# ch11 - CHAPTER 11 POWER SERIES METHODS SECTION 11.1...

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548 Chapter 11 CHAPTER 11 POWER SERIES METHODS SECTION 11.1 INTRODUCTION AND REVIEW OF POWER SERIES The power series method consists of substituting a series y = Σ c n x n into a given differential equation in order to determine what the coefficients { c n } must be in order that the power series will satisfy the equation. It might be pointed out that, if we find a recurrence relation in the form c n +1 = φ ( n ) c n , then we can determine the radius of convergence ρ of the series solution directly from the recurrence relation, 1 1 lim lim . () n nn n c cn →∞ →∞ + == In Problems 1–10 we give first that recurrence relation that can be used to find the radius of convergence and to calculate the succeeding coefficients 123 ,,, ccc ± in terms of the arbitrary constant c 0 . Then we give the series itself 1. 1 ; 1 n n c c n + = + it follows that 0 and lim( 1) ! n n c n →∞ + = . 234 00 0 1 1 26 2 4 1 ! 2 ! 3 ! 4 ! x xxx x yx c x c ce  = + ++++ = + =   ±± 2. 1 4 ; 1 n n c c n + = + it follows that 0 41 and lim !4 n n n c n →∞ + = . 34 2 0 22 33 44 4 32 32 1 4 8 4 4 1 1! 2! 3! 4! x xx c x x x x cc e =+ + +++ = ± ± 3. 1 3 ; 21 n n c c n + =− + it follows that 0 13 2 1 and lim 3 n n n n n c n →∞ −+ = . 23 4 0 39 9 2 7 1 2 8 16 128 x x c + +− ±

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Section 11.1 549 22 33 44 3/ 2 00 234 3 3 1 1!2 2!2 3!2 4!2 x xx x x cc e  =− +− =   ± 4. When we substitute y = Σ c n x n into the equation y' + 2 xy = 0, we find that [] 1 12 0 (2 ) 2 0 . n nn n cn c c x + + = +++ = Hence c 1 = 0 — which we see by equating constant terms on the two sides of this equation — and 2 2 . 2 n n c c n + + It follows that 135 o d d 0 ccc c ==== = ± and 0 2 (1 ) . ! k k c c k = Hence 2 46 246 2 0 () 1 1 23 1 ! 2 ! 3 ! x xxx yx c x c ce + + + + = ±± and . ρ =∞ 5. When we substitute y = Σ c n x n into the equation 2 , y ′ = we find that 1 3 0 2( 3 ) 0 . n n x nc c x + + = ++ + = Hence c 1 = c 2 = 0 — which we see by equating constant terms and x -terms on the two sides of this equation — and 3 . 3 n c c n = + It follows that c 3 k +1 = c 3 k +2 = 0 and 3 . 36 (3) !3 k k c kk == ⋅⋅ ⋅ ± Hence 3 9 3 6 9 (/ 3 ) 0 1 1 . 3 18 162 1!3 2!3 3!3 x x x x x c c =+ + + + + = and . 6. 1 ; 2 n n c c + = it follows that 0 and lim2 2. 2 n n n c c →∞ = 0 1 24 81 6 xx x x c = + ++++ ±
550 Chapter 11 234 00 0 2 1 22 2 2 2 1 2 x xxx c c c x x     =+ ++++ = =         ± 7. 1 2; nn cc + = it follows that 0 11 2a n d l i m . n n n ρ →∞ == = () 23 4 0 1 2 4 8 1 6 yx c x x x x + + ++ ± ()()()() 0 0 12 2 2 2 c cx x x x x = + = ± 8. 1 (2 1) ; n n nc c n + =− + it follows that lim 1. 21 n n n →∞ + 4 0 5 1 2 8 16 128 xx x x c  + +   ± Separation of variables gives 0 1 . c x 9. 1 ) ; 1 n n c n + + = + it follows that 0 (1 ) n cn c and 1 lim 1. 2 n n n →∞ + + 0 3 4 5 c x x x x = + ± Separation of variables gives 0 2 . ) c x = 10. 1 3) ; n n c n + = + it follows that lim 1.

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## This note was uploaded on 04/08/2008 for the course MATH 374 taught by Professor Zhu during the Spring '08 term at Western Michigan.

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ch11 - CHAPTER 11 POWER SERIES METHODS SECTION 11.1...

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