Solve $\frac{f'(x)f'''(x)}{(f''(x))^2}=C$ Let $f$ be twice differentiable almost everywhere and continuous. $C$ is a constant.
Solve for the following differential equation:
$\frac{f'(x)f'''(x)}{(f''(x))^2}=C$ almost everywhere.
If we remove the "almost everywhere", the solution is pretty standard:
$f$ can be a log function, exponential function, or polynom. To be more specific, when $C=1$, it seems that $f(x)=ae^{bx}+c$ where $a,b,c$ are constants. When $C=2$, $f(x)=a\ln(x+b)+c$. When $c\neq 1$ or $2$, $f(x)=a(x+b)^c+d$
Can we get more solutions if we add back the "almost everywhere"?
 A: Updated answer after the OP changed the wording of his question.
$$\frac{f'(x)f'''(x)}{(f''(x))^2}=C$$
$y(x)=f'(x)$
$$\frac{y''}{y'}=C\frac{y'}{y}$$
$$\ln|y'|=C\ln|y[+\text{constant}$$
$$y'=c_1y^C$$
$$\frac{y'}{y^C}=c_1$$
$$\frac{y^{1-C}}{1-C}=c_1x+c_2\quad\text{in case of}\quad C\neq 1$$
$$y=(1-C)(c_1x+c_2)^{1/(1-C)}=f'$$
$$f(x)=\frac{(1-C)^2}{2-C}(c_1x+c_2)^{(2-C)/(1-C)}+c_3\qquad 
\begin{cases}
C\neq 0 \\
C\neq 1 \\
C\neq 2 
\end{cases}$$
or equivalently
$$f(x)=(ax+b)^{(2-C)/(1-C)}+c \qquad 
\begin{cases}
C\neq 0 \\
C\neq 1 \\
C\neq 2 
\end{cases}
\tag 1$$
$$f(x)=ax^2+bx+c \qquad C=0 \tag 2$$
You correctly found the solutions in the cases $C=1$ and $C=2$.
$$f(x)=ae^{bx}+c \qquad C=1 \tag 3$$
$$f(x)=a\ln(x+b)+c\qquad C=2 \tag 4$$
with $a\neq 0$ in all cases so that $f''(x)\neq 0$.
All above holds if $f(x),f'(x),f''(x)$ are differentiable everywhere.
If it is not the case this supposes that the PDE is not valid is some particular points but valid almost everywhere.  Then they are an infinity of continuous piecewise solutions : On each segment between those particular points $f'(x),f''(x),f'''(x)$ exist and $f(x)$ complies to the above equations $(1)$ or $(2)$ or $(3)$ or $(4)$. Two of the parameters can be determined so that there is no discontinuity between adjacent segments. So they are an infinity of continuous piecewise solutions
If no additional condition is specified at each particular point, the third parameter remains arbitrary and $f(x)$ is likely to be multivalued that is with different branches
