# Finding characteristic curves for pde's

A practice question says:

Find and sketch the characteristics for $2u_t + (1+t^2)u_x = 0$

So I found that the vector field for where $u$ is constant is $$(1 + t^2, 2)$$ and so I'm looking for a set of curves such that $$\frac{d}{d\tau}(x(\tau), t(\tau)) = (1+(t(\tau))^2, 2)$$ and I got that $$t(\tau) = 2\tau + t_0$$ and therefore $$x(\tau) = \tau + \frac{1}{6}(2\tau + t_0)^3 + x_0$$ but this seems awfully complicated to sketch?! And so I think I may have done something wrong?

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Just remove $\tau$ from both equations and you get
$$x(t)=x_0+\frac{1}{2}(t-t_0)+\frac{1}{6}t^3$$
and this is just a set of cubic curves parametrized by the two constants $t_0$ and $x_0$.
how can you just remove $\tau$ from both equations? –  user26069 May 12 '12 at 13:17
Invert $t(\tau)=2\tau+t_0$ to $\tau=\frac{1}{2}(t-t_0)$, put this into $x(\tau)$, and you are done. This gives the characteristic curves. –  Jon May 12 '12 at 13:20
In other words, the solution is the one-parameter family $(x_c)_c$ defined by $x_c(t)=\frac16t^3+\frac12t+c$. –  Did May 13 '12 at 8:32