Lecture10 - carrying capacity P(t) = 0 constant in time -...

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Unformatted text preview: carrying capacity P(t) = 0 constant in time - separable -- would use partial fractions to solve solution curve is on (x, y) plane where y*=4 is the equilibria approached by incr and decr curves increases as y<4 use y'''>4 y"' = 5 f(5) = 5(4-5)<0 method: phase line diagram --- similar to cobwebbing technique of DTDS y = y(x) decreases as function of x is y>4 soln curves moving away from eqm is unstable, vice versa for moving towards 2< y^ <3, f(y^) <0 then, dy/dx < 0, and therefore, y(x) decreases 0 <y^ < 2 , f(y^) > 0 , then dy/dx >0 and therefore y(x) increases y>3, f(y).0, dy/dx >0 -> y(x) is increasing as a function of x soln curve will move away from the eqm r- 2rk/k don't want to use separation don't want to use phase line soo, take the derivative of RHS sub in the eqm values ...
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Lecture10 - carrying capacity P(t) = 0 constant in time -...

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