Steps, Solved Example of Frobenius Method - Byju's

The Frobenius method is an approach to identify an infinite series solution to a second-order ordinary differential equation. Generally, the Frobenius ... CheckoutJEEMAINS2022QuestionPaperAnalysis: CheckoutJEEMAINS2022QuestionPaperAnalysis: × DownloadNow Previous Next FrobeniusMethod TheFrobeniusmethodisanapproachtoidentifyaninfiniteseriessolutiontoasecond-orderordinarydifferentialequation.Generally,theFrobeniusmethoddeterminestwoindependentsolutionsprovidedthatanintegerdoesnotdividetheindicialequation’sroots. Considerthesecond-orderordinarydifferentialequationgivenbelow: Here,a(x),b(x),andc(x)are“suitablefunctions”. Letusassumethatthesearethesuitablerationalfunctionsthatmeanspolynomialsdividedbypolynomials,suchasinBessel’sequationoforder1/2. (d2y/dx2)+(1/x)(dy/dx)+[1–(1/4x2)]y=0 Forthisequation,wecanillustratetheFrobeniusmethod.TheprimaryapproachofFrobeniusaimsatsolutionsintheformofpowerseriesaroundsomegivenpointx0multipliedby(x−x0)toacertainpower.Thiscanbeexpressedas: Here,akandraretheconstantsthatcanbeidentifiedthroughtheFrobeniusmethod. Learn:OrdinaryDifferentialEquations Let’sstartwiththeFrobeniusmethodtosolvethesecond-orderordinarydifferentialequation. Step1:Chooseasuitablevalueforx0.Thiscanbedoneintwoways: (i)Ifconditionsaregivenfory(x)atsomepoint,wecanusethatforx0. (ii)Ifnoconditionsaregivenfory(x),wemustchoosex0asperourconvenience.Generally,weprefertochoosex0=0. Step2:Ifthegivendifferentialequationisoftheforma(x)(d2y/dx2)+b(x)(dy/dx)+c(x)y=0,thenconvertthis,asmentionedabove. i.e.,(d2y/dx2)+(1/x)(dy/dx)+[1–(1/4x2)]y=0………(1) Thenmultiplytheequationby4x2sothatwecanavoidfractionstomakethesimplificationeasy. 4x2(d2y/dx2)+4x(dy/dx)+(4x2–1)y=0 (or) 4x2y′′+4xy′+(4x2–1)y=0……..(2) Step3:Letusassumethatthesolutionisoftheform\(\begin{array}{l}y=y(x)=(x-x_0)^r\sum_{k=0}^{\infty}a_k(x-x_0)^k\end{array}\),whereakisanarbitraryconstantsuchthatak≠0. Step3:Now,bringthefactor(x–x0)rinsidethesummation.Thatmeans, \(\begin{array}{l}y=y(x)=\sum_{k=0}^{\infty}a_k(x-x_0)^{k+r}\end{array}\) Step4:Fromtheassumedseriesofy,weneedtocalculatetherespectivemodifiedpowerseriesfory′andy′′bydifferentiating“term-by-term”. Forthissubstitutex0=0.Thus,weget; \(\begin{array}{l}y=y(x)=\sum_{k=0}^{\infty}a_k\x^{k+r}\end{array}\) Then,bydifferentiatingy,weget; Step5:Substitutetheexpressionsfory,y′andy′′inequation(2).Onsimplificationofthisweget; Step6:Foreachseriesobtainedintheaboveequation,weneedtochangetheindexsothateachserieswillbeoftheform: \(\begin{array}{l}\sum_{n=something}^{\infty}[Term\not\containing\x](x-x_0)^n\end{array}\) Step7:Convertthesumofseriesintheobtainedequationintoonebigseries.Afewtermswilllikelyhavetobewrittenseparately,soweneedtosimplifytoapossibleextent. Step8:Now,thefirsttermoftheobtainedseriesmaybeoftheform; a0[formulaofr](x–x0)something Also,rememberthateachtermoftheseriesis0. Fromthis,wecanwritetheformulaofr=0 Thisyieldsaquadraticequationinr. Solvingthisequation,wegettworoots,r1andr2. Step9:Bysubstitutingr1inthelastseriesequation,weget; Fromthis,wecanget; nthformulaofak’s=0forn0≤n Then,solvethisforhighestindex=formulaofnandlowerindexedak’s Tosimplifythetermsatleastalittle,performthechangeofindicessothattherecursionformulawillbederivedandcanberewrittenas; ak=formulaofkandlower-indexedcoefficients Step10:Usingtherecursionformulaoranycorrespondingformulastothelower-orderterms,weneedtofindalltheak’sintermsofa0and,maybe,oneotheram. Step11:Usingr=r1andtheformulas,wehavederivedthecoefficients.Now,writeouttheresultantseriesfory.Simplifyitandfactoroutthearbitraryconstant(s)ifpossible. Step12:Repeatsteps9to11forsubstitutingr=r2. Finally,thelaststepmayyieldyasanarbitrarylinearcombinationoftwodistinctseries.Therefore,thisisreferredtoasthegeneralsolutiontothegivendifferentialequation. Thiscanbefurthersimplifiedas: Readmore: Series Infiniteseriesformula Infiniteseriescalculator FrobeniusMethodSolvedExample Example: Findthesolutionof4xy′′+2y′+y=0byFrobeniusMethod. Solution: Givendifferentialequationis: 4xy′′+2y′+y=0….(1) Let\(\begin{array}{l}y=y(x)=\sum_{k=0}^{\infty}a_k\x^{k+r}\end{array}\)bethesolutionequation. So, Substitutingtheexpressionsofy,y′andy′′inequation(1),weget; \(\begin{array}{l}4x\left[\sum_{k=0}^{\infty}a_k(k+r)(k+r-1)x^{k+r-2}\right]+2\left[\sum_{k=0}^{\infty}a_k(k+r)x^{k+r-1}\right]+\sum_{k=0}^{\infty}a_kx^{k+r}=0\end{array}\) Thiscanbewrittenas: \(\begin{array}{l}\left[\sum_{k=0}^{\infty}a_k4(k+r)(k+r-1)x^{k+r-1}\right]+\left[\sum_{k=0}^{\infty}a_k2(k+r)x^{k+r-1}\right]+\sum_{k=0}^{\infty}a_kx^{k+r}=0\end{array}\)…..(2) Dividingtheaboveequationbyxr-1,weget; \(\begin{array}{l}\left[\sum_{k=0}^{\infty}a_k4(k+r)(k+r-1)x^{k}\right]+\left[\sum_{k=0}^{\infty}a_k2(k+r)x^{k}\right]+\sum_{k=0}^{\infty}a_kx^{k+1}=0\end{array}\) Changingtheindicesofthebases,weget; \(\begin{array}{l}https://latex.codecogs.com/svg.image?\left[\sum_{n=0}^{\infty}a_n4(n+r)(n+r-1)x^{n}\right]+\left[\sum_{n=0}^{\infty}a_n2(n+r)x^{n}\right]+\sum_{n=1}^{\infty}a_{n-1}x^{n}=0\end{array}\) Now,weneedtoexpandthesummationtomaketheindicesequal.Thiscanbedoneasfollows. \(\begin{array}{l}a_04(0+r)(0+r-1)x^0+\left[\sum_{n=1}^{\infty}a_n4(n+r)(n+r-1)x^{n}\right]+a_02(0+r)x^0\left[\sum_{n=1}^{\infty}a_n2(n+r)x^{n}\right]+\sum_{n=1}^{\infty}a_{n-1}x^{n}=0\end{array}\) Nowconsiderthetermwhichisoftheforma0[termofr]=0. i.e.,a0[4r(r–1)+2r]=0 4r2–4r+2r=0 4r2–2r=0 r2–(½)r=0 r[r–(½)]=0 Thus,r=½,r=0. Now,bychangingtheindicesofequation(2),weget; Fork≥0,weget; 4(k+r+1)(k+r)ak+1+2(k+r+1)ak+1+ak=0 Fromthis,wecanwriteak+1as: ak+1=-ak/[(2k+2r+2)(2k+2r+1)];k=0,1,2,3,etc.,………..(3) Substituter=r1=½inequ(3). ak+1=-ak/[(2k+3)(2k+2)] Substitutingk=0,1,2,3,andsoon,intheaboveequation,weget; a1=-a0/3.2=-a0/3! a2=-a1/5.4=a0/5! a3=-a2/7.6=-a0/7! Asweknow,a0isanarbitraryconstant.So,leta0=1. Therefore,ak(r1)=(-1)k/(2k+1)!;k=0,1,2,3,…. Hence, Similarly,bysubstitutingr=r2=0inequ(3),weget; Thus,weobtainedthesolutionsofthegivendifferentialequation. PracticeProblems Solvexy′′+2y′+xy=0byFrobeniusMethod. Solve:5x2y′′+x(1+x)y′−y=0 Considerthedifferentialequationxy”+y’+2xy=0.ObtainthesolutionbyFrobeniusmethod. MATHSRelatedLinks IntroductionToExponents BisectionMethod DifferentialEquationsWorksheets ProbabilityForClass12 TrigonometricEquations Is101APrimeNumber? InverseSine TableOf25 Is1APrimeNumber? MultiplicationOfFractions LeaveaCommentCancelreplyYourMobilenumberandEmailidwillnotbepublished.Requiredfieldsaremarked* * SendOTP DidnotreceiveOTP? RequestOTPon VoiceCall * * Website * Grade/Exam Class1 Class2 Class3 Class4 Class5 Class6 Class7 Class8 Class9 Class10 Class11 Class12 IAS CAT BankExam * * PostComment CBSESamplePapers CBSESamplePapersClass8Maths CBSESamplePapersClass9Maths CBSESamplePapersClass10Maths CBSESamplePapersClass11Maths CBSESamplePapersClass12Maths CBSEPreviousYearQuestionPapers CBSEPreviousYearQuestionPapersClass12Maths CBSEPreviousYearQuestionPapersClass10Maths ICSESamplePapers ICSESamplePapersClass8Maths ICSESamplePapersClass9Maths ICSESamplePapersClass10Maths ISCSamplePapersClass11Maths ISCSamplePapersClass12Maths ICSEPreviousYearQuestionPapers ICSEPreviousYearQuestionPapersClass10 ISCPreviousYearQuestionPapersClass12Maths CBSESamplePapers CBSESamplePapersClass8Maths CBSESamplePapersClass9Maths CBSESamplePapersClass10Maths CBSESamplePapersClass11Maths CBSESamplePapersClass12Maths CBSEPreviousYearQuestionPapers CBSEPreviousYearQuestionPapersClass12Maths CBSEPreviousYearQuestionPapersClass10Maths ICSESamplePapers ICSESamplePapersClass8Maths ICSESamplePapersClass9Maths ICSESamplePapersClass10Maths ISCSamplePapersClass11Maths ISCSamplePapersClass12Maths ICSEPreviousYearQuestionPapers ICSEPreviousYearQuestionPapersClass10 ISCPreviousYearQuestionPapersClass12Maths JoinBYJU'SLearningProgram Grade/Exam Class1 Class2 Class3 Class4 Class5 Class6 Class7 Class8 Class9 Class10 Class11 Class12 IAS CAT BankExam GATE Submit × RegisterNow Share Share Share CallUs RegisterwithBYJU'S&DownloadFreePDFs * SendOTP * * * * * Grade Class1 Class2 Class3 Class4 Class5 Class6 Class7 Class8 Class9 Class10 Class11 Class12 IAS CAT BankExam DownloadNow FREE Signup DOWNLOAD AppNOW