EigenFunction for $\frac{\partial f}{\partial t}+f\frac{\partial f}{\partial x} =\frac{2f^2}{x}$ When studying a computer vision problem I end up with a function $f(x,t)$ that satisfying  $\frac{\partial f}{\partial t}+f\frac{\partial f}{\partial x} =\frac{2f^2}{x}$.
My question includes two parts:


*

*What are the solutions to the above equation in general?  I worked out three specific solutions:  (1) $ f(x,t)= \frac{x}{k-t}$. (2) $ f(x,t)= kx^2$. (3) $ f(x,t)= \frac{kx^2-x}{b+t}$ (k and b are constants). I have no idea whether there is other solutions.

*To resolve the equation in another point view,  we can define an operator $\Omega f=\frac{\partial f}{\partial x} - \frac{2f}{x}$. Suppose we can find the eigenfunctions of the operator $\Omega$, namly $\Omega f=\lambda f $ ($\lambda$ is constant), so that any solution $f$ can be represented by linear sum of these eigenfunctions (very similar to Schrödinger equation). 
Unfortunately, no more functions other than the three above are found, though I expect that there should be some more complicated eigenfunctions, such as Fourier series. Now I am looking into wavelet theories to find a new clue.
Any suggestions on the general solution to $\frac{\partial f}{\partial t}+f\frac{\partial f}{\partial x} =\frac{2f^2}{x}$, or to the eigenfunctions for operator $\Omega f=\frac{\partial f}{\partial x} - \frac{2f}{x}$?
 A: Set $f(x,t)=X(x)T(t)$, then 
$$\frac{\partial f}{\partial t}+f\frac{\partial f}{\partial x} =\frac{2f^2}{x}\tag{1}$$
becomes
$$X(x)\frac{\partial T(t)}{\partial t}+T^2(t)X(x)\frac{\partial X(x)}{\partial x} =2T^2(t)\frac{X^2(x)}{x}\tag{2}$$
Divide $T^2(t)X(x)$ on both sides of (2) leads to:
$$T^{-2}(t)\frac{\partial T(t)}{\partial t}+\frac{\partial X(x)}{\partial x} =2\frac{X(x)}{x}\tag{3}$$
or 
$$T^{-2}(t)\frac{\partial T(t)}{\partial t}=\lambda=-\frac{\partial X(x)}{\partial x} +2\frac{X(x)}{x}\tag{3}$$
The solutions to (3) are
$$T(t)=-\frac{1}{\lambda t+a}\tag{4}$$
$$X(x)=\lambda x+b x^2\tag{5}$$
where $a,b$ are integration constants.
A: $\dfrac{\partial f}{\partial t}+f\dfrac{\partial f}{\partial x}=\dfrac{2f^2}{x}$
$\dfrac{x}{f}\dfrac{\partial f}{\partial t}+x\dfrac{\partial f}{\partial x}=2f$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{ds}=x$ , letting $x(0)=1$ , we have $x=e^s$
$\dfrac{df}{ds}=2f$ , letting $f(0)=f_0$ , we have $f=f_0e^{2s}=f_0x^2$
$\dfrac{dt}{ds}=\dfrac{x}{f}=\dfrac{e^{-s}}{f_0}$ , we have $t=F(f_0)-\dfrac{e^{-s}}{f_0}=F\left(\dfrac{f}{x^2}\right)-\dfrac{x}{f}$
A: Thanks to doraemonpaul and mike, I have applied the method of characteristics to the equation as follows:
$$\frac{\partial f}{\partial t}+f\frac{\partial f}{\partial x} =\frac{2f^2}{x}$$
$$\frac{x}{f}\frac{\partial f}{\partial t}+x\frac{\partial f}{\partial x} =2f$$
with the initial condition $f(x,0)=\Phi(x)$ (i.e. $t(r,0)=0$, $x(r,0)=r$, $f(r,0)= \Phi(r)$), then 
$$\frac{\partial t}{\partial s}(r,s)=\frac{x}{f}$$
$$\frac{\partial x}{\partial s}(r,s)=x$$
$$\frac{\partial f}{\partial s}(r,s)=2f$$
the solutions are:
$$t(r,s)=\frac{r}{\Phi(r)}(1-\mathrm{e}^{-s})$$
$$x(r,s)=r\mathrm{e}^{s}$$
$$f(r,s)=\Phi(r)\mathrm{e}^{2s}$$
canceling $s$ leads to
$$\Phi(r)tx-rx+r^2=0$$
$$f=\Phi(r)\frac{x^2}{r^2}$$
My observation is that 
1) the relationship between $f$,  $x$,  $t$ is implicitly established, while the variable $r$ cannot be easily reduced to give an explicit equation. 
2) this result is insistent with the result $f(x,t)=-\frac{\lambda x+bx^2}{\lambda t+a}$ by assuming $f=T(t)X(x)$, because the couple of equations above lead to the same form by set $\Phi(r) = ar+br^2$. 
Doraemonpaul, I am not very sure whether this result is the same as yours, since I cannot understand the term $F()$ in your result.
