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Please show how to solve? I am stack with lie bracket. Thank you.

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By definition, we have $$[X,Y]=XY-YX.$$ Therefore, we have $$\left[-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y},\frac{\partial}{\partial x}\right]= \left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right)\frac{\partial}{\partial x} -\frac{\partial}{\partial x}\left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right)$$ $$=\left(-y\frac{\partial^2}{\partial x^2}+x\frac{\partial^2}{\partial y\partial x}\right) -\left(-y\frac{\partial^2}{\partial x^2}\right) -\left(\frac{\partial}{\partial y}+x\frac{\partial^2}{\partial y\partial x}\right)=-\frac{\partial}{\partial y}$$ because $$\frac{\partial}{\partial x}\left(-y\frac{\partial}{\partial x}\right)= \frac{\partial}{\partial x}(-y)\frac{\partial}{\partial x}-y\frac{\partial^2}{\partial x^2}=-y\frac{\partial^2}{\partial x^2}$$ and $$\frac{\partial}{\partial x}\left(x\frac{\partial}{\partial y}\right)= \frac{\partial}{\partial x}(x)\frac{\partial}{\partial y}+x\frac{\partial^2}{\partial x\partial y}=\frac{\partial}{\partial y}+x\frac{\partial^2}{\partial y\partial x}.$$

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  • $\begingroup$ Hmm you are just applying a formula related to this topic. Thank you for help. You are the best. :) $\endgroup$
    – 1190
    Apr 12 '13 at 11:44

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