Rudin Chapter 5 #29, getting started

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Homework Statement



Specialize exercise 28 by considering the system

\ y&#039;= y_{j+1} j=(1,...,k-1)<br /> y&#039;_{k}= f(x)-\sum g_{j}(x)y_{j}

where the summation runs from j=1 to j=k, and \g_{j} and f are continuous real functions on [a,b], and derive a uniqueness theorem for solutions of the equation

\ y^{k}+g_{k}(x)y^{k-1}+...+g_{2}y&#039;+g_{1}(x)y = f(x)

subject to initial conditions

\ y(a)=c_{1}, y&#039;(a)= c_{2}, y^{k-1}(a) = c_{k}.

here the kth power denotes the "kth" derivative.

Homework Equations



The problem is supposed to be similar to 26, 27, and 28 in Rudin which we also worked through on this problem set. The idea of those problems is to show the uniqueness of a solution to a system of differentiable equations, by defining some other function as the difference between two possible solutions and applying the result we proved in 26, if f is differentiable on [a,b], f(a)=0, and there exists a real number A such that \left|f&#039;(x)\right| \leq A\left|f(x)\right|, then f(x)=0.

The Attempt at a Solution



I'm sort of confused on exactly what this problem is asking... like, what does the polynomial expression mean here, and what might it's solutions look like? Like if I'm understanding the question we basically have a vector where each component is one of the derivative of that before it except for the final one; so I think we should be able to apply a component wise version of the uniqueness theorem proved in 27/28 (28 being the extension to vectors by operating on each component), and saying if there is an A that bounds two solution vectors like this, the result follows.

What I'm not understanding is how this relates to the final expression and why I care about the g(x)'s.
 
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Ok, so I think I might have an idea of how to the problem. Can someone explain to me where that last polynomial expression comes from though?
 


Grr... this is confusing me a lot. Have I left out any important information someone needs to explain or something?
 
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