<s>[INST] 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.
 [/INST] $$ {dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$ 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 $a$?
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