Differential Equation solve differential equation  $$  \frac{dy}{ dx} =\frac{(3x-y-6)}{(x+y+2)}$$
I tried to do this but it´s first order and posible is separable variables
 A: From $\frac{dy}{ dx} =\frac{(3x-y-6)}{(x+y+2)}$, we have
$$(x+y+2)dy - (3x -y - 6)dx = 0$$
We look for $F(x, y)$ such that
$$F_y = x+y+2 $$
$$F_x = - (3x -y - 6)$$
Solve for F,
$F(x,y) = xy + \frac{y^2}{2} + 2y-\frac{3x^2}{2} + 6x$
The solution is F(x,y) = Constant.
A: We can solve the differential equation given by 
$$\frac{dy}{dx}=\frac{3x-y-6}{x+y+2} \tag 1$$
in a straightforward way.  Rearranging $(1)$ reveals
$$x\,dy+y\,dx+(y+2)\,dy+(6-3x)\,dx=0\tag 2$$
Next, we integrate $(2)$ and write
$$\int (x\,dy+y\,dx)\,+\int (y+2)\,dy\,+\int (6-3x)\,dx=C \tag 3$$
Noting that $(x\,dy+y\,dx)=d(xy)$, and evaluating the integrals in $(3)$, we obtain
$$\bbox[5px,border:2px solid #C0A000]{xy+\frac12 y^2+2y+6x-\frac32 x^2=C}$$
And we are done!
A: Another, albeit more tedious method: Let u = x+y+2. Let v = -3x+y+6. Solve for x and y in terms u and v. Take the differentials of u and v and use the result to express both dx and dy in terms of u,v,du,and dv. Swap the results into the original equation and you now get a homogeneous equation now amenable to associated procedures. 
