Weird differential equation - Jacobi? In class we had differential equations of the type
$$y'=\frac{\left(Ax+By \right)y+ \alpha x + \beta y}{\left(Ax+By \right) x+ax+by},$$
where $A,\alpha,a,B,\beta,b$ are constants. The names of the constants were chosen in a way, so that the 3 different constants for each variable look a bit an a and b.
For example $$y'=\frac{(x-y)y-x-y}{(x-y)x+x+y}$$
I did not fully understand the method of solution. In class the professor called them "Jacobi Differtial calculus" (translated). I did not find anything suitable on the web, except Wolfram Mathworld http://mathworld.wolfram.com/JacobiDifferentialEquation.html There is at least a differential equation named Jacobi. But this does not seem to be the same as in my question. Any help?
My question: 
Q1 What are DE's like those called? 
Q2 How to solve them?
 A: For $y'=\dfrac{(x-y)y-x-y}{(x-y)x+x+y}$ ,
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{(x-xu)xu-x-xu}{(x-xu)x+x+xu}$
$x\dfrac{du}{dx}=\dfrac{(1-u)xu-1-u}{(1-u)x+1+u}-u$
$x\dfrac{du}{dx}=\dfrac{(x-1)u-xu^2-1}{(1-x)u+x+1}-u$
$x\dfrac{du}{dx}=\dfrac{xu^2-(x-1)u+1}{(x-1)u-x-1}-u$
$x\dfrac{du}{dx}=\dfrac{xu^2-(x-1)u+1-(x-1)u^2+(x+1)u}{(x-1)u-x-1}$
$x\dfrac{du}{dx}=\dfrac{u^2+2u+1}{(u-1)x-u-1}$
$\dfrac{dx}{du}=\dfrac{(u-1)x^2-(u+1)x}{(u+1)^2}$
$\dfrac{dx}{du}+\dfrac{x}{u+1}=\dfrac{(u-1)x^2}{(u+1)^2}$
Luckily this becomes a Bernoulli ODE.
Let $v=\dfrac{1}{x}$ ,
Then $x=\dfrac{1}{v}$
$\dfrac{dx}{du}=-\dfrac{1}{v^2}\dfrac{dv}{du}$
$\therefore-\dfrac{1}{v^2}\dfrac{dv}{du}+\dfrac{1}{(u+1)v}=\dfrac{(u-1)}{(u+1)^2v^2}$
$\dfrac{dv}{du}-\dfrac{v}{u+1}=-\dfrac{(u-1)}{(u+1)^2}$
I.F. $=e^{-\int\frac{du}{u+1}}=e^{-\ln(u+1)}=\dfrac{1}{u+1}$
$\therefore\dfrac{d}{du}\left(\dfrac{v}{u+1}\right)=-\dfrac{(u-1)}{(u+1)^3}$
$\dfrac{v}{u+1}=-\int\dfrac{(u-1)}{(u+1)^3}du$
$\dfrac{1}{(u+1)x}=\int\left(-\dfrac{1}{(u+1)^2}+\dfrac{2}{(u+1)^3}\right)du$
$\dfrac{1}{(u+1)x}=\dfrac{1}{u+1}-\dfrac{1}{(u+1)^2}+c$
$\dfrac{1}{x}=\dfrac{u}{u+1}+c(u+1)$
$\dfrac{1}{x}=\dfrac{\dfrac{y}{x}}{\dfrac{y}{x}+1}+c\left(\dfrac{y}{x}+1\right)$
$\dfrac{xy}{x+y}+c(x+y)=1$
$\dfrac{xy}{x+y}-1=C(x+y)$
Now for $y'=\dfrac{(Ax+By)y+\alpha x+\beta y}{(Ax+By)x+ax+by}$ ,
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{(Ax+Bxu)xu+\alpha x+\beta xu}{(Ax+Bxu)x+ax+bxu}$
$x\dfrac{du}{dx}=\dfrac{(A+Bu)xu+\alpha+\beta u}{(A+Bu)x+a+bu}-u$
$x\dfrac{du}{dx}=\dfrac{Bxu^2+(Ax+\beta)u+\alpha}{(Bx+b)u+Ax+a}-u$
$x\dfrac{du}{dx}=\dfrac{Bxu^2+(Ax+\beta)u+\alpha-(Bx+b)u^2-(Ax+a)u}{(Bx+b)u+Ax+a}$
$x\dfrac{du}{dx}=\dfrac{-bu^2+(\beta-a)u+\alpha}{(A+Bu)x+a+bu}$
$\dfrac{dx}{du}=-\dfrac{(A+Bu)x^2+(a+bu)x}{bu^2+(a-\beta)u-\alpha}$
Which also luckily that this becomes a Bernoulli ODE.
