Solving $\frac{dy}{dx} = \frac{xy+3x-2y+6}{xy-3x-2y+6}$ I'm stuck with this problem...
$$\frac{\operatorname{d}y}{\operatorname{d}x} = \frac{xy+3x-2y+6}{xy-3x-2y+6}$$
I have tried separating variables in the following way:
$$\frac{\operatorname{d}y}{\operatorname{d}x} = \frac{x(y+3)2(-y+3)}{y(x-2)3(-x+2)}$$
$$\left(\frac{y+3 \cdot 2(-y+3)}{y}\right)\operatorname{d}y = \left(\frac{x-2 \cdot 3(-x+2)}{x}\right)\operatorname{d}x$$
$$\left(\frac{-y+9}{y}\right) \operatorname{d}y = \left(\frac{-2x+4}{x}\right) \operatorname{d}x$$
Then integrating both sides I end up with
$$y=2x+\ln\left(\frac{y^9}{x^4}\right)+C$$
I'd appreciate very much if someone give me the right answer because I don't know if my solution is correct or not. Thanks in advance
 A: $\begin{align}\frac{xy+3x-2y+6}{xy-3x-2y+6}&=\frac{6x+xy-3x-2y+6}{xy-3x-2y+6}\\
&=\frac{6x+x(y-3)-2(y-3)}{x(y-3)-2(y-3)}\\
&=\frac{6x+(x-2)(y-3)}{(x-2)(y-3)}\\
&=\frac{6x}{(x-2)(y-3)}+1\\
\end{align}$
So, your non-linear ODE is
$$y'=\frac{6x}{(x-2)(y-3)}+1$$
or
$$(y-3)(y'-1)=\frac{6x}{x-2}.$$
If we make a change of variable $u=y-3$ then $y'=u'$ and hence we have
$$uu'-u=\frac{6x}{x-2}.$$
I don't now how to solve it. WA couldn't solve too. https://www.wolframalpha.com/input?i=yy%27-y%3D6x%2F%28x-2%29
I need to unlock pro-wa?
A: Ok so I corrected my algebra and ended up with $\left(\frac{-y+3}{y+3}\right) \operatorname{d}y = \left(\frac{x}{x-2}\right)-\left(\frac{2}{x-2}\right) \operatorname{d}x $ which is one on the x side
Then I integrated both sides and my answer was $y=x-6ln\left({y+3}\right)+C$
I think that is the final answer because I plugged it in the simplified equation and it equals 1, and the derivative of my answer is also 1.
Correct me if I'm wrong, please, and thanks for your answer
