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3f-fall2010-exam_1_solutions

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

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Math 3F — Exam #1Solutions 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) = - 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 0 , where y 0 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

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f ( y ) y - 1 1 0 (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: xdx + ye - x dy = 0 , y (0) = 1 .
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3f-fall2010-exam_1_solutions - Math 3F Exam#1 Solutions...

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