Solution to this type of ODE? Is there a know solution to this ODE:
$$
x= f(x) +f'(x)a(x) + f''(x)b(x);\qquad f(0)=0?
$$
Even in the case where $a,b$ are constant; I don't know how to classify or solve this type of object.  (I have very little background in ODEs).
Thanks.
 A: If you are fine with power series expansions, it shall be solvable quite readily as both differentiation and multiplication with known power series are linear operations on polynomial spaces. 
But I don't know the name of this particular differential equation.
Assume we have a truncated power series expansion for $a(x), b(x)$.
The coefficients for $a(x) : c_{a0},c_{a1},\cdots c_{aN}$
The coefficients for $b(x) : c_{b0},c_{b1},\cdots c_{bN}$


*

*Build the Toeplitz matrices which represent multiplication by these. Call them $\bf M_a,M_b$

*Build the differential operator matrix $\bf D$. If you enumerate from $c_0$ and upwards this shall be: One off diagonal $[1,2,3,4,5\cdots,N]$

*Express everything as a linear least squares problem in this polynomial basis.
$$\min_{\bf v}\| ({\bf I+M_aD+M_b D^2)v} - {\bf d} \|_2^2$$
Where $\bf d$ is the power series expansion for the left hand side. With the vectorization of polynomial coefficients described above will be $[0,1,0,0,\cdots,0]^t$ 
(in other words this $\bf d$ represents the left hand side function $x = 0\cdot 1 +1\cdot x + 0\cdot x^2 + \cdots+ 0\cdot x^N$)
Now one thing remains, to ensure $f(0) = 0$, maybe you can finish by adding it?
