Need help with a proof involving nonlinear differential equations

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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Well, what does it mean to be homogeneous? Once you see that, the answer becomes evident. – Raskolnikov Jun 15 '11 at 21:26
Well, I'm guessing this is a 0 degree homogeneous equation so $F( \sigma [\frac{as + by}{ds+ey}]) = F(\frac{as +by}{ds+ey})$, which is true for this equation, right? – Bananas Jun 15 '11 at 21:32
What is $\sigma$ ? – Raskolnikov Jun 15 '11 at 21:32
The reason $ae \neq bd$ is important is that it keeps the matrix of coefficients non-singular and therefore invertible. – Ross Millikan Jun 15 '11 at 22:09

With $\sigma$ I meant to check if F is homogeneous of degree 0, since I already know it can be written as $$\frac{dy}{ds} = F(y, s)$$. I just define $$G(y, s) = \frac{as + by}{ds+ey}$$ and look at $$F(G(y,s)) = F(\frac{as + by}{ds+ey}) = \frac{dy}{ds}$$. – Bananas Jun 15 '11 at 21:52
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. – Raskolnikov Jun 16 '11 at 9:02