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à 7.4èPower Series Solutions Near a Regular Sïgular Poït
äè Classify ê poït about which ê series is ë be
expåed as Ordïary, Regular Sïgular or Irregular Sïgular
â è Forè xìy»» + (1+x)y» + 3xy = 0èabout x = 0.èAs ê
coefficient ç y»» is zero at x = 0, it is NOT an ORDINARY
poït.èDividïg by xì givesèy»» + [(1+x)/x]y» + [3/x]y = 0
asèèèèè1+xèèèèèèèèèèè 3
èèlimèx ─────è= 1èåè limèx║ ───è= 0
èèx¥0èèèxèèèèèèè x¥0èèèx
are both fïite, x = 0 is a REGULAR SINGULAR POINT
éS è For ê general, lïear second order differential equation,
P(x)y»» + Q(x)y» + R(x)y = 0,
it is assumed that êre are no common facërs as êy can be
cancelled.
èèA poït x╠ is called an ORDINARY POINT if P(x╠) ƒ 0.
If P(x╠) = 0, ên x╠ is called a SINGULAR POINT.
èèWith a differential equation with sïgular poïts, êre
are usually just a few sïgular poïts,so it is reasonable ë
ask why it is important ë attempt ë produce a solution at
a sïgular poït?èIt turns out that, ï many cases, ê
most ïterestïg phenomena occur near sïgular poïts.èSuch
behavior ïcludes large functional values (tendïg ë un-
bounded solutions) å rapidly oscillatïg functions.
èèThere is a class ç sïgular poïts for which a modified
power series solution can be obtaïed via a method due ë
FROBENIUS.èIt can be used when ê poït is a REGULAR
SINGULAR POINT.èThis means that ê poït x╠ satisfies
1) P(x╠) = 0
2) èèèèèèQ(x)
lim (x-x╠) ────── is fïite
x¥0èèèè P(x)
3) èèèèèè R(x)
lim (x-x╠)ì ────── is fïite
x¥0èèèèèP(x)
A sïgular poït that does not satisfy eiêr 2) or 3) is
an IRREGULAR SINGULAR POINT
1 xìy»» + 2xy» + y = 0èabout x = 0
A) Ordïary
B) Regular Sïgular
C) Irregular Sïgular
ü The coefficient ç y»», xì is 0 for x = 0 so this
is not an ORDINARY poït.
èèèèèè 2x
èè limèx ────è=è2è
èè x¥0èè xì
è èèèèèèè 1
èè limèxì ────è=è1
èè x¥0èèèx║
Both ç êse limits are fïite so x = 0 is a REGULAR
SINGULAR POINT
ÇèB
2 (xì-1)y»» + (x+2)y» + xy = 0 about x = 0
A) Ordïary
B) Regular Sïgular
C) Irregular Sïgular
ü è The coefficït ç y»» isèxì-1.èThis expression evaluated
at x = 0 is not zero å hence x = 0 is an ORDINARY poït.
ÇèA
3 (xì-1)y»» + (x+2)y» + xy = 0 about x = 1
A) Ordïary
B) Regular Sïgular
C) Irregular Sïgular
ü The coefficient ç y»», xì - 1 is 0 for x = 1 so this
is not an ORDINARY poït.
èèèèèèèè x+2èèèè
èè limè(x-1) ────è
èè x¥1èèèèxì-1
èèèèèèèèèèx+2
è = limè(x-1) ──────────
èè x¥1èèèè(x-1)(x+1)
è èèèèè x+2èèè3
è = limè─────è= ───
èè x¥1è x+1èèè2
èèèèèèèèè x
èè limè(x-1)ì ─────
èè x¥1èèèèèxì-1
èèèèèèèèèèèèèèèx
è = limè(x-1)ì ──────────
èè x¥1èèèè (x-1)(x+1)
èèèèèèèèè(x-1)xèèè(-1)(1)èèèè1
è = lim ────────è=è───────è=è- ───
èè x¥1èèx+1èèèèè2èèèèè 2
Both ç êse limits are fïite so x = 1 is a REGULAR
SINGULAR POINT
ÇèB
4 (x-1)ìy»» + 2xy» + (x-1)yè=è0èabout x = 1
A) Ordïary
B) Regular Sïgular
C) Irregular Sïgular
ü The coefficient ç y»», (x-1)ì is 0 for x = 1 so this
is not an ORDINARY poït.
èèèèèèèèè2xèèèè
èè limè(x-1) ──────è
èè x¥1èèèè(x-1)ì
è èèèèèè2xè
è = limè─────è which is undefïed
èè x¥1è x-1è
Also
èèèèèx-1
èè limè(x-1)ì ──────
èè x¥1èèèè (x-1)ì
èèèèèèèèèè
è = limèx-1è=è0
èè x¥1èè
èèè
One ç êse limits is fïite but ê oêr does not exits,
so x = 1 is an IRREGULAR SINGULAR POINT
ÇèC
äè Solve ê followïg differential equations about a
regular sïgular poït.è
âè2xìy»» + (xì-x)y» + y = 0 about x = 0. Assume a solution ç
ê formèΣ a┬xⁿó¡, substitute ïë ê equation å ê
INDICIAL EQUATION becomesè2mì - 3m + 1 = 0 which has ê
two solution m = 1/2, 1.èSubstitutïg m = 1/2 ïë ê
recursion relation gives ê solution
xî»ì[ 1 - 1/4 x + 3/80 xì - ∙∙∙] while ê m = 1 solution is
x[1 - 1/3 x + 1/15 xì - ∙∙∙ ]
éSèThe method used ë solve a lïear, second order differential
equation NEAR a REGULAR SINGULAR POINT is a combïation ç
ê methods described ï ê previous two sections.èSection
7.2 used a power series solution ç ê form
èèèèè▄
èèèèèΣèa┬xⁿ
èèèè n=0
ë fïd a solution about an ORDINARY POINT.èThe assumed
solution was differentiated å was substituted ïë ê
differential equation.èThe RECURSION RELATION was determïed
å subsequently ê coefficients a┬ were found ï terms ç
a╙ å a¬.
èèIn Section 7.3, ê EULER type differential equation,
which has x = 0 as a REGULAR SINGULAR POINT, assumed a
solution ç ê formèx¡èwhere m can be any real number.
Differentiatïg ê assumed solution å substitutïg it
ïë ê differential equation produces a quadratic equation
for m.èThere are three distïct classes ç solutions ë ê
Euler type differential equation based on ê three distïct
types ç solutions ë a quadratic equation
1)èèdistïct real roots
2)èèrepeated real roots
3)èècomplex conjugate roots
èèThe solution ç a lïear, second order differential
equation about a regular sïgular poït is assumed ë be ç
ê form
èèèè ▄èèèèèè▄
y =èx¡èΣèa┬xⁿè =èΣèa┬xⁿó¡
èèèèn=0èèèèèn=0
where m can be an arbitrary real number, not just an ïteger.
èèDifferentiation yields
èèè▄
y» =èΣè(n+m)a┬xⁿó¡úî
èè n=0
èèè ▄
y»» =èΣè(n+m)(n+m-1)a┬xⁿó¡úì
èèèn=0
èèEverythïg is substituted ïë ê differential equation
P(x)y»» + Q(x)y» + R(x)yè=è0
èèForè 2xìy»» + (xì-x)y» + y = 0èabout x = 0è
èèèè ▄èèèèèèèèèèèèèèèèè ▄
èè2xìèΣè(n+m)(n+m-1)a┬xⁿó¡úìè+è(xì-x)èΣ (n+m)a┬xⁿó¡úî
èèèèn=0èèèèèèèèèèèèèèèè n=0
èèèèèèèè ▄
èèèèèèè+èΣèa┬xⁿó¡è=è0
èèèèèèèèn=0
orè ▄èèèèèèèèèèèèè▄
èè Σè2(n+m)(n+m-1)a┬xⁿó¡è+èΣè(n+m)a┬xⁿó¡óî
èèn=0èèèèèèèèèèèèn=0
èèèè ▄èèèèèèèèè▄
èèè-èΣè(n+m)a┬xⁿó¡è+èΣèa┬xⁿó¡è=è0
èèèèn=0èèèèèèèèn=0
As ê second sum's exponent is different from ê rest, it
will be re-ïdexed so that all exponents are ê same.
èè ▄èèèèèèèèèèèèè▄
èè Σè2(n+m)(n+m-1)a┬xⁿó¡è+èΣè(n+m-1)a┬▀¬xⁿó¡
èèn=0èèèèèèèèèèèèn=1
èèèè ▄èèèèèèèèè▄
èèè-èΣè(n+m)a┬xⁿó¡è+èΣèa┬xⁿó¡è=è0
èèèèn=0èèèèèèèèn=0
As ê second sum starts at n = 1 while ê oêrs start at
n = 0, ê first term will be isolated å ê remaïïg
terms combïed ïë one sum
0è=è[ 2m(m-1) - m + 1 ] a╠x¡
è ▄
+èΣè{ 2(n+m)(n+m-1)a┬ + (n+m-1)a┬▀¬ - (n+m)a┬ + a┬ }xⁿó¡
èn=1
è For a differential equation ë have a non-trivial solution,
ê coefficient ç a╠x¡ must be zero.èSettïg it ë zero
yields ê INDICIAL EQUATION
2m(m-1) - m + 1è=è0
Solvïg ê ïdicial equation which will be quadratic ï m
will produce 2 roots which are called ê EXPONENTS OF THE
SINGULARITY.èThese values determïe ê behavior ç ê
solution near ê poït ç sïgularity.
è As usual with quadratic equations, êre are three cases
as ë ê types ç solutions produced.èThese, however, are
slightly different than ê usual classifications.
CASE I is for DISTINCT ROOTS ç ê ïdicial equation that
DO NOT DIFFER BY AN INTEGER.èSay êse roots are l å g.
Reconsider
0è=è[ 2m(m-1) - m + 1 ] a╠x¡
è ▄
+èΣè{ 2(n+m)(n+m-1)a┬ + (n+m-1)a┬▀¬ - (n+m)a┬ + a┬ }xⁿó¡
èn=1
The ïdicial equation is
2m(m-1) - m+ 1è=è0
2mì - 2m - m + 1 = 0
2mì - 3m + 1 = 0
This facërs ë
(2m - 1)(m - 1) = 0
The solutions are
m =è1/2,è1
As l å g are solutions ç ê ïdical equation, ê first
term ç ê solution will be zero, so for m = l or g this
reduces ë
▄
Σ { 2(n+m)(n+m-1)a┬ + (n+m-1)a┬▀¬ - (n+m)a┬ + a┬ }xⁿó¡è=è0
n=1
For this sum ë be zero, it is sufficient that every term ï
ê brace be zero.èSettïg each ë zero produces ê
RECURSION RELATION which for this example will be
èè 2(n+m)(n+m-1)a┬ + (n+m-1)a┬▀¬ - (n+m)a┬ + a┬è=è0
As ê roots for m are distïct å do not differ by an
ïteger, subsitutïg l å ên g ïë ê recursion relation
will produce two LINEARLY INDPENDENT power series solutions
èèFirst, let m =1/2 å substitute ïë ê recursion
relation
èè2(n+1/2)(n-1/2)a┬ + (n-1/2)a┬▀¬ - (n+1/2)a┬ + a┬ = 0
Multiplyïg through by 2 ë get rid ç ê fractions gives
(2n+1)(2n-1)a┬ + (2n-1)a┬▀¬ - (2n-1)a┬ + 2a┬ = 0
Combïïg like terms å rearrangïg yields
[ (2n+1)(2n-1) - (2n-1) + 2 ] a┬è=è- (2n-1)a┬▀¬
orèèèèèèèèèèè (2n-1)
èèèa┬ =è- ────────────────────────────èa┬▀¬è n ≥ 1
èèèèèèè (2n+1)(2n-1) - (2n-1) + 2n
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = -(1)/[3(1)-1+2] a╠ = -1/4 a╠
2èèè a½ = -(3)/[5(3)-3+2] a¬ = -3/14 a¬ = 3/56 a╠
è Now let m = 1, substitutïg ïë ê recusion relation gives
è2(n+m)(n+m-1)a┬ + (n+m-1)a┬▀¬ - (n+m)a┬ + a┬è=è0
orè 2(n+1)na┬ + na┬▀¬ - (n+1)an + a┬ = 0
Combïïg like terms å rearrangïg yields
èè[ 2n(n+1) - n ]a┬è=è-na┬▀¬
èè[ n(2n+1) ]a┬è=è-na┬▀¬
orèèèèèèèè nèèèèèèèèè 1
èèèa┬è=è- ───────── a┬▀¬è=è- ────── a┬▀¬èè n ≥ 1
èèèèèèèèn(2n+1)èèèèèèè2n+1
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = -1/[2(1)+1] a╠ = -1/3 a╠
2èèè a½ = -1/[2(2)+1] a¬ = -1/5 a¬ = 1/15 a╠
Thus ê general solution is
èC¬ xî»ì [ 1 - 1/4 xè+è3/56 xìè- ∙∙∙ ]
+ C½ x [ 1 - 1/3 xè+ 1/15 xì - ∙∙∙ ]
CASE IIèReal roots that eiêr are repeated ç differ by an
ïteger.èThis case corresponds ë ê repeated roots case
ç ê EULER differential equation.èIn particular, one series
solution, correspondïg ë ê LARGER root (if different) will
be found.èThe oêr solution will contaï a facër ïvolvïg
ê NATURAL LOGARITHM function ë make it LINEARLY INDEPENDENT.
Its specific form can be found usïg Reduction ï Order,
Section 4.2 or by methods not covered ï this program.
CASE IIIèComplex conjugate roots.èHere ê roots do not
differ by an ïteger as êy are complex conjugates ç each
oêr so êre will be two power series solutions.èHowever,
sïce ê exponents are complex, a technique similar ë ê
one used ï ê similar Euler differential equation must be
employed å ên ê Euler formula is used ë convert an
imagïary exponential ë equivalent trig functions.èAs ï
ê Euler differential equation case, natural logarithmic
functions will be produced.
5 2xìy»» + xy» + (x-1)y = 0èabout x = 0
A)è C¬xúî»ì[1 + xè+ 1/2 xì] + C½x[1 + 1/6 xè+ 1/90 xì]
B)è C¬xúî»ì[1 + xè+ 1/2 xì] + C½x[1 - 1/6 xè+ 1/90 xì]
C)è C¬xúî»ì[1 + xè- 1/2 xì] + C½x[1 + 1/6 xè+ 1/90 xì]
D)è C¬xúî»ì[1 + xè- 1/2 xì] + C½x[1 - 1/6 xè+ 1/90 xì]
ü è Assume a solution ç ê form
èèèèèèèè ▄èèèèèè▄
y =èx¡èΣèa┬xⁿè =èΣèa┬xⁿó¡
èèèèn=0èèèèèn=0
where m can be an arbitrary real number, not just an ïteger.
èèDifferentiation yields
èèè▄
y» =èΣè(n+m)a┬xⁿó¡úî
èè n=0
èèè ▄
y»» =èΣè(n+m)(n+m-1)a┬xⁿó¡úì
èèèn=0
Substitute ïëèè2xìy»» + xy» + (x-1)y = 0èë yield
èèèè ▄èèèèèèèèèèèèèè ▄
èè2xìèΣè(n+m)(n+m-1)a┬xⁿó¡úìè+ xèΣ (n+m)a┬xⁿó¡úî
èèèèn=0èèèèèèèèèèèèè n=0
èèèèèèèè ▄
èèèè+ (x-1)èΣèa┬xⁿó¡è=è0
èèèèèèèèn=0
orè ▄èèèèèèèèèèèèè▄
èè Σè2(n+m)(n+m-1)a┬xⁿó¡è+èΣè(n+m)a┬xⁿó¡ +
èèn=0èèèèèèèèèèèèn=0
èèèè ▄èèèèèèèèè▄
èèèèΣèa┬xⁿó¡óîè-èΣèa┬xⁿó¡è=è0
èèèèn=0èèèèèèèèn=0
As ê third sum's exponent is different from ê rest, it
will be re-ïdexed so that all exponents are ê same.
èè ▄èèèèèèèèèèèèè▄
èè Σè2(n+m)(n+m-1)a┬xⁿó¡è+èΣè(n+m)a┬xⁿó¡
èèn=0èèèèèèèèèèèèn=0
èèèè ▄èèèèèèè ▄
èèè+èΣèa┬▀¬xⁿó¡è-èΣèa┬xⁿó¡è=è0
èèèèn=0èèèèèè n=0
As ê second sum starts at n = 1 while ê oêrs start at
n = 0, ê first term will be isolated å ê remaïïg
terms combïed ïë one sum
0è=è[ 2m(m-1) + m - 1 ] a╠x¡
è ▄
+èΣè{ 2(n+m)(n+m-1)a┬ + (n+m)a┬ + a┬▀¬ - a┬ }xⁿó¡
èn=1
è For differential equation ë have a non-trivial solution,
ê coefficient ç a╠x¡ must be zero.èSettïg it ë zero
yields ê INDICIAL EQUATION
2m(m-1) + m - 1è=è0
orèèè2mì - m - 1è=è0
This facërs ë
(2m + 1)(m - 1) = 0
The solutions are
m = -1/2, 1
èèIf eiêr ç êse two solutions ç ê ïdical equation,
are subsituted ïë ê assumed solution, ê first term ç
ê solution will be zero, so for m = -1/2 or 1 this
reduces ë
▄
Σ { 2(n+m)(n+m-1)a┬ + (n+m)a┬ + a┬▀¬ - a┬ }xⁿó¡è=è0
n=1
For this sum ë be zero, it is sufficient that every term ï
ê braces is zero.èSettïg each ë zero produces ê
RECURSION RELATION which for this example will be
èè 2(n+m)(n+m-1)a┬ + (n+m)a┬ + a┬▀¬ - a┬è=è0
As ê roots for m are distïct å do not differ by an
ïteger, subsitutïg -1/2 å ên 1 ïë ê recursion relation
will produce two LINEARLY INDPENDENT power series solutions.
èèFirst, let m = -1/2 å substitute ïë ê recursion
relation
èè2(n-1/2)(n-3/2)a┬ + (n-1/2)a┬ + a┬▀¬ - a┬ = 0
Multiplyïg through by 2 ë get rid ç ê fractions gives
èè(2n-1)(2n-3)a┬ + (2n-1)a┬▀¬ + 2a┬ - 2a┬ = 0
Combïïg like terms å rearrangïg yields
èè[ (2n-1)(2n-3) + (2n-1) - 2 ] a┬è=è- 2a┬▀¬
orèèèèèèèèèèè 2
èèèa┬ =è- ───────────────────────èa┬▀¬è n ≥ 1
èèèèèèè (2n-1)(2n-3)+(2n-1)-2
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = -2/[1(-1)+1-2] a╠ =èa╠
2èèè a½ = -2/[3(1)+3-2] a¬ = -1/2 a¬ = -1/2 a╠
è Now let m = 1, substitutïg ïë ê recusion relation gives
èèè2(n+m)(n+m-1)a┬ + (n+m)a┬ + a┬▀¬ - a┬è=è0
orèè2(n+1)na┬ + (n+1)a┬ + a┬▀¬ - a┬ = 0
Combïïg like terms å rearrangïg yields
èè[ 2n(n+1) + n+1 - 1 ]a┬è=è-a┬▀¬
èè[ n(2n+1) + n ]a┬è=è-a┬▀¬
èè[ (2n+1)(n+1) ]a┬è=è-a┬▀¬
Rearrangïg
èèèèèèèèèè 1èèèèèè
èèèa┬è=è- ───────────── a┬▀¬èèèn ≥ 1
èèèèèèèè(2n+1)(n+1)èèè
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = -1/[3(2)] a╠ = -1/6 a╠
2èèè a½ = -1/[5(3)] a¬ = -1/15 a¬ = 1/90 a╠
Thus ê general solution is
èC¬ xúî»ì [ 1 + xè- 1/2 xìè- ∙∙∙ ]
+ C½ x [ 1 - 1/6 xè+ 1/90 xì - ∙∙∙ ]
Ç D
6è 3xìy»» + xy» + xyè=è0èabout x = 0
A)è C¬[1 + x + 1/8xì ] + C½x[1 + 1/5 xè+ 1/80 xì]
B)è C¬[1 + x + 1/8xì ] + C½x[1 - 1/5 xè+ 1/80 xì]
C)è C¬[1 - x + 1/8xì ] + C½x[1 + 1/5 xè+ 1/80 xì]
D)è C¬[1 - x + 1/8xì ] + C½x[1 - 1/5 xè+ 1/80 xì]
ü è Assume a solution ç ê form
èèèèèèèèèèèè ∞èèè
èèèèy =èx¡èΣèa┬xⁿè =èΣèa┬xⁿó¡
èèèèn=0èèèèèn=0
where m can be an arbitrary real number, not just an ïteger.
èèDifferentiation yields
èèè▄
y» =èΣè(n+m)a┬xⁿó¡úî
èè n=0
èèè ▄
y»» =èΣè(n+m)(n+m-1)a┬xⁿó¡úì
èèèn=0
Substitute ïëè3xìy»» + xy» + xyè=è0èë yield
èèèè ▄èèèèèèèèèèèèèè ▄
èè3xìèΣè(n+m)(n+m-1)a┬xⁿó¡úìè+ xèΣ (n+m)a┬xⁿó¡úî
èèèèn=0èèèèèèèèèèèèè n=0
èèèèèè▄
èèèè+ x Σèa┬xⁿó¡è=è0
èèèèè n=0
As ê third sum's exponent is different from ê rest, it
will be re-ïdexed so that all exponents are ê same.
orè ▄èèèèèèèèèèèèè▄
èè Σè3(n+m)(n+m-1)a┬xⁿó¡è+èΣè(n+m)a┬xⁿó¡
èèn=0èèèèèèèèèèèèn=0
èèèè ▄èèèèèèèèè
èèè+èΣèa┬xⁿó¡óîè =è0
èèèèn=0
As ê third sum's exponent is different from ê rest, it
will be re-ïdexed so that all exponents are ê same.
èè ▄èèèèèèèèèèèèè▄
èè Σè3(n+m)(n+m-1)a┬xⁿó¡è+èΣè(n+m)a┬xⁿó¡
èèn=0èèèèèèèèèèèèn=0
èèèè ▄èèèèèèèèè
èèè+èΣèa┬▀¬xⁿó¡è =è0
èèèèn=1
As ê third sum starts at n = 1 while ê oêrs start at
n = 0, ê first term will be isolated å ê remaïïg t
erms combïed ïë one sum
0è=è[ 3m(m-1) + m ] a╠x¡
è ▄
+èΣè{ 3(n+m)(n+m-1)a┬ + (n+m)a┬ + a┬▀¬ }xⁿó¡
èn=1
è For differential equation ë have a non-trivial solution,
ê coefficient ç a╠x¡ must be zero.èSettïg it ë zero
yields ê INDICIAL EQUATION
3m(m-1) + mè=è0
orèèè3mì - 2mè=è0
This facërs ë
m(3m - 2) = 0
The solutions are
m = 0, 2/3
èèIf eiêr ç êse two solutions ç ê ïdical equation,
are subsituted ïë ê assumed solution, ê first term ç
ê solution will be zero, so for m = 0 or 2/3 this
reduces ë
▄
Σ { 3(n+m)(n+m-1)a┬ + (n+m)a┬ + a┬▀¬ }xⁿó¡è=è0
n=1
For this sum ë be zero, it is sufficient that every term ï
ê braces is zero.èSettïg each ë zero produces ê
RECURSION RELATION which for this example will be
èè 3(n+m)(n+m-1)a┬ + (n+m)a┬ + a┬▀¬è=è0
As ê roots for m are distïct å do not differ by an
ïteger, subsitutïg 0 å ên 2/3 ïë ê recursion
relation will produce two LINEARLY INDPENDENT power series
solutions.
èèFirst, let m = 0 å substitute ïë ê recursion
relation
èè3(n+m)(n+m-1)a┬ + (n+m)a┬ + a┬▀¬è=è0
èèèè3n(n-1)a┬ + na┬ + a┬▀¬ = 0
Combïïg like terms å rearrangïg yields
èè[ 3n(n-1) + n ] a┬è=è- a┬▀¬
orèèèèèèèè1
èèèa┬ =è- ─────────èa┬▀¬è n ≥ 1
èèèèèèè n(3n-2)
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = -1/[1(1)] a╠ =è-a╠
2èèè a½ = -1/[2(4)] a¬ = -1/8 a¬ = 1/8 a╠
è Now let m = 2/3, substitutïg ïë ê recusion relation
gives
èèè3(n+m)(n+m-1)a┬ + (n+m)a┬ + a┬▀¬è=è0
orèè3(n+2/3)(m-1/3)a┬ + (n+2/3)a┬ + a┬▀¬è= 0
Multiplyïg by 3 ë clear ê fractions gives
èèè(3n+2)(3n-1)a┬ + (3n+2)a┬ + 3a┬▀¬è=è0
Combïïg like terms å rearrangïg yields
èè[ (3n+2)(3n-1) + (3n+2) ]a┬è=è- 3a┬▀¬
èè[ 3n(3n+2) ]a┬è=è-3a┬▀¬
Rearrangïg
èèèèèèèèè 1èèèèèè
èèèa┬è=è- ───────── a┬▀¬èèèn ≥ 1
èèèèèèèèn(3n+2)èèè
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = -1/[1(5)] a╠ = -1/5 a╠
2èèè a½ = -1/[2(8)] a¬ = -1/16 a¬ = 1/80 a╠
Thus ê general solution is
èC¬ [ 1 - xè+ 1/8 xìè- ∙∙∙ ]
+ C½ x [ 1 - 1/5 xè+ 1/80 xì - ∙∙∙ ]
Ç D
7è 9xìy»» + 9xy» + (9x-4)yè=è0è about x = 0
A)è C¬xúì»Ä[1 + 3x + 9/4xì] + C½xì»Ä[1 + 3/7x + 3/91 xì]
B)è C¬xúì»Ä[1 + 3x + 9/4xì] + C½xì»Ä[1 - 3/7x + 3/91 xì]
C)è C¬xúì»Ä[1 + 3x - 9/4xì] + C½xì»Ä[1 + 3/7x + 3/91 xì]
D)è C¬xúì»Ä[1 + 3x - 9/4xì] + C½xì»Ä[1 - 3/7x + 3/91 xì]
ü è Assume a solution ç ê form
èèèèèèèè ▄èèèèèè▄
y =èx¡èΣèa┬xⁿè =èΣèa┬xⁿó¡
èèèèn=0èèèèèn=0
where m can be an arbitrary real number, not just an ïteger.
èèDifferentiation yields
èèè▄
y» =èΣè(n+m)a┬xⁿó¡úî
èè n=0
èèè ▄
y»» =èΣè(n+m)(n+m-1)a┬xⁿó¡úì
èèèn=0
Substitute ïëè9xìy»» + 9xy» + (9x-4)yè=è0èë yield
èèèè ▄èèèèèèèèèèèèèèè▄
èè9xìèΣè(n+m)(n+m-1)a┬xⁿó¡úìè+ 9xèΣ (n+m)a┬xⁿó¡úî
èèèèn=0èèèèèèèèèèèèèèn=0
èèèèèèèè ▄
èèèè+ (9x-4) Σèa┬xⁿó¡è=è0
èèèèèèèèn=0
orè ▄èèèèèèèèèèèèè▄
èè Σè9(n+m)(n+m-1)a┬xⁿó¡è+èΣ 9(n+m)a┬xⁿó¡ +
èèn=0èèèèèèèèèèèèn=0
èèèè ▄èèèèèèè▄
èèèèΣè9a┬xⁿó¡óî -èΣè4a┬xⁿó¡è =è0
èèèèn=0èèèèèèn=0
As ê third sum's exponent is different from ê rest, it
will be re-ïdexed so that all exponents are ê same.
èè ▄èèèèèèèèèèèèè▄
èè Σè9(n+m)(n+m-1)a┬xⁿó¡è+èΣè9(n+m)a┬xⁿó¡
èèn=0èèèèèèèèèèèèn=0
èèèè ▄èèèèèèèè▄
èèè+èΣè9a┬╪¬xⁿó¡è-èΣè4a┬xⁿó¡è=è0
èèèèn=1èèèèèèèn=0
As ê third sum starts at n = 1 while ê oêrs start at
n = 0, ê first term will be isolated å ê remaïïg
terms combïed ïë one sum
0è=è[ 9m(m-1) + 9m - 4 ] a╠x¡
è ▄
+èΣè{ 9(n+m)(n+m-1)a┬ + 9(n+m)a┬ + 9a┬▀¬ - 4a┬ }xⁿó¡
èn=1
è For differential equation ë have a non-trivial solution,
ê coefficient ç a╠x¡ must be zero.èSettïg it ë zero
yields ê INDICIAL EQUATION
9m(m-1) + 9m - 4è=è0
orèèè9mì - 4è=è0
This facërs ë
(3m + 2)(3m - 2) = 0
The solutions are
m = -2/3, 2/3
èèIf eiêr ç êse two solutions ç ê ïdical equation,
are subsituted ïë ê assumed solution, ê first term ç
ê solution will be zero, so for m = -2/3 or 2/3 this
reduces ë
▄
Σ { 9(n+m)(n+m-1)a┬ + 9(n+m)a┬ + 9a┬▀¬ - 4a┬ }xⁿó¡è=è0
n=1
For this sum ë be zero, it is sufficient that every term ï
ê braces is zero.èSettïg each ë zero produces ê
RECURSION RELATION which for this example will be
èè 9(n+m)(n+m-1)a┬ + 9(n+m)a┬ + 9a┬▀¬ - 4a┬è=è0
As ê roots for m are distïct å do not differ by an
ïteger, subsitutïg -2/3 å ên 2/3 ïë ê recursion
relation will produce two LINEARLY INDPENDENT power series
solutions.
èèFirst, let m = 2/3 å substitute ïë ê recursion
relation
èè 9(n+m)(n+m-1)a┬ + 9(n+m)a┬ + 9a┬▀¬ - 4a┬è=è0
èè 9(n-2/3)(n-5/3)a┬ + 9(n-2/3)a┬ + 9a┬▀¬ - 4a┬ = 0
èè
èè (3n-2)(3n-5)a┬ + 3(3n-2)a┬ + 9a┬▀¬ - 4a┬è=è0
Combïïg like terms å rearrangïg yields
èè[ (3n-2)ì - 4 ] a┬è=è- 9a┬▀¬
orèèèèèèèèè9
èèèa┬ =è- ───────────èa┬▀¬è n ≥ 1
èèèèèèè (3n-2)ì-4
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = -9/[1ì-4] a╠ =è3a╠
2èèè a½ = -9/[4ì-4] a¬ = -3/4 a¬ = -9/4 a╠
è Now let m = 2/3, substitutïg ïë ê recusion relation
gives
èèè9(n+m)(n+m-1)a┬ + 9(n+m)a┬ + 9a┬▀¬ - 4a┬è=è0
orèè9(n+2/3)(n-1/3)a┬ + 9(n+2/3)a┬ + 9a┬▀¬ - 4a┬è= 0
orèè(3n+2)(3n-1)a┬ + 3(3m+2)a┬ + 9a┬▀¬ -4a┬è= 0
Combïïg like terms å rearrangïg yields
èè[ (3n+2)(3n-1) + 3(3n+2) - 4 ]a┬è=è- 9a┬▀¬
èè[ (3n+2)ì - 4 ]a┬è=è-9a┬▀¬
Rearrangïg
èèèèèèèèè 9
èèèa┬è=è- ───────── a┬▀¬èèèn ≥ 1
èèèèèèè (3n+2)ì-4èèè
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = -9/21 a╠ = -3/7 a╠
2èèè a½ = -9/117 a¬ = -1/13 a¬ = 3/91 a╠
Thus ê general solution is
èC¬ xúì»Ä [ 1 + 3xè- 9/4 xìè- ∙∙∙ ]
+ C½ xì»Ä [ 1 - 3/7 xè+ 3/91 xì - ∙∙∙ ]
Ç D
8 xìy»» + (xì-3x)y» + 2xyè=è0
A)è C¬ xÄ [ 1 + 5/9 xè+ 10/51 xìè- ∙∙∙ ]
B)è C¬ xÄ [ 1 + 5/9 xè- 10/51 xìè- ∙∙∙ ]
C)è C¬ xÄ [ 1 - 5/9 xè+ 10/51 xìè- ∙∙∙ ]
D)è C¬ xÄ [ 1 - 5/9 xè- 10/51 xìè- ∙∙∙ ]
ü è Assume a solution ç ê form
èèèèèèèè ▄èèèèèè▄
y =èx¡èΣèa┬xⁿè =èΣèa┬xⁿó¡
èèèèn=0èèèèèn=0
where m can be an arbitrary real number, not just an ïteger.
èèDifferentiation yields
èèè▄
y» =èΣè(n+m)a┬xⁿó¡úî
èè n=0
èèè ▄
y»» =èΣè(n+m)(n+m-1)a┬xⁿó¡úì
èèèn=0
Substitute ïëèxìy»» + (xì-3x)y» + 2xyè=è0èë yield
èèèè▄èèèèèèèèèèèèèèèèè ▄
èèxìèΣè(n+m)(n+m-1)a┬xⁿó¡úìè+ (xì-3x)èΣ (n+m)a┬xⁿó¡úî
èèèèn=0èèèèèèèèèèèèèèn=0
èèèèèèèè ▄
èèèè+ 2xy Σèa┬xⁿó¡è=è0
èèèèèèèèn=0
orè ▄èèèèèèèèèèèè ▄
èè Σè(n+m)(n+m-1)a┬xⁿó¡è+èΣ (n+m)a┬xⁿó¡óî
èèn=0èèèèèèèèèèè n=0
èèèè ▄èèèèèèè▄
èèè-èΣè3a┬xⁿó¡ +è Σè2a┬xⁿó¡óîè =è0
èèèèn=0èèèèèèn=0
As ê second å fourth sum's exponent is different from ê
rest, it will be re-ïdexed so that all exponents are ê
same.
èè ▄èèèèèèèèèèèè ▄
èè Σè(n+m)(n+m-1)a┬xⁿó¡è+èΣè(n+m-1)a┬▀¬xⁿó¡
èèn=0èèèèèèèèèèè n=1
èèèè ▄èèèèèèè▄
èèè-èΣè3a┬xⁿó¡è+èΣè2a┬▀¬xⁿó¡è=è0
èèèèn=0èèèèèèn=1
As ê second å fourth sums starts at n = 1 while ê oêrs
start at n = 0, ê first term will be isolated å ê
remaïïg terms combïed ïë one sum
0è=è[ m(m-1) - 2m ] a╠x¡
è ▄
+èΣè{ (n+m)(n+m-1)a┬ + (n+m-1)a┬▀¬ - 3a┬ + 2a┬▀¬ }xⁿó¡
èn=1
è For differential equation ë have a non-trivial solution,
ê coefficient ç a╠x¡ must be zero.èSettïg it ë zero
yields ê INDICIAL EQUATION
m(m-1) - 2mè=è0
orèèèmì - 3mè=è0
This facërs ë
m(m - 3) = 0
The solutions are
m = 0, 3
èèIf eiêr ç êse two solutions ç ê ïdical equation,
are subsituted ïë ê assumed solution, ê first term ç
ê solution will be zero, so for m = 0 or 3 this
reduces ë
▄
Σ { (n+m)(n+m-1)a┬ + (n+m-1)a┬▀¬ - 3a┬ + 2a┬▀¬ }xⁿó¡è=è0
n=1
For this sum ë be zero, it is sufficient that every term ï
ê braces is zero.èSettïg each ë zero produces ê
RECURSION RELATION which for this example will be
èè (n+m)(n+m-1)a┬ + (n+m-1)a┬▀¬ - 3a┬ + 2a┬▀¬è=è0
As ê roots for m are distïct å DIFFER BY AN INTEGER,
this technique will only one solution will be produced by
this technique.èIt will come from substitutïg ê larger
solution, m = 3, back ïë ê recursion relation which is
èè (n+m)(n+m-1)a┬ + (n+m-1)a┬▀¬ - 3a┬ + 2a┬▀¬è=è0
ë yield
èè(n+3)(n+2)a┬ + (n+2)a┬▀¬ - 3a┬ + 2a┬▀¬è= 0
Combïïg like terms å rearrangïg yields
èè[ (n+3)(n+2) - 3 ]a┬è=è- [ (n+2) + 2 ] a┬▀¬
Rearrangïg
èèèèèèèèèèn+4
èèèa┬è=è- ───────────── a┬▀¬èèèn ≥ 1
èèèèèèè (n+3)(n+2)-3èèè
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = -5/[4(3)-3] a╠ = -5/9 a╠
2èèè a½ = -6/[5(4)-3] a¬ = -6/17 a¬ = 10/51 a╠
Thus one solution is
èC¬ xÄ [ 1 - 5/9 xè+ 10/51 xìè- ∙∙∙ ]
Ç B
9 4xìy»» + 4xy» + (3x-1)yè=è0
A)èèC¬ xî»ì [ 1 + 3/4 xè+ 9/80 xìè- ∙∙∙ ]
B)èèC¬ xî»ì [ 1 + 3/4 xè- 9/80 xìè- ∙∙∙ ]
C)èèC¬ xî»ì [ 1 - 3/4 xè+ 9/80 xìè- ∙∙∙ ]
D)èèC¬ xî»ì [ 1 - 3/4 xè- 9/80 xìè- ∙∙∙ ]
ü è Assume a solution ç ê form
èèèèèèèè ▄èèèèèè▄
y =èx¡èΣèa┬xⁿè =èΣèa┬xⁿó¡
èèèèn=0èèèèèn=0
where m can be an arbitrary real number, not just an ïteger.
èèDifferentiation yields
èèè▄
y» =èΣè(n+m)a┬xⁿó¡úî
èè n=0
èèè ▄
y»» =èΣè(n+m)(n+m-1)a┬xⁿó¡úì
èèèn=0
Substitute ïëè4xìy»» + 4xy» + (3x-1)yè=è0èë yield
èèèè ▄èèèèèèèèèèèèèèè▄
èè4xìèΣè(n+m)(n+m-1)a┬xⁿó¡úìè+ 4xèΣ (n+m)a┬xⁿó¡úî
èèèèn=0èèèèèèèèèèèèèèn=0
èèèèèèèè ▄
èèèè+ (3x-1) Σèa┬xⁿó¡è=è0
èèèèèèèèn=0
orè ▄èèèèèèèèèèèèè▄
èè Σè4(n+m)(n+m-1)a┬xⁿó¡è+èΣ 4(n+m)a┬xⁿó¡
èèn=0èèèèèèèèèèèèn=0
èèèè ▄èèèèèèèè▄
èèè+èΣè3a┬xⁿó¡óî -è Σèa┬xⁿó¡è =è0
èèèèn=0èèèèèèèn=0
As ê third sum's exponent is different from ê rest, it
will be re-ïdexed so that all exponents are ê same.
èè ▄èèèèèèèèèèèèè▄
èè Σè4(n+m)(n+m-1)a┬xⁿó¡è+èΣè4(n+m-1)a┬▀¬xⁿó¡
èèn=0èèèèèèèèèèèèn=0
èèèè ▄èèèèèèèèè▄
èèè+èΣè3a┬▀¬xⁿó¡óîè-èΣèa┬xⁿó¡è=è0
èèèèn=1èèèèèèèèn=0
As ê third sum starts at n = 1 while ê oêrs start at
n = 0, ê first term will be isolated å ê remaïïg
terms combïed ïë one sum.
0è=è[ 4m(m-1) + 4m - 1 ] a╠x¡
è ▄
+èΣè{ 4(n+m)(n+m-1)a┬ + 4(n+m-1)a┬ + 3a┬▀¬ - a┬ }xⁿó¡
èn=1
è For differential equation ë have a non-trivial solution,
ê coefficient ç a╠x¡ must be zero.èSettïg it ë zero
yields ê INDICIAL EQUATION
4m(m-1) + 4m - 1è=è0
orèèè4mì - 1è=è0
This facërs ë
(2m - 1)(2m + 1) = 0
The solutions are
m = -1/2, 1/2
èèIf eiêr ç êse two solutions ç ê ïdical equation,
are subsituted ïë ê assumed solution, ê first term ç
ê solution will be zero, so for m = 0 or 3 this
reduces ë
▄
Σ {è4(n+m)(n+m-1)a┬ + 4(n+m-1)a┬ + 3a┬▀¬ - a┬ }xⁿó¡è=è0
n=1
For this sum ë be zero, it is sufficient that every term ï
ê braces is zero.èSettïg each ë zero produces ê
RECURSION RELATION which for this example will be
èèè4(n+m)(n+m-1)a┬ + 4(n+m-1)a┬ + 3a┬▀¬ - a┬è=è0
As ê roots for m are distïct å DIFFER BY AN INTEGER,
this technique will only one solution will be produced by
this technique.èIt will come from substitutïg ê larger
solution, m = 1/2, back ïë ê recursion relation which is
èè 4(n+m)(n+m-1)a┬ + 4(n+m-1)a┬ + 3a┬▀¬ - a┬è=è0
ë yield
èè4(n+1/2)(n-1/2)a┬ + 4(n-1/2)a┬ + 3a┬▀¬ - a┬è= 0
orèè(2n+1)(2n-1)a┬ + 2(2n-1) + 3a┬▀¬ - a┬è=è0
Combïïg like terms å rearrangïg yields
èè[ (2n+1)(2n-1) + 2(2n-1) - 1 ]a┬è=è- 3 a┬▀¬
Rearrangïg
èèèèèèèèèèè3
èèèa┬è=è- ────────────── a┬▀¬èèèn ≥ 1
èèèèèèè (2n+3)(2n-1)-1èèè
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = -3/[5(1)-1] a╠ = -3/4 a╠
2èèè a½ = -3/[7(3)-1] a¬ = -3/20 a¬ = 9/80 a╠
Thus one solution is
èC¬ xî»ì [ 1 - 3/4 xè+ 9/80 xìè- ∙∙∙ ]
Ç B
10 xìy»» - 3xy» + (4-x)yè=è0
A)è C¬ xì [ 1 + 1/4 xè+ 1/36 xìè+ ∙∙∙ ]
B)è C¬ xì [ 1 + 1/4 xè- 1/36 xìè+ ∙∙∙ ]
C)è C¬ xì [ 1 - 1/4 xè+ 1/36 xìè+ ∙∙∙ ]
D)è C¬ xì [ 1 - 1/4 xè- 1/36 xìè+ ∙∙∙ ]
ü è Assume a solution ç ê form
èèèèèèèè ▄èèèèèè▄
y =èx¡èΣèa┬xⁿè =èΣèa┬xⁿó¡
èèèèn=0èèèèèn=0
where m can be an arbitrary real number, not just an ïteger.
èèDifferentiation yields
èèè▄
y» =èΣè(n+m)a┬xⁿó¡úî
èè n=0
èèè ▄
y»» =èΣè(n+m)(n+m-1)a┬xⁿó¡úì
èèèn=0è
Substitutïgèxìy»» - 3xy» + (4-x)yè=è0èyields
èèèè▄èèèèèèèèèèèèèèè▄
èèxìèΣè(n+m)(n+m-1)a┬xⁿó¡úìè+ 5xèΣ (n+m)a┬xⁿó¡úî
èèè n=0èèèèèèèèèèèèèèn=0
èèèèèèèè ▄
èèèè+ (4-x) Σèa┬xⁿó¡è=è0
èèèèèèèèn=0
orè ▄èèèèèèèèèèèè ▄
èè Σè(n+m)(n+m-1)a┬xⁿó¡è-èΣ 3(n+m)a┬xⁿó¡
èèn=0èèèèèèèèèèè n=0
èèèè ▄èèèèèèèè▄
èèè+èΣè4a┬xⁿó¡ -è Σèa┬xⁿó¡óîè =è0
èèèèn=0èèèèèèèn=0
As ê fourth sum's exponent is different from ê rest, it
will be re-ïdexed so that all exponents are ê same.
èè ▄èèèèèèèèèèèè ▄
èè Σè(n+m)(n+m-1)a┬xⁿó¡è-èΣè3(n+m-1)a┬▀¬xⁿó¡
èèn=0èèèèèèèèèèè n=0
èèèè ▄èèèèèèè▄
èèè+èΣè4a┬xⁿó¡è-èΣèa┬▀¬xⁿó¡óîè=è0
èèèèn=0èèèèèèn=1
As ê fourth sum starts at n = 1 while ê oêrs start at
n = 0, ê first term will be isolated å ê remaïïg
terms combïed ïë one sum.
0è=è[ m(m-1) - 3m + 4 ] a╠x¡
è ▄
+èΣè{ (n+m)(n+m-1)a┬ - 3(n+m-1)a┬ + 4a┬ - a┬▀¬ }xⁿó¡
èn=1
è For differential equation ë have a non-trivial solution,
ê coefficient ç a╠x¡ must be zero.èSettïg it ë zero
yields ê INDICIAL EQUATION
m(m-1) - 3m + 4è=è0
orèèèmì - 4m + 4è=è0
This facërs ë
(m-2)ì = 0
The solutions are
m = 2, 2
èèIf êse two solutions ç ê ïdical equation are
subsituted ïë ê assumed solution, ê first term ç
ê solution will be zero, so for m = 2 this reduces ë
▄
Σ { (n+m)(n+m-1)a┬ - 3(n+m-1)a┬ + 4a┬ - a┬▀¬ }xⁿó¡è=è0
n=1
For this sum ë be zero, it is sufficient that every term ï
ê braces is zero.èSettïg each ë zero produces ê
RECURSION RELATION which for this example will be
èèè(n+m)(n+m-1)a┬ - 3(n+m-1)a┬ + 4a┬ - a┬▀¬è=è0
As ê roots for m are REPEATED, this technique will only one
solution will be produced by this technique.èIt will come
from substitutïg ê solution, m = 2, back ïë ê
recursion relation which is
èè (n+m)(n+m-1)a┬ - 3(n+m-1)a┬ + 4a┬ - a┬▀¬è=è0
ë yield
èè(n+2)(n+3)a┬ - 3(n+3)a┬ + 4a┬ - a┬▀¬è= 0
Combïïg like terms å rearrangïg yields
èè[ (n+3)(n-1) + 4 ]a┬è=è a┬▀¬
Rearrangïg
èèèèèèèèèè1
èèèa┬è=è────────────── a┬▀¬èèèn ≥ 1
èèèèèèè(n+3)(n-1)+4èèè
The first few terms are
nèèèèè a┬
───èèèè ────
1èèè a¬ = 1/[4(0)+4] a╠ = 1/4 a╠
2èèè a½ = 1/[5(1)+4] a¬ = 1/9 a¬ = 1/36 a╠
Thus one solution is
èC¬ xì [ 1 + 1/4 xè+ 1/36 xìè+ ∙∙∙ ]
Ç A