Question

consider the two equations below, each having the critical points a, b, and c, suppose that a < b < c. for one of these equations, only the critical point b is stable; for the other equation, b is the only unstable critical point. construct phase diagrams for the two equations to determine which is which. without attempting to solve either equation explicitly, make rough sketches of typical solution curves for each. you should see two funnels and a spout in one case, two spouts and a funnel in the other.
\\(\frac{dx}{dt}=(x - a)(x - b)(x - c)\\) \\(\frac{dx}{dt}=(a - x)(b - x)(c - x)\\)

without attempting to solve it, make rough sketches of typical solution curves for \\(\frac{dx}{dt}=(x - a)(x - b)(x - c)\\). choose the correct answer below.
a. image of solution - curve sketch a
b. image of solution - curve sketch b
c. image of solution - curve sketch c
d. image of solution - curve sketch d

Explanation:

Step1: Analyze sign of $\frac{dx}{dt}$ for $\frac{dx}{dt}=(x - a)(x - b)(x - c)$

Let's consider the intervals $x\lt a$, $a\lt x\lt b$, $b\lt x\lt c$ and $x > c$. When $x\lt a$, all of $(x - a)$, $(x - b)$ and $(x - c)$ are negative, so $\frac{dx}{dt}<0$. When $a\lt x\lt b$, $(x - a)>0$, $(x - b)<0$ and $(x - c)<0$, so $\frac{dx}{dt}>0$. When $b\lt x\lt c$, $(x - a)>0$, $(x - b)>0$ and $(x - c)<0$, so $\frac{dx}{dt}<0$. When $x > c$, all of $(x - a)$, $(x - b)$ and $(x - c)$ are positive, so $\frac{dx}{dt}>0$. Critical - point $a$ is unstable (since $\frac{dx}{dt}$ changes sign from negative to positive as $x$ increases through $a$), $b$ is stable (since $\frac{dx}{dt}$ changes sign from positive to negative as $x$ increases through $b$) and $c$ is unstable (since $\frac{dx}{dt}$ changes sign from negative to positive as $x$ increases through $c$). The solution curves will have two funnels (towards $b$) and a spout (away from $a$ and $c$).

Step2: Analyze sign of $\frac{dx}{dt}$ for $\frac{dx}{dt}=(a - x)(b - x)(c - x)$

Consider the intervals $x\lt a$, $a\lt x\lt b$, $b\lt x\lt c$ and $x > c$. When $x\lt a$, $(a - x)>0$, $(b - x)>0$ and $(c - x)>0$, so $\frac{dx}{dt}>0$. When $a\lt x\lt b$, $(a - x)<0$, $(b - x)>0$ and $(c - x)>0$, so $\frac{dx}{dt}<0$. When $b\lt x\lt c$, $(a - x)<0$, $(b - x)<0$ and $(c - x)>0$, so $\frac{dx}{dt}>0$. When $x > c$, $(a - x)<0$, $(b - x)<0$ and $(c - x)<0$, so $\frac{dx}{dt}<0$. Critical - point $a$ is stable (since $\frac{dx}{dt}$ changes sign from positive to negative as $x$ increases through $a$), $b$ is unstable (since $\frac{dx}{dt}$ changes sign from negative to positive as $x$ increases through $b$) and $c$ is stable (since $\frac{dx}{dt}$ changes sign from positive to negative as $x$ increases through $c$). The solution curves will have two spouts (away from $b$) and a funnel (towards $a$ and $c$).

For $\frac{dx}{dt}=(x - a)(x - b)(x - c)$, the solution curves have two funnels and a spout.

Answer:

The option that shows two funnels and a spout for the equation $\frac{dx}{dt}=(x - a)(x - b)(x - c)$ (where $b$ is the stable critical - point) should be chosen. Without seeing the actual visual details of the options A, B, C, D, we know that the correct option will have solution curves with the behavior described above for this equation. If we assume the correct visual representation based on our analysis, the solution curves for $\frac{dx}{dt}=(x - a)(x - b)(x - c)$ will have solution curves that approach $x = b$ (funnel) and move away from $x=a$ and $x = c$ (spout).