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Unformatted text preview: Name: MA 366, Spring 2010, Quiz 3 (1) (10 points) Determine the solution y ( t ) of y ′ = t y + 1 , y (1) = 1 . Determine for which values of t the solution exists. Solution: We write the equation as ( y + 1) y ′ = t and see that the equation is separable. We integrate to find 1 2 y 2 + y = 1 2 t 2 + C. Since y (1) = 1, we have 1 2 + 1 = 1 2 + C , so C = 2 and y 2 + 2 y + t 2 4 = 0. Solving for y gives y = 2 + radicalbig 2 2 4 × 1 × ( t 2 4) 2 = 1 + radicalbig 5 t 2 , where we need to take the + in the quadratic formula because y (1) = 1 > 0. So a solution exists as long as 5 t 2 ≥ 0, i.e., √ 5 ≤ t ≤ √ 5. Another way to see where the solution exists is to complete the square: y 2 + 2 y + 1 = 5 t 2 and the left hand side is ( y + 1) 2 ≥ 0, so we need 5 t 2 ≥ 0. (2) Assume a sphere of mass 1 gm and diameter d = 1 cm is held still in the center of a glass of water. Assume that at time t = 0 the sphere is released, and that the force due to the viscosity of the liquid on the moving sphere is given by...
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This note was uploaded on 01/18/2012 for the course MATH 366 taught by Professor Edraygoins during the Fall '09 term at Purdue University.
 Fall '09
 EdrayGoins

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