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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.

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Hint
Divide top and bottom by $x$ to get
$$ y^\prime = \frac{1-\frac{y}{x}}{1+\frac{y}{x}}$$

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  • $\begingroup$ 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. $\endgroup$ – Bitrex Jun 19 '11 at 6:39
  • $\begingroup$ take $v$ to the right hand side. simplify and solve by separating variables. Let me know if you need further assistance. $\endgroup$ – Nana Jun 19 '11 at 6:43
  • $\begingroup$ I see it now; I shouldn't do this when I'm tired! For some reason I thought it wasn't separable. $\endgroup$ – Bitrex Jun 19 '11 at 6:45
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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 $y'(x+y)=x-y$.

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