Solve $f ' (x) + f '' (x)/2 = \sqrt f(x)$ How to solve the differential equation
$$f ' (x) + f '' (x)/2 = \sqrt {f(x)}$$

Edit
My efforts
Assume $f(x) = a x^2 + b x + c$.
Then we plug this into the differential equation
$2 a x + b + a = \sqrt (a x^2 + b x + c)$.
I assume that $b = 2 \sqrt a \sqrt c$ So that i can simplify the RHS.
Therefore 
$2 a x + a + 2 \sqrt a \sqrt c = \sqrt a x + \sqrt c$.
When i solve this system of equations i get
$a= c = 0$
( trivial solution )
Or
$a = 1/4 $ and $c = \infty $
A pretty useless result.
( if $ a = 1/4 $ then $a + \sqrt c = \sqrt c $ hence the weird result ).
This attempt was not succesful.
 A: You should already know from my answer to your other question that the equation that does not contain independent variable admits a reduction of order. So once again, setting $\frac{df}{dx}=s$, $f=t$, we get $\frac{d^2f}{dx^2}=\frac{ds}{dt}\cdot\frac{dt}{dx}=s\frac{ds}{dt}$ and the equation becomes
$$\frac12 s\frac{ds}{dt}+s=\sqrt t$$
Further substitution $s=1/u$ gives
$$\frac{du}{dt}=2u^2\left(1-\sqrt t\, u\right).$$
This is a particular case of Abel's equation of the first kind. You can try to compute Abel's invariant with the hope to proceed by Chini method, although this does not seem promising. There are many papers on the web claiming having solved general Abel equation but none of them look serious to me.
A: Let $u=\dfrac{df}{dx}$ ,
Then $\dfrac{d^2f}{dx^2}=\dfrac{du}{dx}=\dfrac{du}{df}\dfrac{df}{dx}=u\dfrac{du}{df}$
$\therefore\dfrac{u}{2}\dfrac{du}{df}+u=\sqrt f$
This belongs to an Abel equation of the second kind.
Let $u=-2v$ ,
Then $\dfrac{du}{df}=-2\dfrac{dv}{df}$
$\therefore2v\dfrac{dv}{df}-2v=\sqrt f$
$v\dfrac{dv}{df}-v=\dfrac{\sqrt f}{2}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf
