I tried to write formulas (part 7)

＜Linear Differential Equation part 2＞
　Linear differential equation can be solved as below.(線形微分方程式は下記のように解くことができる。) \begin{eqnarray}\prod_{i=0}^k\left(\frac{d}{dx}-\lambda_i\right)y&=&0\notag \\ \left(\frac{d}{dx}-\lambda_0\right)\prod_{i=1}^k\left(\frac{d}{dx}-\lambda_i\right)y&=&0\notag\\ \left(\frac{d}{dx}-\lambda_0\right)z_0&=&0\left(\because z_0=\prod_{i=1}^k\left(\frac{d}{dx}-\lambda_i\right)y\right)\notag \\z_0&=&C_0e^{\lambda_0 x} \\\notag\\\prod_{i=1}^k\left(\frac{d}{dx}-\lambda_i\right)y&=&C_0e^{\lambda_0 x}\notag\\\left(\frac{d}{dx}-\lambda_1\right)\prod_{i=2}^k\left(\frac{d}{dx}-\lambda_i\right)y&=&C_0e^{\lambda_0 x}\notag\\\left(\frac{d}{dx}-\lambda_1\right)z_1&=&C_0e^{\lambda_0 x}\left(z_1=\prod_{i=2}^k\left(\frac{d}{dx}-\lambda_i\right)y\right)\notag \\e^{\lambda_1 x}\left(\left(\frac{d}{dx}+\lambda_1\right)-\lambda_1\right)e^{-\lambda_1 x}z_1&=&C_0e^{\lambda_0 x}\notag \\e^{\lambda_1 x}\frac{d}{dx}e^{-\lambda_1 x}z_1&=&C_0e^{\lambda_0 x}\notag \\\frac{d}{dx}e^{-\lambda_1 x}z_1&=&C_0e^{\left(\lambda_0-\lambda_1\right) x}\notag \\e^{-\lambda_1 x}z_1&=&\frac{C_0}{\left(\lambda_0-\lambda_1\right)}e^{\left(\lambda_0-\lambda_1\right) x}+C_1\notag \\z_1&=&\frac{C_0}{\left(\lambda_0-\lambda_1\right)}e^{\lambda_0 x}+C_1e^{\lambda_1 x}\\\notag \\\prod_{i=2}^k\left(\frac{d}{dx}-\lambda_i\right)y&=&\frac{C_0}{\left(\lambda_0-\lambda_1\right)}e^{\lambda_0 x}+C_1e^{\lambda_1 x}\notag \\\left(\frac{d}{dx}-\lambda_2\right)\prod_{i=3}^k\left(\frac{d}{dx}-\lambda_i\right)y&=&\frac{C_0}{\left(\lambda_0-\lambda_1\right)}e^{\lambda_0 x}+C_1e^{\lambda_1 x}\notag \\\left(\frac{d}{dx}-\lambda_2\right)z_2&=&\frac{C_0}{\left(\lambda_0-\lambda_1\right)}e^{\lambda_0 x}+C_1e^{\lambda_1 x}\left(\because z_2= \prod_{i=3}^k\left(\frac{d}{dx}-\lambda_i\right)y\right)\notag \\e^{\lambda_2 x}\left(\left(\frac{d}{dx}+\lambda_2\right)-\lambda_2\right)e^{-\lambda_2 x}z_2&=&\frac{C_0}{\left(\lambda_0-\lambda_1\right)}e^{\lambda_0 x}+C_1e^{\lambda_1 x}\notag \\e^{\lambda_2 x}\frac{d}{dx}e^{-\lambda_2 x}z_2&=&\frac{C_0}{\left(\lambda_0-\lambda_1\right)}e^{\lambda_0 x}+C_1e^{\lambda_1 x}\notag \\ \frac{d}{dx}e^{-\lambda_2 x}z_2&=&\frac{C_0}{\left(\lambda_0-\lambda_1\right)}e^{\left(\lambda_0-\lambda_2\right) x}+C_1e^{\left(\lambda_1-\lambda_2\right)x}\notag \\e^{-\lambda_2 x}z_2&=&\frac{C_0}{\left(\lambda_0-\lambda_1\right)\left(\lambda_0-\lambda_2\right)}e^{\left(\lambda_0-\lambda_2\right) x}+\frac{C_1}{\left(\lambda_1-\lambda_2\right)}e^{\left(\lambda_1-\lambda_2\right)x}+C_2\notag \\z_2&=&\frac{C_0}{\left(\lambda_0-\lambda_1\right)\left(\lambda_0-\lambda_2\right)}e^{\lambda_0 x}+\frac{C_1}{\left(\lambda_1-\lambda_2\right)}e^{\lambda_1 x}+C_2e^{\lambda_2 x} \\\vdots\notag \\z_{k{-1}}=\left(\frac{d}{dx}-\lambda_k\right)y&=&C_0e^{\lambda_0\space x}+\cdots+C_{k{-1}}e^{\lambda_{k{-1}}\space x}\notag\\e^{\lambda_k x}\left(\left(\frac{d}{dx}+\lambda_k\right)-\lambda_k\right)e^{-\lambda_k x}y&=&C_0e^{\lambda_0\space x}+\cdots+C_{k{-1}}e^{\lambda_{k{-1}}x}\notag\\e^{\lambda_k x}\frac{d}{dx}e^{-\lambda_k x}y&=&C_0e^{\lambda_0\space x}+\cdots+C_{k{-1}}e^{\lambda_{k{-1}}\space x}\notag\\\frac{d}{dx}e^{-\lambda_k x}y&=&C_0e^{(\lambda_0-\lambda_k)\space x}+\cdots+C_{k{-1}}e^{(\lambda_{k{-1}}-\lambda_k)x}\notag\\ e^{-\lambda_k x}y&=&\frac{C_0}{(\lambda_0-\lambda_k)}e^{(\lambda_0-\lambda_k)\space x}+\cdots+\frac{C_{k{-1}}}{(\lambda_{k{-1}}-\lambda_k)}e^{(\lambda_{k{-1}}-\lambda_k)x}+C_k\notag \\y&=&\frac{C_0}{(\lambda_0-\lambda_k)}e^{\lambda_0\space x}+\cdots+\frac{C_{k{-1}}}{(\lambda_{k{-1}}-\lambda_k)}e^{\lambda_{k{-1}}\space x}+C_ke^{\lambda_k x}\end{eqnarray}

Here, for the sake of simplicity, the coefficients is expressed as follows.(ここで、簡単のために係数を次のように表す。)

\begin{eqnarray}y&=&C_0e^{\lambda_0\space x}+\cdots+C_{k{-1}}e^{\lambda_{k{-1}}\space x}+C_ke^{\lambda_k x} =\sum_{i=0}^{k}C_{i}e^{\lambda_i x} \end{eqnarray}