#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
