The slope field for \(\dfrac{dy}{dt} = y^2 - 4y = y(y-4)\) shows horizontal stripes (characteristic of autonomous equations) with slopes determined solely by the \(y\)-value. Slopes are zero at \(y=0\) and \(y=4\text{.}\)
Equilibrium solutions: Set \(y(y-4) = 0\) to get \(y = 0\) and \(y = 4\text{.}\) To determine stability, test regions: For \(y < 0\text{,}\) try \(y=-1\text{:}\)\(f(-1) = (-1)(-5) = 5 > 0\) (upward). For \(0 < y < 4\text{,}\) try \(y=1\text{:}\)\(f(1) = (1)(-3) = -3 < 0\) (downward). For \(y > 4\text{,}\) try \(y=5\text{:}\)\(f(5) = (5)(1) = 5 > 0\) (upward). Classification: \(y = 0\) is a sink (stable) since arrows point toward it from both sides. \(y = 4\) is a source (unstable) since arrows point away from it on both sides.
Phase line: Place dots at \(y=0\) and \(y=4\text{.}\) Draw upward arrows in regions \(y < 0\) and \(y > 4\text{,}\) and a downward arrow in region \(0 < y < 4\text{.}\) Long-term behavior: Solutions below \(y=0\) increase toward \(y=0\text{;}\) solutions between \(0\) and \(4\) decrease toward \(y=0\text{;}\) solutions above \(y=4\) increase without bound.
Test regions: For \(y < -\sqrt{3}\text{,}\) try \(y=-2\text{:}\)\(f(-2) = -8 + 6 = -2 < 0\) (downward). For \(-\sqrt{3} < y < 0\text{,}\) try \(y=-1\text{:}\)\(f(-1) = -1 + 3 = 2 > 0\) (upward). For \(0 < y < \sqrt{3}\text{,}\) try \(y=1\text{:}\)\(f(1) = 1 - 3 = -2 < 0\) (downward). For \(y > \sqrt{3}\text{,}\) try \(y=2\text{:}\)\(f(2) = 8 - 6 = 2 > 0\) (upward). Phase line: Place dots at \(y = -\sqrt{3}, 0, \sqrt{3}\text{.}\) Arrows: down for \(y < -\sqrt{3}\text{,}\) up for \(-\sqrt{3} < y < 0\text{,}\) down for \(0 < y < \sqrt{3}\text{,}\) up for \(y > \sqrt{3}\text{.}\) Stability: \(y = -\sqrt{3}\) is a sink (stable), \(y = 0\) is a source (unstable), \(y = \sqrt{3}\) is a sink (stable).
This equation exhibits bistability: two stable equilibria (\(y = \pm\sqrt{3}\)) separated by an unstable equilibrium (\(y = 0\)). Systems starting near \(y = -\sqrt{3}\) approach that stable state, while those near \(y = \sqrt{3}\) approach that stable state. The unstable equilibrium at \(y=0\) acts as a threshold: small perturbations determine which stable state the system settles into. This could model systems like ecological populations with alternative stable states, chemical reactions with multiple equilibria, or bistable mechanical systems.
a) \(y = 0, \pm\sqrt{3}\text{.}\) b) \(y = -\sqrt{3}\) (sink), \(y = 0\) (source), \(y = \sqrt{3}\) (sink). c) Bistable system with two stable states separated by an unstable threshold.
This system does not exhibit chaotic behavior. Chaos typically requires at least three dimensions and sensitivity to initial conditions. This is a one-dimensional autonomous ODE, which cannot exhibit chaos. The system’s behavior is predictable and depends smoothly on initial conditions.
Test regions: For \(y < 1\text{,}\) try \(y=0\text{:}\)\(f(0) = 1 > 0\) (upward). For \(y > 1\text{,}\) try \(y=2\text{:}\)\(f(2) = 4 - 4 + 1 = 1 > 0\) (upward). Phase line: Place a dot at \(y=1\) with upward arrows on both sides. Stability: \(y = 1\) is a semi-stable equilibrium (node). Long-term behavior: Solutions below \(y=1\) increase toward \(y=1\) but never quite reach it (approach asymptotically); solutions above \(y=1\) increase away from \(y=1\) without bound.
a) \(y = 1\text{.}\) b) No chaos; this is a predictable 1D system. c) \(y = 1\) is semi-stable (node); solutions below approach it, solutions above grow unbounded.
Set \(rP\left(1 - \frac{P}{K}\right) = 0\text{.}\) Since \(r \neq 0\text{,}\) we have \(P = 0\) or \(1 - \frac{P}{K} = 0\text{,}\) giving equilibria: \(P = 0\) and \(P = K\text{.}\)
Assuming \(r > 0\) and \(K > 0\text{:}\) For \(P < 0\text{,}\) the term \(1 - \frac{P}{K} > 1\text{,}\) so \(f(P) < 0\) (downward). For \(0 < P < K\text{,}\) try \(P = K/2\text{:}\)\(f(K/2) = r(K/2)(1/2) > 0\) (upward). For \(P > K\text{,}\) try \(P = 2K\text{:}\)\(f(2K) = r(2K)(1 - 2) = -2rK < 0\) (downward). Phase line: Place dots at \(P=0\) and \(P=K\text{.}\) Arrows: down for \(P < 0\text{,}\) up for \(0 < P < K\text{,}\) down for \(P > K\text{.}\) Stability: \(P = 0\) is a source (unstable), \(P = K\) is a sink (stable).
This logistic model predicts that: (1) A population starting near zero will initially grow; (2) As the population increases, growth slows due to limited resources; (3) The population stabilizes at the carrying capacity \(K\text{,}\) which represents the maximum sustainable population; (4) If the population temporarily exceeds \(K\) (e.g., due to migration), it will decline back to \(K\text{.}\) This helps ecologists predict long-term population levels and understand the effects of changing environmental conditions (varying \(r\) or \(K\)).
a) \(P = 0\) and \(P = K\text{.}\) b) \(P = 0\) is unstable (source), \(P = K\) is stable (sink). c) Populations stabilize at carrying capacity \(K\text{;}\) model predicts long-term equilibrium and response to perturbations.
