# 定数係数線形微分方程式 の解の導出

${y}^{\left(n\right)}+{A}_{n-1}{y}^{\left(n-1\right)}+\cdots +{A}_{1}{y}^{\prime }+{A}_{0}y$$=F\left(x\right)$

$F\left(x\right)$$r$ 次の多項式であるとする． ${A}_{0}\ne 0$ ならば，これは $r$ 次の多項式の特殊解をもつ．

## ■導出

${y}^{\left(n\right)}+{A}_{n-1}{y}^{\left(n-1\right)}+\cdots +{A}_{1}{y}^{\prime }+{A}_{0}y$$=f\left(D\right)y$

$F\left(x\right)$$={C}_{r}{x}^{r}+{C}_{r-1}{x}^{r-1}+\cdots +{C}_{1}x+{C}_{0}$

とおく．

$f\left(D\right)y=F\left(x\right)$  より

$y$$=\frac{1}{f\left(D\right)}F\left(x\right)$

$=\frac{1}{f\left(D\right)}\left({C}_{r}{x}^{r}+{C}_{r-1}{x}^{r-1}+\cdots +{C}_{1}x+{C}_{0}\right)$

$=\frac{1}{f\left(D\right)}{C}_{r}{x}^{r}+\frac{1}{f\left(D\right)}{C}_{r-1}{x}^{r-1}+\cdots +\frac{1}{f\left(D\right)}{C}_{1}x+\frac{1}{f\left(D\right)}{C}_{0}$

$={C}_{r}\frac{1}{f\left(D\right)}{x}^{r}+{C}_{r-1}\frac{1}{f\left(D\right)}{x}^{r-1}+\cdots +{C}_{1}\frac{1}{f\left(D\right)}x+{C}_{0}\frac{1}{f\left(D\right)}1$

$\frac{1}{f\left(D\right)}{x}^{i}$$=\frac{1}{\left(D-{a}_{n}\right)\left(D-{a}_{n-1}\right)\cdots \left(D-{a}_{1}\right)}{x}^{i}$

$=\frac{1}{\left(D-{a}_{n}\right)\left(D-{a}_{n-1}\right)\cdots \left(D-{a}_{2}\right)}{e}^{{a}_{1}x}\int {e}^{-{a}_{1}x}{x}^{i}dx$　･･････(1)

$\int {e}^{-{a}_{1}x}{x}^{i}dx={I}_{i}$  とおく

$\int {e}^{-{a}_{1}x}{x}^{i}dx$$=\int {\left(-\frac{1}{{a}_{1}}{e}^{-{a}_{1}x}\right)}^{\prime }{x}^{i}dx$

$=-\frac{1}{{a}_{1}}{e}^{-{a}_{1}x}{x}^{i}-\int \left(-\frac{1}{{a}_{1}}{e}^{-{a}_{1}x}\right)i{x}^{i-1}dx$

$=-\frac{1}{{a}_{1}}{e}^{-{a}_{1}x}{x}^{i}+\frac{i}{{a}_{1}}\int {e}^{-{a}_{1}x}{x}^{i-1}dx$

${I}_{i}=-\frac{1}{{a}_{1}}{e}^{-{a}_{1}x}{x}^{i}+\frac{i}{{a}_{1}}{I}_{i-1}$

${I}_{i-1}=-\frac{1}{{a}_{1}}{e}^{-{a}_{1}x}{x}^{i-1}+\frac{i-1}{{a}_{1}}{I}_{i-2}$

${I}_{i-2}=-\frac{1}{{a}_{1}}{e}^{-{a}_{1}x}{x}^{i-2}+\frac{i-2}{{a}_{1}}{I}_{i-3}$

${I}_{0}=\int {e}^{-{a}_{1}x}dx=-\frac{1}{{a}_{1}}{e}^{-{a}_{1}x}$

${I}_{i}=\left\{-\frac{1}{{a}_{1}}{x}^{i}+{\left(-\frac{1}{{a}_{1}}\right)}^{2}i{x}^{i-1}+{\left(-\frac{1}{{a}_{1}}\right)}^{3}i\left(i-1\right){x}^{i-2}+\cdots +{\left(-\frac{1}{{a}_{1}}\right)}^{i+1}i!\right\}{e}^{-{a}_{1}x}$

${e}^{{a}_{1}x}\int {e}^{-{a}_{1}x}{x}^{i}dx$$=-\frac{1}{{a}_{1}}{x}^{i}+{\left(-\frac{1}{{a}_{1}}\right)}^{2}i{x}^{i-1}+{\left(-\frac{1}{{a}_{1}}\right)}^{3}i\left(i-1\right){x}^{i-2}+\cdots +{\left(-\frac{1}{{a}_{1}}\right)}^{i+1}i!$$={g}_{1i}\left(x\right)$

とおく．${g}_{1i}\left(x\right)$$i$ 次の多項式となる．

(1)に代入すると

$\frac{1}{f\left(D\right)}{x}^{i}$$=\frac{1}{\left(D-{a}_{n}\right)\left(D-{a}_{n-1}\right)\cdots \left(D-{a}_{2}\right)}{g}_{1i}\left(x\right)$

この計算より$\frac{1}{D-{a}_{i}}$ の逆演算を行っても多項式の最高次数保存されることがわかる．

したがって

$y=\frac{1}{f\left(D\right)}F\left(x\right)$  は$r$ 次の多項式となる．

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