t1_s2_sln - YORK UNIVERSITY Faculty of Pure and Applied...

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Unformatted text preview: YORK UNIVERSITY Faculty of Pure and Applied Science AS/SC/MATH 1014 3.0 M June - August 2003 Term Test 1 SOLUTIONS 1. Solve the differential equations: (a) (4 points) √ y cos x + dy dx = 0 Answer: cos xdx + dy √ y = 0 Z cos xdx + Z dy √ y = C- sin x + 2 √ y = C y = ( C- sin x ) 2 4 . (b) (6 points) dy dx + 2 x ( y- 1) = 0 , if y = 3 when x = 0 . Answer: dy dx + 2 xy = 2 x Integrating factor: e R 2 xdx . d dx ( ye x 2 ) = 2 xe x 2 ye x 2 = Z 2 xe x 2 dx = e x 2 + C. So, y = 1 + Ce x 2 . From the initial condition y (0) = 3 follows 1 + C = 3 , that is C = 2 . Hence, y = 1 + 2 e x 2 . 2. (7 points) Find the area of the region trapped between y = ( x + 1) 2 , x = sin πy and y = 0 , (0 ≤ y ≤ 1) . Answer: y = ( x + 1) 2 √ y = x + 1 x = √ y- 1 . So, we have to find the area of the region bounded by the graphs of the functions: x = √ y- 1 , x = sin πy and y = 0 on the interval 0 ≤ y ≤ 1 . But ∀ y ∈ [0 , 1] sin πy ≥ √ y- 1 and only the point of intersection is the point (0...
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t1_s2_sln - YORK UNIVERSITY Faculty of Pure and Applied...

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