Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 32D / a / Solution

The transformations obey $g_1^s{g}_1^r=g_1^{s+r}$ and $g_1^0$ is the identity. If $\tilde{x}(\tilde{t})=x(t)+st$, then $d^2\tilde{x}/d{\tilde{t}}^2=d^{2}x/d{t}^2$. Thus solutions of the free-particle equation are mapped to solutions, so this is a one-parameter group of Lie symmetries.