# 逆演算子の計算順序変更

## ■計算結果が一致する場合

$\frac{1}{D-\alpha }F\left(x\right)={e}^{\alpha x}\int {e}^{-\alpha x}F\left(x\right)dx$  詳細

$\int f\left(x\right){g}^{\prime }\left(x\right)dx$$=f\left(x\right)g\left(x\right)-\int {f}^{\prime }\left(x\right)g\left(x\right)dx$

などを使って次の2種類の計算を行う．

•  $\frac{1}{\left(D+3\right)\left(D-2\right)}x$ $=\frac{1}{D+3}{e}^{2x}\int x{e}^{-2x}dx$ $=\frac{1}{D+3}{e}^{2x}\int x{\left(-\frac{1}{2}{e}^{-2x}\right)}^{\prime }dx$ $=\frac{1}{D+3}{e}^{2x}\left\{-\frac{1}{2}x{e}^{-2x}-\int -\frac{1}{2}{e}^{-2x}dx\right\}$ $=\frac{1}{D+3}{e}^{2x}\left(-\frac{1}{2}x{e}^{-2x}-\frac{1}{4}{e}^{-2x}\right)$ $=\frac{1}{D+3}\left(-\frac{1}{2}x-\frac{1}{4}\right)$ $=-\frac{1}{2}\frac{1}{D+3}\left(x+\frac{1}{2}\right)$ $=-\frac{1}{2}{e}^{-3x}\int {e}^{3x}\left(x+\frac{1}{2}\right)dx$ $=-\frac{1}{2}{e}^{-3x}\int {\left(\frac{1}{3}{e}^{3x}\right)}^{\prime }\left(x+\frac{1}{2}\right)dx$ $=-\frac{1}{2}{e}^{-3x}\left\{\frac{1}{3}{e}^{3x}\left(x+\frac{1}{2}\right)-\int \frac{1}{3}{e}^{3x}dx\right\}$ $=-\frac{1}{2}{e}^{-3x}\left\{\frac{1}{3}{e}^{3x}\left(x+\frac{1}{2}\right)-\frac{1}{9}{e}^{3x}\right\}$ $=-\frac{1}{6}\left\{\left(x+\frac{1}{2}\right)-\frac{1}{3}\right\}$ $=-\frac{1}{6}\left(x+\frac{1}{6}\right)$
•  $\frac{1}{\left(D-2\right)\left(D+3\right)}x$ $=\frac{1}{D-2}{e}^{-3x}\int x{\left(\frac{1}{3}{e}^{3x}\right)}^{\prime }dx$ $=\frac{1}{D-2}{e}^{-3x}\int x{e}^{3x}dx$ $=\frac{1}{D-2}{e}^{-3x}\left(\frac{1}{3}x{e}^{3x}-\int \frac{1}{3}{e}^{3x}dx\right)$ $=\frac{1}{D-2}{e}^{-3x}\left(\frac{1}{3}x{e}^{3x}-\frac{1}{9}{e}^{3x}\right)$ $=\frac{1}{D-2}\left(\frac{1}{3}x-\frac{1}{9}\right)$ $=\frac{1}{3}\frac{1}{D-2}\left(x-\frac{1}{3}\right)$ $=\frac{1}{3}{e}^{2x}\int {e}^{-2x}\left(x-\frac{1}{3}\right)dx$ $=\frac{1}{3}{e}^{2x}\int {\left(-\frac{1}{2}{e}^{-2x}\right)}^{\prime }\left(x-\frac{1}{3}\right)dx$ $=\frac{1}{3}{e}^{2x}\left\{-\frac{1}{2}{e}^{-2x}\left(x-\frac{1}{3}\right)-\int -\frac{1}{2}{e}^{-2x}dx\right\}$ $=\frac{1}{3}{e}^{2x}\left\{-\frac{1}{2}{e}^{-2x}\left(x-\frac{1}{3}\right)-\frac{1}{4}{e}^{-2x}\right\}$ $=-\frac{1}{6}\left\{\left(x-\frac{1}{3}\right)+\frac{1}{2}\right\}$ $=-\frac{1}{6}\left(x+\frac{1}{6}\right)$

この場合は順序を逆にしても答えが一致する．次の場合も答えは一致する．

•  $\frac{1}{\left(D+3\right)\left(D-2\right)}{e}^{x}$ $=\frac{1}{D+3}{e}^{2x}\int {e}^{-2x}{e}^{x}dx$ $=\frac{1}{D+3}{e}^{2x}\int {e}^{-x}dx$ $=\frac{1}{D+3}{e}^{2x}\left(-{e}^{-x}\right)$ $=-\frac{1}{D+3}{e}^{x}$ $=-{e}^{-3x}\int {e}^{3x}{e}^{x}dx$ $=-{e}^{-3x}\int {e}^{4x}dx$ $=-{e}^{-3x}\left(\frac{1}{4}{e}^{4x}\right)$ $=-\frac{1}{4}{e}^{x}$
•  $\frac{1}{\left(D-2\right)\left(D+3\right)}{e}^{x}$ $=\frac{1}{D-2}{e}^{-3x}\int {e}^{3x}{e}^{x}dx$ $=\frac{1}{D+3}{e}^{2x}\int {e}^{-x}dx$ $=\frac{1}{D-2}{e}^{-3x}\left(\frac{1}{4}{e}^{4x}\right)$ $=\frac{1}{4}\frac{1}{D-2}{e}^{x}$ $=\frac{1}{4}{e}^{2x}\int {e}^{-2x}{e}^{x}dx$ $=\frac{1}{4}{e}^{2x}\int {e}^{-x}dx$ $=\frac{1}{4}{e}^{2x}\left(-{e}^{-x}\right)$ $=-\frac{1}{4}{e}^{x}$

## ■計算結果が一致しない場合

•  $\frac{1}{\left(D+3\right)\left(D-2\right)}{e}^{2x}$ $=\frac{1}{D+3}{e}^{2x}\int {e}^{-2x}{e}^{2x}dx$ $=\frac{1}{D+3}{e}^{2x}\int dx$ $=\frac{1}{D+3}x{e}^{2x}$ $={e}^{-3x}\int {e}^{3x}x{e}^{2x}dx$ $={e}^{-3x}\int x{e}^{5x}dx$ $={e}^{-3x}\int x{\left(\frac{1}{5}{e}^{5x}\right)}^{\prime }dx$ $={e}^{-3x}\left(\frac{1}{5}x{e}^{5x}-\int \frac{1}{5}{e}^{5x}dx\right)$ $={e}^{-3x}\left(\frac{1}{5}x{e}^{5x}-\frac{1}{25}{e}^{5x}\right)$ $=\frac{1}{5}{e}^{2x}\left(x-\frac{1}{5}\right)$ $=\frac{1}{5}x{e}^{2x}-\frac{1}{25}{e}^{2x}$
•  $\frac{1}{\left(D-2\right)\left(D+3\right)}{e}^{2x}$ $=\frac{1}{D-2}{e}^{-3x}\int {e}^{3x}{e}^{2x}dx$ $=\frac{1}{D-2}{e}^{-3x}\int {e}^{5x}dx$ $=\frac{1}{D-2}{e}^{-3x}\frac{1}{5}{e}^{5x}$ $=\frac{1}{5}\frac{1}{D-2}{e}^{2x}$ $=\frac{1}{5}{e}^{2x}\int {e}^{-2x}{e}^{2x}dx$ $=\frac{1}{5}{e}^{2x}\int dx$ $=\frac{1}{5}x{e}^{2x}$

$-\frac{1}{25}{e}^{2x}$  は$\left(D+3\right)\left(D-2\right)y=0$ の一般解に含まれる．

よって特殊解としては$-\frac{1}{25}{e}^{2x}$ を省略することができる．

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