{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

SampleFinal1Bsol - c n 2 = n 2 n 2 n 1 c n = 1 n 1 c n n =...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 = 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 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 coefficients): 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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(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
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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 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...
View Full Document

{[ snackBarMessage ]}