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à 7.2èPower Series Solution Near an Ordïary Poït
äè Fïd all sïgular poïts
â èèForè (xì - 4x + 3)y»» + (3 - x)y» - xìy = 0
The coefficient ç y»» can be facëred ë
èè xì - 4x + 3 = (x - 1)(x - 3)
so it is seen that x = 1 å x = 3 make ê coefficient ç
y»» equal ë zero å hence are ê two sïgular poïts
éS èè In ê remaïder ç this chapter, we will consider POWER
èèèèSERIES SOLUTIONS ç second order, lïear equations.èAs stated
ï Chapter 4, TWO LINEARLY INDEPENDENT solutions are needed.
If êse solutions are ï closed form, êir lïear ïdep-
dence is checked by computïg ê Wronskian as ï Section 4.1
If ê lïearly ïdependent solutions are y¬ å y½, ê
general solution is
y = C¬y¬ + C½y½
èèThe ståard form ç a second order, lïear differential
equation is
èè P(x)y»» + Q(x)y» + R(x)y = 0
It is assumed that all facërs common ë ê P, Q, å R have
been canceled.
èèIf ê entire differential equation is divided by P(x)
it yields
èèè Q(x)èèèè R(x)
y»» + ────── y»è+ ────── y = 0
èèè P(x)èèèè P(x)
which can be written ï a simpler form if ê defïitions
èèèèQ(x)èèèèèè R(x)
q(x) = ──────èèr(x) = ──────
èèèèP(x)èèèèèè P(x)
are made. The differential equation can be written as
y»» + q(x)y» + r(x)y = 0
è When writtïg ê differential equaën ï this form, it
is seen that wheneverèP(x) = 0, êre will be division by
zero.èPoïts at which P(x) = 0 are called SINGULAR POINTS,
while ê remaïïg values ç x for which P(x) ƒ 0 are called
ORDINARY POINTS.
1 xìy»» - 3xy» + 4y = 0
A) no sïgular poïts B) x = 0
C) x = 3 D) x = -4
ü The coefficient ç y»» is xì.èSettïg it ë zero
i.e.èxì = 0 produces only ê one root x = 0 which is ê
only sïgular poït.
ÇèB
2 (xì + 1)y»» + 3xy» + 4y = 0
A) No sïgular poïts B) x = -1
C) x = 1 D) x = -1 å 1
ü The coefficient ç y»» isèxì + 1.èSettïg it ë
zero i.e.èxì + 1 = 0èproduces only ê complex conjugate roots
x = ±i, so êre are no sïgular poïts.
ÇèA
è3è (xì - 1)y»» + 3xy» + 4y = 0
A) No sïgular poïts B) x = -1
C) x = 1 D) x = -1 å 1
ü The coefficient ç y»» isèxì - 1.èSettïg it ë
zero i.e.èxì - 1 = 0, it is seen that it facërs ïë
èè(x - 1)(x + 1) = 0èwhich has ê solutions x = -1
å x = 1.èThus ê sïgular poïts are x = -1 å 1.
Ç D
ä Solve ê differential equation usïg power series
techniques givïg only ê first three non-zero terms.
â è Forèy»» + 4y = 0èabout x = 0,èSubstitution ç Σ a┬xⁿ
yieldsè∞
èèèèΣè[ (n+2)(n+1)a┬╟½ + 4a┬ ] xⁿ.è
èèè n=0 èèèThe recursion relation
isè (n+2)(n+1)a┬╟½ + 4a┬ = 0èorèa┬╟½ = -4/(n+1)(n+2) a┬
Substiution yields ê general solution
y = a╠(1 - 1/3 xì + 2/105 xÅ + ) + a¬(x - 2/3 xÄ + 2/15 xÉ - )
éS èFor ê lïear, second order differential equation,
P(x)y»» + Q(x)y» + R(x)y = 0,
if x╠ is an ORDINARY POINT i.e. P(x╠) ƒ 0, ên êre is an
ïterval about x╠ for whichèq(x) = Q(x)/P(x) å
r(x) = R(x)/P(x) will be contïuous.èThus, ï this ïterval
ê Initial Value Problem
y»» + q(x)y» + r(x)y = 0
y(x╙)è=èy╙
y»(x╙) =èy»╙
will be guaranteed a unique solution.èThus if we can fïd
a general solution by POWER SERIES techniques, it can be
used ë get a specific solution ë ê ïitial value problem.
èèThe technique will assume a power series solution about
x = 0 which simplifies ê computation.èWhen ê expansion
is about x = x╙, ê SUBSTITUTION v = x - x╙èi.e. x = v + x╙
will convert ê problem ë one about x = 0.èWhen this
problem is solved, convert back ë x as ê variable ë get
ê general solution ï terms ç ê desired variable.èThis
is illustrated ï Problems 4 å 6.
èèThe method for fïdïg a POWER SERIES SOLUTION ABOUT AN
ORDINARY POINT is as follows:
1) Assume a solution ç ê form
èèèèèèè∞
yè=èΣèa┬xⁿ
èèèèèè n=0
2) Differentiate term-by-term twice
èèè∞
y» =èΣèna┬xⁿúî
èè n=1è
èèè∞
y»» = Σèn(n-1)a┬xⁿú²
èè n=2
3) Substitute êse expressions for y, y», y»» ïë
ê differential equation
4) Adjust ê summation ïdices ç ê sums (see
Section 7.1) so that êy all start at zero.èThis
produces a sum ç ê form
èèèè∞
èèèèΣè[ expression ] xⁿè= 0
èèè n=0
For this summatation ë be valid for all x, ê
expression is brackets must equal zero.èThis
produces ê RECURSION RELATION.èThis recursion
relation can ên be rearranged ë yield ê value
ç a coefficient a┬ ï terms ç one or two ç its
predecessors back ï ê sequence.èThey, ï turn,
depend on êir predecessors until ê process
eventully sëps ï terms ç a╠ or a¬.è
5) These will serve as ê arbitrary constants needed
ï a general solution ç a second order differential
equation which can be written as
y = a╠y¬ + a¬y½
where y¬ å y½ are ê power series associated with
each constant.
6) The ïterval ç convergence ç ê general
solution will be ê smaller ç ê ïtervals for
y¬ å y½.èIf term n can be written as a function
ç n alone, êse ïtervals can be calculated as ï
Section 7.1
Consider
èèy»» + 4y = 0èexpåed about x = 0
The solution å its derivatives are
èèè∞
yè=èΣèa┬xⁿ
èèèèèè n=0
èèè∞
y» =èΣèna┬xⁿúî
èè n=1è
èèè∞
y»» = Σèn(n-1)a┬xⁿú²
èè n=2
Substitutïg ïè y»» + 4y = 0è
èèè∞ èèèèèèèèèè ∞
èèèΣèn(n-1)a┬xⁿú²è+è4èΣ a┬xⁿ = 0
èè n=2èèèèèèèèèèn=0
Changïg ê summation ïdex on ê first sum yields
èèè∞ èèèèèèèèèèèè ∞
èèèΣè(n+2)(n+1)a┬╟½xⁿè+è4èΣ a┬xⁿ = 0
èè n=0èèèèèèèèèèèèn=0
Or
èèè∞ í èèèè┐
èèèΣè▒ (n+2)(n+1)a┬╟½ + 4a┬ ▒xⁿ = 0
èè n=0 └ èèèè ┘
Settïg ê expression ï ê bracket ë zero
(n+2)(n+1)a┬╟½ + 4a┬ = 0
Rearrangïg gives ê RECURSION RELATION
èèèèèèè 4
a┬╟½ = - ──────────── a┬èèn ≥ 0
èèèèè(n+2)(n+1)
Solvïg for ê first coefficients
èèèènèèèèèèèèa┬
èèè ───èèèèèèè────
èèèè0 a½ = - 4/(2)(1) a╠ = - 2 a╠
èèèè1 a¼ = - 4/(3)(2) a¬ = - 2/3 a¬
èèèè2 a« = - 4/(4)(3) a½ = - 1/3 a½ = + 2/3 a╠
èèèè3 a╘ = - 4/(5)(4) a¼ = - 1/5 a¼ = 2/15 a¬
èèThus, through 3 non-zero terms ç each series, ê general
solution is
y =è a╙ [ 1 - 2 xì + 2/3 xÅ - ∙∙∙ ]
èè+ a¬ [ x - 2/3 xÄ + 2/15 xÉ - ∙∙∙ ]
èè This solution can be rewritten as
y =è C¬ [ 1 - (2x)ì/2! + (2x)Å/4! - ∙∙∙ ]
èè+ C½ [ (2x) - (2x)Ä/3! + (2x)É/5! - ∙∙∙ ]
These are known Taylor series so ê general solution is
y = C¬cos[2x] + C½sï[2x]
which is ê same as would be found by ê lïear, constant
coefficient method ç Section 3.3.
4 y»» - 4y = 0èabout x = 0
A)è a╙[ 1 + 2 xì + 2/3 xÅ] + a¬[ x + 2/3 xÄ + 2/15 xÉ]
B)è a╙[ 1 + 1/2 xì + 1/24 xÅ] + a¬[ x - 1/6 xÄ + 1/120 xÉ]
C)è a╙[ 1 - 1/2 xì + 1/24 xÅ] + a¬[ x + 1/6 xÄ + 1/120 xÉ]
D)è a╙[ 1 - 1/2 xì + 1/24 xÅ] + a¬[ x - 1/6 xÄ + 1/120 xÉ]
ü è The solution å its derivatives are
èèè∞
yè=èΣèa┬xⁿ
èèèèèè n=0
èèè∞
y» =èΣèna┬xⁿúî
èè n=1è
èèè∞
y»» = Σèn(n-1)a┬xⁿú²
èè n=2
Substitutïg ïè y»» - 4y = 0è
èèè∞ èèèèèèèèèè ∞
èèèΣèn(n-1)a┬xⁿú²è-è4èΣ a┬xⁿ = 0
èè n=2èèèèèèèèèèn=0
Changïg ê summation ïdex on ê first sum yields
èèè∞èèèèèèèèèèèèè∞
èèèΣè(n+2)(n+1)a┬╟½xⁿè-è4èΣ a┬xⁿ = 0
èè n=0èèèèèèèèèèèèn=0
Or
èèè∞ íèèèèèèèèèèè┐
èèèΣè▒ (n+2)(n+1)a┬╟½ - 4a┬ ▒xⁿ = 0
èè n=0 └èèèèèèèèèèè┘
Settïg ê expression ï ê bracket ë zero
(n+2)(n+1)a┬╟½ - 4a┬ = 0
Rearrangïg gives ê RECURSION RELATION
èèèèèèè4
a┬╟½ =è──────────── a┬èèn ≥ 0
èèèè (n+2)(n+1)
Solvïg for ê first coefficients
èèèènèèèèèèèèa┬
èèè ───èèèèèèè────
èèèè0 a½ =è4/(2)(1) a╠ =è2 a╠
èèèè1 a¼ =è4/(3)(2) a¬ =è2/3 a¬
èèèè2 a« =è4/(4)(3) a½ =è1/3 a½ = + 2/3 a╠
èèèè3 a╘ =è4/(5)(4) a¼ =è1/5 a¼ = 2/15 a¬
èèThus, through 3 non-zero ç each series, ê general
solution is
y =è a╙ [ 1 + 2 xì + 2/3 xÅ - ∙∙∙ ]
èè+ a¬ [ x + 2/3 xÄ + 2/15 xÉ - ∙∙∙ ]
ÇèA
5 y»» + xìy = 0
A)è a╠[1 + 1/12 xÅ + 1/672 xô] + a¬[x + 1/20 xÉ + 1/1440 xö]
B)è a╠[1 + 1/12 xÅ + 1/672 xô] + a¬[x - 1/20 xÉ + 1/1440 xö]
C)è a╠[1 - 1/12 xÅ + 1/672 xô] + a¬[x + 1/20 xÉ + 1/1440 xö]
D)è a╠[1 - 1/12 xÅ + 1/672 xô] + a¬[x - 1/20 xÉ + 1/1440 xö]
ü è The solution å its derivatives are
èèè∞
yè=èΣèa┬xⁿ
èèèèèè n=0
èèè∞
y» =èΣèna┬xⁿúî
èè n=1è
èèè∞
y»» = Σèn(n-1)a┬xⁿú²
èè n=2
Substitutïg ïè y»» + xìy = 0è
èèè∞ èèèèèèèèèè ∞
èèèΣèn(n-1)a┬xⁿú²è+èxì Σ a┬xⁿ = 0
èè n=2èèèèèèèèèèn=0
Changïg ê summation ïdex on ê two sums so ê power
ç x will be n yields
èèèè
èèè∞èèèèèèèèèèè ∞
èèèΣè(n+2)(n+1)a┬╟½xⁿè+èΣ a┬▀½xⁿ = 0
èè n=0èèèèèèèèèè n=2
Or
èèè èèèèè ∞èí èèèèèèèèèèè┐
èè 2a½ + 6a¼x +èΣè▒ (n+2)(n+1)a┬╟½ + a┬▀½ ▒xⁿ = 0
èè èèèèèn=2 └èèèèèèèèèèè ┘
The first two terms must be zero for all x which requires
that
a½ = 0è åè a¼ = 0
Settïg ê expression ï ê bracket ë zero
(n+2)(n+1)a┬╟½ + a┬▀½ = 0
Rearrangïg gives ê RECURSION RELATION
èèèèèèè1
a┬╟½ = - ──────────── a┬▀½èèn ≥ 2
èèèèè(n+2)(n+1)
Solvïg for ê first coefficients
èèèènèèèèèèèèa┬
èèè ───èèèèèèè────
èèèè2 a« =è- 1/(4)(3) a╠ = - 1/12 a╠
èèèè3 a╘ =è- 1/(5)(4) a¬ = - 1/20 a¬
èèèè4 a╒ =è- 1/(6)(5) a½ = - 1/30 a½ = 0
èèèè5 a7 =è- 1/(7)(6) a¼ = - 1/42 a¼ = 0
èèèè6 a8 =è- 1/(8)(7) a« = - 1/56 a« = 1/672 a╠
èèèè7 a9 =è- 1/(9)(8) a╘ = - 1/72 a╘ = 1/1440 a1
èèThus, through 3 non-zero terms ç each series, ê
general solution is
y =è a╙ [ 1 - 1/12 xÅ + 1/672 xô - ∙∙∙ ]
èè+ a¬ [ x - 1/20 xÉ + 1/1440 xö - ∙∙∙ ]
ÇèD
6 (1 + xì) y»»è+èxy»è-èyè=è0
A) a╠[ 1 + 1/2 xì + 1/8 xÅ] + a¬[x + 1/4 xÄ + 1/16 xÉ]
B)è a╠[ 1 - 1/2 xì + 1/8 xÅ] + a¬[x - 1/4 xÄ + 1/16 xÉ]
C) a╠[ 1 + 1/2 xì - 1/8 xÅ] + a¬[x + 1/4 xÄ - 1/16 xÉ]
D) a╠[ 1 + 1/2 xì - 1/8 xÅ] + a¬x
ü è The solution å its derivatives are
èèè∞
yè=èΣèa┬xⁿ
èèèèèè n=0
èèè∞
y» =èΣèna┬xⁿúî
èè n=1è
èèè∞
y»» = Σèn(n-1)a┬xⁿú²
èè n=2
Substitutïg ïè (1 + xì)y»» + xy» - y = 0è
èèèè ∞èèèèèèèèè ∞èèèèè ∞
(1 + xì) Σèn(n-1)a┬xⁿú² + x Σ na┬xⁿúî - Σ a┬xⁿ = 0
èèèèn=2èèèèèèèè n=1èèèè n=0
Changïg ê summation ïdices å multiplyïg yields
èèè∞ èèèèèèèèèè▄èèèèèèè∞èèèèè▄
èèèΣè(n+2)(n+1)a┬╟½xⁿ + Σ n(n-1)a┬xⁿ + Σ na┬xⁿ -èΣ a┬xⁿ = 0
èè n=0èèèèèèèèè n=2èèèèèèn=1èèèèn=0
Takïg out all terms with ïdices less than 2 leaves
è
èè èè2a½ + 6a¼x + a¬x - a╠ - a¬
è▄
+ Σ [ (n+2)(n+1)a┬╟½ + n(n-1)a┬ + na┬ - a┬ ] xⁿ = 0
n=2
Simplifyïg
èèèèèèèèè▄
2a½ - a╙ + 6a¼x + Σ [ (n+2)(n+1)a┬╟½ + (nì+1)a┬ ] xⁿ = 0
èèèèèèèè n=2
The constant terms must add ë zero å ê coefficient ç
ç x must be zero for all x which requires
that
2a½ - a╠ = 0è i.e. a½ = 1/2 a╠
åè6a¼ = 0è i.e. a¼ = 0
Settïg ê expression ï ê bracket ë zero
(n+2)(n+1)a┬╟½ + (nì+1)a┬ = 0
Rearrangïg gives ê RECURSION RELATION
èèèèèè nì-1
a┬╟½ = - ──────────── a┬èèn ≥ 2
èèèèè(n+2)(n+1)
Facërïg å cancellïg yields
èè èn-1
a┬╟½ = - ───── a┬èèn ≥ 2
èèèèèn+2
Solvïg for ê first coefficients
èèèènèèèèèèèèa┬
èèè ───èèèèèèè────
èèèè2 a« =è- 1/4 a½ = -1/8 a╠
èèèè3 a╘ =è- 2/5 a¼ = 0
èèèè4 a╒ =è- 3/6 a« = 1/16 a╠
èèThus, one series termïated after ê first term å
through 3 non-zero terms ç each series, ê general
solution is
y =è a╙ [ 1 + 1/2 xì - 1/8 xÅ + ∙∙∙ ] + a¬x
ÇèD
7 y»» + (x+1)y» + y = 0èabout x = -2
A)è a╙[1 + 1/2 xì + 1/6 xÄ] + a¬[x + 1/2 xì + 1/6 xÄ]
B)è a╙[1 - 1/2 xì - 1/6 xÄ] + a¬[x - 1/2 xì + 1/6 xÄ]
C)è a╙[1 + 1/2(x+2)ì + 1/6(x+2)Ä]
+ a¬[x+2 + 1/2(x+2)ì + 1/6(x+2)Ä]
D)è a╙[1 - 1/2(x+2)ì - 1/6(x+2)Ä]
+ a¬[x+2 - 1/2(x+2)ì + 1/6(x+2)Ä]
ü è As ê solution is ë be about x = -2, all values must
be written ï terms çèx + 2è.èFirst convert ë ê
variableèv = x + 2 .èThen ê coefficient ç y» x + 1 is
converted ë x + 1 = x + 2 - 1 = v - 1
The solution å its derivatives are
èèè∞
yè=èΣèa┬vⁿ
èèèèèè n=0
èèè∞
y» =èΣèna┬vⁿúî
èè n=1è
èèè∞
y»» = Σèn(n-1)a┬vⁿú²
èè n=2
Substitutïg ïèy»» + (x+1)y» + y = 0 å convertïg ë
ê variable v
∞èèèèèèèèèèèè ∞èèèèè ∞
èèèè Σèn(n-1)a┬vⁿú² + (v - 1) Σ na┬vⁿúî + Σ a┬vⁿ = 0
èèèèn=2èèèèèèèèèèè n=1èèèè n=0
Changïg ê summation ïdices å multiplyïg yields
èèè∞èèèèèèèèèè ▄èèèè ∞èèèèèèèè▄
èèèΣè(n+2)(n+1)a┬╟½xⁿ + Σ na┬vⁿ - Σ (n+1)a┬╟¬vⁿ +èΣ a┬vⁿ = 0
èè n=0èèèèèèèèè n=0èèè n=0èèèèèèèn=0
Or
èè ▄
èè Σ [ (n+2)(n+1)a┬╟½ + na┬ - (n+1)a┬╟¬ + a┬ ] vⁿ = 0
èèn=0
Simplifyïg
èè▄
èèΣ [ (n+2)(n+1)a┬╟½ + (n+1)[a┬ - a┬╟¬] ] vⁿ = 0
èèèèè n=0
The coefficient ç x must be zero for all x which requires
that settïg ê expression ï ê bracket ë zero
(n+2)(n+1)a┬╟½ + (n+1)[a┬ - a┬╟¬] = 0
Rearrangïg gives ê RECURSION RELATION
èèèèèè n+1
a┬╟½ = - ──────────── [a┬ - a┬╟¬]èèn ≥ 0
èèèèè(n+2)(n+1)
Cancellïg yields
èèèèè 1
a┬╟½ = - ───── [a┬ - a┬╟¬]èèn ≥ 0
èèèèèn+2
Solvïg for ê first coefficients
èèèènèèèèèèèèa┬
èèè ───èè èèèè────
èèèè0 a½ =è- 1/2 [a╠ - a¬] = -1/2 a╠ + 1/2 a¬
èèèè1 a¼ =è- 1/3 [a¬ - a½] = -1/3 a¬ + 1/3 a½
è =è-1/3 a¬ - 1/6 a╠ + 1/6 a¬
è =è-1/6 a╠ - 1/6 a¬
èèIn this case, ê recursion relation has two predecessor
so, ê computation is messier. Through 3 non-zero terms
ï each series, ê general solution ï v is
y =è a╙ [ 1 - 1/2 vì - 1/6 vÄ + ∙∙∙ ]
èè+ a¬ [ v + 1/2 vì - 1/6 vÄ + ∙∙∙ ]
èè This must be converted back ë ê origïal variable
x by ê transformation v = x + 2 which gives ê general
solution
y =è a╙ [ 1 - 1/2 (x+2)ì - 1/6 (x+2)Ä + ∙∙∙ ]
èè+ a¬ [ x+2 + 1/2 (x+2)ì - 1/6 (x+2)Ä + ∙∙∙ ]
ÇèD
äè Solve ê ïitial value problem.
â è Forèy»» + 4y = 0èabout x = 0,èy(0) = 3; y»(0) = -2
Substitutïg ç Σ a┬xⁿ yields ê recursion relation
èa┬╟½ = -4/(n+1)(n+2) a┬è The general solution is
èèè y = a╠[1 - 1/3 xì + 2/105 xÅ -] + a¬[x - 2/3 xÄ + 2/15 xÉ - ]
èèèèy» = a╙[ - 2/3 x + 8/105 xÄ - ] + a¬[ 1 - 2xì + 2/3 xÅ - ]
y(0) = a╠ = 3 å y»(0) = a¬ = -2 so ê specific solution is
è y = 3[1 - 1/3 xì + 2/105 xÅ - ] - 2[x - 2/3 xÄ + 2/15 xÉ - ]
éS èTo solve an Initial Value Problem
P(x)y»» + Q(x)y» + R(x)y = 0è
y(x╠) = y╠ ; y»(x╠) = y»╠
has two stages.
1) Fïd a general solution ç ê differential equation.
As this is a second order, differential equation,
ê general solution will have TWO ARBITRARY CONSTANTS
2) Substitute ê INITIAL VALUE ç ê ïdependent
variable ïë ê general solution å its deriviative
å set êm equal ë ê TWO INITIAL CONDITIONS.èThis
produces two lïear equations ï two unknowns (ê
arbitrary constants).èSolvïg this system yields ê
value ç ê constants å ê solution ç ê ïitial
value problem.èIt should be noted that as ê soluën
will probably be a power series solution, ê differ-
entiation must be done term-by-term å that ê
ïitial values ç ê ïdependent variable need ë
be at x╠ ë make all but a few terms ç ê power
seriesèzero i.e. if ê expansion is ï terms
ç (x-x╠)ⁿ evaluatïg at x╠ will make êse terms 0.
8 y»» - 4y = 0
y(0)è=è2
y»(0) = -4
A)è 2[ 1 + 1/6 xÄ + 1/180 xæ ] + 4[ x - 1/12 xÅ - 1/504 xÆ ]
B)è 2[ 1 + 1/6 xÄ + 1/180 xæ ] - 4[ x - 1/12 xÅ - 1/504 xÆ ]
C)è -2[ 1 + 1/6 xÄ + 1/180 xæ ] + 4[ x - 1/12 xÅ - 1/504 xÆ]
D)è -2[ 1 + 1/6 xÄ + 1/180 xæ ] - 4[ x - 1/12 xÅ - 1/504 xÆ]
ü è The solution å its derivatives are
èèè∞
yè=èΣèa┬xⁿ
èèèèèè n=0
èèè∞
y» =èΣèna┬xⁿúî
èè n=1è
èèè∞
y»» = Σèn(n-1)a┬xⁿú²
èè n=2
Substitutïg ïè y»» - xy = 0è
èèè∞ èèèèèèèèè ∞
èèèΣèn(n-1)a┬xⁿú²è- x Σ a┬xⁿ = 0
èè n=2èèèèèèèèèn=0
Changïg ê summation ïdex on ê sums so that ê power
ç x will be n yields
èèè∞èèèèèèèèèèè ∞
èèèΣè(n+2)(n+1)a┬╟½xⁿè-èΣ a┬▀¬xⁿ = 0
èè n=0èèèèèèèèèè n=1
Or
èèèèèè ∞èíèèèèèèèèèèè ┐
èè 2a½è+èΣè▒ (n+2)(n+1)a┬╟½ - a┬▀¬ ▒xⁿ = 0
èèèèèèèèèèn=1 └èèèèèèèèèèè ┘
The first terms must be zero which requires that
a½ = 0è
Settïg ê expression ï ê bracket ë zero
(n+2)(n+1)a┬╟½ - a┬▀¬ = 0
Rearrangïg gives ê RECURSION RELATION
èèèèèèè1
a┬╟½ =è──────────── a┬▀¬èèn ≥ 1
èèèèè(n+2)(n+1)
Solvïg for ê first coefficients
èèèènèèèèèèèèa┬
èèè ───èèèèèèè────
èèèè1 a¼ =è1/(3)(2) a╠ = 1/6 a╠
èèèè2 a« =è1/(4)(3) a¬ = 1/12 a¬
èèèè3 a╘ =è1/(5)(4) a½ = 1/20 a½ = 0
èèèè4 a╒ =è1/(6)(5) a¼ = 1/30 a¼ = 1/180 a╠
èèèè5 a7 =è1/(7)(6) a« = 1/42 a« = 1/504 a¬
èèThus, through 3 non-zero terms ç each series, ê
general solution is
y =è a╙ [ 1 + 1/6 xÄ + 1/180 xæ - ∙∙∙ ]
èè+ a¬ [ x + 1/12 xÅ + 1/504 xÆ - ∙∙∙ ]
To evaluate ê constants, differentiate ê general solution
term-by-term
y» =è a╙ [ 1/2 xì + 1/30 xÉ - ∙∙∙ ]
èè + a¬ [ 1 + 1/3 xÄ + 1/72 xæ - ∙∙∙ ]
y(0)è= a╙ = 2
y»(0) = a¬ = -4
Thus ê specific solution is
y =è 2 [ 1 + 1/6 xÄ + 1/180 xæ - ∙∙∙ ]
èè- 4 [ x + 1/12 xÅ + 1/504 xÆ - ∙∙∙ ]
ÇèB
9 y»» - (x+1)y» = 0
y(0) = -3
y»(0) = 7
A) 3 + 7 [ (x+1) + 1/6 (x+1)Ä + (x+1)É + ]
B) 3 - 7 [ (x+1) + 1/6 (x+1)Ä + (x+1)É + ]
C) -3 + 7 [ (x+1) + 1/6 (x+1)Ä + (x+1)É + ]
D) -3 - 7 [ (x+1) + 1/6 (x+1)Ä + (x+1)É + ]
ü è As ê solution is ë be about x = -1, all values must
be written ï terms çèx + 1è.èThe transformation
variable isèv = x + 1 .èThe coefficient ç y» x + 1 is
already ï ê proper form.
The solution å its derivatives are
èèè∞
yè=èΣèa┬vⁿ
èèèèèè n=0
èèè∞
y» =èΣèna┬vⁿúî
èè n=1è
èèè∞
y»» = Σèn(n-1)a┬vⁿú²
èè n=2
Substitutïg ïèy»» - (x+1)y» = 0 å convertïg ë
ê variable v
∞èèèèèèèèè ∞è
èèèè Σèn(n-1)a┬vⁿú² - v Σ na┬vⁿúîè=è0
èèèèn=2èèèèèèèè n=1 èèè
Changïg ê summation ïdices å multiplyïg yields
∞èèèèèèèèèè ▄èèè
èèèè Σè(n+2)(n+1)a┬╟½xⁿ - Σ na┬vⁿè= 0
èè n=0èèèèèèèèè n=1èèè
Simplifyïg
èè è▄
è 2a╠ +èΣ [ (n+2)(n+1)a┬╟½ - na┬ ] vⁿ = 0
èèèèèèèè n=0
As this equation must be true for all v, ê first term must
be zero
èè2a╠ = 0èi.e.èa½ = 0
The quantity ï ê bracket must be zero for all x which
requires
(n+2)(n+1)a┬╟½è- na┬è=è0
Rearrangïg gives ê RECURSION RELATION
èèèèèèèn
a┬╟½ =è──────────── a┬èè n ≥ 1
èèèè (n+2)(n+1)
Solvïg for ê first coefficients
èèèènèèèèèèèèa┬
èèè ───èè èèèè────
èèèè1 a¼è=è1/(3)(2) a¬è=è1/6 a¬
èèèè2 a«è=è2/(4)(3) a½è=è1/6 a½è=è0
èèèè3èèè a╘è=è3/(5)(4) a¼è=è3/20 a¼è=è1/40 a¬
èèIn this case, one ç ê two series termïates after one
term so ê general polynomial is a polynomial plus a
constant.èThrough 3 non-zero terms ï ê series,
ê general solution ï v is
y =è a╙è+ a¬ [ v + 1/6 vÄ + 1/40 vÉ + ∙∙∙ ]
èè This must be converted back ë ê origïal variable
x by ê transformation v = x + 1 which gives ê general
solution
y =è a╙ + a¬ [ x+1 + 1/6 (x+1)Ä + 1/40 (x+1)É + ∙∙∙]
To evaluate ê constants, differentiate ê general solution
term-by-term
y» =èa¬ [ 1 + 1/2 (x+1)ì + 1/8 (x+1)Å - ∙∙∙ ]
y(0)è= a╙ = -3
y»(0) = a¬ = 7
Thus ê specific solution is
y =è -3è+è7 [ x+1 + 1/6(x+1)Ä + 1/40(x+1)É - ∙∙∙ ]
ÇèC