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I’m in need of some help to understand an elementary proof involving the Wronskian of $n$ vector functions and the following existence and uniqueness theorem.

Existence and Uniqueness

Suppose $A(t)$ and $f(t)$ are continuous on an open interval $I$ that contains the point $t_0$. Then, for any choice of initial vector $x_0$, there exists a unique solution $x(t)$ on the whole interval $I$ to the initial value problem

$$x’(t) = A(t)x(t) + f(t) \space \space , \space \space x(t_0) = x_0$$

The proof I’m reading is about how if $x_1, \space \ldots \space, x_n$ are linearly independent solutions on $I$ then the Wronskian is never zero. So I understand that if the Wronskian is zero this implies that the vector functions are linearly dependent which implies that $$c_1 x_1(t_0) \space + \space \ldots \space + \space c_n x_n(t_0) = 0$$

At some point $t_0$ (ie a non-trivial solution exists at some point). Now here is what my textbook goes on to say:

… However, $c_1 x_1(t_0) \space + \space \ldots \space + \space c_n x_n(t_0) = 0$ and the vector function $z(t) \equiv 0$ are both solutions to $x’ = Ax$ on I, and they agree at the point $t_0$. So these solutions must be identical on $I$ according to the existence and uniqueness theorem (What!?).

So this may be obvious to some, but I’m having a hard time understanding what the existence and uniqueness theorem means in this context and how it proves that it is zero for all $t$. Any explanation would be appreciated.

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You have two solutions, one is the linear combination $\sum_i c_ix_i$ the other is the nulfunction $z\equiv 0$. Both of these solutions are equal to $0$ for the point $t_0$.

Uniqueness of the solution then tells you that both have to be the same otherwise there would be two different solutions satisfying the same differential equation and initial condition in $t_0$.

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    $\begingroup$ This makes sense, and I like the explanation, thanks! $\endgroup$ – Amateur Math Guy Mar 21 '13 at 11:54

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