# Ch6 series solutions algebra

14 Oct 2015
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### Ch6 series solutions algebra

• 1. CHAPTER 6 (Power Series Method) Review of Power Series     2 210 0 )()()( axcaxccaxc n n n is a power series in power of (x-a), where ,, 10 cc are real (complex ) constant and “a” is the centre of the series.     2 210 0 xcxccxc n n n is a power series in x with centre at 0 (Maclaurin series).    N n n nN axcxS 0 )()( is called nth partial sum.  If   LxSN N )(lim series is convergent otherwise divergent.  Each Series is always convergent at its centre.  If    0 )( n n n axc converges for all points in Rax  then i. The region Rax  is called region of convergence. R is called radius of convergence. ii. If R = 0 then series converges only at its centre. If R =  then series converges everywhere or x . In other situation R can be any finite positive real number. iii. Rax  is called circle or interval of convergence. x- Real or x-Complex NOTE: If a series converges for Rax  then it must be divergent for Rax  and for Rax  , it may converge at every point or on some points or may not converge at all.
• 2. HOW TO FIND RADIUS OF CONVERGENCE R? Ratio Test: Find L c c axc axc n n n n n n n n         1 1 1 limlim )( )( i. The radius of convergence is L R 1  ii. If Rax  || then the test is inclusive iii. If Rax  || so series diverges iv. If Rax  || so series converges Example: Find radius and region of convergence for the series (i)   1 2 n n n x n (ii)     1 2 )3( )!( )!2( n n x n n Sol: (i) 1 2 , 2 1 1     n c n c n n n n 2 11 22 1 2 2)1( 2 limlim 1        L RL n n n n n n n n and region of convergence is 2 1 x (ii) 212 ))!1(( )22( , )!( )!2(     n n c n n c nn 4 11 4 )1( )12)(22( )!2())!1(( )!()!22( 22 2 limlim        L R n nn nn nn nn and region of convergence is 4 1 3 x  Within region of convergence, series converges absolutely and uniformly. It can be differentiated and integrated term by term and resulting series has the same radius as well as region of convergence. In case R > 0 then the corresponding power series always represents an analytic function.  A function f is analytic at a point a if it can be represented by a power series in )( ax  with a positive or infinite radius of convergence.  Term wise addition:   0 )( n n n axa     0 )( n n n axb =    0 ))(( n n nn axba  Term wise multiplication:
• 3.   0 )( n n n axa     0 )( n n n axb = n nnn n n axbabababa )}({ 02211 0 0       Series addition: in order o perform addition   ln n n axc )(     km m m axb )( given series should satisfy nmkl  and, If the series correspond to a function say 0 0( ) ( )n n n f x c x x      Then the radius of convergence is the distance of the closest singular point from the centre of the series. Example: 11 1 1 0 2      Rxxx x n n POWER SERIES METHOD (Basic method to solve LDE)  Let a differential equation be in standard form ) 0( ( )y yP x Q x y    .  A point “x0” is called ordinary point if P(x) and Q(x) are analytic at “x0”, otherwise the point “x0” will be called singular (Regular, Irregular). Examples: Standard form ) 0( ( )y yP x Q x y    1. '' ' s n 0ix y y ye x   2. '' ' ln 0x y y ye x    THEORM 6.1 Existence of power series solutions: If 0xx  is an ordinary point of the differential equation 0)()("  yxQyxPy , We can always find two linearly independent solutions in the form of power series centered at x0 that is     0 0 )( n n n xxcy .
• 4. A series solution converges at least on some interval Rxx  0 , where R is the distance from 0x to the closest singular point.  A solution of the form     0 0 )( n n n xxcy is said to be a solution about the ordinary point x0 Example: Using Power series method solve 0 yy Sol: i. Assume     0n n n xcy is a solution then      1 1 n n nnxcy ,      2 2 )1( n n n xnncy . ii. 0)1( 2 0 2         n n n n n n xcxnncyy iii. Change k = n-2 in 1st term, k = n in 2nd term        0)1)(2( 0 0 2 k k k k k k xcxkkc        ,2,1,0 )2)(1( 0)2)(1( 2 2 k kk c c cckk k k kk Recurrence formula iv. Let thencandc 01 10  using Recurrence formula 0 !4 1 0, !2 1 54 32     cc cc  4 2 1 !4 1 !2 1 x x y Next y2 :
• 5. Let thencandc 10 10  using Recurrence formula 11 c 6.5.4 1 0 3.2 1 ,0 54 32    cc cc  5 3 2 6.5.4 1 3.2 x x xy Hence, 2211 yCyCy  that is basically xCxCy sincos 21  Example: Using Power series method solve 022  yyxy Sol: i. Assume     0n n n xay is a solution then      1 1 n n nnxcy ,      2 2 )1( n n n xnncy 022)1(22 2 1 0 12            n n n n n n n n n xcnxcxxnncyyxy 022)1( 2 1 0 2            n n n n n n n n n xcnxcxnnc Change k = n-2 in 1st term, k = n in 2nd and 3rd terms           022)1)(2( 0 1 0 2 k k k k k k k k k xckxcxkkc            022)1)(2(1.2 1 1 1 022 k k k k k k k k k xcckxcxkkcc      0)22)1)(2(({22 1 202 k k kkk xckckkccc 0202 022 cccc  (1) ,...3,2,1 2 2 0)22()1)(2( 2 2        kc k c ckckk kk kk (2) Relation (2) is known as Recurrence Formula.
• 6. ii. Let thencandc 01 10  using Recurrence formula 0 7 2 6 1 6 1 0 5 2 2 1 2 1 0 2 3 ,1 5746 3524 1302    cccc cccc cccc ... 6 1 2 1 1 642 1  xxxy Now, let thencandc 10 10  using Recurrence formula, we get 2211 53 2 35 4 3 2 yCyCyxxxy  Example: Find two series solution at x = 0 for 0 yyey x i. Assume     0n n n xay is a solution then      1 1 n n nnxcy ,      2 2 )1( n n n xnncy . ii. 0) !2 1()1( 2 1 0 2 2            n n n n n n n n n xcnxc x xxnnc               0......432 ... !4!3!2 1...201262 3 3 2 210 3 4 2 321 432 3 5 2 432 xcxcxccxcxcxcc xxx xxcxcxcc              tcoefficiencomparingxcxcxcc xcccxccc xcxcxcc 0... ...)32 2 1 ()2( ...201262 3 3 2 210 3 321211 3 5 2 432 )(0123 2 1 )(026 )(02 4321 1213 012 iiicccc iicccc iccc   
• 7. Let    ,0, 6 1 , 2 1 01 43210 ccccandc  32 1 6 1 2 1 1 xxy Now let    , 24 1 , 6 1 , 2 1 10 43210 ccccandc  432 2 24 1 6 1 2 1 xxxxy Hence, 2211 yCyCy 
• 8. Section 6.2: Solution about Singular Points Given ) 0( ( )y yP x Q x y     The point x0 is called regular singular point if the functions p(x) = (x – x0) P(x) and 2 0( ) ( )) (q x x x Q x  are both analytic at x = x0. Otherwise irregular singular point Definition A function, f(x), is called analytic at x= x0 if the Taylor series for f(x) about x= x0 has a positive radius of convergence (series exist), if x0 = 0 we have Maclaurin series.  The point x = x0 is a regular singular point if (x – x0) has at most power 1 in the denominator of P(x) and at most power 2 in the denominator of Q (x). ) 0( ( )y yP x Q x y    Example: 3 2 3 2 3 3 3 3 0 first put theEq.instander form 0 0 0 poin 4 3 4 3 4 . 3 t y y y y y y y y y x x x x x x x x x x irregular               ………………………………….. .int3,0 0)3( poregularx yyxx   …………………………………. iii. irregularxregularx yxyxyxx 5&0 0)25(4)5( 222   NOTE: The power series method is applicable at ordinary points but fails at regular singular points. At regular singular points, Frobenius’ method,
• 9. which is basically an extension of power series method, helps to find the solution.  It possible to obtain the indicial equation in advance of substituting      0n rn n xay into the differential equation, the general indicial equation is 0)1( 00  qrprr where 00 qandp are given by the definition of p(x) and q(x). Example Find the indicial equation and indicial roots of the differential equation 3xy'' + y' - y = 0. Sol: Since x = 0 is a regular singular point of the differential equation 3xy’’ + y’ – y =0 We try a solution of the form      0n rn nxcy . Or Then xp(x) = 3 1 and x2 q(x) = - 3 1 x. Hence p0 = 3 1 and q0 = 0. Hence the indicial equation is r(r – 1) + 3 1 r + 0 = 0  r (3r-2) = 0 NOTE: If r1 and r2 are the roots of indicial equation then Case 1: 2121 rrandrr  is negative integer then solution will be                   0 2 0 1 21 n n n r n n n r xbxcxaxcy Case 2: 2121 rrandrr  is a positive integer then solution will be                   0 12 0 11 21 ln n n n r n n n r xbxxCyyandxaxcy 𝑦 = 𝑐1 𝑦1 + 𝑐2 𝑦2
• 10. Case 3: rrr  21 then                     1 12 0 0 1 ln 0, n n n r n n n r xbxxyy axaxy 𝑦 = 𝑐1 𝑦1 + 𝑐2 𝑦2 NOTE: If 1y is known, method of reduction of order may be used to get     2 1 12 y e yy pdx Example : 0)13()1(  yyxyxxSolve Sol: Assume        00 n rn n n n n r xaxaxy is a solution. n r n n r n y a (x r)x & y a (n r)(n r )x               1 2 1 x(x )y ( x )y y     1 3 1 0        0)()(3 )1)(()1)(( 1 1 rn n rn n rn n rn n rn n xaxarnxarn xarnrnxarnrn     )(0)()1)(( 1)(3)1)(( 1 Axarnrnrn xarnrnrn rn n rn n      
• 11.     )(0)()1)(( 1)(3)1)(( 1 Axarnrnrn xarnrnrn rn n rn n       Indicial Equation: Take n = 0 and select coefficient (except na ) of the smallest power of x we get 0)1(  rrr the indicial equation. Roots of indicial equation are r = 0 , 0 (double root) Now we need Recurrence Formula: In (A), take r = 0      )(0)()1)(( 1)(3)1)(( 1 Axarnrnrn xarnrnrn rn n rn n           0)()1)((1)(3)1)(( 1   n n n n xannnxannn Taking n = k in 1st term and n -1 = k in 2nd term, we get    k k k k(k)(k ) (k) a x (k )(k) (k ) a x         11 3 1 1 1 0 k k a a a a a     1 1 0 1 2 r n n y x a x x ( x x x ) if x x             0 2 3 1 1 1 1 1 Now ?2 y Use the reduction formula:
• 12.     2 1 12 y e yy pdx x(x )y ( x )y y     1 3 1 0        2 )1)(ln( 1 21 )1( 13 )( xxpdx xxxx x xP x x dx xx dx xx x x dx x e xy e yy xxpdx                 1 ln1 1 1 )1( )1( 1 1 )1( 11 1 2 2 2 )1)(ln( 2 1 12 2 2211 ycycy 