- Draw the phase portrait for the logistic equation
$$
\frac{d y}{d x}=y(a-b y) \quad a, b0 .
$$
and find the limit of the population $y(x)$ as the time $x$ goes to infinity.

Question

- Draw the phase portrait for the logistic equation
$$
\frac{d y}{d x}=y(a-b y) \quad a, b>0 .
$$
and find the limit of the population $y(x)$ as the time $x$ goes to infinity.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ The given logistic equation is $\frac{d y}{d x} = y(a - b y)$ where $a, b > 0$. ‖ ‖ step 2 ⋮ To draw the phase portrait, we need to identify the equilibrium points by setting $\frac{d y}{d x} = 0$. ‖ ‖ step 3 ⋮ Solving $y(a - b y) = 0$ gives the equilibrium points $y = 0$ and $y = \frac{a}{b}$. ‖ ‖ step 4 ⋮ The phase portrait can be drawn by plotting these equilibrium points on the $y$-axis and analyzing the behavior of $y$ around these points. ‖ ‖ step 5 ⋮ For $y < 0$, $\frac{d y}{d x} > 0$ and $y$ increases. For $0 < y < \frac{a}{b}$, $\frac{d y}{d x} > 0$ and $y$ increases. For $y > \frac{a}{b}$, $\frac{d y}{d x} < 0$ and $y$ decreases. ‖ ‖ step 6 ⋮ The limit of the population $y(x)$ as $x$ goes to infinity can be found by considering the equilibrium points. ‖ ‖ step 7 ⋮ As $x o \infty$, $y(x)$ approaches the stable equilibrium point, which is $y = \frac{a}{b}$. ‖ ∻Answer∻ ⚹ The phase portrait shows that $y = 0$ is an unstable equilibrium and $y = \frac{a}{b}$ is a stable equilibrium. The limit of the population $y(x)$ as $x$ goes to infinity is $\frac{a}{b}$. ⚹ ∻Key Concept∻ ⚹Logistic Equation Equilibrium Points⚹ ∻Explanation∻ ⚹The logistic equation $\frac{d y}{d x} = y(a - b y)$ has equilibrium points at $y = 0$ and $y = \frac{a}{b}$. The population $y(x)$ will approach the stable equilibrium point $\frac{a}{b}$ as time $x$ goes to infinity.⚹◊What is the general solution to the logistic equation dy/dx = y(a-by)?⍭ Generate me a similar question◊

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