MATH 1005 Midterm: Exam-MATH1005-2003April
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2: a di erential equation which is satis ed by the orthogonal trajectories to the curves x = ky3 is (a) (b) (c) (d) (e) y0 = y. 3x y0 = y y0 = 3ky2. 4 (b) (c) (d) y(2) = sin 2 y(2) = sin 2 c sin 2 1 y(2) = 4 (e) y(2) = sin 2 c. 3: consider the bernoulli di erential equation y0 + 2xy = xe x2. The transformation used to obtain the general solution is y3. (a) u = 1 y (b) u = y2 (c) u = 1 y2 (d) u = y3 (e) u = 1 y3: let f (x; y) = (2x3y2 y)4. 24: let z = 2x3 + y2, x = s et, y = s2 cos t. When s = 1 and t = 0 , @s is equal to (a) (b) (c) (d) (e) 4: let xyz = cos(x + y + z).
