2nd order nonlinear ODE question I am looking for help to solve the following $F(x,y(x),y'(x),y''(x))=0$ equation:
$$
xy''(x)-y'(x)-(x^2)y(x)y'(x)=0
$$
Very much appreciated.
 A: The DE has an integrating factor $1/x^2$. That is, after multiplying by $1/x$, 
we get 
$$  \dfrac{y''}{x} - \dfrac{y'}{x^2} - y y' = 0 $$
which can be written as
$$ \dfrac{d}{dx} \left( \dfrac{y'}{x} - \dfrac{y^2}{2}\right) = 0$$ 
Integrate this:
$$ \dfrac{y'}{x} - \dfrac{y^2}{2} = C $$
 Now this is a separable equation:
$$ \dfrac{dy}{y^2 + 2C} = \dfrac{x \; dx}{2} $$
You might want to do the three cases $C > 0$, $C < 0$, $C = 0$ separately.
EDIT: How to find the integrating factor, you ask?  Actually I used Maple, but here's how one might do it by hand.
In general it's hard to find integrating factors, but if you assume the integrating factor is a function of $x$ alone, it's not too bad. The derivative of $a(x) y' + b(x,y)$ is $a(x) y'' + a'(x) y' + b_x(x,y) + b_y(x,y) y'$.
If this is to be $\mu(x) (x y'' - y' - x^2 y y')$, we need
$a(x) = \mu(x) x$, $b_x = 0$ (so $b(x,y) = b(y)$ and $b_y(x,y) = b'(y)$) and $\mu(x) (-1 + x^2 y) = a'(x) + b'(y)$.  From the last equation $b'(y) = c y$
where $\mu(x) x^2 = c$.  That gives  us the integrating factor up to an arbitrary constant. 
Of course most second order DE's won't have an integrating factor of this form, but enough do that this is a useful addition to one's bag of tricks.  
A: Solution to the DE;
$$-x^2 y'y+xy''-y'=0 \stackrel{\cdot \frac{1}{x^2}}{\iff} \frac{y''}{x}-\frac{y'}{x^2}-y'y=0 \stackrel{\mathrm{\int \mathrm{d}x}}{\iff}\frac{y'}{x}-\frac{y^2}{2}=c \iff \frac{2y'}{2c+y^2}=x \;\;\;\stackrel{\mathrm{\int \mathrm{d}x ,} \; s=\frac{y}{\sqrt{2c}}}{\iff}\;\;\; \frac{\sqrt{2} \tan^{-1} \left (\frac{y}{\sqrt{2c}}\right )}{\sqrt{c}}=\frac{x^2}{2}+c_1 \iff y=\sqrt{2c} \tan \left (\frac{\sqrt{c}(x^2+c_1)}{2\sqrt{2}}\right )$$
