Alligator population Hints only. I feel like I am so close. Population growth: The time rate change of an alligator population $P$ in a swamp is proportional to the square of $P$. The swamp contained a dozen alligators in $1988$ and $2$ dozen in $1998$
$$\frac{dp}{dt} = kp^2$$
$$\int \frac{dp}{p^2} = \int k$$
$$ \frac{p^{-2+1}}{-2+1} = tk + C$$
$$ -\frac{1}{p} = tk+C$$
Use $P(0) = 12$ to solve for C.
$$-\frac{1}{12} = C$$ 
$$p(t) = -\frac{1}{tk - \frac{1}{12}}$$
Use $P(10)=24$ to solve for k:
$$24 = -\frac{1}{10k-\frac{1}{12}}$$
$$10k -\frac{1}{12} = -\frac{1}{24}$$
$$k = \frac{1}{240}$$
Now solve for $t$ with $P(t)=48$
$$48=-\frac{1}{\frac{t}{240}-\frac{1}{12}}$$
$$\frac{t}{240} - \frac{1}{12} = -\frac{1}{48}$$
$$\frac{t}{20}-1=-3$$
$$\frac{t}{20}=-2$$
$$t=-40$$
Which is obviously wrong. I assume that t should in-fact represent 1 year even though our information is given in increments of 10. I am very sure my set-up is correct and I integrated correctly...

You have correctly derived that
$$P(t) = -\frac{1}{tk - \frac{1}{12}}\tag{1}$$
and
$$k = \frac{1}{240}\tag{2}$$
so that
$$P(t) = -\frac{1}{\frac{t}{240} - \frac{1}{12}}\tag{3}$$
Your mistake happens when solving for $P(t)=48$. Starting off at 
$$48=-\frac{1}{\frac{t}{240}-\frac{1}{12}}\implies\frac{t}{240} - \frac{1}{12} = -\frac{1}{48}$$
is correct. However
$$\frac{t}{240} - \frac{1}{12} = -\frac{1}{48}\implies \frac{t}{20}-1=-3$$
is incorrect. Multiplying all three terms by $12$ would produce 
$$\frac{t}{240} - \frac{1}{12} = -\frac{1}{48}\implies \frac{t}{20}-1=-\frac{1}{4}$$
which you could then multiply again by $20$ to form
$$\frac{t}{20}-1=-\frac{1}{4}\implies t-20=5$$
so that $t=15$.
