Question

the differential equation $\frac{dx}{dt}=\frac{1}{16}x(16 - x)-h$ models a logistic population with harvesting at rate h. determine the dependence of the number of critical points on the parameter h, and then construct a bifurcation diagram. the differential equation has two critical points for h < 4, one critical point for h = 4 and no critical points for h > 4 (use integers or fractions for any numbers in the expressions ) write an equation relating the value of critical point(s) c to h. $c^{2}-16c + 16h=0$ (type an equation using c and h as the variables. use integers or fractions for any numbers in the expression ) construct the bifurcation diagram in the ch - plane. choose the correct answer below

Explanation:

Step1: Recall critical - point definition

Critical points occur when $\frac{dx}{dt}=0$. So, set $\frac{1}{16}x(16 - x)-h = 0$. Multiply through by 16 to get $x(16 - x)-16h=0$. Expand to $16x-x^{2}-16h = 0$, or $x^{2}-16x + 16h=0$. Let $c$ be the critical - point value, so $c^{2}-16c + 16h=0$.

Step2: Analyze the discriminant of the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (in our case, $a = 1$, $b=-16$, $c = 16h$), the discriminant $\Delta=b^{2}-4ac$. Here, $\Delta=(-16)^{2}-4\times1\times16h=256 - 64h$.

• When $\Delta>0$, there are two real roots. Solving $256 - 64h>0$ gives $h < 4$.
• When $\Delta = 0$, there is one real root. Solving $256 - 64h=0$ gives $h = 4$.
• When $\Delta<0$, there are no real roots. Solving $256 - 64h<0$ gives $h>4$.

Step3: Analyze the graph of $h=\frac{c^{2}-16c}{-16}=\frac{1}{16}c(16 - c)$

This is a quadratic function of $c$ with $a=-\frac{1}{16}$, $b = 1$, $c = 0$. The vertex of the parabola $y = ax^{2}+bx + c$ is at $c=-\frac{b}{2a}$. Here, $c = 8$, and $h=\frac{1}{16}\times8\times(16 - 8)=4$. The parabola opens downwards.

Answer:

The equation relating the value of critical - point(s) $c$ to $h$ is $c^{2}-16c + 16h=0$. The differential equation has two critical points for $h < 4$, one critical point for $h = 4$ and no critical points for $h>4$. The correct bifurcation diagram is a parabola opening downwards with vertex at $(c = 8,h = 4)$. Without seeing the exact details of the options, if the parabola opens downwards and has its vertex at $(8,4)$ in the $ch$ - plane, that is the correct choice.