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.

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).

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

a) To sketch the phase line for the differential equation dy/dt = 1/((y-2)(y+1)), we need to identify the critical points or equilibrium points. These are the points where dy/dt = 0.

In this case, the equation becomes 1/((y-2)(y+1)) = 0. Since the denominator cannot be zero, we conclude that there are no critical points or equilibrium points in the phase line.

Based on your observation, for values of y > 2 or y < -1, dy/dt > 0, indicating that the solution is increasing. Similarly, for values of -1 < y < 2, dy/dt < 0, indicating that the solution is decreasing.

When y(0) = 1/2, the initial condition falls into the range -1 < y < 2, where dy/dt < 0. This means that the solution will approach values below 1/2 as t increases.

b) To apply analytic techniques to the initial-value problem dy/dt = 1/((y-2)(y+1)), y(0) = 1/2, we need to solve the differential equation explicitly.

We begin by separating variables: (y-2)(y+1) dy = dt. Integrating both sides, we get ∫(y-2)(y+1) dy = ∫dt.

Expanding the left side, we have ∫(y^2 - y - 2) dy = t + C, where C is the constant of integration.

Integrating, we get (1/3)y^3 - (1/2)y^2 - 2y = t + C.

At this point, we can try to solve for y explicitly, but it will be difficult due to the cubic term. Instead, we can consider the behavior of the solution qualitatively.

From the phase line analysis in part (a), we already know that the solution will approach values below 1/2 as t increases. This matches our initial condition.

Comparing the results from part (b) with the phase line analysis in part (a), we find that they align. The analytic technique confirms that the solution will approach values below 1/2 as t increases, consistent with part (a).

a) To sketch the phase line for the given differential equation, you need to determine the critical points and their behavior. The critical points are the values of y for which dy/dt equals zero or is undefined. In this case, dy/dt can be written as:

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

To find the critical points, set dy/dt equal to zero and solve:

1/((y-2)(y+1)) = 0

This equation has no solution. Therefore, there are no critical points for this differential equation.

Next, you can determine the sign of dy/dt for different regions of the y-axis. Based on the given equation, you can see that dy/dt is positive when (y-2)(y+1) is negative, and it is negative when (y-2)(y+1) is positive.

To determine the sign of (y-2)(y+1), you can analyze the intervals on the y-axis. Start by considering values less than -1, between -1 and 2, and greater than 2.

For y < -1, both factors (y-2) and (y+1) are negative. Therefore, their product (y-2)(y+1) is positive, indicating that dy/dt will be negative in this interval.

For -1 < y < 2, (y-2) becomes negative while (y+1) becomes positive. Their product (y-2)(y+1) is negative, indicating that dy/dt will be positive in this interval.

For y > 2, both factors (y-2) and (y+1) become positive. Therefore, their product (y-2)(y+1) is positive, indicating that dy/dt will be negative again in this interval.

Based on this analysis, you can sketch the phase line by drawing arrows indicating the direction of dy/dt for each interval: negative below -1 and above 2, and positive between -1 and 2.

Regarding the initial condition y(0) = 1/2, you mentioned that the solution is negative. Since the phase line indicates that dy/dt is positive for y between -1 and 2, as y approaches those values, the solution will move towards positive values. However, since -1 and 2 are not equilibrium points, the behavior of the solution at those values cannot be precisely determined with the phase line analysis alone.

b) In part (b), you are asked to apply analytic techniques to solve the initial-value problem and compare the results with your discussion in part (a). Analytic techniques involve finding the exact solution to the differential equation.

To solve the initial-value problem dy/dt = 1/((y-2)(y+1)), y(0) = 1/2, you can start by rewriting the equation as:

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

Integrating both sides with respect to y:

∫(y-2)(y+1)dy = ∫dt

This gives:

(y^2 - y - 2) / 2 = t + C

where C is the constant of integration.

Now, substitute the initial condition y(0) = 1/2 into the equation:

(1/4 - 1/2 - 2) / 2 = 0 + C

(-7/4) / 2 = C

Simplifying, you find C = -7/8.

Substituting this value back into the equation:

(y^2 - y - 2) / 2 = t - 7/8

Now, you can solve this quadratic equation for y. To make it easier, you can multiply through by 2:

y^2 - y - 2 = 2t - 7/4

y^2 - y - 2 - 2t + 7/4 = 0

Next, you can use the quadratic formula to find the values for y:

y = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 1, b = -1, and c = -2 - 2t + 7/4.

By solving this quadratic equation, you will find the exact solution for y(t) with the given initial condition.

Finally, compare the results obtained analytically with the behavior discussed in part (a). The behavior of the solution at y = 2 or y = -1 cannot be precisely determined from the phase line analysis alone. However, the analytic solution will provide a more detailed and accurate understanding of the behavior as the solution approaches those values.