公式の証明

${\displaystyle f\left(D\right)=D^n+a_{n-1}{D}^{n-1}+\cdots +a_2{D}^2+a_{1}D+a_0}$

${\displaystyle f\left(D\right)e^{\alpha x}}$${\displaystyle =\left(D^n+a_{n-1}{D}^{n-1}+\cdots +a_2{D}^2+a_{1}D+a_0\right)e^{\alpha x}}$

${\displaystyle =D^n{e}^{\alpha x}+a_{n-1}{D}^{n-1}{e}^{\alpha x}+\cdots +a_2{D}^2{e}^{\alpha x}+a_{1}D{e}^{\alpha x}+a_0{e}^{\alpha x}}$

${\displaystyle ={\alpha }^n{e}^{\alpha x}+a_{n-1}{\alpha }^{n-1}{e}^{\alpha x}+\cdots +a_2{\alpha }^2{e}^{\alpha x}+a_1\alpha e^{\alpha x}+a_0{e}^{\alpha x}}$

${\displaystyle =\left({\alpha }^n+a_{n-1}{\alpha }^{n-1}+\cdots +a_2{\alpha }^2+a_1\alpha +a_0\right)e^{\alpha x}}$

${\displaystyle =f\left(\alpha \right)e^{\alpha x}}$