Proposition 10
Tag 08D9, permanent. Cite it as [Bourbaki, Tag 08D9].
Functions of a Real Variable, chapter IV, § 2. LINEAR DIFFERENTIAL EQUATIONS
Let g and h be two relatively prime polynomials such that . The subspace of solutions of (42) is the direct sum of the subspaces of solutions of the two equations
Now, by the Bezout identity (Alg., VII. 2, th. 1) there exist two polynomials and such that . For every solution of (42) one can write , where and ; and then and . On the other hand, if both and , one deduces that
which completes the proof.
With the preceding notation one can write
and prop. 10, by induction on , shows that the subspace of solutions of (42) is the direct sum of the subspaces of solutions of the equations
Now, for every complex number one has
so (43) is equivalent to
and thus has as solutions the functions , where runs through the set of polynomials of degree ; thus one recovers prop. 9 of IV, p. 194.
Assuming that the homogeneous equation (41) has been solved (that is, assuming that the characteristic roots have been found), one knows that the method of variation of parameters allows one to find the solutions of the inhomogeneous equation
where is an arbitrary regulated function (IV, p. 182); we note that if is indefinitely differentiable on an interval J then all the integrals of (45) are indefinitely differentiable on J. In the particular case , where is a polynomial (with complex coefficients) and is an arbitrary complex number, one obtains an integral of (45) more simply by the following method. Put ; the equation
can, by (44), be written
or, by Taylor’s formula applied to the polynomial ,
Let be the degree of the polynomial ; if (that is, if is not a characteristic root), there exists one and only one polynomial of degree which is a solution of (46), for the coefficients are determined by the system of linear equations
which clearly admits one and only one solution. If, on the other hand, is a characteristic root, and is its multiplicity, the preceding calculation shows that there is one and only one polynomial of degree such that every solution of is an integral; in other words, every polynomial solution of (46) is then of degree ("resonance").
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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