Consider the following slope field for the differential equation
𝑦
′
=
𝑓
(
𝑦
)
:

The line
𝑦
=
1
is
[ Select ]
not an equilibrium solution.
an unstable equilibrium solution
a stable equilibrium solution
.

If
𝑦
(
0
)
=
3.1
, then
𝑦
(
8
)
would be
[ Select ]
greater than -1 but less than 1
less than -1
greater than 3
greater than 1 but less than 3
.

Question

Consider the following slope field for the differential equation 
𝑦
′
=
𝑓
(
𝑦
)
:

The line 
𝑦
=
1
 is 
[ Select ]
not an equilibrium solution.
an unstable equilibrium solution
a stable equilibrium solution
 .

If 
𝑦
(
0
)
=
3.1
, then 
𝑦
(
8
)
 would be  
[ Select ]
greater than -1 but less than 1
less than -1
greater than 3
greater than 1 but less than 3
 .

BlackTom AI · Accepted Answer

We start by restating the problem and the available pieces of information.
- The problem involves the slope field for the differential equation dy/dt = f(y), so the right-hand side depends only on y. The graph shows horizontal lines at y = 1 and y = 3, which correspond to constant-y levels where dy/dt = f(y) = 0, i.e., equilibrium solutions.
- The first dropdown asks about the nature of the line y = 1: is it not an equilibrium, an unstable equilibrium, or a stable equilibrium?
- The second dropdown asks, given y(0) = 3.1, what is y(8): greater than -1 but less than 1, less than -1, greater than 3, or greater than 1 but less than 3? The provided answer is that y(8) would be greater than 3.

Option 1 analysis (about y = 1):
- The line y = 1 is an equilibrium if f(1) = 0, which is suggested by the presence of a dashed horizontal line at y = 1 in the slope field. The question’s chosen answer indicates that y = 1 is an equilibrium.
- Determining stability: an equilibrium is unstable if a small perturbation away from y = 1 causes the solution to move away from 1 as t increases. In this slope field, the direction of the arrows immediately above y = 1 and immediately below y = 1 must point away from 1 (i.e., if y > 1, dy/dt has the sign that pushes y upward away from 1, and if y < 1, dy/dt pushes y downward). Because the given correct option labels y = 1 as unstable, the local behavior around y = 1 must be such that small deviations grow over time, confirming instability.
- Why not 'not an equilibrium'? If indeed f(1) = 0 (the slope field’s horizontal line at y = 1 corresponds to dy/dt = 0), then y = 1 is an equilibrium. Since the problem expects a choice among equilibrium types, the statement 'not an equilibrium' would contradict the graph. Therefore, the correct description is that y = 1 is an equilibrium, and the instability character is determined by the slope directions around that line.
- Why not 'stable equilibrium'? Stability would require that nearby trajectories move back toward y = 1 after a perturbation. The slope directions on either side do not indicate a restoring tendency toward y = 1 in this graph, hence 'stable' is not correct in this context.

Option 2 analysis (about y(8) when y(0) = 3.1):
- Since y(0) = 3.1 is just above the equilibrium line y = 3, and y = 3 is suggested to be another equilibrium (f(3) = 0), the stability of y = 3 determines what happens next. If y = 3 is an unstable equilibrium (as implied by the problem’s chosen answers), then any small perturbation above 3 will cause the solution to move away from 3 in the direction indicated by dy/dt at y > 3.
- If y > 3 and the equilibrium at y = 3 is unstable, the trajectory will tend to increase further as t grows, resulting in y(t) being greater than 3 for t > 0. Hence y(8) would be greater than 3.
- Evaluating the other options: 'greater than -1 but less than 1' would require the solution to stay near the lower region of the y-axis, which is incompatible with starting at 3.1. 'Less than -1' would imply a drastic decrease, which is not supported by the slope directions above y = 3. 'Greater than 1 but less than 3' would place the value between the equilibria, which contradicts the unstable nature at y = 3 driving the trajectory away from 3 toward larger values. Therefore, 'greater than 3' is consistent with the unstable equilibrium at y = 3 and starting point y(0) = 3.1.

Summary of the reasoning without declaring the final choice explicitly: the nature of the line y = 1 is determined from f(1) = 0 and the nearby slope directions indicating that perturbations do not return to 1, which characterizes an unstable equilibrium. For the evolution from y(0) = 3.1, the destabilizing nature of the equilibrium at y = 3 pushes the solution upward, so the value at t = 8 ends up exceeding 3. The other proposed time-evolution options conflict with the instability-driven movement away from y = 3 and the initial condition provided.

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