The supplied identity says
$$
[L,A]\psi
=-3i\left(p_x\psi_x+(p_x\psi)_x\right)
+\left[q_{xxx}-4(q^2)_x\right]\psi.
$$
Expanding it and differentiating $L$ with respect to $t$ gives
$$
[L,A]\psi
=-6ip_x\psi_x
+\left[-3ip_{xx}+q_{xxx}-4(q^2)_x\right]\psi,
$$
$$
L_t\psi=-2iq_t\psi_x+(p_t-iq_{tx})\psi.
$$
Equating the coefficient of $\psi_x$ in <Lax's equation> $L_t=[L,A]$ yields
$$
q_t=3p_x.
$$
The imaginary scalar terms then cancel because $q_{tx}=3p_{xx}$, leaving
$$
p_t=q_{xxx}-4(q^2)_x.
$$
Differentiate the first equation in $t$ and the second in $x$ to eliminate $p$. The <third-order Boussinesq Lax pair> therefore produces
$$
q_{tt}=3(p_t)_x
=3q_{xxxx}-12(q^2)_{xx}.
$$
Thus the requested <Boussinesq equation> has
$$
\boxed{a=3,
\qquad b=-12.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 33A / b / ii / Solution

{title2=$L=iD^3-i(qD+Dq)+p,\quad A=3iD^2-4iq$}

The Lax equation $L_t=[L,A]$ for these operators is equivalent to
$$
q_t=3p_x,
\qquad
p_t=q_{xxx}-4(q^2)_x.
$$
Eliminating $p$ gives the <Boussinesq equation>
$$
q_{tt}=3q_{xxxx}-12(q^2)_{xx}.
$$

Third-order Boussinesq Lax pair

{c}
{wiki}

The Boussinesq equation is a nonlinear dispersive wave equation. One integrable normalization is
$$
q_{tt}=3q_{xxxx}-12(q^2)_{xx}.
$$


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