Im trying to solve the following differential equation: $\frac{dP}{dt} = 8P-(6+ \frac{P}{2500})P$.
I think this might be possible using partial fractions somehow but my attempt so far is:
$$\begin{align}\frac{dP}{dt} &= 2P- \frac{P^2}{2500} \\ \frac{dP}{2P + \frac{P^2}{2500}} &= t+c \\ \ln\left|2P + \frac{P^2}{2500}\right| &= t+c \end{align}$$ - this is about as far as i can get
Any help will be appreciated
edit: I tried a partial fractions approach but didn't get very far
$$\frac{dy}{dx} = 2y - \frac{y^2}{2500}$$
There are a few issues with the attempted solution: (1) The coefficients of $P, P^2$ in the simplification of the given differential equation are wrong. (2) The integration of the differential in $dP$ is incorrect.
In any case, we can write the differential equation in the form $$\frac{dP}{dt} = r P + s P^2$$ for some constants $r, s$, and integrating gives $$\int \frac{dP}{r P + s P^2} = \int dt .$$
To apply the method of partial fractions, we factor the denominator: $r P + s P^2 = P (r + s P)$. Then, (for $r \neq 0$) we can write the integrand of the integral w.r.t. $dP$ as $$\frac{1}{r P + s P^2} = \frac{A}{P} + \frac{B}{r + sP}$$ for constants $A, B$ we can find explicitly in terms of $r, s$.