hints on solving DE How to solve this DE?
$$ {dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$
From the first part, I get $y = c_1x$. How to find the other solution?
The answer according to answer sheet is $ z + \sqrt{x^2 + y^2 + z^2} = c_2$. Thank you for help.

Let 
\begin{equation}
{\frac{dx}{x}} = {\frac{dy}{y}} = {\frac{dz}{z - a \sqrt{x^2+y^2+z^2}}} = K
\end{equation}
\begin{equation}
{\frac{2xdx}{2x^{2}}} = {\frac{2ydy}{y^{2}}} = {\frac{2zdz}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}}} = K
\end{equation}
Then
\begin{equation}
\frac{dx^{2}}{2x^{2}} = \frac{dy^{2}}{2y^{2}} = \frac{dz^{2}}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}} = K
\end{equation}
Adding all the three terms, we get
\begin{equation}
\frac{dx^{2} + dy^{2} + dz^{2}}{2x^{2} + 2y^{2} + 2z^{2} - 2az \sqrt{x^2+y^2+z^2}} = K
\end{equation}
\begin{equation}
\frac{dw^{2}}{2w^{2} - 2az \sqrt{w^{2}}} = {\frac{2zdz}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}}}
\end{equation}
Hence,
\begin{equation}
\frac{dw^{2}}{2w^{2} - 2az \sqrt{w^{2}}} = {\frac{dz}{z - a \sqrt{w^2}}}
\end{equation}
And then, 
\begin{equation}
\frac{2w dw}{2w^{2} - 2az w} = {\frac{dz}{z - a w}}
\end{equation}
And so,
\begin{equation}
\frac{dw}{dz}= {\frac{w - az}{z - a w}}
\end{equation}
