<s>[INST] Does $a\ln(x^2 +y^ 2 )+b$ satisfy Laplace’s equation? I can't verify that $F(x,y) = a\ln(x^2 +y^ 2 )+b$ satisfies Laplace’s equation ($F_{xx}+F_{yy}=0$). Here is what I did:
\begin{align*}
F_x &= \frac{2ax}{x^2 + y^2}
&F_y &= \frac{2ay}{x^2+y^2}
\\
F_{xx} &= \frac{2a}{x^2+y^2} - \frac{4ax^2}{(x^2+y^2)^2}
&
F_{yy} &= \frac{2a}{x^2+y^2} - \frac{4ay^2}{(x^2+y^2)^2}
\end{align*}
and $F_{yy}\neq-F_{xx}$. So what should I do?
 [/INST] Let $F(x,y) = a \ln(x^2 + y^2) + b $. Then, Using quotient rule:
$$ F_x = \frac{ 2ax}{x^2 + y^2} \iff F_{xx} = \frac{2a(x^2+y^2)-2x(2ax)}{(x^2+y^2)^2} = \frac{2ay^2 - 2ax^2}{(x^2+y^2)^2}$$
Similarly,
$$ F_y = \frac{ 2ay}{x^2+y^2} \implies F_{yy} = \frac{2a(x^2+y^2) -2ay(2y)}{(x^2+y^2)^2} = \frac{2ax^2 - 2ay^2}{(x^2+y^2)^2}$$
Hence,
$$ F_{xx} + F_{yy} = 0 $$
</s>