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Show that the function $u=\ln(x^2+y^2)$ is a solution of the two dimensional Laplace equation $$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0.$$

I have found an answer $$\begin{align*} \frac{\partial u}{\partial x} &=\frac{2x}{x^2+y^2}\\ \frac{\partial^2 u}{\partial x^2}&=\frac {2(x^2+y^2)-4x^2}{(x^2+y^2)^2} \end{align*}$$ and $$\frac{\partial^2 u}{\partial y^2}=\frac {2(x^2+y^2)-4y^2}{(x^2+y^2)^2}$$ and added the equations and got $0$.

My question is: is the way I proved it correct or not, or should I use a different method? If so, please mention.

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    It looks correct to me. Your last derivative should be wrt y, not x though.2012-08-12
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    yeah you are right i will edit it2012-08-12
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    @enzotib thanks enzotib for editing i am not aware of using the correct format using the mathjax can you please tell me where i can find some examples of math text so i can easily figure it out2012-08-12
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    For some basic information about writing math at this site see e.g. [here](http://meta.stackexchange.com/questions/68388/there-should-be-universal-latex-mathjax-guide-for-sites-supporting-it/70559#70559), [here](http://meta.math.stackexchange.com/questions/1773/do-we-have-an-equation-editing-howto) and [here](http://math.stackexchange.com/editing-help#latex).2012-08-12
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    You already did a great job, I only made some minor improvements.2012-08-12
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    @MartinSleziak thanks martin that helped a lot2012-08-12

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