= Solution
Since $Eu'+Fv'=\gamma'\cdot r_u$, differentiation gives
$$
\frac d{ds}(Eu'+Fv')=\gamma''\cdot r_u+\frac12E_u(u')^2+F_uu'v'+\frac12G_u(v')^2.
$$
The corresponding calculation with $\gamma'\cdot r_v=Fu'+Gv'$ gives
$$
\frac d{ds}(Fu'+Gv')=\gamma''\cdot r_v+\frac12E_v(u')^2+F_vu'v'+\frac12G_v(v')^2.
$$
Thus the two requested equations hold exactly when $\gamma''$ is perpendicular to both tangent <basis> vectors, which is precisely the <geodesic> condition.
On expanding the left sides, the coefficient matrix of $(u'',v'')$ is the positive-definite metric matrix. It is invertible, so the equations form a smooth second-order <ordinary differential equation> in local coordinates. The local existence and uniqueness theorem gives a unique solution for each initial point and coordinate velocity. For an arbitrary initial tangent vector $w$ this is an affine parameter of constant speed $|w|$; it is arclength when $|w|=1$. For $w=0$ the solution is the constant curve. This specifies how the arbitrary-velocity statement relates to the preceding unit-speed convention.
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