A differential equation introduced from a physics problem Try to solve the equation
\[
c_1 \sqrt{f(x)} + c_2f'(x) = c_3 \sqrt{f(x)} f''(x)
\]
holds for all $x \ge 0$.
There might be another condition: $f(0) = 0$.
It is introduced from a high school physics exam problem on $s, v, a$. The answer to the problem makes a hypothesis that the motion is uniformly accelerated motion and checks and says that it is true. It is equivalent to only check when $f(x) = (\alpha x + \beta)^2$ where $\alpha, \beta \ge 0$, then the equation becomes
\[
c_1 (\alpha x + \beta) + 2c_2 \alpha (\alpha x + \beta) = 2c_3 \alpha (\alpha x + \beta)
\]
and find a solution with $\alpha, \beta \ge 0$, saying proved. I don't think it's a rigorous proof.
I wonder whether the equation can be solved rigorously?
Thanks for help.
 A: The suggested solution is the right answer. This can be seen in the following way. The given equation
\[
c_1 \sqrt{f(x)} + c_2f'(x) = c_3 \sqrt{f(x)} f''(x)
\]
can be written down as
\[
c_1  + c_2\frac{d}{dx}\sqrt{f(x)} = c_3 f''(x)
\]
and so
\[
c_1  =\frac{d}{dx}\left[ c_3 f'(x)- c_2\sqrt{f(x)}\right]
\]
that can be immediately integrated to give
\[
c_1x+c_0  =c_3 f'(x)- c_2\sqrt{f(x)}.
\]
It is straightforward matter to verify that the solution to this equation has the form $f(x)=(\alpha x+\beta)^2$ as stated.
A: I'm no expert on proving uniqueness of solutions to nonlinear ODEs, but consider the following:
Your equation can be trivially integrated if either $c_2=0$ or $c_3=0$, so we will assume that neither is the case.  Then if we define
$$
h(x)=\frac{c_3}{c_2}\sqrt{f(x)}
$$
we may rewrite the equation as
$$
\phantom{(*)}\qquad\qquad\frac{d^2}{dx^2}\left[h(x)\right]^2-2\frac{d}{dx}h(x)-k=0\qquad\qquad(*)
$$
where 
$$
k=\frac{c_1c_3}{c_2^2}.
$$
If we now assume that $h(x)$ is analytic at $x=0$ so that we may write
$$
h(x)=h_0+h_1x+h_2x^2+h_3x^3+\ldots
$$
then we can prove inductively, by repeatedly differentiating (*) and evaluating at $x=0$, that


*

*if $h_0=0$ then $h_2=h_3=h_4=\ldots=0$ and $\displaystyle h_1=\frac{1\pm\sqrt{1+2k}}{2}$,

*if $h_0\ne0$ and $h_2=0$ then $h_3=h_4\ldots=0$ and, again, $\displaystyle h_1=\frac{1\pm\sqrt{1+2k}}{2}$.


Therefore in either of these situations $h(x)$ is a polynomial of degree 1 and $f(x)$ is the square of this polynomial.
If $h_0\ne0$ then we may compute $h_2$, $h_3$, $h_4,\ldots$ iteratively in terms of $h_0$ and $h_1$:
$$
\begin{aligned}
h_2&=\frac{2h_1^2-2h_1-k}{4h_0}\\
h_3&=\frac{(2h_1^2-2h_1-k)(3h_1-1)}{12h_0^2}=h_2\cdot\frac{3h_1-1}{3h_0}\\
h_4&=\frac{(2h_1^2-2h_1-k)(30h_1^2-20h_1+2-3k)}{96h_0^3}=h_2\cdot\frac{30h_1^2-20h_1+2-3k}{24h_0^2}\\
&\vdots
\end{aligned}
$$
These computations are consistent with point (2) above.
