= Solution
Use $\dot{\mathbf x}=\dot b\,\mathbf u$ and $\varphi=Cb\psi/a$, where the growing <linear growth factor> obeys $\ddot b+2H\dot b=Cb/a^3$. The Poisson relation in part (a) immediately becomes
$$
\boxed{\nabla_x^2\psi=\delta/b.}
$$
Along a fluid trajectory,
$$
\ddot{\mathbf x}=\ddot b\,\mathbf u+\dot b^2D_b\mathbf u,\qquad
D_b=\partial_b+\mathbf u\cdot\nabla_x.
$$
Substitute this into the pressureless equation of motion to obtain
$$
\dot b^2D_b\mathbf u+(\ddot b+2H\dot b)\mathbf u
=-\frac{Cb}{a^3}\nabla_x\psi.
$$
The growth equation cancels the second coefficient. By the definition of the logarithmic growth rate, $\dot b=Hbf$, while $C/a^3=(3/2)H^2\Omega(t)$. Therefore
$$
\boxed{D_b\mathbf u=-\frac{3\Omega(t)}{2bf^2}(\mathbf u+\nabla_x\psi).}
$$
This is the printed $d\mathbf u/db$ when interpreted along the particle. Dividing the Eulerian continuity equation by $\dot b$ instead gives
$$
\boxed{\partial_b\delta+\nabla_x\cdot[(1+\delta)\mathbf u]=0.}
$$
The density derivative in this conservative equation must be Eulerian. If both derivatives were read as material derivatives, the density equation would double-count advection; its material form is $D_b\delta+(1+\delta)\nabla_x\cdot\mathbf u=0$. These distinctions make the <growth-time form of cosmological dust equations> valid beyond the first-order terms, up to the single-stream pressureless-fluid limit. The reparameterization assumes $\dot b\ne0$.
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