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. –  Daryl Aug 12 '12 at 9:07
yeah you are right i will edit it –  Biju jose Aug 12 '12 at 9:09
@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 out –  Biju jose Aug 12 '12 at 10:25
For some basic information about writing math at this site see e.g. here, here and here. –  Martin Sleziak Aug 12 '12 at 10:37
You already did a great job, I only made some minor improvements. –  enzotib Aug 12 '12 at 11:20
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