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Let $u_1, \ldots, u_n, u, v$ be functions. If $W(u, u_1, \ldots, u_n)=W(v, u_1, \ldots, u_n)$, is $u=v$? Here $W(u_1, u_2, u_3)$ for example is defined by \pmatrix{ u_1& u_1'& u_1''\\ u_2& u_2'& u_2''\\u_3& u_3'& u_3''}.

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    I don't think so (assuming you're talking about the determinants of those matrices). If all $u_i$ are $0$, then this obviously false, othertwise, pick $u_{i_0}$ a non zero element we have the $W(u+u_{i_0}, u_1, \ldots u_n) = W(u, u_1, \ldots u_n)$.2012-03-01

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No. Here's a counter-example: $W(2x+1, x) = \left| \begin{matrix} 2x+1 & 2 \\ x & 1 \end{matrix} \right| = 2x + 1 - 2x = 1,$ and $W(x+1, x) = \left| \begin{matrix} x+1 & 1 \\ x & 1 \end{matrix} \right| = x + 1 - x = 1.$ But $2x+1 \neq x+1.$

At least for $n =1$, W(u,u_1) = W(v,u_1) \Rightarrow u_1' (u-v)= u_1 ({u'-v'}), but not necessarily $u = v.$