# Homogeneous(?) Differential equation

I was looking over some old notes and I found that the differential equation $y' = \frac{x-y}{x+y}$ is supposedly a homogeneous equation...for some reason I'm blanking on how to batter it into a form where it can be solved with the substitution $u = \frac{y}{x}$. I think I solved it previously by making the substitution $u = x + y$. Any advice would be appreciated.

Divide top and bottom by $x$ to get
$$y^\prime = \frac{1-\frac{y}{x}}{1+\frac{y}{x}}$$
• Heh, I did manage to get that far, but the equation that I get, $v +x\frac{dv}{dx} = \frac{1-v}{1+v}$ doesn't seem to be going anywhere. – Bitrex Jun 19 '11 at 6:39
• take $v$ to the right hand side. simplify and solve by separating variables. Let me know if you need further assistance. – Nana Jun 19 '11 at 6:43
You may also want to consider $\frac{d}{{dx}}\frac{{(y + x)^2 }}{2} = 2x$.
Adding details: $$\frac{d}{{dx}}\frac{{(y + x)^2 }}{2} = \frac{d}{{dx}}\bigg(\frac{{y^2 }}{2} + xy + \frac{{x^2 }}{2}\bigg) = yy' + y + xy' + x = y'(x + y) + y + x.$$ Hence $\frac{d}{{dx}}\frac{{(y + x)^2 }}{2} = 2x$ can be written $y'(x+y)=x-y$.