Am I using the right substitution to solve this differential equation? I want to find the general solution of the following:
$$\frac{dy}{dx} = \frac{y}{x+y+2}$$
I substitute $u = x + y+2$ and got the following $du/dx = 1 + dy/dx$.
Then I got an expression as such:
$$\frac{du}{dx} = 2 - \frac{x+2}{u}$$
But from this expression,  I cannot seem to separate it further. Using an integrating factor also doesn't seem to work. It also doesn't have the form of a bernoulli's equation, due to the $+2$.
How should I further proceed?
 A: Hint...you can write the differential equation upside down i.e. $$\frac {dx}{dy}=\frac xy+1+\frac 2y$$
Now use an integrating factor since you have $$\frac{dx}{dy}-\frac 1yx=1+\frac 2y$$
A: You can work it through as transforming the given ODE to an exact equation :
$$\frac{dy}{dx} = \frac{y}{x+y+2} \Rightarrow y'(x)(x+y+2) -y = 0$$
If you let $R(x,y) = -y$ and $S(x,y) = x+y+2$, then you can see that this is not an exact equation, since :
$$\frac{\partial R}{\partial y} \neq \frac{\partial S}{\partial x}$$
Let's find an integrating factor $μ(y)$, such that the ODE : 
$$\frac{\partial μ R}{\partial y} + \frac{\partial μS}{\partial x} = 0$$
is exact. 
This means, that : 
$$\frac{\partial μR}{\partial y} = \frac{\partial μS}{\partial x}\Rightarrow-\bigg(y\frac{dμ(y)}{dy}\bigg)-μ(y) =μ(y) \Rightarrow \frac{\frac{\partial μ(y)}{\partial y}}{μ(y)} = -\frac{2}{y} \Rightarrow \ln(μ(y))=-2\ln(y) \Rightarrow μ(y) = \frac{1}{y^2}$$
So, now multiply both sides of the initial expression by $μ(y)$ :
$$-\frac{1}{y(x)} + \frac{(x+y(x)+2)\frac{dy(x)}{dx}}{y^2(x)}=0 \Rightarrow \frac{dg(y)}{dy} = \frac{y+2}{y^2} \Rightarrow g(y) = \int \frac{y+2}{y^2}dy = -\frac{2}{y} +\ln(y)$$
And finally, solving for $y$, will yield a not that common result, as the Product-log function will take part : 
$$y(x) = \frac{x+2}{W\bigg(\frac{x+2}{e^{c_1}}\bigg)} \Rightarrow y(x) = \frac{x+2}{W(c_1(x+2))}$$
Of course, if you are not familiar with the $W$ function, you may as well stop at the integration step, which is still considered a solution to the given ODE and is of course 100% correct.
Sampling a bit over initial values $y(0)$ for some trajectories, will yield you the following trajectory spectre for the family of the solutions : 
$\quad \quad \quad \quad \quad \quad \quad \quad$
A: To solve
$$
\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y}{x+y+2}
$$
we can substitute $u=x+2$:
$$
\frac{\mathrm{d}y}{\mathrm{d}u}=\frac{y}{u+y}
$$
Then simple manipulation gives $y\,\mathrm{d}u-u\,\mathrm{d}y=y\,\mathrm{d}y$; therefore,
$$
\begin{align}
\mathrm{d}\frac uy
&=\frac{y\,\mathrm{d}u-u\,\mathrm{d}y}{y^2}\\
&=\frac{\mathrm{d}y}y\\[6pt]
&=\mathrm{d}\log(y)
\end{align}
$$
Solving yields
$$
x+2=y(c+\log(y))
$$
