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Unformatted text preview: Name: MA 366, Spring 2010, Quiz 4 (1) (8 points) Determine all critical (equilibrium) points of the differential equation y = 4 y 2- y 4 . Determine whether each critical point is stable, unstable, or semi-stable. Solution: I should really draw a graph, but its not so easy with what Im using to prepare the solutions. At any rate, Ill write y = f ( y ) = 4 y 2- y 4 = (4- y 2 ) y 2 = (2- y )(2 + y ) y 2 , so the equilibrium points where f ( y ) = 0 are 0,- 2, and 2. We have y = 4 y 2- y 4 < , y <- 2 , > ,- 2 < y < , > , < y < 2 , < , 2 < y. So y =- 2 is unstable, y = 0 is semi-stable, and y = 2 is stable (2) (8 points) Determine the solution y ( x ) of the initial-value problem (2 xy + cos x ) dy dx + y 2- y sin x- x = 0 , y (0) = 1 . Solution: If we write this equation as M ( x,y )+ N ( x,y ) y = 0, then M ( x,y ) = y 2- y sin x- x and N ( x,y ) = 2 xy + cos x and we have M y = 2 y- sin x = N x , so the equation is exact. Soso the equation is exact....
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- Fall '09
- Critical Point