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. (type an equation using c and h as the variables. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Recall critical - point definition

Critical points occur when $\frac{dx}{dt}=0$. So we set $\frac{1}{16}x(16 - x)-h = 0$.

Step2: Rearrange the equation

Multiply through by 16 to get $x(16 - x)-16h=0$, which expands to $16x - x^{2}-16h = 0$. Rearranging gives $x^{2}-16x + 16h=0$.

Step3: Use the quadratic formula

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=-16$, $c = 16h$), the solutions are $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substituting the values, we have $x=\frac{16\pm\sqrt{256 - 64h}}{2}=8\pm\sqrt{64 - 16h}$.

Step4: Analyze the number of solutions based on the discriminant

The discriminant $\Delta=b^{2}-4ac=256 - 64h$. When $\Delta>0$, i.e., $256 - 64h>0$ or $h < 4$, there are two solutions. When $\Delta = 0$, i.e., $h = 4$, there is one solution. When $\Delta<0$, i.e., $h>4$, there are no real - valued solutions.

Step5: Write the equation relating critical points and h

Since $x^{2}-16x + 16h=0$ and we use $c$ as the critical - point variable, the equation is $c^{2}-16c + 16h=0$.

Answer:

$c^{2}-16c + 16h=0$