First order ODE: $y^2+2yy'x+2xy'+y=0$ $$y^2+2yy'x+2xy'+y=0$$
I really have no idea how to do this, I cant fit it into any of the schemes I already know. Also nothing factors out. Maybe should I try differentiating both sides?
 A: Hint; 
$$ y(y + 1) + 2xy'(y + 1) = 0 \iff (y + 1)( y + 2xy' ) = 0  $$ 
Another Hint: 
Linear Equations. 
A: Here's an answer from group theory.
Use a Lie group $$G(x,y)=(\lambda X,\lambda^\beta Y)\lambda_o=1$$
Since $x=\lambda X$, $y=\lambda^\beta Y$, and $y'=\frac{dy}{dx}=\frac{\lambda^\beta Y}{\lambda X}=\lambda^{\beta -1}\frac{dY}{dX}=\lambda^{\beta -1}Y'$, we substitute these into the original ODE and we get
$$
\lambda^{2\beta} Y^2+2\lambda^\beta Y \lambda^{\beta -1} Y' \lambda X+2\lambda X \lambda^{\beta -1}Y' +\lambda^\beta Y=0
$$For invariance, the $\lambda$ coefficients must cancel out.  Therefore, 
$2\beta=\beta=0$, or $\beta=0$.  So the invariant group to your DEQ is 
$x=\lambda X$ and $y=Y$.  Now take the derivative W.r.t. $\lambda$ and set $\lambda=\lambda_o=1$ and you get
$$
\frac{\partial x}{\partial \lambda}\bigg|_{\lambda=1}=X
$$
$$
\frac{\partial y}{\partial \lambda}\bigg|_{\lambda=1}=0
$$
$$
\frac{\partial y'}{\partial \lambda}\bigg|_{\lambda=1}=-Y
$$If we drop the capital letters, the characteristic equation is
$$
\frac{dx}{x}=\frac{dy}{0}=\frac{dy'}{-y}
$$Two independent integrals of this equation are the stabilizers of our invariant group.
$$
\mu=y
$$
$$
\nu=xy'
$$Rewriting the ODE in terms of the stabilizers,
$$
\mu^2+2\nu \mu+2\nu+\mu=0
$$This simplifies to 
$$
\nu=-\frac{\mu}{2}
$$Now we note that 
$$
x\frac{d\mu}{dx}=\nu=-\frac{\mu}{2}
$$which separates to
$$
\frac{d\mu}{\mu}=-\frac{1}{2}\frac{dx}{x}
$$and this integrates to 
$$
ln\mu=lnx^{-\frac{1}{2}}+lnC
$$or
$$
\mu=y=\frac{C}{\sqrt{x}}
$$This is your answer.  Also, note that there is a special answer to be found by using the autodiffeomorphism between the singularities and saddle points in the direction field of the stabilizers where the relation is $\nu=\beta \mu=0$.  Substituting this into the DEQ yields
$$
\mu^2+\mu=0
$$Eliminating the trivial solution of $\mu=y=0$, we have $\mu=-1$ which is also a solution.
Of course, there are easier ways of finding these solutions, but very few people know anything about how to solve DEQ's using group theory so I like to "beat the drum" when I can.
A: If you write equation in this form it is more clear:
$2x(1+y)dy+(y+y^{2})dx=0$
$\frac{(1+y)}{y(1+y)}dy+\frac{1}{2x}dx=0$
$lny+\frac{1}{2}lnx=C$
$y=\frac{A}{\sqrt{x}}$
$y=-1$ could be a particular solution.
