次の各式を求めよ。(ただし、任意定数は省略してよい。)
\begin{align}
(1) & \quad \frac{1}{D^4 + D^2} \cos 2 x \\
(2) & \quad \frac{1}{D^4 – D^2 + 3} \sin x \\
(3) & \quad \frac{1}{D^2 + 9} \cos 3 x \\
(4) & \quad \frac{1}{(D^2 + 9)(D^2 + 2)} \sin 3 x
\end{align}
以下、適宜これまでに導いた結果、微分演算子の性質(3),微分演算子の性質(4)を使う。
(1)
\begin{align}
\frac{1}{D^4 + D^2} cos 2 x &= \frac{\cos 2 x}{(-2^2)^2 + (-2^2)} \\
&= \frac{1}{12} \cos 2x
\end{align}
(2)
\begin{align}
\frac{1}{D^4 – D^2 + 3} \sin x &= \frac{\sin x}{(-1^2)^2 – (-1^2) + 3} \\
&= \frac{1}{5} \sin x
\end{align}
(3)
\begin{align}
\frac{1}{D^2 + 9} \cos 3 x &= \frac{x \sin 3 x}{2 \cdot 3} \\
&= \frac{1}{6} \sin 3 x
\end{align}
(4)
\begin{align}
\frac{1}{(D^2 + 9)(D^2 + 2)} \sin 3 x &= \frac{1}{D^2 + 9} \bigg(\frac{1}{D^2 + 2} \sin 3 x\bigg) \\
&= \frac{1}{D^2 + 9} \frac{\sin 3x}{- 3^2 + 2} \\
&= – \frac{1}{7} \frac{1}{D^2 + 9} \sin 3 x \\
&= – \frac{1}{7} \bigg(- \frac{x \cos 3 x}{2 \cdot 3}\bigg) \\
&= \frac{1}{42} x \cos 3 x
\end{align}
