Hw 4 Solution

Hw 4 Solution - 162 Chapter 5. Series Solutions of Second...

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Unformatted text preview: 162 Chapter 5. Series Solutions of Second Order Linear Equations The series converges absolutely for |x| < 1 . Term-by-term differentiation results in ∞ n2 xn−1 = 1 + 4x + 9x2 + 16 x3 + . . . y= n=1 ∞ n2 (n − 1) xn−2 = 4 + 18x + 48x2 + 100x3 + . . . y= n=2 Shifting the indices, we can also write ∞ ∞ (n + 1)2 xn y= and (n + 2)2 (n + 1) xn . y= n=0 n=0 Sec 5.1, Pg 249 20. Shifting the index in the second series, that is, setting n = k + 1 , ∞ ∞ ak xk+1 = an−1 xn . n=1 k=0 Hence ∞ ∞ ak+1 xk + k=0 ∞ ak xk+1 = k=0 ∞ ak+1 xk + k=0 ak −1 x k k=1 ∞ (ak+1 + ak−1 )xk+1 . = a1 + k=1 21. Shifting the index by 2 , that is, setting m = n − 2 , ∞ ∞ n(n − 1)an xn−2 = n=2 (m + 2)(m + 1)am+2 xm m=0 ∞ (n + 2)(n + 1)an+2 xn . = n=0 22. Shift the index down by 2 , that is, set m = n + 2 . It follows that ∞ ∞ ∞ an xn+2 = n=0 am−2 xm = m=2 an−2 xn . n=2 24. Clearly, ∞ ∞ (1 − x2 ) ∞ n(n − 1)an xn−2 = n=2 n(n − 1)an xn−2 − n=2 n(n − 1)an xn . n=2 Shifting the index in the first series, that is, setting k = n − 2 , ∞ ∞ n(n − 1)an xn−2 = n=2 (k + 2)(k + 1)ak+2 xk k=0 ∞ (n + 2)(n + 1)an+2 xn . = n=0 162 Chapter 5. Series Solutions of Second Order Linear Equations The series converges absolutely for |x| < 1 . Term-by-term differentiation results in ∞ n2 xn−1 = 1 + 4x + 9x2 + 16 x3 + . . . y= n=1 ∞ n2 (n − 1) xn−2 = 4 + 18x + 48x2 + 100x3 + . . . y= n=2 Shifting the indices, we can also write ∞ ∞ (n + 1)2 xn y= and (n + 2)2 (n + 1) xn . y= n=0 n=0 20. Shifting the index in the second series, that is, setting n = k + 1 , ∞ ∞ ak xk+1 = an−1 xn . n=1 k=0 Hence ∞ ∞ ak+1 xk + k=0 ∞ ak xk+1 = k=0 ∞ ak+1 xk + k=0 ak −1 x k k=1 ∞ (ak+1 + ak−1 )xk+1 . = a1 + k=1 21. Shifting the index by 2 , that is, setting m = n − 2 , ∞ ∞ n(n − 1)an xn−2 = n=2 (m + 2)(m + 1)am+2 xm m=0 ∞ (n + 2)(n + 1)an+2 xn . = n=0 22. Shift the index down by 2 , that is, set m = n + 2 . It follows that ∞ ∞ ∞ an xn+2 = n=0 am−2 xm = m=2 an−2 xn . n=2 Sec 5.1, Pg 249 24. Clearly, ∞ ∞ (1 − x2 ) ∞ n(n − 1)an xn−2 = n=2 n(n − 1)an xn−2 − n=2 n(n − 1)an xn . n=2 Shifting the index in the first series, that is, setting k = n − 2 , ∞ ∞ n(n − 1)an xn−2 = n=2 (k + 2)(k + 1)ak+2 xk k=0 ∞ (n + 2)(n + 1)an+2 xn . = n=0 5.2 163 Hence ∞ ∞ (1 − x2 ) ∞ n(n − 1)an xn−2 = (n + 2)(n + 1)an+2 xn − n=2 n=0 n(n − 1)an xn . n=2 Note that when n = 0 and n = 1 , the coefficients in the second series are zero. So ∞ (1 − x2 ) ∞ n(n − 1)an xn−2 = n=2 [(n + 2)(n + 1)an+2 − n(n − 1)an ] xn . n=0 26. Clearly, ∞ ∞ ∞ nan xn−1 + x n=1 ∞ an xn = n=0 nan xn−1 + n=1 an xn+1 . n=0 Shifting the index in the first series, that is, setting k = n − 1 , ∞ ∞ nan xn−1 = n=1 (k + 1)ak+1 xk . k=0 Shifting the index in the second series, that is, setting k = n + 1 , ∞ ∞ an xn+1 = n=0 ak−1 xk . k=1 Combining the series, and starting the summation at n = 1 , ∞ ∞ ∞ nan xn−1 + x n=1 an xn = a1 + n=0 [(n + 1)an+1 + an−1 ] xn . n=1 27. We note that ∞ ∞ ∞ n(n − 1)an xn−2 + x n=2 ∞ an xn = n=0 n(n − 1)an xn−1 + n=2 an xn . n=0 Shifting the index in the first series, that is, setting k = n − 1 , ∞ ∞ ∞ n(n − 1)an xn−1 = n=2 k (k + 1)ak+1 xk = k=1 k (k + 1)ak+1 xk , k=0 since the coefficient of the term associated with k = 0 is zero. Combining the series, ∞ ∞ n(n − 1)an xn−2 + x n=2 ∞ an xn = n=0 [n(n + 1)an+1 + an ] xn . n=0 5.2 1.(a,b,d) Let y = a0 + a1 x + a2 x2 + . . . + an xn + . . . . Then ∞ ∞ n(n − 1)an xn−2 = y= n=2 (n + 2)(n + 1)an+2 xn . n=0 5.2 169 Note that for n = 2 , a4 = 0 . Since the indices differ by two, we also have a2k = 0 for k = 2, 3, . . . . On the other hand, for k = 1, 2, . . . , (2k − 3)a2k−1 (2k − 5)(2k − 3)a2k−3 −a1 = = ... = k . 4(2k + 1) 42 (2k − 1)(2k + 1) 4 (2k − 1)(2k + 1) a2k+1 = Therefore the general solution is ∞ y = a0 + a1 x − a0 x2 x2n+1 − a1 . n (2n − 1)(2n + 1) 4 4 n=1 Hence the linearly independent solutions are y1 (x) = 1 − x2 /4 and ∞ y2 (x) = x − x5 x7 x3 x2n+1 − − − ... = x − . 12 240 2240 4n (2n − 1)(2n + 1) n=1 (c) The Wronskian at 0 is 1. 2 n Sec 5.2, Pg 260 11.(a,b,d) Let y = a0 + a1 x + a2 x + . . . + an x + . . . . Then ∞ ∞ nan xn−1 = y= n=1 and (n + 1)an+1 xn n=0 ∞ ∞ n(n − 1)an xn−2 = y= n=2 (n + 2)(n + 1)an+2 xn . n=0 Substitution into the ODE results in ∞ (3 − x2 ) ∞ ∞ (n + 2)(n + 1)an+2 xn − 3x n=0 (n + 1)an+1 xn − n=0 an xn = 0 . n=0 Before proceeding, write ∞ x2 ∞ (n + 2)(n + 1)an+2 xn = n=0 n(n − 1)an xn n=2 and ∞ ∞ (n + 1)an+1 xn = x n=0 n an x n . n=1 It follows that 6a2 − a0 + (−4a1 + 18a3 )x+ ∞ [3(n + 2)(n + 1)an+2 − n(n − 1)an − 3n an − an ] xn = 0. + n=2 We obtain a2 = a0 /6 , 2a3 = a1 /9 , and 3(n + 2)an+2 = (n + 1)an , n = 0, 1, 2, . . . . The indices differ by two, so for k = 1, 2, . . . a2k = (2k − 3)(2k − 1)a2k−4 3 · 5 . . . (2k − 1) a0 (2k − 1)a2k−2 = = ... = k 2 (2k − 2)(2k ) 3(2k ) 3 3 · 2 · 4 . . . (2k ) 170 Chapter 5. Series Solutions of Second Order Linear Equations and a2k+1 = (2k )a2k−1 (2k − 2)(2k )a2k−3 2 · 4 · 6 . . . (2k ) a1 =2 = ... = k . 3(2k + 1) 3 (2k − 1)(2k + 1) 3 · 3 · 5 . . . (2k + 1) Hence the linearly independent solutions are ∞ x4 5x6 x2 3 · 5 . . . (2n − 1) x2n + + + ... = 1 + y1 (x) = 1 + 6 24 432 3n · 2 · 4 . . . (2n) n=1 ∞ y2 (x) = x + 2x3 8x5 16x7 2 · 4 · 6 . . . (2n) x2n+1 + + + ... = x + . 9 135 945 3n · 3 · 5 . . . (2n + 1) n=1 (c) The Wronskian at 0 is 1. 12.(a,b,d) Let y = a0 + a1 x + a2 x2 + . . . + an xn + . . . . Then ∞ ∞ nan xn−1 = y= n=1 (n + 1)an+1 xn n=0 and ∞ ∞ n(n − 1)an xn−2 = y= n=2 (n + 2)(n + 1)an+2 xn . n=0 Substitution into the ODE results in ∞ ∞ (n + 2)(n + 1)an+2 xn + x (1 − x) n=0 ∞ (n + 1)an+1 xn − n=0 an xn = 0 . n=0 Before proceeding, write ∞ ∞ (n + 2)(n + 1)an+2 xn = x n=0 (n + 1)n an+1 xn n=1 and ∞ ∞ (n + 1)an+1 xn = x n=0 n an xn . n=1 It follows that ∞ [(n + 2)(n + 1)an+2 − (n + 1)n an+1 + n an − an ] xn = 0. 2a2 − a0 + n=1 We obtain a2 = a0 /2 and (n + 2)(n + 1)an+2 − (n + 1)n an+1 + (n − 1)an = 0 5.2 171 for n = 0, 1, 2, . . . . Writing out the individual equations, 3 · 2 a3 − 2 · 1 a2 4 · 3 a4 − 3 · 2 a3 + a2 5 · 4 a5 − 4 · 3 a4 + 2 a3 6 · 5 a6 − 5 · 4 a5 + 3 a4 =0 =0 =0 =0 . . . The coefficients are calculated successively as a3 = a0 /(2 · 3), a4 = a3 /2 − a2 /12 = a0 /24, a5 = 3a4 /5 − a3 /10 = a0 /120, . . .. We can now see that for n ≥ 2 , an is proportional to a0 . In fact, for n ≥ 2 , an = a0 /(n!) . Therefore the general solution is a0 x3 a0 x4 a0 x2 y = a0 + a1 x + + + + ... . 2! 3! 4! Hence the linearly independent solutions are y2 (x) = x and ∞ y1 (x) = 1 + xn = ex − x. n! n=2 (c) The Wronskian is ex (1 − x). At x = 0 it is 1. Sec 5.2, Pg 260 13.(a,b,d) Let y = a0 + a1 x + a2 x2 + . . . + an xn + . . . . Then ∞ ∞ nan xn−1 = y= n=1 and (n + 1)an+1 xn n=0 ∞ ∞ n(n − 1)an xn−2 = y= n=2 (n + 2)(n + 1)an+2 xn . n=0 Substitution into the ODE results in ∞ ∞ (n + 2)(n + 1)an+2 xn + x 2 n=0 ∞ (n + 1)an+1 xn + 3 n=0 First write ∞ n=0 ∞ (n + 1)an+1 xn = x an xn = 0 . n=0 n an x n . n=1 We then obtain ∞ [2(n + 2)(n + 1)an+2 + n an + 3an ] xn = 0 . 4a2 + 3a0 + n=1 It follows that a2 = −3a0 /4 and 2(n + 2)(n + 1)an+2 + (n + 3)an = 0 172 Chapter 5. Series Solutions of Second Order Linear Equations for n = 0, 1, 2, . . . . The indices differ by two, so for k = 1, 2, . . . (2k + 1)a2k−2 (2k − 1)(2k + 1)a2k−4 =2 = ... 2(2k − 1)(2k ) 2 (2k − 3)(2k − 2)(2k − 1)(2k ) (−1)k 3 · 5 . . . (2k + 1) a0 . = 2k (2k )! a 2k = − and (2k )(2k + 2)a2k−3 (2k + 2)a2k−1 =2 = ... 2(2k )(2k + 1) 2 (2k − 2)(2k − 1)(2k )(2k + 1) (−1)k 4 · 6 . . . (2k )(2k + 2) = a1 . 2k (2k + 1)! a2k+1 = − Hence the linearly independent solutions are ∞ 3 5 76 (−1)n 3 · 5 . . . (2n + 1) 2n y1 (x) = 1 − x2 + x4 − x + ... = x 4 32 384 2n (2n)! n=0 ∞ 1 17 (−1)n 4 · 6 . . . (2n + 2) 2n+1 1 x + ... = x + x . y2 (x) = x − x3 + x5 − 3 20 210 2n (2n + 1)! n=1 (c) The Wronskian at 0 is 1. 15.(a) From Problem 2, we have ∞ y1 (x) = x2n 2n n! n=0 ∞ and y2 (x) = 2n n! x2n+1 . (2n + 1)! n=0 Since a0 = y (0) and a1 = y (0) , we have y (x) = 2 y1 (x) + y2 (x) . That is, 1 1 1 1 y (x) = 2 + x + x2 + x3 + x4 + x5 + x6 + . . . . 3 4 15 24 The four- and five-term polynomial approximations are p4 = 2 + x + x2 + x3 /3, and p5 = 2 + x + x2 + x3 /3 + x4 /4 . (b) 174 Chapter 5. Series Solutions of Second Order Linear Equations (b) (c) The four-term approximation p4 appears to be reasonably accurate (within 10%) on the interval |x| < 0.9 . 20. Two linearly independent solutions of Airy’s equation (about x0 = 0) are ∞ y1 (x) = 1 + x 3n 2 · 3 . . . (3n − 1)(3n) n=1 ∞ y2 (x) = x + x3n+1 . 3 · 4 . . . (3n)(3n + 1) n=1 Applying the ratio test to the terms of y1 (x) , 2 · 3 . . . (3n − 1)(3n) x3n+3 1 3 = lim |x| = 0 . n → ∞ |2 · 3 . . . (3n + 2)(3n + 3) x3n | n → ∞ (3n + 1)(3n + 2)(3n + 3) lim Similarly, applying the ratio test to the terms of y2 (x) , 3 · 4 . . . (3n)(3n + 1) x3n+4 1 3 = lim |x| = 0 . n → ∞ |3 · 4 . . . (3n + 3)(3n + 4) x3n+1 | n → ∞ (3n + 2)(3n + 3)(3n + 4) lim Hence both series converge absolutely for all x . Sec 5.2, Pg 260 21. Let y = a0 + a1 x + a2 x2 + . . . + an xn + . . . . Then ∞ ∞ nan xn−1 = y= n=1 and (n + 1)an+1 xn n=0 ∞ ∞ n(n − 1)an xn−2 = y= n=2 (n + 2)(n + 1)an+2 xn . n=0 Substitution into the ODE results in ∞ ∞ (n + 2)(n + 1)an+2 xn − 2x n=0 ∞ (n + 1)an+1 xn + λ n=0 an xn = 0 . n=0 5.2 175 First write ∞ ∞ (n + 1)an+1 xn = x n=0 n an x n . n=1 We then obtain ∞ [(n + 2)(n + 1)an+2 − 2n an + λ an ] xn = 0 . 2a2 + λ a0 + n=1 Setting the coefficients equal to zero, it follows that (2n − λ) an (n + 1)(n + 2) an+2 = for n = 0, 1, 2, . . . . Note that the indices differ by two, so for k = 1, 2, . . . (4k − 4 − λ)a2k−2 (4k − 8 − λ)(4k − 4 − λ)a2k−4 = = ... (2k − 1)2k (2k − 3)(2k − 2)(2k − 1)2k λ . . . (λ − 4k + 8)(λ − 4k + 4) = (−1)k a0 . (2k )! a2k = and (4k − 2 − λ)a2k−1 (4k − 6 − λ)(4k − 2 − λ)a2k−3 = = ... 2k (2k + 1) (2k − 2)(2k − 1)2k (2k + 1) (λ − 2) . . . (λ − 4k + 6)(λ − 4k + 2) = (−1)k a1 . (2k + 1)! a2k+1 = Hence the linearly independent solutions of the Hermite equation (about x0 = 0) are λ λ(λ − 4) 4 λ(λ − 4)(λ − 8) 6 y1 (x) = 1 − x2 + x− x + ... 2! 4! 6! y2 (x) = x − λ − 2 3 (λ − 2)(λ − 6) 5 (λ − 2)(λ − 6)(λ − 10) 7 x+ x− x + ... . 3! 5! 7! (b) Based on the recurrence relation an+2 = (2n − λ) an , (n + 1)(n + 2) the series solution will terminate as long as λ is a nonnegative even integer. If λ = 2m, then one or the other of the solutions in part (b) will contain at most m/2 + 1 terms. In particular, we obtain the polynomial solutions corresponding to λ = 0, 2, 4, 6, 8, 10 : λ=0 λ=2 λ=4 λ=6 λ=8 λ = 10 y1 (x) = 1 y2 (x) = x y1 (x) = 1 − 2x2 y2 (x) = x − 2x3 /3 y1 (x) = 1 − 4x2 + 4x4 /3 y2 (x) = x − 4x3 /3 + 4x5 /15 5.3 179 That is, a3 = 1/6 , a4 = 1/12 , a5 = 1/24 , a6 = 1/45 , . . . . Hence the series solution of the initial value problem is 1 1 1 1 13 7 y (x) = x + x3 + x4 + x5 + x6 + x + ... . 6 12 24 45 1008 5.3 Sec 5.3, Pg 265 2. Let y = φ(x) be a solution of the initial value problem. First note that y = −(sin x)y − (cos x)y . Differentiating twice, y = −(sin x)y − 2(cos x)y + (sin x)y y (4) = −(sin x)y − 3(cos x)y + 3(sin x)y + (cos x)y . Given that φ(0) = 0 and φ (0) = 1 , the first equation gives φ (0) = 0 and the last two equations give φ (0) = −2 and φ(4) (0) = 0 . 3. Let y = φ(x) be a solution of the initial value problem. First write y =− 1+x 3 ln x y− y. 2 x x2 Differentiating twice, y −1 (x + x2 )y + (3x ln x − x − 2)y + (3 − 6 ln x)y . x3 −1 y (4) = 4 (x2 + x3 )y + (3x2 ln x − 2x2 − 4x)y + x = + (6 + 8x − 12x ln x)y + (18 ln x − 15)y . Given that φ(1) = 2 and φ (1) = 0 , the first equation gives φ (1) = 0 and the last two equations give φ (1) = −6 and φ(4) (1) = 42 . 4. Let y = φ(x) be a solution of the initial value problem. First note that y = −x2 y − (sin x)y . ~t;.~ ~--- q L ('>l - L"'--3) 0 f1 II) '- t '0 . (t YO 'o--Q \."- - L'>L - .J ~ nX):--tl X \....­ ~ L "ll-) ~ "-- . )( -"5 t lt 1 { ~ 3 ~ .: . x. ( yt ~ t I ( \{ - 3) ~ , ("( - ~ ( 't + f) ----a V \ . ,)(::-? .)/ ~~ 1 ~L . ~ (x) ~ ...-I­ "-­ ':.­ 3, - i ..-~ ~+--.~, ~ ~ 11 L\ X~ ~ 4/ ti 3 4 b -I - ~ ~"i~ S­ ~ 2-~ ~.~ !x- 4 1«1- ~~~, ~.r-- (x - l1) '1\ (i,,- ~'"-7 <:: ~. ~ ~h-~ Y ~ - It -te 3 '1 L{ - If t -J ~ .~:l ~~ ~ x~ ~ - 4 I ~ ~ ~ J ' t ~ ~ ~~ ~ ~ ~ f0-.- \x r\.>. t - t't ,,\1\...­ L\:'-""(X-t-Li) ':C::s" , C-< f d ) ~ D D -b 3 t -1 5.4 187 27. We have n (x2 − 1)n = k=0 (−1)n−k n ! 2k x, k !(n − k )! which is a polynomial of degree 2n. Differentiating n times, dn 2 (x − 1)n = dxn n (−1)n−k n ! (2k )(2k − 1) . . . (2k − n + 1)x2k−n , k !(n − k )! k=µ in which the lower index is µ = [n/2] + 1 . Note that if n = 2m + 1, then µ = m + 1 . Now shift the index, by setting k = n − j . Hence dn 2 (x − 1)n = dxn [n/2] j =0 (−1)j n ! (2n − 2j )(2n − 2j − 1) . . . (n − 2j + 1)xn−2j (n − j )!j ! [n/2] (−1)j (2n − 2j )! n−2j x . (n − j )!j !(n − 2j )! = n! j=0 Based on Problem 25, dn 2 (x − 1)n = n! 2n Pn (x). dxn Sec 5.3, Pg 268 29. Since the n + 1 polynomials P0 , P1 , . . ., Pn are linearly independent, and the degree of Pk is k , any polynomial, f , of degree n can be expressed as a linear combination n f (x) = ak Pk (x) . k=0 Multiplying both sides by Pm and integrating, n 1 f (x)Pm (x)dx = −1 1 ak k=0 Pk (x)Pm (x)dx . −1 Based on Problem 28, 1 Pk (x)Pm (x)dx = 2 δkm . 2m + 1 f (x)Pm (x)dx = 2 am . 2m + 1 −1 Hence 1 −1 5.4 1. Substitution of y = xr results in the quadratic equation F (r) = 0, where F (r) = r(r − 1) + 4r + 2 = r2 + 3r + 2 . ...
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