定数係数線形微分方程式 の解の導出

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

${\displaystyle =\frac{1}{f\left(D\right)}\left(C_r{x}^r+C_{r-1}{x}^{r-1}+\cdots +C_{1}x+C_0\right)}$

${\displaystyle =\frac{1}{f\left(D\right)}{C}_r{x}^r+\frac{1}{f\left(D\right)}{C}_{r-1}{x}^{r-1}+\cdots +\frac{1}{f\left(D\right)}{C}_{1}x+\frac{1}{f\left(D\right)}{C}_0}$

${\displaystyle =C_r\frac{1}{f\left(D\right)}{x}^r+C_{r-1}\frac{1}{f\left(D\right)}{x}^{r-1}+\cdots +C_1\frac{1}{f\left(D\right)}x+C_0\frac{1}{f\left(D\right)}1}$

${\displaystyle \frac{1}{f\left(D\right)}{x}^i}$${\displaystyle =\frac{1}{\left(D-a_n\right)\left(D-a_{n-1}\right)\cdots \left(D-a_1\right)}{x}^i}$

${\displaystyle =\frac{1}{\left(D-a_n\right)\left(D-a_{n-1}\right)\cdots \left(D-a_2\right)}{e}^{a_{1}x}{\displaystyle \int e^{-a_{1}x}{x}^{i}dx}}$　･･････(1)

${\displaystyle {\displaystyle \int e^{-a_{1}x}{x}^{i}dx}}$${\displaystyle ={\displaystyle \int {\left(-\frac{1}{a_1}{e}^{-a_{1}x}\right)}^{\prime }{x}^{i}dx}}$

${\displaystyle =-\frac{1}{a_1}{e}^{-a_{1}x}{x}^i-{\displaystyle \int \left(-\frac{1}{a_1}{e}^{-a_{1}x}\right)i{x}^{i-1}dx}}$

${\displaystyle =-\frac{1}{a_1}{e}^{-a_{1}x}{x}^i+\frac{i}{a_1}{\displaystyle \int e^{-a_{1}x}{x}^{i-1}dx}}$

${\displaystyle I_i=\left\{-\frac{1}{a_1}{x}^i+{\left(-\frac{1}{a_1}\right)}^{2}i{x}^{i-1}+{\left(-\frac{1}{a_1}\right)}^{3}i\left(i-1\right)x^{i-2}+\cdots +{\left(-\frac{1}{a_1}\right)}^{i+1}i!\right\}{e}^{-a_{1}x}}$  

${\displaystyle e^{a_{1}x}{\displaystyle \int e^{-a_{1}x}{x}^{i}dx}}$${\displaystyle =-\frac{1}{a_1}{x}^i+{\left(-\frac{1}{a_1}\right)}^{2}i{x}^{i-1}+{\left(-\frac{1}{a_1}\right)}^{3}i\left(i-1\right)x^{i-2}+\cdots +{\left(-\frac{1}{a_1}\right)}^{i+1}i!}$${\displaystyle =g_{1i}\left(x\right)}$  

${\displaystyle \frac{1}{f\left(D\right)}{x}^i}$${\displaystyle =\frac{1}{\left(D-a_n\right)\left(D-a_{n-1}\right)\cdots \left(D-a_2\right)}{g}_{1i}\left(x\right)}$