= Solution
Write $u_j=u^{(j)}$. The transformation $\psi_s:(x,u)\mapsto(\widetilde x,\widetilde u)$ induces a transformation of the $n$th <jet space of a scalar ordinary differential equation> by differentiating $\widetilde u$ with respect to $\widetilde x$ through order $n$. The infinitesimal generator of this induced action is the <prolongation of a vector field>
$$
\boxed{
\operatorname{pr}^{(n)}V
=V+\sum_{j=1}^n\eta_j\frac{\partial}{\partial u_j}.}
$$
To determine its coefficients, put $\eta_0=\eta$ and use the <total derivative operator>
$$
D_x=\frac{\partial}{\partial x}
+u_1\frac{\partial}{\partial u}
+\sum_{j\geq1}u_{j+1}\frac{\partial}{\partial u_j}.
$$
Differentiating transformed derivatives by
$$
\widetilde u_j
=\frac{D_x\widetilde u_{j-1}}{D_x\widetilde x}
$$
and then differentiating with respect to the group parameter at $s=0$ gives the recursion
$$
\boxed{
\eta_j=D_x\eta_{j-1}-u_jD_x\xi,
\qquad j=1,\ldots,n.}
$$
This also proves the displayed form of the prolongation by mathematical induction on the derivative order.
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