Question: Consider the PDE $u_t + u^3 u_x = u^2$, $x \in \mathbb{R}$, $t>0$. Find a parametric solution to the PDE.
My attempt: We use the method of characteristics. The characteristic equations are $\frac{dt}{ds}=1$, $\frac{dx}{ds}=u^3$, $\frac{du}{ds}=u^2$. Hence we obtain $\frac{dx}{du}=\frac{u^3}{u^2}=u$ so $x=\frac{u^2}{2}+c$. Let us solve the PDE with the initial condition $u(x,0)=f(x)$. Then we have $x_0=\xi$, $t_0=0$, $u_0=f(\xi)$ so $x=\frac{u^2}{2}+\xi-\frac{f(\xi)^2}{2}$.
This is where I have reached so far. I am unable to eliminate the parameter $\xi$. My attempts thus far have produced $u^ 2= 2x - 2\xi + f(\xi)^2$ so $\frac{du}{ds} = u^2 = 2x - 2\xi + f(\xi)^2$ but I am unable to proceed further as I am unable to find an expression for $x$ in terms of $s, \xi$, to help me solve this ODE.
Any help would be greatly appreciated! Thank you for your time.
Hint:
\begin{align} \frac{\mathrm d t}{\mathrm du} &= \frac1{u^2} &\implies t &= -\frac1u + a\\ \frac{\mathrm d x}{\mathrm du} &= u& \implies x &= \frac12 u^2 + b\\ \end{align}
So, $$x = \frac12\times\frac1{(a-t)^2} + b$$
Can you finish the solution?