Find the general solution to y′=x+yx−yy' = \frac{x+y}{x-y}y′=x−yx+y.
tan−1(y/x)=ln∣x∣+C\tan^{-1}(y/x) = \ln|x| + Ctan−1(y/x)=ln∣x∣+C
12ln(x2+y2)+tan−1(y/x)=C\frac{1}{2} \ln(x^2+y^2) + \tan^{-1}(y/x) = C21ln(x2+y2)+tan−1(y/x)=C
12ln(1+(y/x)2)+tan−1(y/x)=ln∣x∣+C\frac{1}{2} \ln(1+(y/x)^2) + \tan^{-1}(y/x) = \ln|x| + C21ln(1+(y/x)2)+tan−1(y/x)=ln∣x∣+C
12ln(x2+y2)=C\frac{1}{2} \ln(x^2+y^2) = C21ln(x2+y2)=C