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.

To sketch the phase line for the differential equation and discuss the behavior of the solution with the given initial condition, we can follow these steps:

a) Sketching the Phase Line:
1. Start by finding the critical points or equilibrium points of the differential equation where dy/dt = 0. In this case, the equation is dy/dt = 1/((y-2)(y+1)).

To find the critical points, we need to solve the equation (y-2)(y+1) = 0. Solving this equation, we get y = 2 and y = -1 as the critical points.

2. Now, we can draw a number line and mark the critical points. Place open circles to represent non-equilibrium points.

-1 2

3. Between each pair of critical points, we need to determine the sign of dy/dt. To do this, we can pick any value in each interval and substitute it into the differential equation to see if the solution is positive or negative.

For example, pick y = 0 in the interval (-1, 2). Substituting into the equation, we have dy/dt = 1/((0-2)(0+1)) = -1/2 < 0. Therefore, the solution is negative in this interval.

4. Mark the signs of the derivative in each interval on the number line:

-1 - 2

5. Finally, draw arrows to indicate the direction of the solution. If the sign of dy/dt is positive, draw an arrow pointing upwards, and if the sign is negative, draw an arrow pointing downwards.

-1 ↓ - → 2 ↓

This phase line represents the behavior of the solution for the given differential equation without considering the equilibrium points.

Now let's discuss the behavior of the solution with the initial condition y(0) = 1/2:

Starting at y(0) = 1/2, we can see that the solution is in the negative interval. As the solution moves towards 2 or -1, the denominator of the differential equation approaches zero, and the derivative becomes infinite. This indicates that the solution becomes unbounded (blows up) near these points.

b) Applying Analytic Techniques and Comparing Results:
In part (b), the question is asking you to solve the initial-value problem analytically using appropriate techniques and compare the results with the behavior observed in part (a).

To solve the initial-value problem dy/dt = 1/((y-2)(y+1)), y(0) = 1/2 analytically, you can try techniques such as separation of variables or finding a suitable integrating factor.

Once you find the solution, compare its behavior near the critical points with the behavior observed from the phase line. Specifically, check if the solution blows up (becomes unbounded) near y = 2 or y = -1 as indicated by the phase line.

By comparing the results, you can verify if the behavior observed in the phase line (unboundedness near 2 and -1) is consistent with the solution obtained analytically.

Remember, the phase line is a graphical tool to understand the qualitative behavior of the solution, while the analytic techniques provide the exact solution and allow for a more precise analysis.