已知一个二维函数，画出距离函数2D的ODEs

假设函数为 $f(x,y)$，则距离函数为 $d(x,y)=\sqrt{x^2+y^2}$。距离函数的ODEs为：

$$
\begin{aligned}
\frac{dx}{dt} &= \frac{\partial d}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} \\
\frac{dy}{dt} &= \frac{\partial d}{\partial y} = \frac{y}{\sqrt{x^2+y^2}} \\
\end{aligned}
$$

将函数 $f(x,y)$ 和距离函数的ODEs 结合起来，可以得到以下二阶ODEs：

$$
\begin{aligned}
\frac{d^2f}{dt^2} &= \frac{d}{dt} \left( \frac{dx}{dt} \frac{\partial f}{\partial x} + \frac{dy}{dt} \frac{\partial f}{\partial y} \right) \\
&= \frac{d}{dt} \left( \frac{x}{\sqrt{x^2+y^2}} \frac{\partial f}{\partial x} + \frac{y}{\sqrt{x^2+y^2}} \frac{\partial f}{\partial y} \right) \\
&= \frac{x^2-y^2}{(x^2+y^2)^{3/2}} \frac{\partial^2f}{\partial x^2} + 2\frac{xy}{(x^2+y^2)^{3/2}} \frac{\partial^2f}{\partial x \partial y} + \frac{y^2-x^2}{(x^2+y^2)^{3/2}} \frac{\partial^2f}{\partial y^2}
\end{aligned}
$$

其中，$\frac{\partial^2f}{\partial x^2}$、$\frac{\partial^2f}{\partial x \partial y}$ 和 $\frac{\partial^2f}{\partial y^2}$ 分别为函数 $f(x,y)$ 的二阶偏导数。