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$$\frac{dy}{dx}=\frac{-x+y}{x+y}=\frac{1-\frac{x}{y}}{1+\frac{x}{y}}.$$ Let $z=x/y$, so that $y=x/z$ and $$\frac{dy}{dx}=\frac{z-x\frac{dz}{dx}}{z^2}=\frac{1-z}{1+z}.$$ Then $z-x\frac{dz}{dx}=\frac{z^2-z^3}{1+z} \implies -x\frac{dz}{dx}=-\frac{z^3+z}{z+1} \implies \frac{dx}{x}=\frac{dz}{z^2+1} \implies \arctan(z)=\ln(|x|)+C \implies z=\tan(\ln(|x|)+C)$

Thus $$y = \frac{x}{\tan(\ln(|x|)+C)}$$.

However, my book: *A Modern Introduction to Differential Equations 2nd edition by Henry Ricardo* says "this first-order equation is homogeneous and can be solved implicitly".

They pursued the method of letting $z=y/x$ and obtained the solution as

$$2\arctan(y/x)+\ln(x^2+y^2)-C=0.$$

Why are the two different substitutions, which should both be suitable, giving me two different answers?