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.
Homogeneous(?) Differential equation
4
$\begingroup$
ordinary-differential-equations
2 Answers
2
Hint
Divide top and bottom by $x$ to get
$ y^\prime = \frac{1-\frac{y}{x}}{1+\frac{y}{x}}$
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0I see i$t$ now; I shouldn't do this when I'm tired! For some reason I thought it wasn't separable. – 2011-06-19
2
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.