Toolbox:
Step1
The given equation can be rearranged and written as$ dy/dx = (x+y)/(x-y)$ $F(x,y) = (x+y)/(x-y)$ $F(kx,ky) = (kx+ky)/(kx-ky) = k^0.f(x,y)$ Hence it is a homogenous equation with degree zero. Step2
Using the information in the tool box , let us substitute $y = vx$ and $dy/dx = v + xdv/dx.$ $v + x.dv/dx = (x + vx)/(x-vx)$ Taking $v$ as tha common factor in the RHS and cancelling we get $v + x.dv/dx = (1+v)/(1-v)$ Bringing $v$ from the LHS to RHS we get, $x.dv/dx = (1+v^2)/(1-v)$ Step3:
Separating the variables we get $(1-v)dv/(1+v^2) = dx/x$ $dv/1+v^2 - vdv/1+v^2 = dx/x$ Integrating on both sides we get, $tan^{-1}v - (1/2)log(1+v^2) = logx + C$ Step4:
Substituting back for v as y/x we get $tan^{-1}(y/x) - (1/2)log(x^2+y^2)/x^2 = logx + C$ This can be rewritten as , $tan^{-1}(y/x) - (1/2)x^2+y^2 + (1/2)log x^2 = logx + C$ $tan^{-1}(y/x) = (1/2)log(x^2+y^2) + C$ This is the required solution.