Hello Indie,
We are given to solve the first order ODE:
$$\frac{dy}{dx}=0.3y-0.0001y^2$$
First, let's multiply through by $10000$ to get rid of the decimals:
$$10000\frac{dy}{dx}=3000y-y^2$$
Next, let's separate variables:
$$\frac{1}{y(3000-y)}\,dy=\frac{1}{10000}\,dx$$
Now, the expression on the left may be rewritten using partial fraction decomposition. We may assume the decomposition will take the form:
$$\frac{1}{y(3000-y)}=\frac{A}{y}+\frac{B}{3000-y}$$
Multiplying through by $$y(3000-y)$$ we obtain:
$$1=A(3000-y)+By$$
Since this is true for all $y$, letting $y=0$ we get:
$$1=3000A\implies A=\frac{1}{3000}$$
And letting $y=3000$, we obtain:
$$1=3000B\implies B=\frac{1}{3000}$$
Hence:
$$\frac{1}{y(3000-y)}=\frac{1}{3000}\left(\frac{1}{y}+\frac{1}{3000-y} \right)$$
And so the ODE may be written:
$$\left(\frac{1}{y}-\frac{1}{y-3000} \right)\,dy=\frac{3}{10}\,dx$$
Integrating, we find:
$$\int \frac{1}{y}-\frac{1}{y-3000}\,dy=\frac{3}{10}\int\,dx$$
$$\ln\left|\frac{y}{y-3000} \right|=\frac{3}{10}x+C$$
Converting from logarithmic to exponential form (and rewriting the parameter $C$), we have:
$$\frac{y}{y-3000}=Ce^{\frac{3}{10}x}$$
Now we want to solve for $y$. Multiply through by $y-3000$:
$$y=(y-3000)Ce^{\frac{3}{10}x}$$
$$y\left(Ce^{\frac{3}{10}x}-1 \right)=3000Ce^{\frac{3}{10}x}$$
$$y=\frac{3000Ce^{\frac{3}{10}x}}{Ce^{\frac{3}{10}x}-1}$$
Now, dividing each term in the numerator and denominator by $$Ce^{\frac{3}{10}x}$$ and rewriting the parameter again, we obtain:
$$y(x)=\frac{3000}{1+Ce^{-\frac{3}{10}x}}$$
