Nonlinear ODE: $y'=(1+axy)/(1+bxy)$ Consider the  first order nonlinear ODE problem: 
$$
y'(x)=\frac{1+ay(x)x}{1+by(x)x}, \quad x>0
$$
where $a,  b>0$ are some  constants. I would like to know if these kind of equations were studied somewhere else  (i.e., existence of global solutions, uniqueness  etc). Any references will be helpful. 
Remarks: 
Mathematica does not give me anything. 
I have tried to use Maple and it shows me that $y$ is a solution of an equation involving  some Whittaker functions but I am not sure about this. 
If $a=b$ then $y=x+C$.
If $a=1,  b=2$ then one can take $y(x) = \frac{1}{2}\left(x-\frac{1}{x} \right)$
 A: You might find it useful to make a change of variables to reduce the equation to a more familiar form.  For example, if we assume, as we may, that $a$ and $b$ are not equal, then we can substitute $y = (z+ax)/b$, where $z$ is a new unknown, and then the equation can be written in the form
$$
\frac{dx}{dz} = \frac{(1 + xz + a x^2)}{b-a},
$$ 
which is a Riccati equation for $x$ as a function of $z$.  Consequently, we understand all of the blow-up conditions since, by the usual linearization of Ricatti equations, we can express the solutions as the ratio of linear combinations of two solutions of a linear equation.  Moreover, if we are given a particular, solution, we can now get the general solution by quadrature.
A: Mathematica does give solutions (in terms of the Lambert $W$ function) for specific values of $a, b.$ it's just that it finds different numbers of solutions for different $a, b$ so can't formulate a general answer. for example, for $a=3, b=7,$ it gives:
$$\left\{\left\{y(x)\to \frac{1}{21} \left(-4 W\left(-\frac{7}{4} \sqrt[4]{e^{-9
   c_1-9 x}}\right)-7\right)\right\},\left\{y(x)\to \frac{1}{21} \left(-7-4
   W\left(-\frac{7}{4} i \sqrt[4]{e^{-9 c_1-9
   x}}\right)\right)\right\},\left\{y(x)\to \frac{1}{21} \left(-7-4
   W\left(\frac{7}{4} i \sqrt[4]{e^{-9 c_1-9
   x}}\right)\right)\right\},\left\{y(x)\to \frac{1}{21} \left(-4
   W\left(\frac{7}{4} \sqrt[4]{e^{-9 c_1-9 x}}\right)-7\right)\right\}\right\}$$
A: Global existence (or not) can be clarified without any tools by just taking a good look at your equation. Clearly, the RHS $f(x,y)$ is bounded away from the curve $1+bxy=0$, where it is undefined. So global existence fails for a given solution precisely if this solution approaches $1+bxy=0$ in finite time (in particular, global existence for a solution $y(x)$ is guaranteed as soon as $y(x_0)\ge 0$ for some $x_0$).
This won't happen if $a\le b$, as we see by just checking what the ODE does close to our curve. For example, if $1+bx_0y(x_0)=\epsilon>0$, then $y'(x_0)>0$, so we're moving away from the curve.
If $a>b$, then, by the same argument, there are solutions that reach $1+bxy=0$ in finite time (those with sufficiently negative initial value $y(0)$).
