Multiplier for solving the pde $\frac{dx}{y+x}=\frac{dy}{y-x}=\frac{dz}{z}$ . I did the following for first relation like:
$$\frac{dx}{y+x}=\frac{dy}{y-x}$$
$$or, ydx-xdx=ydy+xdy$$
Now, what should I do to find the relation for $c_1$ & $c_2$ ? It would have been easier for me if 'twas  $ydx+xdy$ . But here it's $ydx-xdy$ . I have recently started solving such questions. Any suggestion is appreciated. Thanks ...
Edit:
Could I do like this:
$$or, ydx-xdy=ydy+xdx$$
$$or, \frac{ydx-xdy}{xy}=\frac{ydy+xdx}{xy}$$
$$or, d[\log(\frac{x}{y})]=\frac{d(xy)}{xy}$$
$$or, \log(\frac{x}{y})=\log(xy)+\log(c_1)$$
$$or, \frac{1}{y^2}=c_1$$
And then I could use it as follows to find $c_2$ :
$$\frac{dx}{\frac{1}{\sqrt{c_1}}+x}=\frac{dz}{z}$$
and then continue ?

By introducing polar coordinates I find that $(x,y)$ follows logarithmic spirals:
$$
\frac{dx}{y+x}=\frac{dy}{y-x} \\
(y-x) \, dx = (y+x) \, dy \\
y\,dx-x\,dy = x\,dx+y\,dy \\
y^2 d(x/y) = \frac12 d(x^2+y^2) \\
r^2 \sin^2\theta \, d(\cot\theta) = 2r\,dr \\
\sin^2\theta \, d(\cot\theta) = 2r^{-1}\,dr \\
\sin^2\theta \, \frac{-d\theta}{\sin^2\theta} = 2r^{-1}\,dr \\
-d\theta = 2r^{-1}\,dr \\
C_1-\theta = 2\ln r \\
r = C_2 e^{-\theta/2} \\
x^2+y^2 = C_3 e^{-\arctan(y/x)}
$$
