Given, x*dx + y*dy = x*dy - y*dx

=> y*dy - x*dy = -x*dx - y*dx

=> (y - x)dy = (-x - y)dx

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

=> 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)

So, f(kx, ky) = f(x, y)

Now, dy/dx = (1 + y/x)/(1 - y/x)

Put y = vx

dy/dx = v + x*dv/dx

=> (1 + y/x)/(1 - y/x) = v + x*dv/dx

=> (1 + v)/(1 - v) = v + x*dv/dx

=> x*dv/dx = -v + (1 + v)/(1 - v)

=> x*dv/dx = {-v(1- v) + (1 + v)}/(1 - v)

=> x*dv/dx = (v² + 1)/(1 - v) 

=> x/dx= (v² + 1)/{(1 - v)*dv}

=> dx/x = {(1 - v)/(v² + 1)}*dv

=> dx/x = {-(v - 1)/(v² + 1)}*dv

=> -dx/x = {(v - 1)/(v² + 1)}*dv

=> -dx/x = {v/(v² + 1)}*dv - {1/(v² + 1)}*dv 

=> -∫dx/x = ∫{v/(v² + 1)}*dv - ∫{1/(v² + 1)}*dv

=> -logx + C =  log(v² + 1)/2 - tan^-1 (v) 

=> log{(y/x)² + 1}/2 - tan^-1 (y/x) = -logx + C