A theorem from algebra says that if \(P_{1}\) and \(P_{2}\) are polynomials with
no common factors then there are polynomials \(Q_{1}\) and \(Q_{2}\) such that
$$
Q_{1} P_{1}+Q_{2} P_{2}=1
$$
This implies that
$$
Q_{1}(D) P_{1}(D) y+Q_{2}(D) P_{2}(D) y=y
$$
for every function \(y\) with enough derivatives for the left side to be
defined.
(a) Use this to show that if \(P_{1}\) and \(P_{2}\) have no common factors and
$$
P_{1}(D) y=P_{2}(D) y=0
$$
then \(y=0\)
(b) Suppose \(P_{1}\) and \(P_{2}\) are polynomials with no common factors. Let
\(u_{1}, \ldots, u_{r}\) be linearly independent solutions of \(P_{1}(D) y=0\) and
let \(v_{1}, \ldots, v_{s}\) be linearly independent solutions of \(P_{2}(D) y=0
.\) Use (a) to show that \(\left\\{u_{1}, \ldots, u_{r}, v_{1}, \ldots,
v_{s}\right\\}\) is a linearly independent set.
(c) Suppose the characteristic polynomial of the constant coefficient equation
$$
a_{0} y^{(n)}+a_{1} y^{(n-1)}+\cdots+a_{n} y=0
$$
has the factorization
$$
p(r)=a_{0} p_{1}(r) p_{2}(r) \cdots p_{k}(r)
$$
where each \(p_{j}\) is of the form
$$
p_{j}(r)=\left(r-r_{j}\right)^{n_{j}} \text { or }
p_{j}(r)=\left[\left(r-\lambda_{j}\right)^{2}+w_{j}^{2}\right]^{m_{j}}
\quad\left(\omega_{j}>0\right)
$$
and no two of the polynomials \(p_{1}, p_{2}, \ldots, p_{k}\) have a common
factor. Show that we can find a fundamental set of solutions \(\left\\{y_{1},
y_{2}, \ldots, y_{n}\right\\}\) of ( \(\mathrm{A}\) ) by finding a fundamental
set of solutions of each of the equations
$$
p_{j}(D) y=0, \quad 1 \leq j \leq k
$$
and taking \(\left\\{y_{1}, y_{2}, \ldots, y_{n}\right\\}\) to be the set of all
functions in these separate fundamental sets.
