#dy/dx=(x+y)/(x-y)#

this is first order linear and **homogeneous** in the sense that when written in the form #dy/dx = f(x,y)# then

#f(kx, ky) = f(x,y)#

so we re-write it as #dy/dx=(1+y/x)/(1-y/x)#

in order to make the **standard** sub #v(x) = (y(x))/x#

because #y = v x#, then #y' = v' x + v#

so we have

#v'x + v = (1+v)/(1-v)# and we may as well now separate it out

#v'x = - v + (1+v)/(1-v) #

#= (- v(1-v)+ 1+v)/(1-v)#

#implies v'x = (v^2+ 1)/(1-v)#

#(v- 1)/(v^2+1) v' = -1/x #

#int \ (v- 1)/(v^2+1) v' \ dx = - int \ 1/x \ dx#

looks absolutely horrible

#int \ (v- 1)/(v^2+1) \ dv = - int \ 1/x \ dx#

#int \ v/(v^2+1) - 1/(v^2+1) \ dv = - int \ 1/x \ dx#

#1/2 ln(v^2+1) - arctan v = - ln x + C#

#1/2 ln((y/x)^2+1) - arctan (y/x) = - ln x + C#

horrible