A step in a proof of linear dependence of ODE solutions

  • #1
Hill
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How do we know that the factors are constant?
Please consider the following step in a proof:

1715999934788.png


After transposing the matrix, its coefficients still are functions of ##x##. Why then the solution ##a, a_1, a_2## is constant?
 
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  • #2
The excerpt above is from the textbook, Leslie Copley, Mathematics for the Physical Sciences.
Does it skip a step showing that ##a, a_1, a_2## are constants, or rather I miss it?
 
  • #3
I think the point is that for any fixed x the functions are just numbers.
 
  • #4
Office_Shredder said:
I think the point is that for any fixed x the functions are just numbers.
That is what I think, too. Thus, for any fixed ##x## there is a non-trivial solution ##a, a_1, a_2##. But these solutions can be different for different ##x##. Then, the functions ##y, y_1, y_2## are not necessarily linearly dependent. This is why I think that this proof does not prove that the functions are linearly dependent.
 
  • #5
I'm not sure this is a proof, or if it is then it requires more explanation than the author has given.

Once you have an existecne and uniqueness proof for the IVP, you can set up a linear map between the space of solutions of the homogenous ODE and the space of initial conditions ([itex](y(x_0), y'(x_0)) \in \mathbb{R}^2[/itex]). The existence and uniqueness theorem implies that the kernel of this map is trivial, and that it has an inverse. Hence the space of solutions has dimension 2. (See Apostol, Calculus, vol II, p. 147.) The result that any solution can be expressed as a linear combination of two particular linearly independent solutions then follows from the general theory of vector spaces.
 
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