I tried to write formulas (part 6)

＜Linear Differential Equation＞

　Linear differential equation can be solved as follows.(線形微分方程式は次のようにして解くことができる。)

\begin{eqnarray}\left(\frac{d}{dx}-\lambda\right)^{k}y&=&0\notag\\\left(\frac{d}{dx}-\lambda\right)\left(\frac{d}{dx}-\lambda\right)^{k{-1}}y&=&0\notag\\\left(\frac{d}{dx}-\lambda\right)z_0&=&0\left(\because z_0 = \left(\frac{d}{dx}-\lambda\right)^{k{-1}}y\right)\notag\\z_0&=&C_0e^{\lambda x}\\&(&\because {\rm the \space result\space in\space the\space case\space of\space} i=1)\notag\\\notag\\\left(\frac{d}{dx}-\lambda\right)^{k{-1}}y&=&C_0e^{\lambda x}\notag\\\left(\frac{d}{dx}-\lambda\right)\left(\frac{d}{dx}-\lambda\right)^{k{-2}}y&=&C_0e^{\lambda x}\notag\\ \left(\frac{d}{dx}-\lambda\right)z_1&=&C_0e^{\lambda x}\left(\because z_1=\left(\frac{d}{dx}-\lambda\right)^{k{-2}}y\right)\notag \\e^{\lambda x}\left(\left(\frac{d}{dx}+\lambda\right)-\lambda\right)e^{-\lambda x}z_1&=&C_0e^{\lambda x}\notag\\e^{\lambda x}\frac{d}{dx}e^{-\lambda x}z_1&=&C_0e^{\lambda x}\notag\\\frac{d}{dx}e^{-\lambda x}z_1&=&C_0\notag\\e^{-\lambda x}z_1&=&C_0x+C_1\notag\\z_1&=&(C_0x+C_1)e^{\lambda x} \\\notag\\ \left(\frac{d}{dx}-\lambda\right)^{k{-2}}y&=&(C_0x+C_1)e^{\lambda x}\notag\\\left(\frac{d}{dx}-\lambda\right)\left(\frac{d}{dx}-\lambda\right)^{k{-3}}y&=&(C_0x+C_1)e^{\lambda x}\notag \\\left(\frac{d}{dx}-\lambda\right)z_2&=&(C_0x+C_1)e^{\lambda x}\left(\because z_2=\left(\frac{d}{dx}-\lambda\right)^{k{-3}}y\right)\notag \\e^{\lambda x}\left(\left(\frac{d}{dx}+\lambda\right)-\lambda\right)e^{-\lambda x}z_2&=&(C_0x+C_1)e^{\lambda x}\notag\\e^{\lambda x}\frac{d}{dx}e^{-\lambda x}z_2&=&(C_0x+C_1)e^{\lambda x}\notag\\\frac{d}{dx}e^{-\lambda x}z_2&=&(C_0x+C_1)\notag\\e^{-\lambda x}z_2&=&C_0x^2+C_1x+C_2\notag\\&(&C_0, C_1, C_2:{\rm arbitrary\space constants})\notag\\z_2&=&(C_0x^2+C_1x+C_2)e^{\lambda x}\\\vdots\notag\\z_{k{-1}}=\left(\frac{d}{dx}-\lambda\right)^{k-k}y&=&\left(C_0x^{k{-1}}+C_1x^{k{-2}}+\cdots+C_{k{-2}}x+C_{k{-1}}\right)e^{\lambda x}\notag\\y&=&\sum_{i=0}^{k{-1}}C_ix^{k{-1}{-i}}e^{\lambda x}\end{eqnarray}