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I encountered this in my homework and am not sure why it is true:

$Y$ is a function on $\theta$ and $\phi$, and $\frac{\partial}{\partial \phi} Y(\theta,\phi)=iY(\theta,\phi)$, then $Y$ takes the form $P(\theta)e^{i\phi}$.

I know this is true assuming $Y$ is separable, but I don't know why this is true in general. Thank you!

orangecat
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    You can shift the term to the left and multiply by the integrating factor to obtain $$\frac{\partial}{\partial \phi}(e^{-i\phi}Y) = 0$$ and do a "partial integration with respect to theta". This yields $$e^{-i\phi}Y = P(\theta)$$ (here, $P$ serves as an arbitrary constant depending on $\theta$). – HK Tan Feb 28 '22 at 21:54
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    The equivalent of the above comment, suppose that $P$ is a function of both $\theta$ and $\phi$. If you plug $Y(\theta,\phi)=P(\theta,\phi)e^{i\phi}$ into the differential equation, you should get $$\frac{\partial}{\partial \phi}P(\theta,\phi)=0$$ – Andrei Feb 28 '22 at 22:00

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