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For $u^{\prime }=F(x,u)$, a point-symmetry generator $V=\xi {\partial }_x+\eta {\partial }_u$ must satisfyIt follows by applying the first prolongation to $u^{\prime }-F=0$ and restricting to its solution manifold.

Let $\Delta (x,u,u^{\prime })=u^{\prime }-F(x,u)$. Applying the first prolongation givesOn the solution manifold $\Delta =0$, one has $u^{\prime }=F$. The infinitesimal symmetry criterion is therefore exactly