# Wronskian for multivariate functions

I've been reading about the wronskian and I got stuck in the following:

Suppose we are given a multivariate function describing e.g. a plane: $z = m_1 x + m_2 y + b$. How is the wronskian computed? We have here to two variables ($x,y$). How is this case dealt with the wronskian?

Best regards

Theorem: Let $$y_{1}$$ and $$y_{2}$$ be functions of two independence variables $$x_{1}$$ and $$x_{2}$$ i.e., $$y_{1} = y_{1}(x_{1} ,x_{2})$$ and $$y_{2} = y_{1}(x_{1} ,x_{2})$$ for which all partial derivatives of $$1^{st}$$ order, $$\frac{\partial y_{1}}{\partial x_{k}}$$, $$\frac{\partial y_{2}}{\partial x_{k}}$$, $$(k = 1,2)$$ exists throughout the region $$A$$. Suppose, farther, that one of the functions, say $$y_{1}$$, vanishes at no point of $$A$$. Then if all the two rowed determinants in the matrix $$\begin{pmatrix} y_{1} & y_{2} \\ \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{2}}{\partial x_{1}} \\ \frac{\partial y_{1}}{\partial x_{2}} & \frac{\partial y_{2}}{\partial x_{2}} \end{pmatrix}$$ vanish identically in $$A$$, $$y_{1}$$ and $$y_{2}$$ are linearly dependent in $$A$$, and in fact $$y_{2}=c y_{1}$$.
• Do you mean the following three determinants? $$\begin{vmatrix} y_{1} & y_{2} \\ \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{2}}{\partial x_{1}} \end{vmatrix} , \quad \begin{vmatrix} y_{1} & y_{2} \\ \frac{\partial y_{1}}{\partial x_{2}} & \frac{\partial y_{2}}{\partial x_{2}} \end{vmatrix} ,\quad \text{and}\quad \begin{vmatrix} \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{2}}{\partial x_{1}} \\ \frac{\partial y_{1}}{\partial x_{2}} & \frac{\partial y_{2}}{\partial x_{2}} \end{vmatrix}.$$ Commented Mar 30, 2022 at 3:51