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, one critical point for h and no critical points for h. (use integers or fractions for any numbers in the expressions.)

Explanation:

Step1: Recall critical - point definition

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

Step2: Rearrange the equation

Expand and rewrite it as $-\frac{1}{16}x^{2}+x - h=0$. Multiply through by - 16 to get $x^{2}-16x + 16h=0$.

Step3: Use the quadratic formula

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

Step4: Determine number of critical points based on the discriminant

The discriminant $\Delta=b^{2}-4ac=256 - 64h$.

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

Answer:

The differential equation has two critical points for $h<4$, one critical point for $h = 4$, and no critical points for $h>4$.