Solve the following differential equations $\frac{dy}{dx}=\frac{x-y}{x+y}$

$\frac{dy}{dx}=\frac{x-y}{x+y}$

$\text{ put }y=\vartheta x$

$\frac{dy}{dx}=\vartheta +x\frac{d\vartheta }{dx}$

$\vartheta +x\frac{d\vartheta }{dx}=\frac{x-\vartheta x}{x+\vartheta x}$

$\vartheta +x\frac{d\vartheta }{dx}=\frac{x(1-\vartheta )}{x(1+\vartheta )}$

$x\frac{d\vartheta }{dx}=\frac{1-\vartheta }{1+\vartheta }-\vartheta$

$=\frac{1-\vartheta -\vartheta -{\vartheta }^2}{1+\vartheta }=\frac{1-2\vartheta -{\vartheta }^2}{1+\vartheta }$

$\Rightarrow -\frac{1}{2}\log \left(1-2\cdot \frac{y}{x}-\frac{y^2}{x^2}\right)=\log cx$

$\Rightarrow \log \frac{\left(x^2-2xy-y^2\right)}{x^2}=-2\log cx$

$\Rightarrow \log \frac{\left(x^2-2xy-y^2\right)}{x^2}=\log (cx{)}^{-2}$

$\text{ solution is }\frac{x^2-2xy-y^2}{x^2}=(cx{)}^{-2}=\frac{1}{c^2{x}^2}$

$\left(x^2-2xy-y^2\right)=\frac{1}{c^2}=K$