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- 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=0
-
- Substitute ïëèü 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
-
-