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3.3 解答

3.3

1. (a)

${\bf X}^{\prime} = \left(\begin{array}{cc} 2&1\\ 3&4 \end{array}\right){\bf X} - \left(\begin{array}{c} e^{t}\\ 7e^{t} \end{array}\right)$よりまず ${\bf X}^{\prime} = A{\bf X}$の解$\Phi$を求める．

$\displaystyle \det(A - \lambda I) = \left\vert\begin{array}{cc} 2-\lambda&1\\ 3&4-\lambda \end{array}\right\vert = \lambda^2 - 6\lambda + 5 = 0$

より固有値 $\lambda = 1,5$を得る．

固有値 $\lambda = 1$に対する固有ベクトル${\bf C}$をGaussの消去法を用いて求める．

$\displaystyle A - I = \left(\begin{array}{cc} 1&1\\ 3&3 \end{array}\right) \longrightarrow \left(\begin{array}{cc} 1&1\\ 0&0 \end{array}\right) $

より $c_{2} = \alpha$とおくと

$\displaystyle {\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right... ...\end{array}\right) = \alpha \left(\begin{array}{c} -1\\ 1 \end{array}\right). $

よって解 ${\bf X}_{1} = \left(\begin{array}{c} -1\\ 1 \end{array}\right)e^{t}$を得る．

固有値 $\lambda = 5$に対する固有ベクトル${\bf C}$は

$\displaystyle A - 5I = \left(\begin{array}{cc} -3&1\\ 3&-1 \end{array}\right) \longrightarrow \left(\begin{array}{cc} 1&-\frac{1}{3}\\ 0&0 \end{array}\right) $

より $c_{2} = 3\beta$とおくと

$\displaystyle {\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right... ...a \end{array}\right) = \beta \left(\begin{array}{c} 1\\ 3 \end{array}\right). $

よって解 ${\bf X}_{2} = \left(\begin{array}{c} 1\\ 3 \end{array}\right)e^{5t}$を得る．これより基本行列$\Phi$は

$\displaystyle \Phi(t) = \left(\begin{array}{cc} -e^{t}&e^{5t}\\ e^{t}&3e^{5t} \end{array}\right) $

で与えられる．

次に $\Phi {\bf U}^{\prime} = {\bf F}$を解き特殊解${\bf U}$を求める．

$\displaystyle \left(\begin{array}{cc} -e^{t}&e^{5t}\\ e^{t}&3e^{5t} \end{array... ...\end{array}\right) = \left(\begin{array}{c} e^{t}\\ 7e^{t} \end{array}\right) $

をCramerの公式を用いて解くと

$\displaystyle u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} e^{t}&e^{5t}\\... ...^{5t}\\ e^{t}&3e^{5t} \end{array}\right\vert} = \frac{-4e^{6t}}{-4e^{6t}} = 1 $

よって

$\displaystyle u_{1} = \int 1 dt = t \ ($特殊解を求めるので積分定数をつけない$\displaystyle )$

また

$\displaystyle u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} -e^{t}&e^{t}\\... ... e^{t}&3e^{5t} \end{array}\right\vert} = \frac{-8e^{2t}}{-4e^{6t}} = 2e^{-4t} $

よって

$\displaystyle u_{2} = \int 2e^{-4t} dt = -\frac{1}{2}e^{-4t} \ ($特殊解を求めるので積分定数をつけない$\displaystyle )$

したがって一般解は
$\displaystyle {\bf X}$ $\displaystyle =$ $\displaystyle \Phi{\bf C} + \Phi {\bf U}$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{cc} -e^{t}&e^{5t}\\ e^{t}&3e^{5t} \end{array... ...array}\right)\left(\begin{array}{c} t\\ -\frac{1}{2}e^{-4t} \end{array}\right)$  

で与えられる． \framebox{終}

(b)

${\bf X}^{\prime} = \left(\begin{array}{cc} 0&4\\ -1&0 \end{array}\right){\bf X} - \left(\begin{array}{c} -3\cos{t}\\ 0 \end{array}\right)$よりまず ${\bf X}^{\prime} = A{\bf X}$の解$\Phi$を求める．

$\displaystyle \det(A - \lambda I) = \left\vert\begin{array}{cc} -\lambda&4\\ -1&-\lambda \end{array}\right\vert = \lambda^2 + 4 = 0$

より固有値 $\lambda = \pm 2i$を得る．

固有値 $\lambda = 2i$に対する固有ベクトル${\bf C}$をGaussの消去法を用いて求める．

$\displaystyle A - 2i I = \left(\begin{array}{cc} -2i&4\\ -1&-2i \end{array}\ri... ...}\right) \longrightarrow \left(\begin{array}{cc} 1&2i\\ 0&0 \end{array}\right)$

より $c_{2} = \alpha$とおくと

$\displaystyle {\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right... ...end{array}\right) = \alpha \left(\begin{array}{c} -2i\\ 1 \end{array}\right). $

よって解 ${\bf X}_{1} = \Re {\bf C}e^{2it}, {\bf X}_{2} = \Im {\bf C}e^{2it}$を得る．ここで
$\displaystyle {\bf C}e^{2it}$ $\displaystyle =$ $\displaystyle \left(\begin{array}{c} -2i\\ 1 \end{array}\right)e^{2it} = \left(\begin{array}{c} -2i\\ 1 \end{array}\right)(\cos{2t} + i\sin{2t})$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{c} 2\sin{2t}\\ \cos{2t} \end{array}\right) + i \left(\begin{array}{c} -2\cos{2t}\\ \sin{2t} \end{array}\right)$  

これより基本行列$\Phi$は

$\displaystyle \Phi(t) = \left(\begin{array}{cc} 2\sin{2t}&-2\cos{2t}\\ \cos{2t}&\sin{2t} \end{array}\right) $

で与えられる．

次に $\Phi {\bf U}^{\prime} = {\bf F}$を解き特殊解${\bf U}$を求める．

$\displaystyle \left(\begin{array}{cc} 2\sin{2t}&-2\cos{2t}\\ \cos{2t}&\sin{2t}... ... \end{array}\right) = \left(\begin{array}{c} -3\cos{t}\\ 0 \end{array}\right) $

をCramerの公式を用いて解くと

$\displaystyle u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} -3\cos{t}&-2\c... ...\sin{2t}\cos{t}}{2\sin^{2}{2t}+2\cos^{2}{2t}} = \frac{-3}{4}(\sin{3t}+\sin{t}) $

よって

$\displaystyle u_{1} = \frac{-3}{4}\int (\sin{3t}+\sin{t}) dt = \frac{3}{4}(\frac{\cos{3t}}{4} + \cos{t}) \ ($積分定数をつけない$\displaystyle )$

また

$\displaystyle u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} 2\sin{2t}&-3\c... ...}\right\vert} = \frac{3\cos{2t}{\sin{t}}}{2} = \frac{3}{4}(\cos{3t} + \cos{t}) $

よって

$\displaystyle u_{2} = \frac{3}{4} \int (\cos{3t} + \cos{t}) dt = \frac{3}{4}(\frac{\sin{3t}}{3} + \sin{t}). $

したがって一般解は
$\displaystyle {\bf X}$ $\displaystyle =$ $\displaystyle \Phi{\bf C} + \Phi {\bf U}$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{cc} 2\sin{2t}&-2\cos{2t}\\ \cos{2t}&\sin{2t}... ...rac{\cos{3t} + 3\cos{t}}{4}\\ \frac{\sin{3t} + 3\sin{t}}{4} \end{array}\right)$  

で与えられる． \framebox{終}

(c)

${\bf X}^{\prime} = \left(\begin{array}{ccc} 1&1&1\\ 0&-1&0\\ -2&-1&-2 \end{array}\right){\bf X} + \left(\begin{array}{c} 1\\ 0\\ 3e^{-t} \end{array}\right)$よりまず ${\bf X}^{\prime} = A{\bf X}$の解$\Phi$を求める．

$\displaystyle \det(A - \lambda I) = \left\vert\begin{array}{ccc} 1-\lambda&1&1\... ...ay}\right\vert = (-1-\lambda)(\lambda^2 + \lambda) = -\lambda(\lambda + 1)^{2} $

より固有値 $\lambda = -1,0$を得る．

固有値 $\lambda = 0$に対する固有ベクトル${\bf C}$をGaussの消去法を用いて求める．

$\displaystyle A - 0 I$ $\displaystyle =$ $\displaystyle \left(\begin{array}{ccc} 1&1&1\\ 0&-1&0\\ -2&-1&-2 \end{array}\... ...grightarrow \left(\begin{array}{ccc} 1&1&1\\ 0&1&0\\ 0&1&0 \end{array}\right)$  
  $\displaystyle \longrightarrow$ $\displaystyle \left(\begin{array}{ccc} 1&1&1\\ 0&1&0\\ 0&0&0 \end{array}\righ... ...grightarrow \left(\begin{array}{ccc} 1&0&1\\ 0&1&0\\ 0&0&0 \end{array}\right)$  

より $c_{3} = \alpha$とおくと，

$\displaystyle {\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{arr... ...array}\right) = \alpha \left(\begin{array}{c} -1\\ 0\\ 1 \end{array}\right). $

これより解 ${\bf X}_{1} = \left(\begin{array}{c} -1\\ 0\\ 1 \end{array}\right)e^{0t} = \left(\begin{array}{c} -1\\ 0\\ 1 \end{array}\right)$を得る．

固有値 $\lambda = -1$に対する固有ベクトル${\bf C}$を求める．

$\displaystyle A + I = \left(\begin{array}{ccc} 2&1&1\\ 0&0&0\\ -2&-1&-1 \end{... ...gin{array}{ccc} 1&\frac{1}{2}&\frac{1}{2}\\ 0&0&0\\ 0&0&0 \end{array}\right) $

より自由度2なので， $c_{2} = \beta, c_{3} = \gamma$とおくと，

$\displaystyle {\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{arr... ...ht) + \gamma \left(\begin{array}{c} -\frac{1}{2}\\ 0\\ 1 \end{array}\right). $

よって解 ${\bf X}_{2} = \left(\begin{array}{c} -\frac{1}{2}\\ 1\\ 0 \end{array}\right)e... ...X}_{3} = \left(\begin{array}{c} -\frac{1}{2}\\ 0\\ 1 \end{array}\right)e^{-t}$を得る．これより基本行列$\Phi$は

$\displaystyle \Phi(t) = \left(\begin{array}{ccc} -1& - \frac{e^{-t}}{2}&- \frac{e^{-t}}{2}\\ 0&e^{-t}&0\\ 1&0&e^{-t} \end{array}\right) $

で与えられる．

次に $\Phi {\bf U}^{\prime} = {\bf F}$より特殊解${\bf U}$を求める．

$\displaystyle \left(\begin{array}{ccc} -1& - \frac{e^{-t}}{2}&- \frac{e^{-t}}{2... ...nd{array}\right) = \left(\begin{array}{c} 1\\ 0\\ 2e^{-t} \end{array}\right) $

をCramerの公式を用いて解くと

$\displaystyle u_{1}^{\prime} = \frac{\left\vert\begin{array}{ccc} 1& - \frac{e^... ...t\vert} = \frac{e^{-2t} + e^{-3t}}{-e^{-2t} + \frac{e^{-2t}}{2}} = -2 -2e^{-t} $

よって

$\displaystyle u_{1} = \int (-2 -2e^{-t}) dt = -2t + 2e^{-t} \ ($特殊解を求めるので積分定数をつけない$\displaystyle )$

$\displaystyle u_{2}^{\prime} = \frac{\left\vert\begin{array}{ccc} -1& 1&- \frac... ...-t}&0\\ 1&0&e^{-t} \end{array}\right\vert} = \frac{0}{-\frac{e^{-2t}}{2}} = 0 $

よって

$\displaystyle u_{2} = \int 0 dt = 0 \ ($特殊解を求めるので積分定数をつけない$\displaystyle )$

$\displaystyle u_{3}^{\prime} = \frac{\left\vert\begin{array}{ccc} -1& - \frac{e... ...{array}\right\vert} = \frac{-2e^{-t}-1}{-\frac{e^{-2t}}{2}} = 4e^{t} + 2e^{2t} $

よって

$\displaystyle u_{3} = \int (4e^{t} + 2e^{2t}) dt = 4e^{t} + e^{2t} \ ($特殊解を求めるので積分定数をつけない$\displaystyle )$

したがって一般解は
$\displaystyle {\bf X}$ $\displaystyle =$ $\displaystyle \Phi{\bf C} + \Phi {\bf U}$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{ccc} -1& - \frac{e^{-t}}{2}&- \frac{e^{-t}}{2... ...\left(\begin{array}{c} -2t + 2e^{-t}\\ 0\\ 4e^{t} + e^{2t} \end{array}\right)$  

で与えられる． \framebox{終}

(d)


$\displaystyle \det(A - \lambda I)$ $\displaystyle =$ $\displaystyle \left\vert\begin{array}{rrrr} -\lambda&1&0&0\\ 0&-\lambda&1&0\\ ... ...&-\lambda&1\\ 1&0&0&-\lambda \end{array}\right\vert = -\lambda(-\lambda^3) - 1$  
  $\displaystyle =$ $\displaystyle \lambda^4 - 1 = (\lambda^2 + 1)(\lambda + 1)(\lambda - 1) = 0$  

より固有値 $\lambda = -1,1,\pm i$を得る．

固有値 $\lambda = -1$に対する固有ベクトル${\bf C}$をGaussの消去法を用いて求める．

$\displaystyle A + I$ $\displaystyle =$ $\displaystyle \left(\begin{array}{rrrr} 1&1&0&0\\ 0&1&1&0\\ 0&0&1&1\\ 1&0&0&... ...begin{array}{rrrr} 1&1&0&0\\ 0&1&1&0\\ 0&0&1&1\\ 0&-1&0&1 \end{array}\right)$  
  $\displaystyle \longrightarrow$ $\displaystyle \left(\begin{array}{rrrr} 1&1&0&0\\ 0&1&1&0\\ 0&0&1&1\\ 0&0&1&... ...\begin{array}{rrrr} 1&1&0&0\\ 0&1&1&0\\ 0&0&1&1\\ 0&0&0&0 \end{array}\right)$  
  $\displaystyle \longrightarrow$ $\displaystyle \left(\begin{array}{rrrr} 1&1&0&0\\ 0&1&0&-1\\ 0&0&1&1\\ 0&0&0... ...begin{array}{rrrr} 1&0&0&1\\ 0&1&0&-1\\ 0&0&1&1\\ 0&0&0&0 \end{array}\right)$  

より $c_{4} = \alpha$とおくと，

$\displaystyle {\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3}\\ c_{4}... ...\right) = \alpha \left(\begin{array}{c} - 1\\ 1\\ -1\\ 1 \end{array}\right) $

よって解 ${\bf X}_{1} = \left(\begin{array}{c} - 1\\ 1\\ -1\\ 1 \end{array}\right)e^{-t}$を得る．

固有値 $\lambda = 1$に対する固有ベクトル${\bf C}$を求める．

$\displaystyle A - I$ $\displaystyle =$ $\displaystyle \left(\begin{array}{rrrr} -1&1&0&0\\ 0&-1&1&0\\ 0&0&-1&1\\ 1&0... ...in{array}{rrrr} 1&-1&0&0\\ 0&-1&1&0\\ 0&0&-1&1\\ 0&1&0&-1 \end{array}\right)$  
  $\displaystyle \longrightarrow$ $\displaystyle \left(\begin{array}{rrrr} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1\\ 0&0... ...gin{array}{rrrr} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1\\ 0&0&0&0 \end{array}\right)$  
  $\displaystyle \longrightarrow$ $\displaystyle \left(\begin{array}{rrrr} 1&-1&0&0\\ 0&1&0&-1\\ 0&0&1&-1\\ 0&0... ...gin{array}{rrrr} 1&0&0&-1\\ 0&1&0&-1\\ 0&0&1&-1\\ 0&0&0&0 \end{array}\right)$  

より $c_{4} = \beta$とおくと，

$\displaystyle {\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3}\\ c_{4}... ...ay}\right) = \beta \left(\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right). $

よって解 ${\bf X}_{2} = \left(\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right)e^{t}$を得る．

固有値 $\lambda = i$に対する固有ベクトル${\bf C}$を求める．

$\displaystyle A - i I$ $\displaystyle =$ $\displaystyle \left(\begin{array}{rrrr} -i&1&0&0\\ 0&-i&1&0\\ 0&0&-i&1\\ 1&0... ...egin{array}{rrrr} 1&i&0&0\\ 0&1&i&0\\ 0&0&1&i\\ 0&-i&0&-i \end{array}\right)$  
  $\displaystyle \longrightarrow$ $\displaystyle \left(\begin{array}{rrrr} 1&i&0&0\\ 0&1&i&0\\ 0&0&1&i\\ 0&0&-1... ...\begin{array}{rrrr} 1&i&0&0\\ 0&1&i&0\\ 0&0&1&i\\ 0&0&0&0 \end{array}\right)$  
  $\displaystyle \longrightarrow$ $\displaystyle \left(\begin{array}{rrrr} 1&i&0&0\\ 0&1&0&1\\ 0&0&1&i\\ 0&0&0&... ...\begin{array}{rrrr} 1&0&0&i\\ 0&1&0&1\\ 0&0&1&i\\ 0&0&0&0 \end{array}\right)$  

より $c_{4} = \gamma$とおくと，

$\displaystyle {\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3}\\ c_{4}... ...\right) = \gamma \left(\begin{array}{c} i\\ -1\\ -i\\ 1 \end{array}\right). $

よって

$\displaystyle {\bf X}_{3} = \Re {\bf C}e^{it} = \left(\begin{array}{c} -\sin{t}... ...\begin{array}{c} \cos{t}\\ -\sin{t}\\ -\cos{t}\\ \sin{t} \end{array}\right) $

これより，基本行列$\Phi$は

$\displaystyle \Phi(t) = \left(\begin{array}{rrrr} -e^{-t}&e^{t}&-\sin{t}&\cos{t... ...{-t}&e^{t}&\sin{t}&-\cos{t}\\ e^{-t}&e^{t}&\cos{t}&\sin{t} \end{array}\right) $

で与えられる．

次に， $\Phi {\bf U}^{\prime} = {\bf F}$より特殊解${\bf U}$を求める．

$\displaystyle \left(\begin{array}{rrrr} -e^{-t}&e^{t}&-\sin{t}&\cos{t}\\ e^{-t... ...d{array}\right) = \left(\begin{array}{c} t^2\\ 0\\ 0\\ 0 \end{array}\right) $

をCramerの公式を用いて解くと

$\displaystyle u_{1}^{\prime} = \frac{\left\vert\begin{array}{rrrr} t^2&e^{t}&-\... ...\\ e^{-t}&e^{t}&\cos{t}&\sin{t} \end{array}\right\vert} = \frac{2t^2 e^t}{-8} $

よって

$\displaystyle u_{1} = -\frac{1}{4} \int t^2 e^{t} dt = -\frac{1}{4}(t^2 e^{t} - 2t e^{t} + 2e^{t}) $

$\displaystyle u_{2}^{\prime} = \frac{\left\vert\begin{array}{rrrr} -e^{-t}&t^2&... ...e^{-t}&e^{t}&\cos{t}&\sin{t} \end{array}\right\vert} = \frac{-2t^2 e^{-t}}{-8} $

よって

$\displaystyle u_{2} = \frac{1}{4} \int t^2 e^{-t} dt = -\frac{1}{4}(t^2 e^{-t} + 2t e^{-t} + 2e^{-t}) $

$\displaystyle u_{3}^{\prime} = \frac{\left\vert\begin{array}{rrrr} -e^{-t}&e^{t... ...e^{-t}&e^{t}&\cos{t}&\sin{t} \end{array}\right\vert} = \frac{4t^2 \sin{t}}{-8} $

よって

$\displaystyle u_{3} = -\frac{1}{2} \int t^2 \sin{t} dt = \frac{1}{2}(t^2 \cos{t} - 2t \sin{t} - 2\cos{t}) $

$\displaystyle u_{4}^{\prime} = \frac{\left\vert\begin{array}{rrrr} -e^{-t}&e^{t... ...^{-t}&e^{t}&\cos{t}&\sin{t} \end{array}\right\vert} = \frac{-4t^2 \cos{t}}{-8} $

よって

$\displaystyle u_{4} = \frac{1}{2} \int t^2 \cos{t} dt = \frac{1}{2}(t^2 \sin{t} - 2t \cos{t} + 2\sin{t}) $

これより一般解は
$\displaystyle {\bf X}$ $\displaystyle =$ $\displaystyle \Phi{\bf C} + \Phi {\bf U}$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{rrrr} -e^{-t}&e^{t}&-\sin{t}&\cos{t}\\ e^{-t... ...ight)\left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3}\\ c_{4} \end{array}\right)$  
  $\displaystyle +$ $\displaystyle \left(\begin{array}{rrrr} -e^{-t}&e^{t}&-\sin{t}&\cos{t}\\ e^{-t... ...\cos{t})\\ \frac{1}{2}(t^2 \sin{t} - 2t \cos{t} + 2\sin{t}) \end{array}\right)$  

で与えられる． \framebox{終}

2.

$y_{1}^{\prime} = y_{2}$とおくと， $y_{2}^{\prime} = y^{\prime\prime} = -4y^{\prime} - 3y + t$．よって

$\displaystyle \left\{\begin{array}{l} y_{1}^{\prime} = y_{2}\\ y_{2}^{\prime} = -4y_{2} - 3y_{1} + t \end{array}\right. $

ここで ${\bf Y} = \left(\begin{array}{c} y_{1}\\ y_{2} \end{array}\right)$とおくと，

$\displaystyle {\bf Y}^{\prime} = \left(\begin{array}{cc} 0&1\\ -3&-4 \end{array}\right){\bf Y} + \left(\begin{array}{c} 0\\ t \end{array}\right). $

と書き直せる．

$\displaystyle \det(A - \lambda I) = \left\vert\begin{array}{cc} -\lambda & 1\\ -3 & -4 - \lambda \end{array}\right\vert = \lambda^2 + 4\lambda + 3 = 0 $

より固有値 $\lambda = -3,-1$を得る．

固有値 $\lambda = -3$に対する固有ベクトル${\bf C}$をGaussの消去法を用いて求める．

$\displaystyle A + 3I = \left(\begin{array}{cc} 3&1\\ -3&-1 \end{array}\right) ... ...rightarrow \left(\begin{array}{cc} 1 & \frac{1}{3}\\ 0 & 0 \end{array}\right) $

より $c_{2} = 3\alpha$とおくと，

$\displaystyle {\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right... ...\end{array}\right) = \alpha \left(\begin{array}{c} -1\\ 3 \end{array}\right). $

よって解 ${\bf Y}_{1} = \left(\begin{array}{c} -1\\ 3 \end{array}\right)e^{-3t}$を得る．

固有値 $\lambda = -1$に対する固有ベクトル${\bf C}$をGaussの消去法を用いて求める．

$\displaystyle A + I = \left(\begin{array}{cc} 1&1\\ -3&-3 \end{array}\right) \longrightarrow \left(\begin{array}{cc} 1 & 1\\ 0 & 0 \end{array}\right) $

より $c_{2} = 3\beta$とおくと，

$\displaystyle {\bf C} = \left(\begin{array}{c} c_{1}\\ c_{2} \end{array}\right... ...\end{array}\right) = \alpha \left(\begin{array}{c} -1\\ 1 \end{array}\right). $

よって解 ${\bf Y}_{2} = \left(\begin{array}{c} -1\\ 1 \end{array}\right)e^{-t}$を得る．これより基本行列$\Phi$は

$\displaystyle \Phi(t) = \left(\begin{array}{cc} -e^{-3t} & -e^{-t}\\ 3e^{-3t} & e^{-t} \end{array}\right) $

で与えられる．

次に， $\Phi {\bf U}^{\prime} = {\bf F}$より特殊解${\bf U}$を求める．

$\displaystyle \left(\begin{array}{cc} -e^{-3t} & -e^{-t}\\ 3e^{-3t} & e^{-t} \... ...{\prime} \end{array}\right) = \left(\begin{array}{c} 0\\ t \end{array}\right) $

よりCramerの公式を用いると

$\displaystyle u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} 0&-e^{-t}\\ t... ...^{-t} \end{array}\right\vert} = \frac{te^{-t}}{2e^{-4t}} = \frac{1}{2}te^{3t}. $

これより

$\displaystyle u_{1} = \frac{1}{2}\int te^{3t} dt = \frac{1}{2}(\frac{te^{3t}}{3} - \frac{e^{3t}}{9}). $

また

$\displaystyle u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} -e^{-3t} & 0\\... ...-t} \end{array}\right\vert} = \frac{-te^{-3t}}{2e^{-4t}} = -\frac{1}{2}te^{t}. $

これより

$\displaystyle u_{2} = -\frac{1}{2}\int te^{t} dt = -\frac{1}{2}(te^{t} - e^{t}). $

よって一般解は
$\displaystyle {\bf Y}$ $\displaystyle =$ $\displaystyle \Phi{\bf C} + \Phi {\bf U}$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{cc} -e^{-3t} & -e^{-t}\\ 3e^{-3t} & e^{-t} \... ...{3t}}{3} - \frac{e^{3t}}{9})\\ -\frac{1}{2}(te^{t} - e^{t}) \end{array}\right)$  

で与えられる． \framebox{終}

3.

(a)

$\displaystyle \left\{\begin{array}{ll} (D-2)x_{1} + (D-4)x_{2} & = e^{t}\\ Dx_{1} + (D-1)x_{2} & = e^{4t} \end{array}\right.$ (A.10)

より

$\displaystyle \left\vert\begin{array}{cc} D-2 & D-4\\ D & D-1 \end{array}\righ... ...\left\vert\begin{array}{cc} e^{t} & D-4\\ e^{4t} & D-1 \end{array}\right\vert $

を得る．これを展開すると
$\displaystyle (D^2 - 3D + 2 - D^2 + 4D)x_{1}$ $\displaystyle =$ $\displaystyle (D-1)e^{t} - (D-4)e^{4t}$  
  $\displaystyle =$ $\displaystyle e^{t} - e^{t} - 4e^{4t} + 4e^{4t} = 0$  

つまり

$\displaystyle (D+2)x_{1} = 0 $

よって $x_{1} = c_{1}e^{-2t}$．同様にして$x_{2}$を求めることができるが，一般に $x_{2}$を求めるには，A.10の式から$Dx_{2}$を消去してそこに$x_{1}$を代入する方が簡単である．実際にやってみるとA.10の最初の式から次の式を引くと， $-2x_{1}-3x_{2} = e^{t} - e^{4t}$を得る．よって

$\displaystyle x_{2} = -\frac{1}{3}(2x_{1} + e^{4t} - e^{t}) = -\frac{1}{3}(2c_{1}e^{-2t} + e^{4t} - e^{t}). $

これより，一般解は

$\displaystyle \left\{\begin{array}{l} x_{1} = c_{1}e^{-2t}\\ x_{2} = -\frac{1}{3}(2c_{1}e^{-2t} + e^{4t} - e^{t}) \end{array}\right. $

で与えられる． \framebox{終}

(b)

$\displaystyle \left\{\begin{array}{ll} (D^2 - 1)x_{1} + (D + 1)x_{2} & = 1\\ (D-1)x_{1} + (D^2 + 1)x_{2} & = 0 \end{array}\right.$ (A.11)

より

$\displaystyle \left\vert\begin{array}{cc} D^2 - 1 & D + 1\\ D-1 & D^2 + 1 \end... ...x_{1} = \left\vert\begin{array}{cc} 1 & D+1\\ 0&D^2 +1 \end{array}\right\vert $

を得る．これを展開すると

$\displaystyle (D^4 - 1 - D^2 + 1)x_{1} = (D^2 + 1)1 - (D+1)0 = 1$

つまり

$\displaystyle D^2(D^2 - 1)x_{1} = 1. $

特性方程式は $m^2(m^2 - 1) = 0$より， $m = -1,0,0,1$． よって余関数は

$\displaystyle x_{1c} = c_{1}e^{-t} + c_{2} + c_{3}t + c_{4}e^{t}. $

で与えられる． 次に特殊解$x_{1p}$を未定係数法で求める． $D1 = 0$より

$\displaystyle D^3(D^2 - 1)x_{1} = D1 = 0.$

よって

$\displaystyle x_{1p} = At^2 + Bt + C $

と置くことができる． $D^2(D^2-1)x_{1p} = 1$より，$-2A = 1$．よって $A = -\frac{1}{2}$．これより，

$\displaystyle x_{1p} = -\frac{1}{2}t^2 $

よって

$\displaystyle x_{1} = x_{1c} + x_{1p} = c_{1}e^{-t} + c_{2} + c_{3}t + c_{4}e^{t} - \frac{1}{2}t^2. $

次に$x_{2}$を求める．A.11の式からまず． $x_{2}^{\prime\prime}$を消去すると

$\displaystyle (D^3 - 2D + 1)x_{1} + Dx_{2} - x_{2} = 0 $

を得る．これからA.11の最初の式を引くと，

$\displaystyle (D^3 - D^2 - 2D + 2)x_{1} -2x_{2} = 0 $

を得る．これより
$\displaystyle x_{2}$ $\displaystyle =$ $\displaystyle \frac{1}{2}(D^3 - D^2 - 2D + 2)x_{1}$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}(D^2 - 2)(D - 1)x_{1}$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}(D^2 - 2)(-2c_{1}e^{-t} -c_{2} + c_{3} - c_{3}t - t + \frac{1}{2}t^2)$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}(-2c_{1}e^{-t} + 1 -2(-2c_{1}e^{-t} -c_{2} + c_{3} - c_{3}t - t + \frac{1}{2}t^2))$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}(2c_{1}e^{-t} + 2c_{2} - 2c_{3} + 2c_{3}t + 2t - t^2).$  

よって一般解は

$\displaystyle \left\{\begin{array}{l} x_{1} = c_{1}e^{-t} + c_{2} + c_{3}t + c_... ...= c_{1}e^{-t} + c_{2} - c_{3} + c_{3}t + t - \frac{1}{2}t^2 \end{array}\right. $

で与えられる． \framebox{終}

(c)

$\displaystyle \left\{\begin{array}{ll} (2D + 1)x_{1} + (D + 5)x_{2} & = 1\\ (D + 2)x_{1} + (D + 2)x_{2} & = 0 \end{array}\right.$ (A.12)

より

$\displaystyle \left\vert\begin{array}{cc} 2D + 1 & D + 5\\ D + 2 & D + 2 \end{... ...x_{1} = \left\vert\begin{array}{cc} 4t & D+5\\ 2&D + 2 \end{array}\right\vert $

を得る．これを展開すると
$\displaystyle (2D^2 + 5D + 2 - D^2 - 7D - 10)x_{1}$ $\displaystyle =$ $\displaystyle (D+2)4t - (D+5)2$  
  $\displaystyle =$ $\displaystyle 4+ 8t -10$  
  $\displaystyle =$ $\displaystyle 8t - 6$  

つまり

$\displaystyle (D^2 - 2D - 8 )x_{1} = 8t - 6. $

特性方程式は $m^2 - 2m - 8 = 0$より，$m = -2, 4$． よって余関数は

$\displaystyle x_{1c} = c_{1}e^{-2t} + c_{2}e^{4t}. $

で与えられる． 次に特殊解$x_{1p}$を未定係数法で求める． $D^2(8t-6) = 0$より

$\displaystyle D^2(D + 2)(D - 4)x_{1} = D^2(8t-6) = 0.$

よって

$\displaystyle x_{1p} = At + B $

と置くことができる． $(D^2 -2D - 8)x_{1p} = 8t-6$より， $A = -1, B = 1$．これより，

$\displaystyle x_{1p} = -t + 1 $

よって

$\displaystyle x_{1} = x_{1c} + x_{1p} = c_{1}e^{-2t} + c_{2}e^{4t} - t + 1. $

次に$x_{2}$を求める．A.12の式からまず． $x_{2}^{\prime}$を消去すると

$\displaystyle (D - 1)x_{1} + 3x_{2} = 4t - 2 $

を得る．これより
$\displaystyle x_{2}$ $\displaystyle =$ $\displaystyle \frac{1}{3}(4t - 2 -(D - 1)x_{1})$  
  $\displaystyle =$ $\displaystyle \frac{1}{3}(4t -2 - Dx_{1} + x_{1})$  
  $\displaystyle =$ $\displaystyle \frac{1}{3}(4t - 2 + 2c_{1}e^{-2t} - 4c_{2}e^{4t} + 1)$  
  $\displaystyle =$ $\displaystyle \frac{1}{3}(3c_{1}e^{-2t} - 3c_{2}e^{4t} + 3t)$  
  $\displaystyle =$ $\displaystyle c_{1}e^{-2t} - c_{2}e^{4t} + t .$  

よって一般解は

$\displaystyle \left\{\begin{array}{l} x_{1} = c_{1}e^{-2t} + c_{2}e^{4t} - t + 1\\ x_{2} = c_{1}e^{-2t} - c_{2}e^{4t} + t \end{array}\right. $

で与えられる． \framebox{終}

(d)

$\displaystyle \left\{\begin{array}{ll} D^{2}x_{1} - Dx_{2} & = t +1\\ (D - 3)x_{1} + (D + 1)x_{2} & = 2t - 1 \end{array}\right.$ (A.13)

より

$\displaystyle \left\vert\begin{array}{cc} D^2 & -D\\ D - 3 & D + 1 \end{array}... ... = \left\vert\begin{array}{cc} t+1 & -D\\ 2t - 1&D + 1 \end{array}\right\vert $

を得る．これを展開すると
$\displaystyle (D^3 + D^2 + D^2 - 3D)x_{1}$ $\displaystyle =$ $\displaystyle (D+1)(t+1) + D(2t-1)$  
  $\displaystyle =$ $\displaystyle 1 + t+1 + 2$  
  $\displaystyle =$ $\displaystyle t + 4$  

つまり

$\displaystyle D(D-1)(D+3)x_{1} = t + 4. $

特性方程式は $m(m-1)(m+3) = 0$より， $m = 0,-3,1$． よって余関数は

$\displaystyle x_{1c} = c_{1} + c_{2}e^{-3t} + c_{3}e^{t}. $

で与えられる． 次に特殊解$x_{1p}$を未定係数法で求める． $D^2(t + 4) = 0$より $D^3(D-1)(D+3)x_{1} = D^2(t + 4) = 0$．このうち， $m = -3,0,1$は余関数に用いられているので，

$\displaystyle x_{1p} = At^2 + Bt $

と置くことができる． $D(D-1)(D+3)x_{1p} = t + 4$より， $-6A = 1, 3B = -\frac{14}{3}$．これより，

$\displaystyle x_{1p} = -\frac{t^2}{6} - \frac{14t}{9} $

よって

$\displaystyle x_{1} = x_{1c} + x_{1p} = c_{1} + c_{2}e^{-3t} + c_{3}e^{t} -\frac{t^2}{6} - \frac{14t}{9} . $

次に$x_{2}$を求める．A.13の式からまず． $x_{2}^{\prime}$を消去すると

$\displaystyle (D^2 + D - 3)x_{1} + x_{2} = 3t $

を得る．これより
$\displaystyle x_{2}$ $\displaystyle =$ $\displaystyle 3t - (D^2 + D - 3)x_{1}$  
  $\displaystyle =$ $\displaystyle 3t - (9c_{2}e^{-3t} + c_{3}e^{t} - \frac{1}{3}$  
  $\displaystyle +$ $\displaystyle -3c_{2}e^{-3t} + c_{3}e^{t} - \frac{t}{3} - \frac{14}{3}$  
  $\displaystyle -$ $\displaystyle 3(c_{1} + c_{2}e^{-3t} + c_{3}e^{t} - -\frac{t^2}{6} - \frac{14t}{9}))$  
  $\displaystyle =$ $\displaystyle 3c_{1} - 3c_{2}e^{-3t} + c_{3}e^{t} - \frac{t^2}{2} - \frac{4t}{3} + \frac{17}{9} .$  

よって一般解は

$\displaystyle \left\{\begin{array}{l} x_{1} = c_{1} + c_{2}e^{-3t} + c_{3}e^{t}... ... + c_{3}e^{t} - \frac{t^2}{2} - \frac{4t}{3} + \frac{17}{9} \end{array}\right. $

で与えられる． \framebox{終}

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