How can I solve $ yy''+y'+1=0$? I know that a particular solution for this equation is $y=-x$.
So, I have tried to find another solution with $y=xf(x)$ but I have encountered an equation even more difficult.
I also have attempted to make $u=y'$ and $u=yy'$ but it was not successful.
 A: $$yy''+y'+1=0$$
$$yy'''+y'y''=-y''$$
$$yy'''=-y''(y'+1)$$
$$yy'''=y''yy''$$
$$y'''=y''^2$$
Let $u=y''$
$$u'=u^2$$
$$\frac{\mathrm dx }{\mathrm du}=\frac1{u^2}$$
$$x=-\frac{1}{u}+b$$
$$u=-\frac{1}{x-b}$$
Can you take it from here?

If no, here's the rest.
$$y''=-\frac{1}{x-b}$$
$$y'=-\ln (x-b)+c$$
$$y=-(x-b)\ln (x-b)+(x-b)+cx+d$$
$$y=-(x-b)\ln (x-b)+Cx+D$$
When we put this back to the original ODE, we have $D=-bC$, thus our final answer is,
$$y=-(x-b)\ln (x-b)+Cx-bC$$
A: Let us introduce a new variable $p= y'$, then $y''= p'$ and our equation becomes $yp'+p+1=0$, so our equation is transformed to the following system:
$$\left\{\matrix{\frac{dy}{dx}&=p\hfill\cr\frac{dp}{dx}&=-\frac{1+p}{y}}\right.$$
Now, if we consider $y$ as function of $p$ we see that
$$\frac{dy}{dp}=\frac{dy}{dx}\left/\frac{dp}{dx}\right.=-\frac{py}{1+p}
$$
And this gives us
$$\frac1y\frac{dy}{dp}=-\frac{p}{1+p}=\frac{1}{1+p}-1$$
Hence
$$ y(p)=k(1+p)e^{-p}\tag1$$
Also we have
$$\frac{dx}{dp}=1\left/\frac{dp}{dx}\right.=-\frac{y}{1+p}=-ke^{-p}$$
Thus
$$x(p)=ke^{-p}+\ell\tag2$$
Eleminating $p$ we get the general solution
$$y(x)=(x-\ell)\ln\frac{k}{x-\ell}+x-\ell$$
