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

I got for the phase line:
up
-2
down
-1
down

a)negative slope:
on the graph -2 and -1 are asymptotes where the line is like an S in that it goes from riding the -2 line down through the y-axis to riding the -1 line.

b) * S = differentiation symbol I cant type)
S (y-2)(y+1)dy = S1dt
S y^2 - y - 2 dy = t+c
[(y^3)/3] - [(y^2)/2] -2y = t+c
...