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**Unformatted text preview: **Math 415 Name: Quiz 3A For the differential equation y = y 2 (1- y 2 ), sketch the graph of f ( y ) versus y , determine the equilibrium points, and classify them. Draw the phase line and several solutions for- < y < (where y (0) = y ). .5-1-0.5 0.5 1 1-1 1 Here we see that y ( t ) =- 1 is an unstable solution, y ( t ) = 0 is semistable, and y ( t ) = 1 is stable. Solutions below the line y =- 1 will drop to- , those between 0 and 1 will asymptotically approach y = 1 as t (and approach y = 0 as t - ), and those between- 1 and 0 will approach 0. Solve the initial value problem y = (3 x 2 + 2 y- 1) dx + (4 y + 2 x ) dy = 0 , y (1) = 0 , and determine (approximately) where the solution is valid. The equation is exact as (3 x 2 +2 y- 1) y = 2 = (4 y +2 x ) x . So there is a function ( x,y ) with x = 3 x 2 +2 y- 1 and y = 4 y + 2 x . Integrating one of these we get ( x,y ) = Z 4 y + 2 xdy = 2 y 2 + 2 xy + h ( x ) ....

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