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I had to find a function for the population of a species that reproduces at a rate proportional to the current population and that dies at a rate proportional to the square root of the current population. Therefore, I assumed that this meant I had to solve

$$ \frac{dP}{dt}=\beta P-\delta \sqrt{P} $$

where $P$ is the population (a function of $t$, time) and $\beta$ and $\delta$ the birth and death rates repectively. Solving this differential equation gave me

$$ P(t)=\left(\frac{e^\left(\frac{\beta t}{2}+C \right)+\delta}{\beta}\right)^2 $$

which is correct, according to Wolfram. The next part of the questionis to show that for some value of $P_0$ (the initial population at $t=0$), the population will be extinct in the long run. However, it is clear that my formula for $P(t)$ is always increasing and that it is never equal to zero. Did I miss anything?Like I said, I checked with Wolfram and it is a solution to my differential equation.

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    Let it be noted that your equation decays when $\beta < 0$, and when $\delta = 0$ your function will decay to zero (i.e. extinction) – 2017-02-01
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    @BrevanEllefsen Can a birth rate be negative? – 2017-02-01
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    depends on your definition of birth rate I suppose. I would generally say no, but I was pointing out that you should explicitly assume that $\beta$ is positive to have your analysis be correct. – 2017-02-01

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With $t=0$ we have $e^C=\beta\sqrt{P_0}-\delta$ and the population after $t$ (usually in years) is $$ P(t)=\left(\frac{(\beta\sqrt{P_0}-\delta)e^\left(\frac{\beta t}{2} \right)+\delta}{\beta}\right)^2 $$ this makes the equation easier.

In these types problems we can't say when population will be extinct, scince as you mention, $P(t)$ is never equal to zero. Instead of it we could say for which times the population will be smaller than an amount for example $P(t)<\dfrac{1}{1000}P(t_0)$.

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However, it is clear that my formula for $P(t)$ is always increasing and that it is never equal to zero.

The problem with your analysis is that your function is NOT always increasing. Whenever $\beta < 0$ your function decays, and when $\delta = 0 $ your function decays to zero (extinction)

All we can is that your function must be non-negative because of the squaring operation

Note: the decay is not monotonic unless $\delta \geq 0,$ although $\lim x \to -\infty$ will still be unbounded towards $+\infty$ and $\lim x \to +\infty$ will still be bounded (assuming still that $\beta < 0$)