how to solve this nonlinear partial differential equation?

How to solve this nonlinear partial differential equation? $$\displaystyle\frac{\partial^2}{\partial x^2} f(x,t) +b \frac{\partial^2}{\partial x^2} f(x,t) \cdot \frac{\partial^2}{\partial t^2} f(x,t) + a = 0,$$ where $a,b$ are constant.

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Can someone experienced in LaTeX edit this to use proper format? – Alex Becker Mar 26 '11 at 5:11
Does $\left(\frac{\partial}{\partial x}\right)^2$ mean the second partial? – Arturo Magidin Mar 26 '11 at 5:14
@Theo: If it is inside $$...$$, it's already in displaystyle... – Arturo Magidin Mar 26 '11 at 5:15
@aniket: I tried to texify your source, I hope I haven't messed it up. @Arturo: my mistake I clicked on submit too early and thanks for the remark about \displaystyle. The original source said (d/dx)^2. – t.b. Mar 26 '11 at 5:16
thanks theo for latex – user8143 Mar 26 '11 at 5:29

Also it is possible to look for traveling waves solutions $f(x,t)=g(x-сt)\,$. It leads to ODE $$g''(x)+bc^2(g''(x))^2+a=0$$ solutions of which can be written down explicitly: $$g(x)=C_1+C_2 x+\frac{1\pm\sqrt{1-4 a b c^2}}{4 b c^2}x^2.$$
You might find some solutions by making the ansatz $f(x,t) = g(x) + h(t)$, i.e., that the solution separates. The equation then reads $$g''(x) + b \, g''(x) h''(t) +a =0,$$ $$g''(x) [1+ b \,h''(t)] = -a.$$ This equation can be solved by setting $g''(x)=c$ and $1+ b \,h''(t) = -a/c$. The solutions read $$g(x) = \frac{c}{2} x^2 + c_1 x + c_2$$ and $$h(t) = -\frac{a+c}{2 b c} t^2 +C_1 t + C_2$$ with $c\neq0$, $c_1$, $c_2$, $C_1$, $C_2$ arbitrary constants.