Consider the differential equation \((D-a)(D-b) y=P_{n}(x),\) where \(P_{n}(x)\)
is a polynomial of degree \(n\). Show that a particular solution of this
equation is given by (6.24) with \(c=0 ;\) that is, \(y_{p}\) is
\(\left\\{\begin{array}{l}\text { a polynomial } Q_{n}(x) \text { of degree } n
\text { if } a \text { and } b \text { are both different from zero; } \\ x
Q_{n}(x) \text { if } a \neq 0, \text { but } b=0 \\ x^{2} Q_{n}(x) \quad
\text { if } a=b=0\end{array}\right.\)
Hint: To show that \(Q_{n}(x)=\sum a_{n} x^{n}\) is a solution of the
differential equation for
a given \(P_{n}=\sum b_{n} x^{n},\) you have only to show that the coefficients
\(a_{n}\) can be found
so that \((D-a)(D-b) Q_{n}(x) \equiv P_{n}(x) .\) Equate coefficients of \(x^{n},
x^{n-1}, \cdots,\) to see that this is always possible if \(a \neq b\). For
\(b=0,\) the differential equation becomes \((D-a) D y=P_{n} ;\) what is \(D y\) if
\(y=x Q_{n} ?\) Similarly, consider \(D^{2} y\) if \(y=x^{2} Q_{n}\).