Consider the differential equation. A detailed direction field is not needed. Find the solutions that are constant, for all (the equilibrium solutions) . In what regions are solutions increasing? Decreasing?
-y'(t) = y(y + 4) ( 6 - y)

Question

Consider the differential equation. A detailed direction field is not needed. Find the solutions that are constant, for all (the equilibrium solutions). In what regions are solutions increasing? Decreasing? -y'(t) = y(y + 4)( 6 - y) A) y = -4, y = 6; increasing for -4 < y < 6; decreasing for y < -4, y > 6 B) y = 0, y = -4, y = 6; increasing for y < -4, 0 < y < 6; decreasing for -4 < y < 0, y > 6 C) y = 0, y = -4, y = 6; increasing for -4 < y < 0; decreasing for y < -4, y > 0 D) y = -4, y = 6; increasing for y < -4, 0 < y < 6; decreasing for -4 < y < 0, y > 6

Quizplus · Accepted Answer

The correct answer is B because the given differential equation y'(t) = y(y + 4)(6 - y) has critical points at y = 0, y = -4, and y = 6. To determine where y'(t) is positive or negative, we analyze intervals around these points. For y < -4, the product is positive (three negative factors); for -4 < y < 0, the product is negative (two negative factors, one positive); for 0 < y < 6, the product is positive again (one negative factor, two positive); and for y > 6, it is negative (three positive factors, but the negative sign from the (6 - y) term). This matches the increasing/decreasing behavior described in option B.

Consider the Differential Equation