SampleFinal1Bsol - c n +2 = ( n + 2) ( n + 2)( n + 1) c n =...

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Solutions to Practice Final 1B 1. Z cos 2 (tan - 1 ( x )) dx = ± t = tan - 1 ( x ) , x = tan t, dx = dt cos 2 t ² = Z cos 2 t · dt cos 2 t = Z dt = t + C = tan - 1 ( x ) + C 2. a n = 1 n 1 n tan - 1 ( n ) + 1 q 1 + 1 n 3 , lim n →∞ a n = lim n →∞ 1 n · lim n →∞ 1 n tan - 1 ( n ) + 1 q 1 + 1 n 3 (since both limits exist) lim n →∞ 1 n = 0, lim n →∞ 1 n tan - 1 ( n ) + 1 q 1 + 1 n 3 = 1 ³ tan - 1 ( n ) π 2 , n ??? ´ lim n →∞ a n = 0. 3. (a) Characteristic equation of the complimentary homogeneouse equation: r 2 - 1 = 0 , r = ± 1 , general solution to the complimentary equation: y c = C 1 e x + C 2 e - x . (b) Particular solution (method of undetermined coeFcients): notice that e - x is a solution to the complimentary homogeneouse equation, therefore try: y p = Ax + B + Cxe - x y 00 p = ( Cxe - x ) 00 = - 2 Ce - x + Cxe - x y 00 p - y p = - Ax - B - 2 Ce - x we should have: y 00 p - y p = x + e - x therefore A = - 1, B = 0, C = - 1 2 .
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(c) The general selection is y = C 1 e x + C 2 e - x - x - 1 2 xe - x 4. y = X n =0 c n x n , xy 0 = X n =1 nc n x n , y 00 = X n =0 c n +2 ( n + 2)( n + 1) x n , substitute this into the equation: X n =0 ( n + 2)( n + 1) c n +2 x n - X n =1 nc n x n - 2 X n =0 c n x n = 0 this gives the equation:
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Unformatted text preview: c n +2 = ( n + 2) ( n + 2)( n + 1) c n = 1 ( n + 1) c n , n = 0 , 1 , 2 . Solution: (a) c 2 k = 1 2 k-1 c 2 k-2 = = 1 (2 k-1)(2 k-3) . . . 3 1 (0 , k ) (b) c 2 k +1 = 1 2 k c 2 k-1 = 1 2 k 1 2 k-2 1 2 c 2 = ( 1 2 ) k k ! c 2 The power series solution: y = X n =0 c n x n = X k =0 c 2 k x 2 k + X k =0 c 2 k +1 x 2 k +1 = c X k =1 x 2 k (2 k-1)(2 k-3) . . . 3 + c + c 1 + c 1 X k =1 1 k ! 1 2 k x 2 k +1 = c X k =1 x 2 k (2 k-1)(2 k-3) . . . 3 + c + c 1 xe x 2 2 , y (0) = c , y (0) = c 1 c = 0 , c 1 = 1 y = xe x 2 2...
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SampleFinal1Bsol - c n +2 = ( n + 2) ( n + 2)( n + 1) c n =...

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