\[
\begin{array}{l}
\frac{d x}{d t}=\beta x\left(1-\frac{x}{k}\right)-c_{1} x y \\
\frac{d y}{d t}=-\alpha y+c_{2} x y
\end{array}
\]
Classify all equilibrium point of the following

Answer

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Answer

\(x = 0, y = 0 \) or \(x = k, y = \frac{\alpha}{c_{2}} \) or \(y= \frac{\beta}{c_{1}}x\left(1 - \frac{x}{k}\right) \)

Steps

Step 1 ：\(0 = \beta x\left(1-\frac{x}{k}\right)-c_{1} xy \)

Step 2 ：\(0 = -\alpha y+c_{2} xy \)

Step 3 ：\(x = 0, y = 0 \) or \(x = k, y = \frac{\alpha}{c_{2}} \) or \(y= \frac{\beta}{c_{1}}x\left(1 - \frac{x}{k}\right) \)