MATH201 Lecture 25: MATH201(LEC25)
10 Jan 2017
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Department
Course
Professor
Frobenius*Method….*(cont’d)*
*
Recap:*Use*the*Frobenius*method*if*𝑥!=0*is*a*regular-singular*point*for*a*
differential*equation.*
We*used*𝑦(𝑥)*for*the*previous*2*examples,*but*here,*the*𝑦𝑥=𝜙!(𝑥)*
𝑦𝑥=𝜙!𝑥=𝑥!!𝑐!𝑥!
∞
!!!
;𝑐!=1*
where*𝑐!*is:**
𝑐!𝑠!=−
1
𝑓(𝑠!+𝑛)𝑘+𝑠!𝑎!!!+𝑏!!!𝑐!
!!!
!!!
*
where*𝑠!*is*the*biggest*root*of*the*equation:*
𝑓𝑠=𝑠!+𝑎!−1𝑠+𝑏!*
and*𝑎!*and*𝑏!*are*the*first*term*of*the*expansion*of*the*functions*of*𝑎𝑥*and*𝑏(𝑥):*
𝑎𝑥=𝑥𝑝 𝑥 & 𝑏𝑥=𝑥!𝑞(𝑥)*
*
The*Second*Solution*
*
Consider:* 𝑥−𝑥!
!𝑦!! +𝑎𝑥𝑦!+𝑏𝑥𝑦=0*
When*𝑥!=0,*then:*𝑥!𝑦!! +𝑥𝑎 𝑥𝑦!+𝑏𝑥𝑦=0*
*
• Case*1:*
Assume*that*the*equation*𝑓𝑠=0*has*two*real*roots,*𝑠!>𝑠!*and*𝑠!−𝑠!∉N*
(is*a*non-integer*number_,*then*the*second*solution*is:*
𝜙!𝑥=𝑥!!𝑐!(𝑠!)𝑥!
∞
!!!
*
where*𝑐!=1*and*𝑠=𝑠!*
𝑐!𝑠!=−
1
𝑓𝑠!+𝑛𝑘+𝑠!𝑎!!!+𝑏!!!
!!!
!!!
𝑐!*
which*is*from*the*recursive*formula*above.*
*
• Case*2:*
Assume*that*𝑓𝑠=0*has*a*repeated*root*𝑠!,*then*the*second*solution*is:*
𝜙!𝑥=𝑦𝑥𝐿𝑛 𝑥+𝑥!!ℎ(𝑥)*
where*h(x)*is*an*analytic*function:*
ℎ𝑥=ℎ!𝑥!
∞
!!!
*
where*ℎ!*is*determined*from*the*recursive*formula:*
ℎ!=−
1
𝑛!𝑘+𝑠!𝑎!!!+𝑏!!!ℎ!+𝑎!!!𝑐!}
!!!
!!!
−
2
𝑛𝑐!*
*
Document Summary
= 0 is a regular-singular point for a differential equation. We used () for the previous 2 examples, but here, the = ! where ! is: 1 where ! is the biggest root of the equation: 1 + ! and ! and ! are the first term of the expansion of the functions of and (): Assume that the equation = 0 has two real roots, ! and ! N (is a non-integer number_, then the second solution is: where ! 1 which is from the recursive formula above: case 2: Assume that = 0 has a repeated root !, then the second solution is: where h(x) is an analytic function: where ! is determined from the recursive formula: Assume = 0 has real roots !, !, where ! = is an integer, then is: where ! = 0 !, 1 is determined below: (i) From p(x) and q(x) both not being analytic, we know that @ !
