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?
To sketch the phase line for the given differential equation, dy/dt = 1/((y-2)(y+1)), we need to find the equilibrium points and determine the behavior of the solution around those points.
To find the equilibrium points, we set dy/dt = 0 and solve for y:
0 = 1/((y-2)(y+1))
For the denominator to be zero, we must have y = 2 or y = -1. These are the equilibrium points.
Now, let's analyze the behavior of the solution based on the sign of dy/dt in different intervals:
1. When y < -1: In this range, both (y-2) and (y+1) are negative. Therefore, dy/dt = 1/((y-2)(y+1)) is positive. This means that the solution is increasing.
2. When -1 < y < 2: In this interval, (y-2) is negative, but (y+1) is positive. As a result, dy/dt = 1/((y-2)(y+1)) is negative. Hence, the solution is decreasing.
3. When y > 2: Both (y-2) and (y+1) are positive, so dy/dt = 1/((y-2)(y+1)) is positive. Again, the solution is increasing.
Now, let's discuss the behavior of the solution with the initial condition y(0) = 1/2.
Starting from y(0) = 1/2, we can see that the solution is initially in the range -1 < y < 2, where the solution is decreasing.
As the solution approaches y = 2, the denominator (y-2) approaches zero, causing dy/dt to approach infinity. This indicates that the solution blows up or becomes unbounded as y approaches 2.
Similarly, as the solution approaches y = -1, the denominator (y+1) approaches zero, resulting in dy/dt approaching negative infinity. Once again, the solution blows up or becomes unbounded as y approaches -1.
Therefore, the solution initially decreases towards the interval between -1 and 2, but it never reaches those values (as they are not equilibrium points) and instead diverges towards infinity as y approaches 2 or -1.
Now, let's proceed to part (b) and apply analytic techniques to the initial-value problem dy/dt = 1/((y-2)(y+1)), y(0) = 1/2, to compare the results with part (a).