Set $z=x-ct$ and $u(t,x)=Q_c(z)$. Substitution into the equation gives
$$
\partial_z(Q_c''-cQ_c+Q_c^p)=0.
$$
Decay at infinity makes the integration constant zero. If
$$
Q_c(x)=c^{1/(p-1)}Q(\sqrt c\,x),
$$
then every term in $Q_c''-cQ_c+Q_c^p$ equals $c^{1+1/(p-1)}$ times the corresponding term in $Q''-Q+Q^p$. Thus the required <Generalized KdV solitary wave> is
$$
\boxed{Q_c(x)=c^{1/(p-1)}Q(\sqrt c\,x)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 154 / 1 / 4 / Solution

{c}

If $Q''-Q+Q^p=0$, then
$$
Q_c(x)=c^{1/(p-1)}Q(\sqrt c\,x)
$$
satisfies $Q_c''-cQ_c+Q_c^p=0$, and $u(t,x)=Q_c(x-ct)$ is a traveling-wave solution of the <Generalized Korteweg–De Vries equation>.

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