統計学にあまりでない数学

連立微分方程式

{\begin{cases} \frac{d x}{d t} = p_{1} x + q_{1} y \\ \frac{d y}{d t} = p_{2} x + q_{2} y \end{cases}

$\begin{equation} \begin{cases} \ds\frac{dx}{dt} = p_{1}x + q_{1}y \\ \\ \ds\frac{dy}{dt} = p_{2}x + q_{2}y \end{cases} \end{equation}$

ここで次の特性方程式を解き，

λ

$\lambda$ を求める．

| \begin{matrix} p_{1} - λ & q_{1} \\ p_{2} & q_{2} - λ \end{matrix} | = 0

$\begin{equation} \begin{vmatrix} p_{1}-\lambda & q_{1} \\ p_{2} & q_{2}-\lambda \end{vmatrix} = 0 \end{equation}$

解が

λ_{1}, λ_{2}

$\lambda_{1},~\lambda_{2}$ であったとする．

$\lambda_{1} \ne \lambda_{2}$ のとき，

$\lambda = \lambda_{1}$ のとき，
$(\begin{matrix} p_{1} - λ_{1} & q_{1} \\ p_{2} & q_{2} - λ_{1} \end{matrix}) (\begin{matrix} α_{1} \\ β_{1} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$
$\begin{equation} \begin{pmatrix} p_{1}-\lambda_{1} & q_{1} \\ p_{2} & q_{2}-\lambda_{1} \end{pmatrix} \begin{pmatrix} \alpha_{1} \\ \beta_{1} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{equation}$
を解いて $\alpha_{1},~\beta_{1}$ が求まり，
$x = α_{1} e^{λ_{1} t}, y = β_{1} e^{λ_{1} t}$
$\begin{equation} x = \alpha_{1}e^{\lambda_{1}t},~~~~y = \beta_{1}e^{\lambda_{1}t} \end{equation}$
$\lambda = \lambda_{2}$ のときも，同様にして，
$x = α_{2} e^{λ_{2} t}, y = β_{2} e^{λ_{2} t}$
$\begin{equation} x = \alpha_{2}e^{\lambda_{2}t},~~~~y = \beta_{2}e^{\lambda_{2}t} \end{equation}$
よって，求める一般解は，
$x = c_{1} α_{1} e^{λ_{1} t} + c_{2} α_{2} e^{λ_{2} t}, y = c_{1} β_{1} e^{λ_{1} t} + c_{2} β_{2} e^{λ_{2} t}$
$\begin{equation} x = c_{1}\alpha_{1}e^{\lambda_{1}t} + c_{2}\alpha_{2}e^{\lambda_{2}t},~~~~y = c_{1}\beta_{1}e^{\lambda_{1}t} + c_{2}\beta_{2}e^{\lambda_{2}t} \end{equation}$
$\lambda_{1} = \lambda_{2}$ (重解)のとき，

一般解は，
$x = (A + B t) e^{λ t}, y = (C + D t) e^{λ t}$
$\begin{equation} x = (A+Bt)e^{\lambda t},~~~~y = (C+Dt)e^{\lambda t} \end{equation}$
あとは，これらを連立微分方程式の第1式に代入し，恒等式を解き， $A,B,C,D$ を求めればよい．