Tuesday

August 23, 2016
Posted by **Erica** on Sunday, February 3, 2013 at 4:53pm.

dy/dt=1/((y-2)(y+1))

and discuss the behavior of the solution with initial condition y(0)=1/2

b) Apply analytic techniques to the initial-value problem

dy/dt=1/((y-2)(y+1))), y(0)=1/2

and compare your results with your discussion in part (a).

I couldn't get the equilibrium points for the equation so I did the phase line without them, and everything above 2 and below -1 was positive and between 2 and -1 is negative.

When y(0)=1/2, the solution is negative but what happens when it gets to 2 or -1 if they are not equilibrium points?