Nonlineary differential equation I have following problem:

let's assume this nonlinear differential equation
$$y''(x)+ay(x)+by'(x)^2=0$$
for unknown function $y(x)$ of one variable $x$ and $a,b \in \mathbb{R}$ are real parameters. Is there any method to find general solution? I can only find solution for $b=0$ as it is then linear ode.

P.S. Does this equation have a special name?
 A: Nonlinear ordinary differential equations of the form $y^{\prime\prime} = f(y,y^{\prime})$ where $x$ does not explicitly appear on the right hand side can sometimes be solved by the following method.
Let $u = y^{\prime}$, then 
$$y^{\prime\prime} = \frac{du}{dx} = \frac{du}{dy}\frac{dy}{dx} = u\frac{du}{dy}$$
Applying this to the above yields the equation
$$u\frac{du}{dy} = f(y,v)$$
which is of first order.
In your particular case this yields the equation
$$u\frac{du}{dy} = ay + bu^2$$
or in another form
$$(ay + bu^2)dy - udu = 0$$
In this case, we are lucky, there is an obvious integrating factor $e^{-2by}$,
multiplying by this factor yields
$$(ay + bu^2)e^{-2by}dy - ue^{-2by}du = 0$$
one can check that this leaves us with an exact equation. From here you can use the known solution to exact differential equations to solve for $u$. Recovering the solution for $y$ then amounts to solving the equation $y^{\prime} = u(y)$, which is separable.
A: Here is a line of attack that is sometimes useful for certain non-linear ODE's with no explcit coordinate dependence.
Put $u = y'$ then by using $y'' = u' = \frac{du}{dy}\frac{dy}{dx}$ we find
$$\frac{du}{dy}u + b u^2 = -ay$$
which has the solution
$$u^2 = -e^{-2by}\int2aye^{2by}dy = \frac{a^2}{b^2}-\frac{2a^2y}{b} + C e^{-2by}$$
where $C$ is an integration constant. Now we are left with
$$y' = \sqrt{\frac{a^2}{b^2}-\frac{2a^2y}{b} + C e^{-2by}}$$
which has the formal implicit solution
$$\int \frac{dy}{\sqrt{\frac{a^2}{b^2}-\frac{2a^2y}{b} + C e^{-2by}}} = x$$
which is doubt can be solved in general. However for certain initial conditions (dictated by the value of $C$ such as $C=0$) there might be nice solution. 
