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.
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
...
a) To sketch the phase line for the differential equation dy/dt = 1/((y-2)(y+1)), we can start by finding the equilibrium points, where dy/dt = 0. In this case, we need to solve the equation 1/((y-2)(y+1)) = 0.
However, note that this equation has no real solutions because the denominator is always nonzero. Therefore, there are no equilibrium points.
Next, we can determine the behavior of the solution with the given initial condition y(0) = 1/2. Since the phase line is positive above 2 and negative below -1, we can see that the solution starts at a value of 1/2, which is below -1. As time progresses, the solution will move towards the line y = -1 and approach it from below.
Regarding what happens when the solution reaches 2 or -1, even though these points are not equilibrium points, it's still important to consider their influence on the solution behavior. When the solution gets close to y = 2, we can observe that the denominator (y-2) approaches zero, making dy/dt very large in magnitude. This suggests that the solution will experience a "jump" or discontinuity near y = 2. Similarly, when the solution gets close to y = -1, the denominator (y+1) becomes small, causing dy/dt to be large in magnitude. This indicates a potential "jump" or discontinuity near y = -1.
b) Analytic techniques refer to solving the given initial-value problem using mathematical methods like separation of variables, integrating factors, or other approaches. In this case, we are asked to apply such techniques to the initial-value problem dy/dt = 1/((y-2)(y+1)), y(0) = 1/2, and compare the results with the discussion in part (a).
To solve the initial-value problem analytically, we can separate variables, integrate, and then solve for the constant of integration using the given initial condition.
First, we can rewrite the differential equation as: (y-2)(y+1) dy = dt
Next, we integrate both sides: ∫(y-2)(y+1) dy = ∫dt
Integrating the left side: (1/3)y^3 - y^2 - 2y = t + C
Now, we can apply the initial condition y(0) = 1/2. Plugging in these values, we get: (1/3)(1/2)^3 - (1/2)^2 - 2(1/2) = 0 + C
Simplifying and solving for C, we find: C = -(13/12)
So the particular solution to the initial-value problem is: (1/3)y^3 - y^2 - 2y = t - (13/12)
To compare these analytic results with part (a), we can evaluate the equation at different values of t and see if the behavior matches the discussion. For example, when t = 0, y = 1/2 as given. As time progresses, we can observe whether the solution indeed approaches y = -1 as described.
Note: The analytic solution provides a mathematical expression for the solution at any given time t, whereas the phase line gives a qualitative representation of the overall behavior of the solution. Both approaches are complementary in understanding the differential equation and its solution.