solution2-2011

# solution2-2011 - Solutions to Math 105 Practice Midterm2,...

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Unformatted text preview: Solutions to Math 105 Practice Midterm2, Spring 2011 1. Short answer questions (1) Let u = arctan x and du = 1 x 2 +1 dx . Then ∫ arctan x x 2 + 1 dx = ∫ u du = 1 2 u 2 + C = 1 2 (arctan x ) 2 + C. So ∫ ∞ 1 arctan x x 2 + 1 dx = lim b →∞ ∫ b 1 arctan x x 2 + 1 dx = lim b →∞ [ 1 2 (arctan x ) 2 ] | b 1 = lim b →∞ 1 2 [(arctan b ) 2 − (arctan 1) 2 ] = 1 2 [( π 2 ) 2 − ( π 4 ) 2 ] = 3 π 2 32 . (2). The differential equation is separable. So we have x x 2 + 5 dx = dt, then ∫ x x 2 + 5 dx = ∫ dt Let u = x 2 + 5 and du = 2 x dx . Then ∫ x x 2 + 5 dx = 1 2 ∫ 1 u du = 1 2 ln | u | + C. So we have 1 2 ln | u | = t + C, ln | u | = 2 t + 2 C, | u | = Ce 2 t , x 2 + 5 = Ce 2 t , x 2 = Ce 2 t − 5 . Since x (0) = 1, we have 1 2 = Ce − 5, that is C = 6. Thus x ( t ) = √ 6 e 2 t − 5 . (We do not take the branch x = − √ 6 e 2 t − 5 because x (0) = 1 > 0.) (3). The function x 2 y + xy 2 y 3 − x 3 is not well-define at (0 , 0) and can not be simplified....
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## This note was uploaded on 01/14/2012 for the course MATH 105 taught by Professor Malabikapramanik during the Fall '10 term at UBC.

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solution2-2011 - Solutions to Math 105 Practice Midterm2,...

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