How to get the Jacobian matrix for this differential equation?

Question

$$ \begin{cases} y_1'(t) = -k_1y_1 + k_2y_2 \\ y_2'(t) = k_1y_1 - k_3y_2^2 \end{cases} $$

let's put $(y_1',y_2')$ as the function part in jacobian matrix, and put $(y_1,y_2)$ as the self change variable part in jacobian matrix, the partial diffrential of $y_1'$ with respect to $y_1$ should be $-k_1$? but there is some relationship between $y_1$ and $y_2$, can we just throw the $k_2y_2$ part?

Answer 1

\begin{equation} \begin{cases} y_1'(t) = -k_1~y_1 + k_2~y_2\\ y_2'(t) = k_1~y_1 - k_3~y_2^2 \end{cases} \end{equation}

$$\text{Jacobian },~J = \begin{bmatrix}\dfrac{\partial y_1'}{\partial y_1} & \dfrac{\partial y_1'}{\partial y_2}\\\dfrac{\partial y_2'}{\partial y_1} & \dfrac{\partial y_2'}{\partial y_2}\end{bmatrix} = \begin{bmatrix}-k_1 & k_2\\ k_1 & -2k_3y_2\end{bmatrix}$$