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I'm trying to solve a problem that stated:

If $ae \neq bd$ prove that you can choose 2 constants, h and k, so that the substitution $t= s - h$ , $ x = y - k $ reduce the following equation to a homogeneous equation. $\frac{dx}{dt} = F \left(\frac{at+bx+c}{dt+ex+f}\right)$

I'm unsure about what I'm supposed to be doing here.
Doing all the substitutions I got $ \frac{dy}{ds} = F\left(\frac {as -ah+by-bk+c}{ds-dh+ey-ek+f}\right) $

From this I gathered that if $\displaystyle k = \frac{fa-dc}{ea-db} $ ( Note that I'm not dividing by 0 since I know $ae\neq bd$) and $ \displaystyle h = \frac{c}{a} - \frac{b}{a} \left[\frac{fa - dc}{ea - db}\right] $ then I would have $ \frac{dy}{ds} = F\left(\frac{as+by}{ds+ey}\right)$ With $ v = \frac{s}{y} $ I can write this as $ G(v) = F(\frac{a+bv}{d+ev}) $

Which if I understand it correctly would make this a 0 order differential equation, right? It seem kind of weird that this works regardless of what F is as long as $ ae \neq bd $.

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    The reason $ae \neq bd$ is important is that it keeps the matrix of coefficients non-singular and therefore invertible.2011-06-15

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Here's a hint.

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    The point is that $G(y,s)$ is invariant for a rescaling of both variables with a same factor. Therefore so is $F\circ G$ and thus the differential equation is homogeneous. And that is all there is to it.2011-06-16