制御と振動の数学/第一類/連立微分方程式の解法/例題による考察/同次微分方程式

(5.1)
${\displaystyle {\begin{cases}{\frac {dx}{dt}}=-y\\{\frac {dy}{dt}}=x\end{cases}}\quad {\begin{cases}x(0)=\alpha \\y(0)=\beta \end{cases}}}$

を解け．

これは普通に解けばよい．${\displaystyle x\sqsupset X,y\sqsupset Y}$ とおいて式 (5.1) Laplace 変換すれば

${\displaystyle {\begin{cases}sX-\alpha &=-Y\\sY-\beta &=X\end{cases}}}$
${\displaystyle {\begin{pmatrix}s&1\\-1&s\end{pmatrix}}{\begin{pmatrix}X\\Y\end{pmatrix}}={\begin{pmatrix}\alpha \\\beta \end{pmatrix}}}$[1]
${\displaystyle \therefore {\begin{pmatrix}X\\Y\end{pmatrix}}={\begin{pmatrix}s&1\\-1&s\end{pmatrix}}^{-1}{\begin{pmatrix}\alpha \\\beta \end{pmatrix}}={\frac {1}{s^{2}+1}}{\begin{pmatrix}s&-1\\1&s\end{pmatrix}}{\begin{pmatrix}\alpha \\\beta \end{pmatrix}}={\begin{pmatrix}{\frac {\alpha s}{s^{2}+1}}+{\frac {-\beta }{s^{2}+1}}\\{\frac {\alpha }{s^{2}+1}}+{\frac {\beta s}{s^{2}+1}}\end{pmatrix}}}$

この原像を求めると，

(5.2)
${\displaystyle {\begin{cases}x(t)=\alpha \cos t-\beta \sin t\\y(t)=\alpha \sin t+\beta \cos t\end{cases}}}$

を得る． ${\displaystyle \diamondsuit }$

1. ^
${\displaystyle s{\begin{pmatrix}X\\Y\end{pmatrix}}+{\begin{pmatrix}Y\\-X\end{pmatrix}}={\begin{pmatrix}\alpha \\\beta \end{pmatrix}}}$
${\displaystyle {\begin{pmatrix}s&0\\0&s\end{pmatrix}}{\begin{pmatrix}X\\Y\end{pmatrix}}+{\begin{pmatrix}0&1\\-1&0\end{pmatrix}}{\begin{pmatrix}X\\Y\end{pmatrix}}={\begin{pmatrix}\alpha \\\beta \end{pmatrix}}}$
${\displaystyle \therefore {\begin{pmatrix}s&1\\-1&s\end{pmatrix}}{\begin{pmatrix}X\\Y\end{pmatrix}}={\begin{pmatrix}\alpha \\\beta \end{pmatrix}}}$

${\displaystyle {\begin{cases}x'=-x+3y\\y'=-2x+4y\end{cases}}\quad {\begin{cases}x(0)=\alpha \\y(0)=\beta \end{cases}}}$

${\displaystyle X\sqsubset x,Y\sqsubset y}$ とおいて，与方程式を Laplace 変換すると，
${\displaystyle {\begin{pmatrix}s+1&-3\\2&s-4\end{pmatrix}}{\begin{pmatrix}X\\Y\end{pmatrix}}={\begin{pmatrix}\alpha \\\beta \end{pmatrix}}}$
${\displaystyle \therefore {\begin{pmatrix}X\\Y\end{pmatrix}}={\begin{pmatrix}s+1&-3\\2&s-4\end{pmatrix}}^{-1}{\begin{pmatrix}\alpha \\\beta \end{pmatrix}}={\frac {1}{s^{2}-3s+2}}{\begin{pmatrix}s-4&3\\-2&s+1\end{pmatrix}}{\begin{pmatrix}\alpha \\\beta \end{pmatrix}}={\frac {1}{(s-1)(s-2)}}{\begin{pmatrix}\alpha (s-4)+3\beta \\-2\alpha +\beta (s+1)\end{pmatrix}}}$
${\displaystyle {\frac {\alpha s-4\alpha }{(s-1)(s-2)}}={\frac {3\alpha }{s-1}}-{\frac {2\alpha }{s-2}}}$…①
${\displaystyle {\frac {3\beta }{(s-1)(s-2)}}={\frac {-3\beta }{s-1}}+{\frac {3\beta }{s-2}}}$…②
①②の原像は，
${\displaystyle x=3\alpha e^{t}-2\alpha e^{2t}-3\beta e^{t}+3\beta e^{2t}}$
${\displaystyle \therefore x=\alpha (3e^{t}-2e^{2t})+3(-e^{t}+e^{2t})\beta .}$

${\displaystyle {\frac {-2\alpha }{(s-1)(s-2)}}=2\alpha \left({\frac {1}{s-1}}-{\frac {1}{s-2}}\right)}$…③
${\displaystyle {\frac {\beta (s+1)}{(s-1)(s-2)}}={\frac {-2\beta }{s-1}}+{\frac {3\beta }{s-2}}}$…④
③④の原像は，
${\displaystyle y=2\alpha e^{t}-2\alpha e^{2t}-2\beta e^{t}+3\beta e^{2t}}$
${\displaystyle \therefore y=2(e^{t}-e^{2t})\alpha +(-2e^{t}+3e^{2t})\beta }$
${\displaystyle \diamondsuit }$

${\displaystyle {\begin{cases}x'-\alpha x-\beta y&=\beta e^{\alpha t}\\y'+\beta x-\alpha y&=0\end{cases}}\quad {\begin{cases}x(0)=0\\y(0)=1\end{cases}}}$

${\displaystyle X\sqsubset x,Y\sqsubset y}$ とおいて，与方程式を Laplace 変換すると，
${\displaystyle {\begin{cases}sX-\alpha X-\beta Y={\frac {\beta }{s-\alpha }}\\sY-1+\beta X-\alpha Y=0\end{cases}}}$

${\displaystyle \therefore {\begin{pmatrix}s-\alpha &-\beta \\\beta &s-\alpha \end{pmatrix}}{\begin{pmatrix}X\\Y\end{pmatrix}}={\begin{pmatrix}{\frac {\beta }{s-\alpha }}\\1\end{pmatrix}}}$
${\displaystyle \therefore {\begin{pmatrix}X\\Y\end{pmatrix}}={\frac {1}{(s-\alpha )^{2}+\beta ^{2}}}{\begin{pmatrix}s-\alpha &\beta \\-\beta &s-\alpha \end{pmatrix}}{\begin{pmatrix}{\frac {\beta }{s-\alpha }}\\1\end{pmatrix}}}$
${\displaystyle ={\frac {1}{(s-\alpha )^{2}+\beta ^{2}}}{\begin{pmatrix}2\beta \\{\frac {-\beta ^{2}}{s-\alpha }}+(s-\alpha )\end{pmatrix}}}$
${\displaystyle ={\mathcal {L}}^{-1}[e^{\alpha t}]}$[1]${\displaystyle \cdot {\frac {1}{s^{2}+\beta ^{2}}}{\begin{pmatrix}2\beta \\{\frac {-\beta }{s}}+s\end{pmatrix}}}$
${\displaystyle X={\mathcal {L}}^{-1}[e^{\alpha t}]\cdot {\frac {2\beta }{s^{2}+\beta ^{2}}}}$
この原像は，
${\displaystyle x=e^{\alpha t}\cdot 2\sin \beta t}$
また，
${\displaystyle Y={\mathcal {L}}^{-1}[e^{\alpha t}]{\frac {1}{s^{2}+\beta ^{2}}}\left({\frac {-\beta ^{2}}{s}}+s\right)}$
${\displaystyle ={\mathcal {L}}^{-1}[e^{\alpha t}]{\frac {1}{s^{2}+\beta ^{2}}}{\frac {s^{2}-\beta ^{2}}{s}}}$
これを部分分数展開すると，
${\displaystyle Y={\mathcal {L}}^{-1}[e^{\alpha t}]\left({\frac {2s}{s^{2}+\beta ^{2}}}-{\frac {1}{s}}\right)}$
この原像は，
${\displaystyle y=e^{\alpha t}(2\cos \beta t-1)}$
${\displaystyle \diamondsuit }$

${\displaystyle {\begin{cases}x'+m(y-z)&=1\\y'+m(z-x)&=1\\z'+m(x-y)&=1\end{cases}}\quad {\begin{cases}x(0)=\alpha \\y(0)=\beta \\z(0)=\gamma \end{cases}}}$

${\displaystyle \diamondsuit }$

1. ^ この書き方は…厳密にはおかしいというべきかもしれない．