30th October, 2020
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = f(x) \]
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = x \left(1 - \frac{x}{10}\right) - p\frac{x^2}{1 + x^2} \]
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = x \left(1 - \frac{x}{10}\right) - p\frac{x^2}{1 + x^2} \]
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = x \left(1 - \frac{x}{10}\right) - p\frac{x^2}{1 + x^2} \]
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = x \left(1 - \frac{x}{10}\right) - p\frac{x^2}{1 + x^2} \]
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = x \left(1 - \frac{x}{10}\right) - p\frac{x^2}{1 + x^2} \]
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = x \left(1 - \frac{x}{10}\right) - p\frac{x^2}{1 + x^2} \]
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = x \left(1 - \frac{x}{10}\right) - p\frac{x^2}{1 + x^2} \]
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = x \left(1 - \frac{x}{10}\right) - p\frac{x^2}{1 + x^2} \]
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = x \left(1 - \frac{x}{10}\right) - p\frac{x^2}{1 + x^2} \]