3f-fall2010-exam_1_solutions

3f-fall2010-exam_1_solutions - Math 3F Exam #1 Solutions...

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Unformatted text preview: Math 3F Exam #1 Solutions Laney College, Fall 2010 Fred Bourgoin 1. (12 points) A direction field for the equation y prime = y ( y- 2)( y- 4) is shown below. (a) Sketch the graphs of the solutions that satisfy the initial conditions: (i) y (0) =- . 3; (ii) y (0) = 1; (iii) y (0) = 3; (iv) y (0) = 4 . 3 . Solution. See above. (b) What are the equilibrium solutions. Solution. The equilibrium solutions are y = 0, y = 2, and y = 4. (c) If the initial condition is y (0) = c , for what values c is lim t y ( t ) finite? Solution. The solutions with finite limits as t are the ones for which c [0 , 4]. All other solutions diverge. 2. (14 points) Consider the initial-value problem dy dt = f ( y ) = y 2 ( y 2- 1) , y (0) = y , where y R . (a) Sketch the graph of f ( y ) versus y , determine the equilibrium solutions, and classify each as stable, unstable, or semistable. Solution. The equilibrium solutions are y =- 1 (stable), y = 0 (semistable), and y = 1 (unstable). See graph on the next page. 1 f ( y ) y- 1 1 (b) Sketch several graphs of solutions in the ty-plane. Solution. Here is what DFIELD gave: 3. (14 points) Find an explicit solution to the initial-value problem: x dx + ye- x dy = 0 , y (0) = 1 ....
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This note was uploaded on 01/14/2011 for the course MATH 3F taught by Professor Williamson during the Fall '09 term at Laney College.

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3f-fall2010-exam_1_solutions - Math 3F Exam #1 Solutions...

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