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**Unformatted text preview: **Math 334 Lecture #7 2.5: Autonomous First Order ODEs Geometric Analysis of y = f ( y ). The zeros of f are called the critical points of the ODE. The critical points correspond to the equilibrium solutions of the ODE. Between consecutive zeros of f , the values of f are of one sign, so that y = f ( y ) are also. Between consecutive zeros of f , solutions are increasing if y = f ( y ) > 0 there, or they are decreasing if y = f ( y ) < 0 there. A graph of the function f reveals the location of the zeros of f and the intervals where f is positive or negative. Here is the graph of a function f for the purpose of this discussion: 5 2 ! 1 3 f(y) 1 ! 1 ! 3 1 4 6 2 3 ! 2 ! 3 y 4 ! 2 Note the zeros of f , and the sign of f between consecutive zeros. The graph of f determines the phase portrait of the ODE y = f ( y ): 2 ! 3 4 2$0 & (&) 5 0$5 3$0 ! 1 1$5 ! 2 1$0 0$0 6 2$5 1 3 An equilibrium is asymptotically stable if every nearby solution converges to it as t . [Identify on the phase portrait all of the equilibria that are asymptotically stable.] [Question: Do any of the solutions near an asymptotically stable equilibrium y = A achieve the value of A in finite time? NO! Because if any did, it would contradict the uniqueness of solutions.] An equilibrium solution is unstable if every nearby solution moves away from it....

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