What method to use for differential equations I'm having doubts as to what method to use in the following ODE: 
$$2t+3x+(x+2)x'=0$$
As this can be changed into: 
$$x'=\frac{-2t-3x}{x+2}$$
I'm thinking it can be solved by using homogeneous equations method, but I'm not sure this applies because for the $x+2$ in the denominator I'm not sure it would be a homogeneous equation (grade 0 and all). 
What would you suggest? Thanks!
 A: $$2t+3x+(x+2)\frac{dx}{dt}=0$$
By inspection one see that : $\quad 2t+3x=2(t-3)+3(x+2)$
$$2(t-3)+3(x+2)+(x+2)\frac{dx}{dt}=0$$
This draw us to the change of variables : $\quad\begin{cases}
X=x+2 \\
T=t-3
\end{cases}\quad$ thus $\quad \frac{dX}{dT}= \frac{dx}{dt}$.
$$2T+3X+X\frac{dX}{dT}=0\quad\implies\quad \frac{dX}{dT}=-\frac{2T+3X}{X}$$
This is an homogeneous ODE. The usual change of function is :
$$X(T)=T\:F(T)$$
$X'=F+TF'=-\frac{2T+3TF}{TF}$
$$T\frac{dF}{dT}=-\frac{(F+1)(F+2)}{F}$$
This ODE is separable. Solving (easy, but rather boring) is for you.
A: Compressing the d'Alembert treatment: Insert $p=x'$, then 
$$
2t+3x+(x+2)p=0.
$$
If $p$ is constant, then the $t$ derivative of this equation gives
$$
2+3p+p^2=0\implies p=-1\text{  or  } p=-2.
$$
In the other cases, locally use $p$ as parameter and use $T(p)$, $X(p)$ as the dependent functions. Then from the chain rule $X'(p)=pT'(p)$ and from the equation
$$
0=2T'+3X'+(X+2)+pX'\implies 0=(2+3p+p^2)X'+p(X+2)
$$
which is separable
$$
\frac{X'}{X+2}=\frac{p}{(p+1)(p+2)}=\frac{2}{p+2}-\frac1{p+1}
\implies
X+2=\frac{C(p+2)^2}{p+1}
$$
Inserting this back into the original equation gives a parametrization of the solution curves with
$$
2T=6-\frac{C(p+3)(p+2)^2}{p+1}.
$$
