I need to find percentage $P$ as a function of time, and I am given the equation $\dfrac{dP}{dt} = 1.3P(70-P)$. At time $t = 0$, the population is $P = 14$.
I have found $P = \dfrac{70\exp(91t)}{\exp(91t)-1} - 21\;$ as my equation, but Maple is telling me that's not correct. If anyone could confirm with me or correct the equation, that would be wonderful.
P.S. Sorry if the equation is painful to read, I don't know any of the display languages.
I can't tell you what exactly you did wrong, since you didn't show your work, but at first glance it looks incorrect. The equation has two equilibria at $P=0$ and $P=70$, so the steady-state solution at $t\to\infty$ should be one of those values. Instead, your solution has $P(t\to\infty) = 49$.
Here's how you solve it using separation of variables
\begin{align} \int\frac{dP}{P(70-P)} &= \int1.3\ dt \\ \frac{1}{70} \int\frac{P + (70-P)}{P(70-P)}dP &= 1.3 t + C \\ \frac{1}{70} \int \left(\frac{1}{70-P} + \frac{1}{P}\right)dP &= 1.3t + C \\ \frac{1}{70} \left(\ln |P| - \ln|70-P|\right) &= 1.3t + C \\ \frac{1}{70} \ln \left| \frac{P}{70-P}\right| &= 1.3t + C \\ \frac{P}{70-P} &= Ae^{91t} \end{align}
The initial condition $P(0)=14$ gives $A = \dfrac{14}{56} = \dfrac{1}{4}$. Solving for $P$ we get
\begin{align} \frac{70-P}{P} &= 4e^{-91t} \\ \frac{70}{P}-1 &= 4e^{-91t} \\ P(t) &= \frac{70}{1+4e^{-91t}} \end{align}
This indeed satisfies $P(\infty) = 70$