Finding Harmonic Function in the form $\phi(x + \sqrt{x^2 + y^2})$ I must find a harmonic function $u(x,y)$ of the form $\phi(x + \sqrt{x^2 + y^2})$
Let $u(x,y) = \phi(x + \sqrt{x^2 + y^2})$.
$$
\phi_x = \phi'(x + \sqrt{x^2 + y^2})((x+\frac{\sqrt(x^2+y^2))}{\sqrt(x^2+y^2)}  )
$$
\begin{align*}
\phi_{xx} &= \left (  \phi'(x + \sqrt{x^2 + y^2}) \right )' \left ((x+\frac{\sqrt(x^2+y^2))}{\sqrt(x^2+y^2)} \right ) + \left (\phi'(x + \sqrt{x^2 + y^2} \right ) \left ( \frac{(x+\sqrt(x^2+y^2))}{\sqrt(x^2+y^2)}\right )'\\
&= \phi''(x + \sqrt{x^2 + y^2})\left (\frac{(x+\sqrt(x^2+y^2))}{\sqrt(x^2+y^2)} \right )^2 + \left (\phi'(x + \sqrt{x^2 + y^2} \right ) \left ( \frac{1}{\sqrt{x^2 + y^2}} - \frac{2x^2}{(x^2 + y^2)^2}\right )
\end{align*}
$$
\phi_y =\left ( \phi'(x + \sqrt{x^2 + y^2})\right ) \left(\frac{y}{\sqrt{x^2 + y^2}} \right )
$$
\begin{align*}
\phi_{yy} &= \left ( \phi'(x + \sqrt{x^2 + y^2})\right )'\left(\frac{y}{\sqrt{x^2 + y^2}} \right ) +  \left ( \phi'(x + \sqrt{x^2 + y^2})\right )\left(\frac{y}{\sqrt{x^2 + y^2}} \right )'\\
&= \phi''(x + \sqrt{x^2 + y^2})\left(\frac{y}{\sqrt{x^2 + y^2}} \right )^2 +  \left ( \phi'(x + \sqrt{x^2 + y^2})\right ) \left (\frac{1}{\sqrt{x^2 + y^2}} - \frac{2y^2}{(x^2 + y^2)^2} \right )
\end{align*}
This is obviously pretty convoluted so when I attempt to solve the DE discovered from the Laplacian Equation, I get lost. How do I make this more simple with a substitution?
 A: I came across this question on here as I am studying Complex Analysis independently and was looking for something not typically found on the books, but see that the answer is still "up in the air". As Massive Jack said, the key is to use polar coordinates to transform the given multi-variable function to one dependent on $r$ and $\theta$ , which yields the ordinary differential equation (ODE) above, dependent only on the variable $z$ after making the substitution $z= r(1+cos(\theta))$. That's all correct and everything, but I just wanted to provide more details into how to obtain the final result after one obtains the ODE. As Complex, Oh did, you may want to proceed by reducing the order of the ODE to a first order ODE, but you'd end up obtaining (after letting $\phi'(z)$ = $y(z)$ and then $\phi''(z)$ = $y'(z)$) :   $2zy'(z)$ + $y(z)$ = $0$. I want to point out that there are no restrictions here on the variable $z$, so it is not always the case that you can divide like it was done above. ($cos(\pi$) = $-1$ ... ). With that small observation, note that then we proceed to solve the ODE , which is really a separable equation and whose solution can be readily be obtained as $y(z) = \frac{c}{\sqrt(z)}$ 
The final step is to put this in terms of $\phi(z)$, and thus we get that $\phi(z) = \int \frac{c}{\sqrt(z)}dz = c{\sqrt(z)}+ d$. You may then substitute back for $z$ as appropriate.
