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)=0,\qquad f(1/2)=a,\, f^{\prime}(1/2)=\sqrt{2\pi b}, \qquad x\in(0,1).
\end{equation}
where $a\in\mathbb{R}$ and $b\in(0,\infty)$ are constants. 
I have tried the considering the following: Rewrite the $f^{\prime}$ term in order to obtain an equation of first order. But I get stuck in the substitutions as I do not know what to make of the $f$ term.
Any help or hint is greatly appreciated.
 A: First, $g(x):=f(x)-a$ gives us a simplification $$g''=g'g/b,\quad g(1/2) =0,\quad g'(1/2)=\sqrt {2\pi b}.$$
Next, change of variables $y=2x-1$, so the interval becomes $(-1,1)$, initial data is taken at zero.
$$h(y):=g((y+1)/2),\quad 4h''(y) = g'' ((y+1)/2),\quad \quad 2h'(y) = g' ((y+1)/2).$$
The equation becomes
$$h''(y)= \frac{h(y)h'(y)}{2b},\quad h(0) =0,\quad h'(0)=\frac{\sqrt {2\pi b}}{2}. $$
We can integrate this equation:
$$h'(y) = \frac{h^2}{4b}+C,\quad h(0) =0,\quad h'(0)=\frac{\sqrt {2\pi b}}{2}.$$
We find $C$ from initial conditions: $C= h'(0)=\frac{\sqrt {2\pi b}}{2}>0$.
At last, we obtain something of the first order. Once we recall that $\tan x'= \tan^2 x+1$, the further direction of reasoning is clear. We make another scaling of variables; if
$$h(y) = A\tan(By),$$then$$   h' (y) =   AB(\tan^2(By)+1)=  \frac BA h^2(y) +AB= \frac{h^2}{4b}+ \frac{\sqrt {2\pi b}}{2}.$$
All we have to do now is to solve $$\frac BA =\frac{1}{4b},$$
$$ AB=   \frac{\sqrt {2\pi b}}{2},$$
plug $A$, $B$ back to $h$ and  revert all our changes of variables and translations back to $f$.
A: If we think of $f$ as the independent variable (this step might take some justification), we can write $f'(x)=g(f)$. Then, by the chain rule, $f''(x) =\frac{dg}{dx} = \frac{dg}{df}\frac{df}{dx}$
Using this substitution, we can rewrite the DE as
$$
\frac{dg}{df}f'(x)-\frac{f-a}{b}f'(x)=0\\
\frac{dg}{df}=\frac{f-a}{b}
$$
This is a separable DE, which we can solve by integration. This gives us a solution for $g$. We can then use this to perform a substitution which will reduce the order of the original DE by 1. The resulting equation will also be separable, and can be solved by integration to find the general solution.
A: substitute values  of $f(\frac{1}{2})$ &$f'(\frac{1}{2})$ on the above equation  and you get $f''(\frac{1}{2}) = 0$.approximating using Taylor expansion for $f(x)$ centered at $x = \frac{1}{2}$ is,
$$f(x) = \frac{f(  \frac{1}{2} )}{0!} + \frac{f'(  \frac{1}{2} )}{1!}x + \frac{f''(  \frac{1}{2} )}{2!}x^2 $$
or $$f(x)_{x = \frac{1}{2}} \approx a + x\sqrt{2πb}$$ 
A: The solution by TZakrevskiy is nice. I will add that after the same change of variables, he suggests, to obtain:
$$ g'' = \frac{g}{2b} g'$$
with boundary conditions of:
$$ g(0)=0$$ and
$$ g'(0)=\frac{\sqrt{2 \pi b}}{2} $$
you can apply the general solution of that nonlinear ODE as given in the handbook by Polyanin and Zaitsev:
For a general equation of the form $g''=f(g)g'$ and $g=g(x)$ the solution is:
$$ \int \frac{dg}{F(g) + C_1}=C_2 + x $$
where $C_1$ and $C_2$ are the integration constants and $F(g)=\int f(g) dg$. Applying this will yield the same $tan$ based solution as noted above.
Cheers,
Paul Safier
