecaseof realrootsworksjustaswellinthiscase,oncew ehave ,rememberequation( ):thegeneralsolutionofy y 0 is( )y x y 0 cosx y 0 sinxIfy x eixis torepresenta solutionofthisdifferentialequation,wehav ey 0 e0 1,andy 0 ie0 i, sowemusthave( )eix cosx isinxNoticethatif wedifferentiatethisexpression,weget( ) sinx icosx i cosx isinx sothisexpressionis consistentwiththedifferentiationrulefort heexponential:( )ddxeix ieix Chapter12 SecondOrderLinearDifferentialEquations17 8 Infact,definingthecomplex exponentialby( )is , if wesubstitutetheMaclaurinseriesforallthef unctionsin( )wegetanidentity:( ) n 0 ix nn!

6 N 0 1 nx2n 2n ! i n 0 1 nx2n 1 2n 1 ! complex number i if wedefinetheexponentialfunctionas( )e i x e x cos x isin x , ofcourse,weareinterestedonlyin shownis thatif i aretherootsoftheequationr2 ar b 0,thenthefunctionse i xsolve thedifferentialequationy ay b ,whichgivesusthedesiredtwo theauxiliaryequationforthedifferentialeq uation( )y ay b 0hasthecomplex roots i , theneverysolutionofthedifferentialequati onis oftheform( )Ae xcos x Be xsin x e x Acos x Bsin x Insolvinginitialvalueproblems,wecanworkw iththecomplex solutionsorsolutionsoftheform( );usuallythelatteris x t ofx 2x 0 Sincetherootsoftheauxiliaryequationr2 2 0 are i , thegeneralsolutionis( )x t Acos t Bsin tIt is easytoseewhatthisfunctionlookslike bydefining( )C A2 B2 arctan B A Then( )becomes( )x t C cos cos t sin sin t Ccos x Ccos x Thusthegraphofx x t is a simplecosinecurve ofamplitudeC, andperiod2 , shiftedtotherightbythephase.

7 ( ). 2 1 5 1 1 0 5 0 5000 50 5111 51 522 y x ofy 2y 5y 0, withtheinitialvaluesy 0 2 y 0 2r 5 0 hasthesolutionsr 1 2i. Thusthegeneralsolutionisy e x Acos 2x Bsin 2x . To solve forAandBusingtheinitialvalueswemustfirst differentiatey:( )y e x Acos 2x Bsin 2x e x 2 Asin 2x 2 Bcos 2x Substitutingtheinitialvaluesgivestheequa tionsA 2 A 2B 1,whichhasthesolutionsA 2 B 1 ( )y e x 2 cos 2x 12sin 2x Caseofa doubleroot. If thediscriminanta2 4b 0, thentheauxiliaryequationhasonerootr, tryy uerx, whereuis a , thedifferentialequationis( )y 2ry r2y 0 Substitutingthisyintheequationwegetto( )y 2ry r2y erx u 2u r ur2 2r u ur r2u erxu 0 Thusu Ax theauxiliaryequationforthedifferentialeq uation( )y ay b 0hasonlytherootr, theneverysolutionis oftheform( ) Ax B 4y 4y 0, withinitialvaluesy 0 2 y 0 Ax B e2x, withderivativey 2 Ax B e2x Ae2x.

8 Substitutingtheinitialconditionsgivesthe Equations ( )2 B 1 2B A ThusA 5 B 2 andtheansweris( )y 5x 2 e2x 17 PSfragreplacements00112345670 20 40 60 20 40 60 500400 300200 1000010020030040050012345610 50 1 0 50 5 ,oneusuallywantstohave ,theyalsodeterminetheirbehavior. Exceptfortheidenticallyzerosolution,alls olutionsgrow Exceptfortheidenticallyzerosolution, negative anda positive root. Allsolutionsgrow exponentially, exceptforthemultiplesoftheexpo-nentialwi ththenegative Thisis thecaseoftheequationwitha 0 andb 0, whichwecanwriteas( )y 2y 0 Aswesaw ,thegeneralsolutioncanbewrittenas( )y x Acos x Bsin x ory x Ccos x anoscillationofperiod2 , amplitudeCandphase ( ).

9 Complexroots. Inthiscasetherootsareoftheform i andthegeneralsolutioncanbewritteninthefo rm(followingthepreviousdiscussion)( )y x Ce xcos x 0, thisgivesanoscillationwithexponentiallyi ncreasingamplitude,andif 0, thisgivesanexponentiallydampedoscillatio n. massmontheendofa verticallyhangingspring,andthensetit inmotion;how canwedescribethesubsequentmotion?Aswehav e ,thespringis s secondlawofmotion,thisforceisma, whereais theaccelerationofthemassm. Lettingxrepresentthedownwarddisplacement fromequilibrium,wehavea x , andif thespringconstantisk, thisgivesustheequation( )mx kx orx kmx 0 Letting k m, thishasthesolution( )( )x t Ccos t whereCand have tobea ,whenmis measuredinkilogramsandxinmeters,thenforc eis measuredinnewtons, smallerscale,mis ingrams,xincentimeters,forceindynes,andk indynes/meter.

10 However, intheBritishsystemit is customarytorefertotheweightwoftheobject( inpounds), w g, whereg 32ft/sec2is theaccelerationdueto gravity. Finally, intheBritishsystem,thespringconstantis massof10g hangsfroma springwithspringconstantk 0 4 extendedanadditional8 ,give 0 x 0 8 x 0 alsohave 0 4 10 0 04 0 ( )x t Ccos 0 2t We get,fromtheinitialconditions( )8 Ccos 0 0 2 Csin so 0 andC 8, andtheequationofmotionis( )x t 8 cos 0 2t We couldhave concludedthismorequickly, byobservingthattheinitialconditionstellu sthatx 8whenthevelocityis 0, sothemaximumextension(theamplitude) configurationalreadyinmotion,andwhenwema ke ourobservation(attimet 0),themassis displaced12cmdownwardandis travelingdownwardat a velocityof1 ,theequationhasthegeneralform( )