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The sterile insect release method for pest control releases a number of sterile insects into a population. If a population n of sterile insects is maintained in a population, a possible simple model for the population of fertile insects N(t) is $\frac{dN}{dt}=[\frac{aN}{N+n}-b]N-kN(N+n)$, s.t. $a>b>0$ and k>0 are constant parameters. How do I determine the critical number of sterile insects $n_{c}$ which would eradicate the pests and show that this is less than a quarter of the environment carrying capacity? I'm having trouble understanding what this is asking of me?

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A stable solution has $dN/dt=0$. Rearrange the equation $dN/dt=0$ into a cubic polynomial.
When $n=0$, the cubic has three solutions $0,\pm p$, where $p$ is the stable population.
As $n$ increases, the cubic moves. The solutions that start at $n=0,n=p$ move towards each other until they meet, and disappear. That is because the cubic's local min moves above the $x$-axis.
It means there is no longer a stable positive solution to the differential equation, so the insects will get wiped out.
So: Find a value of $n$ where the cubic has a positive double-zero. Or at least, show that there are no positive solutions when $n=p/4$

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