Example 7 If real numbers $x, y$ satisfy $x^{2}+y^{2}-2 x+4 y=0$, find the maximum value of $x-2 y$.

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The original equation can be transformed into $(x-1)^{2}+(y+2)^{2}=(\sqrt{5})^{2}$, which represents a circle.
We can set $\left\{\begin{array}{l}x=1+\sqrt{5} \cos \theta, \\ y=-2+\sqrt{5} \sin \theta,\end{array} \quad \theta\right.$ as the parameter.
$$
x-2 y=5+\sqrt{5} \cos \theta-2 \sqrt{5} \sin \theta=5+5 \cos (\theta+\varphi) \leqslant 5+5=10 \text {. }
$$

Thus, the maximum value of $x-2 y$ is 10.