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.
 A: 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}
A: $$ {dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$ You get $y=c_1x$, so put it into the third fraction:
$$ {dx \over x} =  {dz \over z - a \sqrt{x^2+c_1^2x^2+z^2}}$$ $$ {dx \over x} =  {dz \over z - a \sqrt{(1+c_1^2)x^2+z^2}}={dz \over z - a \sqrt{Cx^2+z^2}}$$ which is homogeneous equation: $$(z - a \sqrt{Cx^2+z^2})dx=xdz, x\neq 0$$ by taking $u=\frac{z}{x}$, you get: $${-adx \over x} =  {du \over \sqrt{C+u^2}}$$ then integrating from both sides gives: $$\ln|u+\sqrt{C+u^2}|=-a\ln|x|+c_2$$ or $$\ln|z+\sqrt{x^2+y^2+z^2}|=(1-a)\ln|x|+c_2$$ Are you sure, you don't have any information about that $a$?
