Solve this nonlinear second order ODE: $yy'+xyy''+1=0$ Please could you help me  to solve the differential equation 
$$yy'+xyy''+1=0$$
with initial conditions $y(0)=1, y'(0)=-1$. Thank you in advance.
 A: From a quick glance, one can pretty much figure out that there is no closed form way leading to a result for this IVP. Hence, trying out the power series approach would be a good idea.
Let's assume a solution will be of the form : 
$$y(x) = \sum_{n=0}^\infty a_n(x-x_0)^n$$
In our case, the initial point for which we have to work around to find a solution, is the point $x_0=0$ as we can see from the initial values given. This means that : 
$$y(x) = \sum_{n=0}^\infty a_nx^n$$
Then, we'll have to find the derivatives of our summation, which follow as implicit differentiations :
$$y'(x) = \sum_{n=1}^\infty na_nx^{n-1}$$
$$y''(x) = \sum_{n=2}^\infty n(n-1)a_nx^{n-2}$$
You now need to find an expression or calculation for some of the constants (or all if possible) $a_n$. This will be done by plugging in the expressions above to your initial differential equation, since we have assumed that they are solutions. First, let's reform our initial given expression a bit to not get messed with summation products. For $y\neq 0$, it is :
$$yy' + xyy'' + 1 = 0 \Rightarrow y'+xy'' + y^{-1}=0$$
$$\Rightarrow$$
$$\sum_{n=1}^\infty na_nx^{n-1} + x\sum_{n=2}^\infty n(n-1)a_nx^{n-2} + \sum_{n=0}^\infty a_n^{-1}x^{-n}=0$$
All you have to do now, is reform the summations to match the same starting point ($n=2$ for example) and then, given that you have initial values for $y(x_0=0)=1, \space y'(x_0=0)=1$ plus the fact that $y\neq 0$, you can make conclusions about a part of the constants $a_n$ and thus approach/estimate a solution for your given IVP (left as exercise).
In a different case, sampling over initial values for $x=1$, we get the following trajectories sample plane (similar goes for $x=0$, this case just simplifies the expression and its characteristics a bit while also keeping the second term) :

A: Let me show how to lower the order of this ODE. Note that since the initial conditions are not consistent with the equation (as Daniel Gendin pointed out in the comments), I will consider a more general equation:
$$y \frac{dy}{dx}+xy\frac{d^2 y}{dx^2}+a=0$$
First let us make the change of variable:
$$x=e^t$$
$$\frac{dy}{dx}=e^{-t} \frac{dy}{dt}$$
$$\frac{d^2 y}{dx^2}=e^{-2t} \left( \frac{d^2y}{dt^2}-\frac{dy}{dt} \right)$$
Substituting into the equation, we obtain:
$$e^{-t} y \frac{dy}{dt}+e^{-t} y \left( \frac{d^2y}{dt^2}-\frac{dy}{dt} \right)+a=0$$
Or, after simplifications:
$$y \frac{d^2y}{dt^2}+a e^t=0$$
Now, let us introduce a new function:
$$y(t)=v(t) e^{t/2}$$
$$\frac{d^2y}{dt^2}=\left(\frac{d^2v}{dt^2}+\frac{dv}{dt} +\frac{v}{4} \right)e^{t/2}$$
Substituting into the equation, we obtain:
$$v \left(\frac{d^2v}{dt^2}+\frac{dv}{dt} +\frac{v}{4} \right)+a=0$$
As the variable disappeared from the equation, we can use the old trick:
$$u(v)=\frac{dv}{dt}$$
$$\frac{d^2v}{dt^2}=u \frac{du}{dv}$$
Finally we obtain:
$$v \left(u \frac{du}{dv}+u +\frac{v}{4} \right)+a=0$$
Or, assuming $v \neq 0$:

$$u\frac{du}{dv}=-u-\frac{v}{4}-\frac{a}{v}$$

This is Abel equation for the function "$-u$". See Table 1 in the provided link, which, unfortunately, doesn't deal with this particular case. See also Wikipedia which shows how to transform this into canonical form, and gives a reference to the paper, which constructs the solutions for the general case of Abel equations.
We can try some other ways, either an integrating factor (though I doubt it's simple to find) or the series solution as Rebellos proposes.
