Solution to the differential equation $\frac{f}{f^\prime}=\frac{f^\prime}{f^{\prime\prime}}$

I am wondering if anyone could please post the solution to the following differential equation for the function $f(x)$:

$$\frac{f}{f^\prime}=\frac{f^\prime}{f^{\prime\prime}}$$

Thanks!

-
In the context of differential equations it is extremely useful to memorize the particular chain rule $(\ln f)'=f\,'/f$ both forwards and backwards, because as it turns out you can use it backwards quite a lot. – anon Oct 30 '11 at 4:22
right. thanks again! – Qiang Li Oct 30 '11 at 4:34

$\rm\bf Start$: Multiply through by $f\,''/f$ and integrate with respect to $x$: $$\frac{f\,''}{f\,'}=\frac{f\,'}{f} \implies \ln (f\,')=\ln f+C.$$ Now exponentiate and solve another differential equation similarly...

$\rm\bf Finish$:

$$f\,'=e^Cf=Af\implies f(x)=Be^{Ax}.$$

-
Ah, thanks! Did not know it should be that easy! – Qiang Li Oct 30 '11 at 4:20

$$\frac{f}{f'} = \frac{f'}{f''} \qquad \Longleftrightarrow \qquad \frac{f'}{f} = \frac{f''}{f'} \qquad \Longleftrightarrow \qquad \int \frac{f'}{f} = \int\frac{f''}{f'} + C \qquad \Longleftrightarrow \qquad \ln f = \ln f' + C$$

Hence, taking exponentials on both sides,

$$f = K f' \ ,$$

where $K = e^C$. Renaming $K$ as $\frac{1}{K}$, this is the same as

$$\frac{f'}{f} = K \qquad \Longleftrightarrow \qquad \int \frac{f'}{f} = \int K + C \qquad \Longleftrightarrow \qquad \ln f = Kx + C \qquad \Longleftrightarrow \qquad f(x) = A e^{Kx}$$

where $A = e^C$.

-
There is a second solution with opposite sign for K (and maybe A?). – zyx Oct 30 '11 at 7:34