Pro One: Differential Equations

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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