Difficult nonlinear ODE of second order. I have problems finding out whether this initial value problem has an explicit form solution or if it is possible to grind out a term-by-term representation of this solution using power series expansions.
\begin{equation}f^{\prime\prime}(x)-\frac{f(x)-a}{b}f^{\prime}(x)^2=0,\quad x\in(0,1),\qquad f(1/2)=a,\,f^{\prime}(1/2)=\sqrt{2\pi b}.\end{equation}
where $a\in\mathbb{R}$ and $b>0$ are constants.
Attempt: Find a solution for the easier problem, \begin{equation}f^{\prime\prime}(x)-\frac{f(x)-a}{b}f^{\prime}(x)=0,\quad x\in(0,1),\qquad f(1/2)=a,\,f^{\prime}(1/2)=\sqrt{2\pi b}.\end{equation}
where $a\in\mathbb{R}$ and $b>0$ are constants. 
The solution for the easy problem is\begin{equation}f(x)=a+\tan\left[\frac{\sqrt{2b\sqrt{2\pi b}}(x-1/2)}{2b}\right]\sqrt{2b\sqrt{2\pi b}}
\end{equation}
Now, solve the original system with the $f^{\prime}(x)^2$ replacing the easier $f^{\prime}(x)$. However, I can't make sense of substituting the squared term into my calculations... 
Any help or hint is greatly appreciated.
 A: Let us make the change of variables:
$$g(x)=\frac{f(x)-a}{\sqrt{b}},$$
which implies $g'(x)=f'(x)/\sqrt{b}$, $g''(x)=f''(x)/\sqrt{b}$. The equation for $g(x)$ is
$$g''(x)-g(x)g'(x)^2=0.$$


*

*Dividing this equation by $g'(x)$, we find
$$\frac{g''(x)}{g'(x)}-g(x)g'(x)=\left(\ln g'(x)-\frac12 g(x)^2\right)'=0.$$
Hence
$$\ln g'(x)-\frac12 g(x)^2=\mathrm{const}.\tag{1}$$

*The constant can be fixed using the initial conditions $g(1/2)=0$, $g'(1/2)=\sqrt{2\pi}$. Substituting these values in (1), one finds that
$$\ln g'(x)-\frac12 g(x)^2=\ln\sqrt{2\pi},$$
or, in another form
$$g'(x)=\sqrt{2\pi}\,e^{g(x)^2/2}.\tag{2}$$

*The equation (2) is separable. In particular, it implies that
$$\frac{1}{\sqrt{2\pi}}\int_{g(1/2)}^{g(x)}e^{-h^2/2}dh=\int_{1/2}^xdt.$$
Applying the initial conditions once more, one obtains
$$\frac{1}{\sqrt{2\pi}}\int_{0}^{g(x)}e^{-h^2/2}dh=x-\frac12.$$

*The integral on the left is expressed in terms of the error function:
$$\frac12\operatorname{erf}\frac{g(x)}{\sqrt{2}}=x-\frac12.$$
Hence 
$g(x)=\sqrt{2}\operatorname{erf}^{-1}(2x-1)$,
and finally
$$f(x)=a+\sqrt{2b}\operatorname{erf}^{-1}(2x-1). \tag{3}$$
The answer (3) is given in terms of special function $\operatorname{erf}^{-1}$, but one can, for example, write it in the form of a series (around $x=1/2$) with recursively determined coefficients, see e.g. here.
