Question

Reduce the homogeneous equation to the separated one

(x+y)dx+(y-x)dy=0

Answer 1

$$(x+y)+(y-x)y'=0$$

Let $y(x)=xu(x).$ Then

$$y'=u+xu'$$

$$x+xu+(ux-x)(u+xu')=0$$

$$x(1+u+(u-1)(u+xu')=0$$

$$1+u+u^2-u+xuu'-xu'=0$$

$$x(1-u)u'=1+u^2$$

$$\dfrac{1-u}{1+u^2}du=\dfrac{dx}{x}$$

Integrate

$$\int\dfrac{1-u}{1+u^2}du=\int \dfrac{dx}{x}$$

$$\tan^{-1}(u)-\dfrac{1}{2}\ln(1+u^2)=\ln x+C$$

$$\tan^{-1}(\dfrac{y}{x})-\dfrac{1}{2}\ln(1+\dfrac{y^2}{x^2})=\ln x+C$$