chapter 5 series solutions of second order linear equations

C H A P T E R

5

SeriesSolutionsof SecondOrderLinearEquations

Finding the generalsolutionof a linear differentialequationrestson determiningafundamentalsetof solutionsof the homogeneousequation.So far, we have given asystematicprocedurefor constructingfundamentalsolutionsonly if theequationhasconstantcoefficients.To dealwith themuchlargerclassof equationshaving variablecoefficients it is necessaryto extend our searchfor solutions beyond the familiarelementaryfunctionsof calculus.Theprincipal tool thatweneedis therepresentationof agivenfunctionby apowerseries.Thebasicideais similar to thatin themethodofundeterminedcoefficients:Weassumethatthesolutionsof agivendifferentialequationhave power seriesexpansions,and thenwe attemptto determinethecoefficientssoasto satisfythedifferentialequation.

5.1 Review of Power Series

In this chapterwe discussthe use of power seriesto constructfundamentalsetsofsolutionsof secondorderlineardifferentialequationswhosecoefficientsarefunctionsof theindependentvariable.Webegin by summarizingvery briefly thepertinentresultsaboutpower seriesthatwe need.Readerswho arefamiliar with power seriesmay goonto Section5.2.Thosewhoneedmoredetailsthanarepresentedhereshouldconsultabook oncalculus.

231

232 Chapter 5. Series Solutions of Second Order Linear Equations

1. A power series∞∑

n=0an(x − x0)

n is saidto convergeatapoint x if

limm→∞

m∑

n=0

an(x − x0)n

existsfor thatx. Theseriescertainlyconvergesfor x = x0; it mayconvergefor allx, or it may convergefor somevaluesof x andnot for others.

2. Theseries∞∑

n=0an(x − x0)

n is saidto convergeabsolutelyatapoint x if theseries

∞∑

n=0

|an(x − x0)n| =

∞∑

n=0

|an||x − x0|n

converges.It canbeshown that if theseriesconvergesabsolutely, thentheseriesalsoconverges;however, theconverseis not necessarilytrue.

3. Oneof themostusefultestsfor theabsoluteconvergenceof a power seriesis theratio test.If an �= 0, andif for afixed valueof x

limn→∞

∣

∣

∣

∣

∣

an+1(x − x0)n+1

an(x − x0)n

∣

∣

∣

∣

∣

= |x − x0| limn→∞

∣

∣

∣

∣

an+1

an

∣

∣

∣

∣

= L|x − x0|,

thenthe power seriesconvergesabsolutelyat that valueof x if |x − x0| < 1/L,anddivergesif |x − x0| > 1/L. If |x − x0| = 1/L, thetestis inconclusive.

EXAMPLE

1

For which valuesof x doesthepower series∞

∑

n=1

(−1)n+1n(x − 2)n

converge?To testfor convergenceweusetheratio test.Wehave

limn→∞

∣

∣

∣

∣

∣

(−1)n+2(n + 1)(x − 2)n+1

(−1)n+1n(x − 2)n

∣

∣

∣

∣

∣

= |x − 2| limn→∞

n + 1

n= |x − 2|.

Accordingto statement3 theseriesconvergesabsolutelyfor |x − 2| < 1, or 1 <

x < 3,anddivergesfor |x − 2| > 1.Thevaluesof x correspondingto |x − 2| = 1arex = 1 and x = 3. The seriesdivergesfor eachof thesevaluesof x sincethenth termof theseriesdoesnot approachzeroas n → ∞.

4. If the power series∞∑

n=0an(x − x0)

n convergesat x = x1, it convergesabsolutely

for |x − x0| < |x1 − x0|; and if it divergesat x = x1, it divergesfor |x − x0| >

|x1 − x0|.5. Thereis a nonnegative numberρ, called the radius of convergence, such that

∞∑

n=0an(x − x0)

n convergesabsolutelyfor |x − x0| < ρ anddivergesfor |x − x0| >

ρ. For a seriesthatconvergesonly at x0, we define ρ to bezero;for a seriesthat

5.1 Review of Power Series 233

convergesforall x, wesaythatρ is infinite. If ρ > 0,thentheinterval |x − x0| < ρ

is calledtheinterval of convergence; it i s indicatedby thehatchedlinesin Figure5.1.1.Theseriesmay eitherconvergeor diverge when|x − x0| = ρ.

x

Seriesdiverges

Seriesdiverges

Series convergesabsolutely

x0 x0 +

Series mayconverge or diverge

x0 – ρ ρ

FIGURE 5.1.1 Theinterval of convergenceof apowerseries.

EXAMPLE

2

Determinetheradiusof convergenceof thepowerseries∞

∑

n=1

(x + 1)n

n2n .

Weapplytheratio test:

limn→∞

∣

∣

∣

∣

∣

(x + 1)n+1

(n + 1)2n+1

n2n

(x + 1)n

∣

∣

∣

∣

∣

=|x + 1|

2lim

n→∞

n

n + 1=

|x + 1|2

.

Thustheseriesconvergesabsolutelyfor |x + 1| < 2,or−3 < x < 1,anddivergesfor |x + 1| > 2. Theradiusof convergenceof thepower seriesis ρ = 2. Finally,wechecktheendpointsof theintervalof convergence.At x = 1 theseriesbecomestheharmonicseries

∞∑

n=1

1

n,

whichdiverges.At x = −3 we have∞

∑

n=1

(−3 + 1)n

n2n =∞

∑

n=1

(−1)n

n,

which converges,but doesnot convergeabsolutely. Theseriesis saidto convergeconditionally at x = −3. To summarize,the given power seriesconverges for−3 ≤ x < 1,anddivergesotherwise.It convergesabsolutelyfor −3 < x < 1,andhasa radiusof convergence2.

If∞∑

n=0an(x − x0)

n and∞∑

n=0bn(x − x0)

n convergeto f (x) andg(x), respectively,

for |x − x0| < ρ, ρ > 0, thenthefollowing aretruefor |x − x0| < ρ.6. Theseriescanbeaddedor subtractedtermwiseand

f (x) ± g(x) =∞

∑

n=0

(an ± bn)(x − x0)n.


7. Theseriescanbeformally multiplied and

f (x)g(x) =

[

∞∑

n=0

an(x − x0)n

] [

∞∑

n=0

bn(x − x0)n

]

=∞

∑

n=0

cn(x − x0)n,

wherecn = a0bn + a1bn−1 + · · · + anb0. Further, if g(x0) �= 0, the seriescanbeformally dividedand

f (x)

g(x)=

∞∑

n=0

dn(x − x0)n.

In mostcasesthecoefficientsdn canbemosteasilyobtainedby equatingcoeffi-cientsin theequivalent relation

∞∑

n=0

an(x − x0)n =

[

∞∑

n=0

dn(x − x0)n

] [

∞∑

n=0

bn(x − x0)n

]

=∞

∑

n=0

(

n∑

k=0

dkbn−k

)

(x − x0)n.

Also, in thecaseof division,theradiusof convergenceof theresultingpowerseriesmaybe lessthanρ.

8. The function f is continuousandhasderivatives of all ordersfor |x − x0| < ρ.Further, f ′, f ′′, . . . canbecomputedby differentiatingtheseriestermwise;thatis,

f ′(x) = a1 + 2a2(x − x0) + · · · + nan(x − x0)n−1 + · · ·

=∞

∑

n=1

nan(x − x0)n−1,

f ′′(x) = 2a2 + 6a3(x − x0) + · · · + n(n − 1)an(x − x0)n−2 + · · ·

=∞

∑

n=2

n(n − 1)an(x − x0)n−2,

andso forth,and eachof theseriesconvergesabsolutelyfor |x − x0| < ρ.9. Thevalueof an is given by

an =f (n)(x0)

n!.

TheseriesiscalledtheTaylor1 seriesfor thefunction f aboutx = x0.

10. If∞∑

n=0an(x − x0)

n =∞∑

n=0bn(x − x0)

n for each x, then an = bn for n = 0, 1,

2, 3, . . . . In particular, if∞∑

n=0an(x − x0)

n = 0 for eachx, then a0 = a1 = · · · =

an = · · · = 0.

1BrookTaylor (1685–1731)was theleadingEnglishmathematicianin thegenerationfollowing Newton.In 1715hepublisheda generalstatementof theexpansiontheoremthat is namedfor him, a resultthat is fundamentalinall branchesof analysis.He was also oneof thefoundersof thecalculusof finite differences,and was thefirst torecognizetheexistenceof singularsolutionsof differentialequations.


A function f thathasaTaylor seriesexpansionaboutx = x0,

f (x) =∞

∑

n=0

f (n)(x0)

n!(x − x0)

n,

with a radiusof convergenceρ > 0, is said to be analytic at x = x0. Accordingtostatements6 and7, if f andg areanalyticat x0, then f ± g, f · g, and f/g [providedthatg(x0) �= 0] areanalytic at x = x0.

Shift of Index of Summation. Theindex of summationinaninfiniteseriesisadummyparameterjust as the integrationvariablein a definite integral is a dummyvariable.Thusit is immaterialwhich letteris usedfor theindex of summation.For example,

∞∑

n=0

2nxn

n!=

∞∑

j =0

2 j x j

j !.

Justaswemakechangesof thevariableof integrationin a definiteintegral,wefindit convenientto make changesof summationindicesin calculatingseriessolutionsofdifferentialequations.We illustrateby several exampleshow to shift the summationindex.

EXAMPLE

3

Write∞∑

n=2anxn asaserieswhosefirst termcorrespondsto n = 0 ratherthann = 2.

Let m = n − 2; thenn = m + 2 and n = 2 correspondsto m = 0. Hence∞

∑

n=2

anxn =∞

∑

m=0

am+2xm+2. (1)

By writing out the first few termsof eachof theseseries,you can verify that theycontainpreciselythesameterms.Finally, in theserieson theright sideof Eq. (1), wecanreplacethedummyindex m by n, obtaining

∞∑

n=2

anxn =∞

∑

n=0

an+2xn+2. (2)

In effect,wehaveshiftedtheindex upwardby 2, andcompensatedby startingto countata level 2 lower thanoriginally.

EXAMPLE

4

Write theseries∞

∑

n=2

(n + 2)(n + 1)an(x − x0)n−2 (3)

asaserieswhosegenericterminvolves(x − x0)n ratherthan(x − x0)

n−2.Again, we shift the index by 2 so that n is replacedby n + 2 and startcounting2

lower. Weobtain∞

∑

n=0

(n + 4)(n + 3)an+2(x − x0)n. (4)

Youcanreadilyverify thatthetermsin theseries(3) and(4) areexactly thesame.


EXAMPLE

5

Write theexpression

x2∞

∑

n=0

(r + n)anxr +n−1 (5)

asaserieswhosegenericterminvolvesxr +n.First take thex2 insidethesummation,obtaining

∞∑

n=0

(r + n)anxr +n+1. (6)

Next, shift theindex down by 1andstartcounting1 higher. Thus∞

∑

n=0

(r + n)anxr +n+1 =∞

∑

n=1

(r + n − 1)an−1xr +n. (7)

Again, you caneasilyverify that the two seriesin Eq. (7) areidentical,and thatbothareexactly thesameas theexpression(5).

EXAMPLE

6

Assumethat∞

∑

n=1

nanxn−1 =∞

∑

n=0

anxn (8)

for all x, and determinewhatthis impliesaboutthecoefficientsan.We want to usestatement10 to equatecorrespondingcoefficientsin thetwo series.

In order to do this, we mustfirst rewrite Eq. (8) so that the seriesdisplay the samepowerof x in their genericterms.For instance,in theserieson theleft sideof Eq.(8),wecanreplacen by n + 1 and startcounting1 lower. ThusEq.(8) becomes

∞∑

n=0

(n + 1)an+1xn =∞

∑

n=0

anxn. (9)

Accordingto statement10 weconcludethat

(n + 1)an+1 = an, n = 0, 1, 2, 3, . . .

or

an+1 =an

n + 1, n = 0, 1, 2, 3, . . . (10)

Hence,choosingsuccessivevaluesof n in Eq. (10), we have

a1 = a0, a2 =a1

2=

a0

2, a3 =

a2

3=

a0

3!,

andso forth. In general,

an =a0

n!, n = 1, 2, 3, . . . . (11)

Thustherelation(8) determinesall the following coefficientsin termsof a0. Finally,usingthecoefficientsgivenby Eq.(11),weobtain

∞∑

n=0

anxn = a0

∞∑

n=0

xn

n!= a0ex,

wherewehave followedtheusualconvention that0! = 1.


PROBLEMS In eachof Problems1 through8 determinetheradiusof convergenceof thegivenpowerseries.

1.∞∑

n=0(x − 3)n 2.

∞∑

n=0

n

2n xn

3.∞∑

n=0

x2n

n!4.

∞∑

n=02nxn

5.∞∑

n=1

(2x + 1)n

n2 6.∞∑

n=1

(x − x0)n

n

7.∞∑

n=1

(−1)nn2(x + 2)n

3n 8.∞∑

n=1

n!xn

nn

In eachof Problems9 through16 determinetheTaylor seriesaboutthepoint x0 for thegivenfunction.Also determinetheradiusof convergenceof theseries.

9. sinx, x0 = 0 10. ex, x0 = 0

11. x, x0 = 1 12. x2, x0 = −1

13. ln x, x0 = 1 14.1

1 + x, x0 = 0

15.1

1 − x, x0 = 0 16.

1

1 − x, x0 = 2

17. Given thaty =∞∑

n=0nxn, computey′ andy′′ andwrite out thefirst four termsof eachseries

aswell asthecoefficientof xn in thegeneralterm.

18. Given that y =∞∑

n=0anxn, computey′ and y′′ and write out the first four termsof each

seriesaswell asthe coefficient of xn in the generalterm.Show that if y′′ = y, thenthecoefficientsa0 anda1 arearbitrary, and determinea2 anda3 in termsof a0 anda1. Showthatan+2 = an/(n + 2)(n + 1), n = 0, 1, 2, 3, . . . .

In eachof Problems19and 20 verify thegivenequation.

19.∞∑

n=0an(x − 1)n+1 =

∞∑

n=1an−1(x − 1)n

20.∞∑

k=0ak+1xk +

∞∑

k=0akxk+1 = a1 +

∞∑

k=1(ak+1 + ak−1)x

k

In eachof Problems21 through27 rewrite thegivenexpressionas a sumwhosegenericterminvolvesxn.

21.∞∑

n=2n(n − 1)anxn−2 22.

∞∑

n=0anxn+2

23. x∞∑

n=1nanxn−1 +

∞∑

k=0akxk 24. (1 − x2)

∞∑

n=2n(n − 1)anxn−2

25.∞∑

m=2m(m − 1)amxm−2 + x

∞∑

k=1kakxk−1 26.

∞∑

n=1nanxn−1 + x

∞∑

n=0anxn

27. x∞∑

n=2n(n − 1)anxn−2 +

∞∑

n=0anxn


28. Determinethean sothattheequation

∞∑

n=1

nanxn−1 + 2∞

∑

n=0

anxn = 0

is satisfied.Try to identify thefunctionrepresentedby theseries∞∑

n=0anxn.

5.2 Series Solutions near an Ordinary Point, Part I

In Chapter3wedescribedmethodsof solvingsecondorderlineardifferentialequationswith constantcoefficients.We now considermethodsof solving secondorder linearequationswhen the coefficients are functionsof the independentvariable. In thischapterwe will denotethe independentvariable by x. It is sufficient to considerthehomogeneousequation

P(x)d2y

dx2 + Q(x)dy

dx+ R(x)y = 0, (1)

sincetheprocedurefor thecorrespondingnonhomogeneousequationis similar.A wideclassof problemsin mathematicalphysicsleadsto equationsof theform(1)

having polynomialcoefficients;for example,theBesselequation

x2y′′ + xy′ + (x2 − ν2)y = 0,

whereν is a constant,and theLegendreequation

(1 − x2)y′′ − 2xy′ + α(α + 1)y = 0,

whereα isa constant.For thisreason,aswell astosimplify thealgebraiccomputations,we primarily considerthecasein which the functionsP, Q, and R arepolynomials.However, aswe will see,themethodof solutionis also applicablewhenP, Q, and Raregeneralanalyticfunctions.

For thepresent,then,supposethat P, Q, and R arepolynomials,and thatthey haveno commonfactors.Supposealso that we wish to solve Eq. (1) in the neighborhoodof a point x0. Thesolutionof Eq. (1) in an interval containingx0 is closelyassociatedwith thebehavior of P in thatinterval.

A point x0 suchthat P(x0) �= 0 is calledan ordinary point. Since P is continuous,it follows thatthereisan interval aboutx0 in which P(x) is never zero.In thatintervalwecandivide Eq.(1) by P(x) to obtain

y′′ + p(x)y′ + q(x)y = 0, (2)

where p(x) = Q(x)/P(x) andq(x) = R(x)/P(x) arecontinuousfunctions.Hence,accordingto theexistenceanduniquenessTheorem3.2.1,thereexistsin thatinterval auniquesolutionof Eq. (1) thatalsosatisfiestheinitial conditionsy(x0) = y0, y′(x0) =y′

0 for arbitrary valuesof y0 and y′0. In this and the following sectionwe discussthe

solutionof Eq. (1) in theneighborhoodof anordinarypoint.On the other hand,if P(x0) = 0, thenx0 is calleda singular point of Eq. (1). In

this caseat leastoneof Q(x0) andR(x0) is not zero.Consequently, at leastoneof the

ODE

5.2 Series Solutions near an Ordinary Point, Part I 239

coefficientsp andq in Eq.(2) becomesunboundedasx → x0, andthereforeTheorem3.2.1doesnot apply in this case.Sections5.4 through5.8dealwith finding solutionsof Eq. (1) in theneighborhoodof asingularpoint.

We now take up theproblemof solvingEq. (1) in theneighborhoodof anordinarypoint x0. We look for solutionsof theform

y = a0 + a1(x − x0) + · · · + an(x − x0)n + · · · =

∞∑

n=0

an(x − x0)n, (3)

and assumethat the seriesconverges in the interval |x − x0| < ρ for some ρ > 0.While atfirst sightit mayappearunattractive to seekasolution in theform of apowerseries,thisisactuallya convenientandusefulform for asolution.Within their intervalsof convergencepower seriesbehave very much like polynomialsand are easytomanipulatebothanalyticallyandnumerically. Indeed,even if wecanobtainasolutionin termsof elementaryfunctions,suchas exponentialor trigonometricfunctions,wearelikely to needa power seriesor someequivalent expressionif we want to evaluatethemnumericallyor to plot their graphs.

The most practicalway to determinethe coefficientsan is to substitutethe series(3) and its derivatives for y, y′, and y′′ in Eq. (1). The following examplesillustratethisprocess.Theoperations,suchasdifferentiation,thatareinvolved in theprocedureare justifiedso long as we staywithin the interval of convergence.The differentialequationsin theseexamplesarealsoof considerableimportancein their own right.

EXAMPLE

1

Findaseriessolutionof theequation

y′′ + y = 0, −∞ < x < ∞. (4)

Asweknow, twolinearlyindependentsolutionsof thisequationaresinx andcosx, soseriesmethodsarenotneededto solve thisequation.However, thisexampleillustratestheuseof power seriesin a relatively simplecase.For Eq. (4), P(x) = 1, Q(x) = 0,andR(x) = 1; henceevery point is anordinarypoint.

We look for asolutionin theform of apower seriesaboutx0 = 0,

y = a0 + a1x + a2x2 + · · · + anxn + · · · =∞

∑

n=0

anxn, (5)

andassumethattheseriesconvergesin someinterval |x| < ρ.Differentiatingtermby termyields

y′ = a1 + 2a2x + · · · + nanxn−1 + · · · =∞

∑

n=1

nanxn−1, (6)

y′′ = 2a2 + · · · + n(n − 1)anxn−2 + · · · =∞

∑

n=2

n(n − 1)anxn−2. (7)

Substitutingtheseries(5) and (7) for y andy′′ in Eq. (4) gives

∞∑

n=2

n(n − 1)anxn−2 +∞

∑

n=0

anxn = 0.


To combinethetwo serieswe needto rewrite at leastoneof themsothatbothseriesdisplaythesamegenericterm.Thus,in thefirst sum,weshift theindex of summationby replacingn by n + 2 and startingthesumat 0 ratherthan2. Weobtain

∞∑

n=0

(n + 2)(n + 1)an+2xn +∞

∑

n=0

anxn = 0

or∞

∑

n=0

[(n + 2)(n + 1)an+2 + an]xn = 0.

For this equationto be satisfiedfor all x, the coefficient of eachpower of x mustbezero;hence,weconcludethat

(n + 2)(n + 1)an+2 + an = 0, n = 0, 1, 2, 3, . . . . (8)

Equation(8) is referredto asarecurrence relation. Thesuccessivecoefficientscanbe evaluatedoneby oneby writing the recurrencerelationfirst for n = 0, then forn = 1, andso forth. In this exampleEq. (8) relateseachcoefficient to thesecondonebeforeit. Thustheeven-numberedcoefficients(a0, a2, a4, . . .) andtheodd-numberedones(a1, a3, a5, . . .) aredeterminedseparately. For theeven-numberedcoefficientswehave

a2 = −a0

2 · 1= −

a0

2!, a4 = −

a2

4 · 3= +

a0

4!, a6 = −

a4

6 · 5= −

a0

6!, . . . .

Theseresultssuggestthatin general,if n = 2k, then

an = a2k =(−1)k

(2k)!a0, k = 1, 2, 3, . . . . (9)

WecanproveEq.(9) by mathematicalinduction.Firstobservethatit is truefor k = 1.Next, assumethatit is truefor an arbitrary valueof k andconsiderthecasek + 1. Wehave

a2k+2 = −a2k

(2k + 2)(2k + 1)= −

(−1)k

(2k + 2)(2k + 1)(2k)!a0 =

(−1)k+1

(2k + 2)!a0.

HenceEq.(9) isalsotruefork + 1andconsequentlyit is truefor all positiveintegersk.Similarly, for theodd-numberedcoefficients

a3 = −a1

2 · 3= −

a1

3!, a5 = −

a3

5 · 4= +

a1

5!, a7 = −

a5

7 · 6= −

a1

7!, . . . ,

andin general,if n = 2k + 1, then2

an = a2k+1 =(−1)k

(2k + 1)!a1, k = 1, 2, 3, . . . . (10)

SubstitutingthesecoefficientsintoEq. (5), wehave

y = a0 + a1x −a0

2!x2 −

a1

3!x3 +

a0

4!x4 +

a1

5!x5 +

2Theresultgivenin Eq. (10) andothersimilar formulasin this chaptercanbeproved by an inductionargumentresemblingtheonejustgiven for Eq.(9).We assumethattheresultsareplausibleandomit theinductive argumenthereafter.


+ · · · +(−1)na0

(2n)!x2n +

(−1)na1

(2n + 1)!x2n+1 + · · ·

= a0

[

1 −x2

2!+

x4

4!+ · · · +

(−1)n

(2n)!x2n + · · ·

]

+ a1

[

x −x3

3!+

x5

5!+ · · · +

(−1)n

(2n + 1)!x2n+1 + · · ·

]

= a0

∞∑

n=0

(−1)n

(2n)!x2n + a1

∞∑

n=0

(−1)n

(2n + 1)!x2n+1. (11)

Now thatwehaveformallyobtainedtwo seriessolutionsof Eq.(4),wecantestthemfor convergence.Usingtheratiotest,it iseasytoshow thateachof theseriesinEq.(11)convergesfor all x, and this justifiesretroactively all the stepsusedin obtainingthesolutions.Indeed,we recognizethat the first seriesin Eq. (11) is exactly the Taylorseriesfor cosx aboutx = 0 and thesecondis theTaylor seriesfor sinx aboutx = 0.Thus,as expected,weobtainthesolutiony = a0 cosx + a1 sinx.

Noticethatno conditionsareimposedon a0 anda1; hencethey arearbitrary. FromEqs.(5) and (6) we seethat y and y′ evaluated at x = 0 are a0 anda1, respectively.Sincetheinitial conditionsy(0) andy′(0) canbechosenarbitrarily, it follows thata0anda1 shouldbe arbitrary until specificinitial conditionsarestated.

Figures5.2.1 and 5.2.2 show how the partial sums of the seriesin Eq. (11) ap-proximatecosx andsin x. As thenumberof termsincreases,the interval over whichthe approximationis satisfactorybecomeslonger, andfor eachx in this interval theaccuracy of theapproximationimproves.However, you shouldalwaysrememberthata truncatedpower seriesprovides only a local approximationof the solution in aneighborhoodof theinitial point x = 0; it cannotadequatelyrepresentthesolutionforlarge |x|.

2

2 4 6

1

–1

–2

8 10

y = cos x

y

x

n = 4 n = 8 n = 20n = 16n = 12

n = 2 n = 6 n = 10 n = 14 n = 18

FIGURE 5.2.1 Polynomialapproximationsto cosx. The value of n is the degree of theapproximatingpolynomial.


2

2 4 6

1

–1

–2

8 10

y

x

y = sin x

n = 5 n = 9 n = 13 n = 17 n = 21

n = 3 n = 7 n = 11 n = 15 n = 19

FIGURE 5.2.2 Polynomial approximationsto sinx. The value of n is the degree of theapproximatingpolynomial.

In Example1 we knew from thestartthatsin x andcosx form a fundamentalsetofsolutionsof Eq.(4). However, if wehadnotknown thisandhadsimplysolvedEq.(4)usingseriesmethods,wewouldstill haveobtainedthesolution(11). In recognitionofthefactthatthedifferentialequation(4) oftenoccursin applicationswe might decideto give thetwo solutionsof Eq. (11) specialnames;perhaps,

C(x) =∞

∑

n=0

(−1)n

(2n)!x2n, S(x) =

∞∑

n=0

(−1)n

(2n + 1)!x2n+1.

Then we might ask what propertiesthesefunctions have. For instance,it followsat once from the seriesexpansionsthat C(0) = 1, S(0) = 0, C(−x) = C(x), andS(−x) = −S(x). It is also easy to show that

S′(x) = C(x), C′(x) = −S(x).

Moreover, by calculatingwith theinfiniteseries3 wecanshow thatthefunctionsC(x)

andS(x) have all theusualanalyticalandalgebraicpropertiesof thecosineandsinefunctions,respectively.

Although you probablyfirst saw the sine and cosinefunctionsdefinedin a moreelementarymannerin termsof right triangles,it is interestingthatthesefunctionscanbe definedassolutionsof a certainsimplesecondorder linear differentialequation.To be precise,the function sin x canbe definedasthe uniquesolutionof the initialvalue problemy′′ + y = 0, y(0) = 0, y′(0) = 1; similarly, cosx canbedefinedastheuniquesolutionof the initial valueproblemy′′ + y = 0, y(0) = 1, y′(0) = 0. Manyotherfunctionsthatareimportantin mathematicalphysicsarealsodefinedassolutionsof certaininitial valueproblems.For most of thesefunctionsthereis no simpler ormoreelementaryway to approachthem.

3Suchan analysisis given in Section24 of K. Knopp, Theoryand Applicationsof Infinite Series(New York:Hafner, 1951).


EXAMPLE

2

Findaseriessolution in powersof x of Airy’ s4 equation

y′′ − xy = 0, −∞ < x < ∞. (12)

For this equation P(x) = 1, Q(x) = 0, and R(x) = −x; henceevery point is anordinarypoint.We assumethat

y =∞

∑

n=0

anxn, (13)

andthat the seriesconvergesin someinterval |x| < ρ. The seriesfor y′′ is given byEq.(7); as explainedin theprecedingexample,wecanrewrite it as

y′′ =∞

∑

n=0

(n + 2)(n + 1)an+2xn. (14)

Substitutingtheseries(13) and(14) for y andy′′ in Eq. (12),weobtain∞

∑

n=0

(n + 2)(n + 1)an+2xn = x∞

∑

n=0

anxn =∞

∑

n=0

anxn+1. (15)

Next, weshift theindex of summationin theseriesontheright sideof thisequationbyreplacingn by n − 1 andstartingthesummationat 1 ratherthanzero.Thuswehave

2 · 1a2 +∞

∑

n=1

(n + 2)(n + 1)an+2xn =∞

∑

n=1

an−1xn.

Again, for this equationto be satisfiedfor all x it is necessarythat thecoefficientsoflikepowersof x beequal;hencea2 = 0, andweobtaintherecurrencerelation

(n + 2)(n + 1)an+2 = an−1 for n = 1, 2, 3, . . . . (16)

Sincean+2 is given in terms of an−1, the a’s aredeterminedin stepsof three.Thusa0 determinesa3, which in turn determinesa6, . . . ; a1 determinesa4, which in turndeterminesa7, . . . ; and a2 determinesa5, which in turn determinesa8, . . . . Sincea2 = 0, we immediatelyconcludethata5 = a8 = a11 = · · · = 0.

For thesequencea0, a3, a6, a9, . . . wesetn = 1, 4, 7, 10, . . . in therecurrencerela-tion:

a3 =a0

2 · 3, a6 =

a3

5 · 6=

a0

2 · 3 · 5 · 6, a9 =

a6

8 · 9=

a0

2 · 3 · 5 · 6 · 8 · 9, . . . .

For this sequenceof coefficients it is convenient to write a formula for a3n, n =1, 2, 3, . . . . Theprecedingresultssuggestthegeneralformula

a3n =a0

2 · 3 · 5 · 6 · · · (3n − 4)(3n − 3)(3n − 1)(3n), n = 1, 2, . . . .

For the sequencea1, a4, a7, a10, . . . , we set n = 2, 5, 8, 11, . . . in the recurrencerelation:

a4 =a1

3 · 4, a7 =

a4

6 · 7=

a1

3 · 4 · 6 · 7, a10 =

a7

9 · 10=

a1

3 · 4 · 6 · 7 · 9 · 10, . . . .

4SirGeorgeAiry (1801–1892),anEnglishastronomerandmathematician,wasdirectorof theGreenwichObserva-tory from 1835to 1881.OnereasonAiry’ sequationis of interestis thatfor x negativethesolutionsareoscillatory,similar to trigonometricfunctions,and for x positive they aremonotonic,similar to hyperbolicfunctions.Canyouexplainwhy it is reasonableto expectsuchbehavior?


In thesamewayasbefore wefind that

a3n+1 =a1

3 · 4 · 6 · 7 · · · (3n − 3)(3n − 2)(3n)(3n + 1), n = 1, 2, 3, . . . .

Thegeneralsolutionof Airy’s equationis

y = a0

[

1 +x3

2 · 3+

x6

2 · 3 · 5 · 6+ · · · +

x3n

2 · 3 · · · (3n − 1)(3n)+ · · ·

]

+ a1

[

x +x4

3 · 4+

x7

3 · 4 · 6 · 7+ · · · +

x3n+1

3 · 4 · · · (3n)(3n + 1)+ · · ·

]

= a0

[

1 +∞

∑

n=1

x3n

2 · 3 · · · (3n − 1)(3n)

]

+ a1

[

x +∞

∑

n=1

x3n+1

3 · 4 · · · (3n)(3n + 1)

]

.

(17)

Havingobtainedthesetwoseriessolutions,wecannow investigatetheirconvergence.Becauseof the rapid growth of the denominatorsof the termsin the series(17), wemight expecttheseseriesto have alarge radiusof convergence.Indeed,it is easytousetheratio testto show thatboththeseseriesconvergefor all x; seeProblem20.

Assumingfor themomentthattheseriesdo converge for all x, let y1 andy2 denotethe functionsdefinedby the expressionsin the first and secondsetsof brackets,respectively, in Eq.(17).Then,by first choosinga0 = 1, a1 = 0 andthen a0 = 0, a1 =1, it followsthaty1 andy2 areindividually solutionsof Eq.(12).Noticethaty1 satisfiesthe initial conditionsy1(0) = 1, y′

1(0) = 0 and that y2 satisfiesthe initial conditionsy2(0) = 0, y′

2(0) = 1. Thus W(y1, y2)(0) = 1 �= 0, and consequentlyy1 and y2 arelinearly independent.Hencethegeneralsolutionof Airy’s equationis

y = a0y1(x) + a1y2(x), −∞ < x < ∞.

In Figures5.2.3and 5.2.4,respectively, we show thegraphsof thesolutionsy1 andy2 of Airy’ s equation,aswell asgraphsof several partial sumsof the two seriesinEq. (17). Again, the partial sumsprovide local approximationsto the solutions in aneighborhoodof the origin. While the quality of the approximationimproves as thenumberof termsincreases,nopolynomialcanadequatelyrepresenty1 andy2 for large|x|. A practicalway to estimatetheinterval in whichagiven partial sumis reasonablyaccurateis to comparethe graphsof that partial sum andthe next one,obtainedbyincludingonemoreterm.As soonasthegraphsbegin to separatenoticeably, onecanbeconfidentthattheoriginal partial sumis no longeraccurate.For example,in Figure5.2.3the graphsfor n = 24 andn = 27 begin to separateat about x = −9/2. Thus,beyondthispoint, thepartialsumof degree24 is worthlessasan approximationto thesolution.

Observe that both y1 and y2 are monotonefor x > 0 and oscillatory for x < 0.Onecanalsoseefrom the figuresthat the oscillationsarenot uniform, but decayinamplitudeandincreaseinfrequency asthedistancefromtheoriginincreases.Incontrastto Example1, thesolutionsy1 andy2 of Airy’ s equationare not elementaryfunctionsthatyou have alreadyencounteredin calculus.However, becauseof their importancein somephysicalapplications,thesefunctionshave beenextensively studiedandtheirpropertiesarewell known to appliedmathematiciansandscientists.


2

2

–2

–2–4–6–8

y

x

n = 48 36 24 12

n = 45 33 21 9

39 27 15

3

42 30 18 6

y = y1(x)

n ≥ 6n = 3

–10

FIGURE 5.2.3 Polynomialapproximationsto thesolutiony1(x) of Airy’ sequation.Thevalueof n is thedegreeof theapproximatingpolynomial.

2

2

–2

–2–6–8–10 x

n = 46n ≥ 4

34

40 16 4

10

n = 49

31 19 7

25 13

28

22

43

37

–4

y

y = y2(x)

FIGURE 5.2.4 Polynomialapproximationsto thesolutiony2(x) of Airy’ sequation.Thevalueof n is thedegreeof theapproximatingpolynomial.

EXAMPLE

3

Findasolutionof Airy’s equationin powersof x − 1.Thepoint x = 1 is an ordinarypoint of Eq. (12),and thuswe look for a solutionof

theform

y =∞

∑

n=0

an(x − 1)n,

wherewe assumethattheseriesconvergesin someinterval |x − 1| < ρ. Then

y′ =∞

∑

n=1

nan(x − 1)n−1 =∞

∑

n=0

(n + 1)an+1(x − 1)n,


and

y′′ =∞

∑

n=2

n(n − 1)an(x − 1)n−2 =∞

∑

n=0

(n + 2)(n + 1)an+2(x − 1)n.

Substitutingfor y andy′′ in Eq. (12),weobtain∞

∑

n=0

(n + 2)(n + 1)an+2(x − 1)n = x∞

∑

n=0

an(x − 1)n. (18)

Now to equatethe coefficients of like powers of (x − 1) we must expressx, thecoefficientof y in Eq. (12), in powers of x − 1; thatis,wewrite x = 1 + (x − 1). Notethatthis is preciselytheTaylor seriesfor x aboutx = 1. ThenEq.(18) takes theform

∞∑

n=0

(n + 2)(n + 1)an+2(x − 1)n = [1 + (x − 1)]∞

∑

n=0

an(x − 1)n

=∞

∑

n=0

an(x − 1)n +∞

∑

n=0

an(x − 1)n+1.

Shifting theindex of summationin thesecondserieson theright gives∞

∑

n=0

(n + 2)(n + 1)an+2(x − 1)n =∞

∑

n=0

an(x − 1)n +∞

∑

n=1

an−1(x − 1)n.

Equatingcoefficientsof likepowersof x − 1, weobtain

2a2 = a0,

(3 · 2)a3 = a1 + a0,

(4 · 3)a4 = a2 + a1,

(5 · 4)a5 = a3 + a2,

...

Thegeneralrecurrencerelationis

(n + 2)(n + 1)an+2 = an + an−1 for n ≥ 1. (19)

Solvingfor thefirst few coefficientsan in termsof a0 anda1, we find that

a2 =a0

2, a3 =

a1

6+

a0

6, a4 =

a2

12+

a1

12=

a0

24+

a1

12, a5 =

a3

20+

a2

20=

a0

30+

a1

120.

Hence

y = a0

[

1 +(x − 1)2

2+

(x − 1)3

6+

(x − 1)4

24+

(x − 1)5

30+ · · ·

]

+ a1

[

(x − 1) +(x − 1)3

6+

(x − 1)4

12+

(x − 1)5

120+ · · ·

]

. (20)

In general,whenthe recurrencerelationhasmorethantwo terms,as in Eq. (19),thedeterminationof a formulafor an in termsa0 anda1 will befairly complicated,ifnot impossible.In thisexamplesucha formulaisnot readilyapparent.Lackingsuchaformula,we cannottestthe two seriesin Eq. (20) for convergenceby directmethods


suchastheratio test.However, even without knowing theformulafor an we shallseein Section5.3 thatit is possibleto establishthattheseriesin Eq. (20) convergefor allx, and furtherdefinefunctionsy3 andy4 thatarelinearly independentsolutionsof theAiry equation(12).Thus

y = a0y3(x) + a1y4(x)

is thegeneralsolutionof Airy’sequationfor −∞ < x < ∞.

It is worth emphasizing,aswe saw in Example3, that if we look for a solutionof

Eq. (1) of the form y =∞∑

n=0an(x − x0)

n, thenthecoefficients P(x), Q(x), and R(x)

in Eq. (1) mustalsobeexpressedin powersof x − x0. Alternatively, we canmake thechangeof variablex − x0 = t , obtaininganew differentialequationfor y asafunction

of t , and thenlook for solutionsof this new equationof the form∞∑

n=0antn. When we

havefinishedthecalculations,we replacet by x − x0 (seeProblem19).In Examples2 and 3 we have found twosetsof solutionsof Airy’ s equation.The

functionsy1 andy2 definedby theseriesin Eq. (17)arelinearly independentsolutionsof Eq. (12) for all x, and this is alsotrue for the functionsy3 and y4 definedby theseriesin Eq. (20). Accordingto the generaltheoryof secondorder linear equationseachof thefirsttwo functionscanbeexpressedasalinearcombinationof thelattertwofunctionsandviceversa—aresultthatis certainlynotobviousfrom an examinationoftheseriesalone.

Finally, we emphasizethatit is notparticularlyimportantif, as in Example3,weareunableto determinethegeneralcoefficient an in termsof a0 anda1. Whatis essentialis thatwe candetermineasmanycoefficientsas we want. Thuswe canfind as manytermsin thetwo seriessolutionsaswe want, even if we cannotdeterminethegeneralterm. While the taskof calculatingseveral coefficients in a power seriessolution isnot difficult, it canbetedious.A symbolicmanipulationpackage canbevery helpfulhere;someare ableto find a specifiednumberof termsin a power seriessolutioninresponseto asinglecommand.With asuitablegraphicspackageonecanalsoproduceplotssuchas thoseshown in thefiguresin this section.

PROBLEMS In eachof Problems1 through14 solve the given differentialequationby meansof a powerseriesaboutthegivenpointx0. Find therecurrencerelation;alsofind thefirst four termsin eachof two linearly independentsolutions(unlesstheseriesterminatessooner).If possible,find thegeneraltermin eachsolution.

1. y′′ − y = 0, x0 = 0 2. y′′ − xy′ − y = 0, x0 = 03. y′′ − xy′ − y = 0, x0 = 1 4. y′′ + k2x2y = 0, x0 = 0, k a constant5. (1 − x)y′′ + y = 0, x0 = 0 6. (2 + x2)y′′ − xy′ + 4y = 0, x0 = 07. y′′ + xy′ + 2y = 0, x0 = 0 8. xy′′ + y′ + xy = 0, x0 = 19. (1 + x2)y′′ − 4xy′ + 6y = 0, x0 = 0 10. (4 − x2)y′′ + 2y = 0, x0 = 0

11. (3 − x2)y′′ − 3xy′ − y = 0, x0 = 0 12. (1 − x)y′′ + xy′ − y = 0, x0 = 013. 2y′′ + xy′ + 3y = 0, x0 = 0 14. 2y′′ + (x + 1)y′ + 3y = 0, x0 = 2


In eachof Problems15 through18:(a) Find thefirst fivenonzerotermsin thesolutionof thegiven initial valueproblem.(b) Plot thefour-termandthefive-termapproximationsto thesolutionon thesameaxes.(c) Fromtheplot in part (b) estimatetheinterval in which thefour-termapproximationisreasonablyaccurate.

� 15. y′′ − xy′ − y = 0, y(0) = 2, y′(0) = 1; seeProblem2

� 16. (2 + x2)y′′ − xy′ + 4y = 0, y(0) = −1, y′(0) = 3; seeProblem6

� 17. y′′ + xy′ + 2y = 0, y(0) = 4, y′(0) = −1; seeProblem7

� 18. (1 − x)y′′ + xy′ − y = 0, y(0) = −3, y′(0) = 2; seeProblem1219. By makingthe changeof variablex − 1 = t andassumingthat y is a power seriesin t ,

find two linearly independentseriessolutionsof

y′′ + (x − 1)2y′ + (x2 − 1)y = 0

in powersof x − 1. Show thatyou obtainthesameresultdirectly by assumingthat y is aTaylor seriesin powersof x − 1 and alsoexpressingthe coefficient x2 − 1 in powersofx − 1.

20. Show directly, using the ratio test,that the two seriessolutionsof Airy’ s equationaboutx = 0 converge for all x; seeEq.(17) of thetext.

21. The Hermite Equation. Theequation

y′′ − 2xy′ + λy = 0, −∞ < x < ∞,

whereλ is a constant,is known asthe Hermite5 equation.It is an importantequationinmathematicalphysics.(a) Find thefirst four termsin eachof two linearly independentsolutionsaboutx = 0.(b) Observe that if λ is a nonnegative even integer, thenoneor the other of the seriessolutionsterminatesandbecomesa polynomial.Find thepolynomialsolutionsfor λ = 0,2, 4, 6, 8,and10. Note that eachpolynomial is determinedonly up to a multiplicativeconstant.(c) TheHermitepolynomialHn(x) is definedasthepolynomialsolution of theHermiteequationwith λ = 2n for which thecoefficientof xn is 2n. Find H0(x), . . . , H5(x).

22. Considertheinitial valueproblemy′ =√

1 − y2, y(0) = 0.(a) Show that y = sinx is thesolutionof this initial valueproblem.(b) Look for a solutionof the initial valueproblemin the form of a power seriesaboutx = 0. Find thecoefficientsup to thetermin x3 in thisseries.

In eachof Problems23 through28 plot several partial sumsin a seriessolutionof the giveninitial valueproblemaboutx = 0, therebyobtaininggraphsanalogousto thosein Figures5.2.1through5.2.4.

� 23. y′′ − xy′ − y = 0, y(0) = 1, y′(0) = 0; seeProblem2

� 24. (2 + x2)y′′ − xy′ + 4y = 0, y(0) = 1, y′(0) = 0; seeProblem6

� 25. y′′ + xy′ + 2y = 0, y(0) = 0, y′(0) = 1; seeProblem7

� 26. (4 − x2)y′′ + 2y = 0, y(0) = 0, y′(0) = 1; seeProblem10

� 27. y′′ + x2y = 0, y(0) = 1, y′(0) = 0; seeProblem4

� 28. (1 − x)y′′ + xy′ − 2y = 0, y(0) = 0, y′(0) = 1

5CharlesHermite (1822–1901)was an influential Frenchanalyst and algebraist.He introducedthe Hermitefunctionsin 1864,andshowedin 1873thate is a transcendentalnumber(thatis,e is notarootof any polynomialequationwith rationalcoefficients).His nameis also associatedwith Hermitianmatrices(seeSection7.3), someof whosepropertieshediscovered.

5.3 Series Solutions near an Ordinary Point, Part II 249

5.3 Series Solutions near an Ordinary Point, Part II

In theprecedingsectionwe consideredtheproblemof finding solutionsof

P(x)y′′ + Q(x)y′ + R(x)y = 0, (1)

where P, Q, and R are polynomials,in the neighborhoodof an ordinary point x0.AssumingthatEq. (1) doeshave asolution y = φ(x), and that φ hasaTaylor series

y = φ(x) =∞

∑

n=0

an(x − x0)n, (2)

whichconvergesfor |x − x0| < ρ, whereρ > 0,wefoundthatthean canbedeterminedby directly substitutingtheseries(2) for y in Eq.(1).

Letusnow considerhow wemight justify thestatementthatif x0 is anordinarypointof Eq. (1), thenthereexist solutionsof theform (2). We also considerthequestionoftheradiusof convergenceof sucha series.In doingthis we areledto a generalizationof thedefinition of an ordinarypoint.

Suppose,then,that thereis a solution of Eq. (1) of theform (2). By differentiatingEq.(2) m timesandsettingx equalto x0 it follows that

m!am = φ(m)(x0).

Hence,to computean in theseries(2), we mustshow thatwe candetermineφ(n)(x0)

for n = 0, 1, 2, . . . from thedifferentialequation(1).Supposethat y = φ(x) is a solution of Eq. (1) satisfying the initial conditions

y(x0) = y0, y′(x0) = y′0. Then a0 = y0 and a1 = y′

0. If we are solely interestedinfinding a solution of Eq. (1) without specifyingany initial conditions,thena0 anda1remainarbitrary. To determineφ(n)(x0) andthecorrespondingan for n = 2, 3, . . . , weturn to Eq.(1). Sinceφ is asolutionof Eq. (1), wehave

P(x)φ′′(x) + Q(x)φ′(x) + R(x)φ(x) = 0.

For theinterval aboutx0 for which P is nonvanishingwecanwrite thisequationin theform

φ′′(x) = −p(x)φ′(x) − q(x)φ(x), (3)

where p(x) = Q(x)/P(x) andq(x) = R(x)/P(x). Setting x equalto x0 in Eq. (3)gives

φ′′(x0) = −p(x0)φ′(x0) − q(x0)φ(x0).

Hencea2 is given by

2!a2 = φ′′(x0) = −p(x0)a1 − q(x0)a0. (4)

To determinea3 wedifferentiateEq.(3) andthenset x equalto x0, obtaining

3!a3 = φ′′′(x0) = −[ pφ′′ + (p′ + q)φ′ + q′φ]∣

∣

∣

x=x0

= −2!p(x0)a2 − [ p′(x0) + q(x0)]a1 − q′(x0)a0. (5)

Substitutingfor a2 from Eq. (4) givesa3 in termsof a1 anda0. Since P, Q, and R arepolynomialsandP(x0) �= 0,all of thederivativesof p andq exist at x0. Hence,wecan

ODE


continueto differentiateEq.(3) indefinitely, determiningaftereachdifferentiationthesuccessivecoefficientsa4, a5, . . . by settingx equalto x0.

Notice that the importantproperty that we used in determiningthe an was thatwe couldcomputeinfinitely many derivatives of thefunctionsp andq. It might seemreasonableto relaxourassumptionthatthefunctionsp andq areratiosof polynomials,and simply requirethat they be infinitely differentiablein the neighborhoodof x0.Unfortunately, this conditionis too weakto ensurethatwe canprove theconvergenceof the resultingseriesexpansionfor y = φ(x). What is neededis to assumethat thefunctions p andq areanalytic at x0; that is, they have Taylor seriesexpansionsthatconvergeto themin someinterval aboutthepoint x0:

p(x) = p0 + p1(x − x0) + · · · + pn(x − x0)n + · · · =

∞∑

n=0

pn(x − x0)n, (6)

q(x) = q0 + q1(x − x0) + · · · + qn(x − x0)n + · · · =

∞∑

n=0

qn(x − x0)n. (7)

With this idea in mind we can generalizethe definitionsof an ordinary point anda singular point of Eq. (1) as follows: If the functions p = Q/P andq = R/P areanalytic at x0, then the point x0 is said to be anordinary point of the differentialequation(1); otherwise,it isa singular point.

Now let usturn to thequestionof theinterval of convergenceof theseriessolution.Onepossibility is actuallyto computethe seriessolutionfor eachproblemandthento apply oneof the testsfor convergenceof an infinite seriesto determineits radiusof convergence.However, the questioncanbe answeredat oncefor a wide classofproblemsby thefollowing theorem.

Theorem 5.3.1 If x0 is anordinarypoint of thedifferentialequation(1),

P(x)y′′ + Q(x)y′ + R(x)y = 0,

that is, if p = Q/P andq = R/P areanalytic at x0, then the generalsolution ofEq. (1) is

y =∞

∑

n=0

an(x − x0)n = a0y1(x) + a1y2(x), (8)

wherea0 anda1 arearbitrary, andy1 andy2 arelinearly independentseriessolutionsthat are analytic at x0. Further, the radius of convergencefor eachof the seriessolutionsy1 andy2 is at leastas largeastheminimum of theradii of convergenceoftheseriesfor p andq.

Noticefrom theform of theseriessolution that y1(x) = 1 + b2(x − x0)2 + · · · and

y2(x) = (x − x0) + c2(x − x0)2 + · · · . Hencey1 is the solution satisfying the initial

conditionsy1(x0) = 1, y′1(x0) = 0,andy2 is thesolutionsatisfyingtheinitial conditions

y2(x0) = 0, y′2(x0) = 1. Also note that while the calculationof the coefficients by

successively differentiatingthedifferentialequationisexcellentin theory, it isusuallynotapracticalcomputationalprocedure.Rather, oneshouldsubstitutetheseries(2) for


y in thedifferentialequation(1) anddeterminethecoefficientssothatthedifferentialequationis satisfied,asin theexamplesin theprecedingsection.

We will not prove this theorem,which in a slightly more generalform is due toFuchs.6 What is important for our purposesis that thereis a seriessolution of theform (2), andthat theradiusof convergenceof theseriessolutioncannotbelessthanthesmallerof theradii of convergenceof theseriesfor p andq; hencewe needonlydeterminethese.

Thiscanbedonein oneof two ways.Again,onepossibilityis simplyto computethepower seriesfor p andq, and thento determinetheradii of convergenceby usingoneof the convergencetestsfor infinite series.However, thereis an easierway when P,Q, and R arepolynomials.It is shown in thetheoryof functionsof a complex variablethat theratio of two polynomials,say Q/P, hasa convergentpower seriesexpansionabouta point x = x0 if P(x0) �= 0. Further, if we assumethatany factorscommontoQ and P have beencanceled,thenthe radiusof convergenceof thepower seriesforQ/P aboutthepoint x0 is preciselythedistancefrom x0 to thenearestzeroof P. Indeterminingthis distancewe mustrememberthat P(x) = 0 may have complex roots,andthesemustalsobeconsidered.

EXAMPLE

1

Whatis theradiusof convergenceof theTaylor seriesfor (1 + x2)−1 aboutx = 0?Oneway to proceedis to find theTaylor seriesin question,namely,

1

1 + x2 = 1 − x2 + x4 − x6 + · · · + (−1)nx2n + · · · .

Thenit canbe verified by the ratio testthatρ = 1. Anotherapproachis to notethatthezerosof 1 + x2 arex = ±i . Sincethedistancefrom 0 to i or to −i in thecomplexplaneis 1, theradiusof convergenceof thepower seriesaboutx = 0 is 1.

EXAMPLE

2

Whatis theradiusof convergenceof theTaylorseriesfor (x2 − 2x + 2)−1 aboutx = 0?About x = 1?

First noticethat

x2 − 2x + 2 = 0

hassolutionsx = 1 ± i . Thedistancefrom x = 0 to either x = 1 + i or x = 1 − i inthecomplex planeis

√2;hencetheradiusof convergenceof theTaylor seriesexpansion

∞∑

n=0anxn aboutx = 0 is

√2.

Thedistancefrom x = 1 to either x = 1 + i or x = 1 − i is 1; hencetheradiusof

convergenceof theTaylor seriesexpansion∞∑

n=0bn(x − 1)n aboutx = 1 is 1.

6ImmanuelLazarusFuchs(1833–1902)wasastudentandlateraprofessorat theUniversityof Berlin. Heprovedthe resultof Theorem5.3.1 in 1866. His most importantresearchwas on singular points of linear differentialequations.He recognizedthe significanceof regular singular points (Section5.4), and equationswhoseonlysingularities,includingthepointat infinity, are regularsingularpointsareknown asFuchsianequations.


Accordingto Theorem5.3.1theseriessolutionsof theAiry equationin Examples2 and 3 of theprecedingsectionconverge for all valuesof x andx − 1, respectively,sincein eachproblemP(x) = 1 and henceis never zero.

A seriessolution may converge for a wider rangeof x thanindicatedby Theorem5.3.1,sothetheoremactuallygives only a lower bound ontheradiusof convergenceof the seriessolution. This is illustratedby the Legendrepolynomialsolution of theLegendreequationgivenin thenext example.

EXAMPLE

3

Determinealowerboundfor theradiusof convergenceof seriessolutionsaboutx = 0for theLegendreequation

(1 − x2)y′′ − 2xy′ + α(α + 1)y = 0,

whereα is a constant.Note that P(x) = 1 − x2, Q(x) = −2x, and R(x) = α(α + 1) are polynomials,

andthatthezerosof P, namely, x = ±1, are a distance 1 from x = 0. Hencea series

solutionof the form∞∑

n=0anxn convergesfor |x| < 1 at least,andpossiblyfor larger

values of x. Indeed,it canbe shown that if α is a positive integer, oneof the seriessolutionsterminatesafter a finite numberof termsandhenceconvergesnot just for|x| < 1 but for all x. For example,if α = 1, the polynomial solution is y = x. SeeProblems22 through29 at theend of this sectionfor a more completediscussionoftheLegendreequation.

EXAMPLE

4

Determinea lower bound for the radius of convergenceof seriessolutions of thedifferentialequation

(1 + x2)y′′ + 2xy′ + 4x2y = 0 (9)

aboutthepoint x = 0; aboutthepoint x = −12.

Again P, Q, and R arepolynomials,and P haszerosat x = ±i . Thedistancein the

complex planefrom 0 to ±i is 1, andfrom −12 to ±i is

√

1 + 14 =

√5/2. Hencein the

first casetheseries∞∑

n=0anxn convergesat leastfor |x| < 1, andin thesecondcasethe

series∞∑

n=0bn(x + 1

2)n convergesat leastfor |x + 1

2| <√

5/2.

An interestingobservation thatwe canmake aboutEq. (9) follows from Theorems3.2.1and 5.3.1.Supposethat initial conditionsy(0) = y0 and y′(0) = y′

0 aregiven.Since1 + x2 �= 0 for all x, we know from Theorem3.2.1 that thereexists a uniquesolutionof the initial valueproblemon −∞ < x < ∞. On theotherhand,Theorem

5.3.1only guaranteesa seriessolution of the form∞∑

n=0anxn (with a0 = y0, a1 = y′

0)

for −1 < x < 1. The uniquesolutionon the interval −∞ < x < ∞ may not have apower seriesaboutx = 0 thatconvergesfor all x.

EXAMPLE

5

Canwedetermineaseriessolutionaboutx = 0 for thedifferentialequation

y′′ + (sinx)y′ + (1 + x2)y = 0,

andif so, whatis theradiusof convergence?


For thisdifferentialequation,p(x) = sinx andq(x) = 1 + x2. Recallfrom calculusthatsinx hasaTaylor seriesexpansionaboutx = 0 thatconvergesfor all x. Further, qalsohasaTaylor seriesexpansionaboutx = 0,namely, q(x) = 1 + x2, thatconverges

for all x. Thus there is a seriessolution of the form y =∞∑

n=0anxn with a0 and a1

arbitrary, andtheseriesconvergesfor all x.

PROBLEMS In eachof Problems1 through4 determineφ′′(x0), φ′′′(x0), andφ iv(x0) for thegiven point x0 if

y = φ(x) is asolutionof thegiven initial valueproblem.

1. y′′ + xy′ + y = 0; y(0) = 1, y′(0) = 02. y′′ + (sinx)y′ + (cosx)y = 0; y(0) = 0, y′(0) = 13. x2y′′ + (1 + x)y′ + 3(ln x)y = 0; y(1) = 2, y′(1) = 04. y′′ + x2y′ + (sinx)y = 0; y(0) = a0, y′(0) = a1

In eachof Problems5 through8 determinealowerboundfor theradiusof convergenceof seriessolutionsabouteachgiven point x0 for thegiven differentialequation.

5. y′′ + 4y′ + 6xy = 0; x0 = 0, x0 = 46. (x2 − 2x − 3)y′′ + xy′ + 4y = 0; x0 = 4, x0 = −4, x0 = 07. (1 + x3)y′′ + 4xy′ + y = 0; x0 = 0, x0 = 28. xy′′ + y = 0; x0 = 1

9. Determinea lowerboundfor theradiusof convergenceof seriessolutionsaboutthegivenx0 for eachof thedifferentialequationsin Problems1 through14of Section5.2.

10. The Chebyshev Equation. TheChebyshev7 differentialequationis

(1 − x2)y′′ − xy′ + α2y = 0,

whereα is a constant.(a) Determinetwo linearly independentsolutionsin powersof x for |x| < 1.(b) Show thatif α is anonnegative integern, thenthereis apolynomialsolutionof degreen. Thesepolynomials,whenproperlynormalized,arecalledtheChebyshev polynomials.They are very useful in problemsrequiring a polynomial approximationto a functiondefinedon−1 ≤ x ≤ 1.(c) Find apolynomialsolution for eachof thecasesα = n = 0, 1, 2,and3.

For eachof thedifferentialequationsin Problems11through14find thefirst four nonzerotermsin eachof two linearly independentpowerseriessolutionsabouttheorigin. Whatdo youexpecttheradiusof convergenceto befor eachsolution?

11. y′′ + (sinx)y = 0 12. ex y′′ + xy = 013. (cosx)y′′ + xy′ − 2y = 0 14. e−x y′′ + ln(1 + x)y′ − xy = 0

15. Supposethatyou aretold thatx andx2 aresolutionsof a differentialequationP(x)y′′ +Q(x)y′ + R(x)y = 0. Can you say whetherthe point x = 0 is an ordinary point or asingularpoint?Hint: UseTheorem3.2.1,andnotethevaluesof x andx2 at x = 0.

7Pafnuty L. Chebyshev (1821–1894),professorat Petersburg University for 35 yearsand the most influentialnineteenthcenturyRussianmathematician,foundedtheso-called“Petersburg school,” whichproducedalonglineof distinguishedmathematicians.Hisstudyof Chebyshev polynomialsbeganabout1854aspartof an investigationof theapproximationof functionsby polynomials.Chebyshev is alsoknown for his work in numbertheoryandprobability.

chapter 5 series solutions of second order linear equations

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