Let \(a, b, c,\) and \(\omega\) be constants, with \(a \neq 0\) and \(\omega>0\), and
let
$$
P(x)=p_{0}+p_{1} x+\cdots+p_{k} x^{k} \quad \text { and } \quad
Q(x)=q_{0}+q_{1} x+\cdots+q_{k} x^{k}
$$
where at least one of the coefficients \(p_{k}, q_{k}\) is nonzero, so \(k\) is
the larger of the degrees of \(P\) and \(Q\).
(a) Show that if \(\cos \omega x\) and \(\sin \omega x\) are not solutions of the
complementary equation
$$
a y^{\prime \prime}+b y^{\prime}+c y=0
$$
then there are polynomials
$$
A(x)=A_{0}+A_{1} x+\cdots+A_{k} x^{k} \quad \text { and } \quad
B(x)=B_{0}+B_{1} x+\cdots+B_{k} x^{k}
$$
such that
$$
\begin{aligned}
\left(c-a \omega^{2}\right) A+b \omega B+2 a \omega B^{\prime}+b A^{\prime}+a
A^{\prime \prime} &=P \\
-b \omega A+\left(c-a \omega^{2}\right) B-2 a \omega A^{\prime}+b B^{\prime}+a B^{\prime \prime} &=Q
\end{aligned}
$$
where \(\left(A_{k}, B_{k}\right),\left(A_{k-1}, B_{k-1}\right),
\ldots,\left(A_{0}, B_{0}\right)\) can be computed successively by solving the
systems
$$
\begin{aligned}
\left(c-a \omega^{2}\right) A_{k}+b \omega B_{k} &=p_{k} \\
-b \omega A_{k}+\left(c-a \omega^{2}\right) B_{k} &=q_{k}
\end{aligned}
$$
and, if \(1 \leq r \leq k,\)
$$
\begin{aligned}
\left(c-a \omega^{2}\right) A_{k-r}+b \omega B_{k-r} &=p_{k-r}+\cdots \\
-b \omega A_{k-r}+\left(c-a \omega^{2}\right) B_{k-r} &=q_{k-r}+\cdots
\end{aligned}
$$
where the terms indicated by "..." depend upon the previously computed
coefficients with subscripts greater than \(k-r\). Conclude from this and
Exercise \(36(\mathbf{b})\) that
$$
y_{p}=A(x) \cos \omega x+B(x) \sin \omega x
$$
is a particular solution of
$$
a y^{\prime \prime}+b y^{\prime}+c y=P(x) \cos \omega x+Q(x) \sin \omega x .
$$
(b) Conclude from Exercise \(36(\mathbf{c})\) that the equation
$$
a\left(y^{\prime \prime}+\omega^{2} y\right)=P(x) \cos \omega x+Q(x) \sin
\omega x
$$
does not have a solution of the form (B) with \(A\) and \(B\) as in (A). Then show
that there are polynomials
$$
A(x)=A_{0} x+A_{1} x^{2}+\cdots+A_{k} x^{k+1} \quad \text { and } \quad
B(x)=B_{0} x+B_{1} x^{2}+\cdots+B_{k} x^{k+1}
$$
such that
$$
\begin{aligned}
a\left(A^{\prime \prime}+2 \omega B^{\prime}\right) &=P \\
a\left(B^{\prime \prime}-2 \omega A^{\prime}\right) &=Q
\end{aligned}
$$
where the pairs \(\left(A_{k}, B_{k}\right),\left(A_{k-1}, B_{k-1}\right),
\ldots,\left(A_{0}, B_{0}\right)\) can be computed successively as follows:
$$
\begin{aligned}
A_{k} &=-\frac{q_{k}}{2 a \omega(k+1)} \\
B_{k} &=\frac{p_{k}}{2 a \omega(k+1)},
\end{aligned}
$$
and, if \(k \geq 1\),
$$
A_{k-j}=-\frac{1}{2 \omega}\left[\frac{q_{k-j}}{a(k-j+1)}-(k-j+2)
B_{k-j+1}\right]
$$
$$
B_{k-j}=\frac{1}{2 \omega}\left[\frac{p_{k-j}}{a(k-j+1)}-(k-j+2)
A_{k-j+1}\right]
$$
for \(1 \leq j \leq k\). Conclude that (B) with this choice of the polynomials
\(A\) and \(B\) is a particular solution of \((\mathrm{C})\)
