Question

a) Sketch the phase line for the differential equation
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? And I don't really understand what they are asking for in part b.

Answer 1

The page at

http://www.sosmath.com/diffeq/first/phaseline/phaseline.html

has quite a lengthy and clear discussion of phase lines and equilibrium points.

Part (b) wants you to solve the equation analytically and compare the solution with your qualitative analysis in part (a).

Answer 2

I read the info, but it doesnt talk about functions with no equilibrium point.