5.5 解答

5.5

1.

(a) 両辺にラプラス変換を施すと

$${\cal L}\{y^{\prime\prime} + 4y^{\prime} - 5y\} = {\cal L}\{e^{t}\}$$

微分法則を用いて変形すると

$$s^{2}Y(s) - sy(0) - y^{\prime}(0) + 4(sY(s) - y(0)) - 5Y(s) = \frac{1}{s-1}$$

初期値 $y(0) = 0 , y^{\prime}(0) = 1$を代入すると

$$s^2 Y(s) - 1 + 4sY(s) - 5Y(s) = \frac{1}{s-1}$$

これを$Y(s)$について解くと

$$Y(s)(s^2 + 4s -5) = \frac{1}{s-1} + 1 = \frac{s}{s-1}$$

より

$$Y(s) = \frac{s}{(s-1)(s^2+4s-5)}$$

ここで$Y(s)$の逆変換を求めると$y(t)$が得られる．つまり

$$y(t) = {\cal L}^{-1}\{\frac{s}{(s-1)(s^2+4s-5)}\}$$

部分分数分解を用いて

$$\frac{s}{(s-1)(s^2+4s-5)} = \frac{s}{(s-1)^{2}(s+5)} = \frac{A}{s-1} + \frac{B}{(s-1)^{2}} + \frac{C}{s+5}$$

$$B = (s-1)^{2}(\frac{s}{(s-1)^{2}(s+5)})\mid_{s=1} = \frac{1}{6}, A = (\frac{s}{s+5})^{\prime}\mid_{s=1} = -\frac{5}{36}$$

$$C = (s+5)(\frac{s}{(s-1)^{2}(s+5)})\mid_{s=-5} = -\frac{5}{36}$$

よって

$$y(t) = {\cal L}^{-1}\{\frac{5/36}{s-1} + \frac{1/6}{(s-1)^2} + \f... ...\} = \frac{5e^{t}}{36} + \frac{te^{t}}{6} - \frac{5e^{-5t}}{36} \ \framebox{終}$$

(b) 両辺にラプラス変換を施すと

$${\cal L}\{y^{\prime\prime} + 4y^{\prime} + 4y\} = {\cal L}\{e^{-t}\}$$

微分法則を用いて変形すると

$$s^{2}Y(s) - sy(0) - y^{\prime}(0) + 4(sY(s) - y(0)) + 4Y(s) = \frac{1}{s+1}$$

初期値 $y(0) = 1, y^{\prime}(0) = 1$を代入すると

$$s^2 Y(s) - s - 1 + 4sY(s) - 4 + 4Y(s) = \frac{1}{s+1}$$

これを$Y(s)$について解くと

$$Y(s)(s^2 + 4s + 4) = \frac{1}{s+1} + s + 5 = \frac{1+s^2 + 6s + 5}{s+1}$$

より

$$Y(s) = \frac{s^2 + 6s + 6}{(s+1)(s^2+4s+4)}$$

ここで$Y(s)$の逆変換を求めると$y(t)$が得られる．つまり

$$y(t) = {\cal L}^{-1}\{\frac{s^2 + 6s + 6}{(s+1)(s^2+4s+4)}\}$$

部分分数分解を用いて

$$\frac{s^2 + 6s + 6}{(s+1)(s^2+4s+4)} = \frac{s^2 + 6s + 6}{(s+1)(s+2)^{2}} = \frac{A}{s+1} + \frac{B}{s+2} + \frac{C}{(s+2)^2}$$

$$A = (s+1)( \frac{s^2 + 6s + 6}{(s+1)(s+2)^{2}})\mid_{s=-1} = 1$$

$$C = (s+2)^{2}(\frac{s^2 + 6s + 6}{(s+1)(s+2)^{2}})\mid_{s=-2} = 2$$

$$B = (\frac{s^2 + 6s + 6}{s+1})^{\prime}\mid_{s= -2} = \frac{(2s+6)(s+1)-(s^2 +6s +6)}{(s+1)^{2}}\mid_{s=-2} = 0$$

よって

$$y(t) = {\cal L}^{-1}\{\frac{1}{s+1} + \frac{2}{(s+2)^{2}}\} = e^{-t} + 2te^{-2t} \ \framebox{終}$$

(c) 両辺にラプラス変換を施すと

$${\cal L}\{y^{\prime\prime} + 4y^{\prime} - 5y\} = {\cal L}\{f(t)\}$$

微分法則を用いて変形すると

$$s^{2}Y(s) - sy(0) - y^{\prime}(0) + 4(sY(s) - y(0)) - 5Y(s) = {\cal L}\{2u_{0}(t) + \cos({t} -2)u_{\pi}(t)\}$$

初期値 $y(0) = 0 , y^{\prime}(0) = 0$を代入すると

$$s^2 Y(s) + 4sY(s) - 5Y(s) = \frac{2}{s} + e^{-\pi s}{\cal L}\{\cos{(t+\pi)} - 2\}$$

これを$Y(s)$について解くと

$$Y(s)(s^2 + 4s -5) = \frac{2}{s} + e^{-\pi s}(\frac{-s}{s^2 + 1} - \frac{2}{s})$$

よって

$$Y(s) = \frac{1}{s^2 + 4s -5}(\frac{2}{s} - e^{-\pi s}( \frac{3s^2 + 2}{s(s^2 + 1)})$$

ここで$Y(s)$の逆変換を求めると$y(t)$が得られる．つまり

$$y(t) = 2{\cal L}^{-1}\{\frac{1}{s(s^2 + 4s -5)}\} - {\cal L}^{-1}\{e^{-\pi s}\frac{3s^2 + 2}{s(s^2 + 1)(s^2 + 4s -5)} \}$$

ここで注意することは

$$f(t) = {\cal L}^{-1}\{\frac{3s^2 + 2}{s(s^2 + 1)(s^2 + 4s -5)}\}$$

とおくと第2移動法則より

$${\cal L}^{-1}\{e^{-\pi s}\frac{3s^2 + 2}{s(s^2 + 1)(s^2 + 4s -5)} \} = u_{\pi}(t)f(t - \pi)$$

で表わせることである． そこで，部分分数分解を用いると

$$\frac{3s^2 + 2}{s(s^2 + 1)(s^2 + 4s -5)} = \frac{3s^2 + 2}{s(s^2 ... ...)(s+5)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 1} + \frac{D}{s-1} + \frac{E}{s+5}$$

ここで

$$A = s( \frac{3s^2 + 2}{s(s^2 + 1)(s-1)(s+5)})\mid_{s=0} = -\frac{2}{5}$$

$$Bs +C\mid_{s=i} = \frac{3s^2 + 2}{s(s-1)(s+5)}\mid_{s=i} = \frac{-1}{i(i-1)(i+5)} = \frac{1}{4+6i} = \frac{4-6i}{52}$$

よって

$$B = \frac{-6}{52} = -\frac{3}{26} , C = \frac{4}{52} = \frac{1}{13}$$

$$D = \frac{3s^2 + 2}{s(s^2 + 1)(s+5)}\mid_{s=1} = \frac{5}{12} , E = \frac{3s^2 + 2}{s(s^2 + 1)(s-1)}\mid_{s=-5} = \frac{77}{780}$$

これより

$$f(t) = {\cal L}^{-1}\{\frac{3s^2}{s(s^2 + 1)(s^2 + 4s -5)} \}$$

$$= {\cal L}^{-1}\{\frac{-2/5}{s} + \frac{-3s/26 + 1/13}{s^2 + 1} + \frac{5/12}{s-1} + \frac{77/780}{s+5}\}$$

$$= -\frac{2}{5} - \frac{3\cos{t}}{26} + \frac{\sin{t}}{13} + \frac{5e^{t}}{12} + \frac{77e^{-5t}}{780}$$

よって

$${\cal L}^{-1}\{e^{-\pi s}\frac{3s^2 + 2}{s(s^2 + 1)(s^2 + 4s -5)} \}$$

$$= u_{\pi}(t)(-\frac{2}{5} - \frac{3\cos{(t-\pi)}}{26} + \frac{\sin{(t-\pi)}}{13} + \frac{5e^{t-\pi}}{12} + \frac{77e^{-5(t-\pi)}}{780})$$

また

$$\frac{1}{s(s^2 + 4s -5)} = \frac{1}{s(s-1)(s+5)} = \frac{A}{s} + \frac{B}{s-1} + \frac{C}{s+5}$$

$$A = s(\frac{1}{s(s-1)(s+5)})\mid_{s=0} = -\frac{1}{5}, B = \frac{... ...(s+5)}\mid_{s=1} = \frac{1}{6}, C = \frac{1}{s(s-1)}\mid_{s=-5} = \frac{1}{30}$$

よって

$$f(t) = {\cal L}^{-1}\{\frac{1}{s(s^2 + 4s -5)}\} = {\cal L}^{-1}\... ...1} + \frac{1/30}{s+5}\} = -\frac{1}{5} + \frac{e^{t}}{6} + \frac{e^{-5t}}{30}.$$

これより

$$2{\cal L}^{-1}\{\frac{1}{s(s^2 + 4s -5)}\} = 2(-\frac{1}{5} + \frac{e^{t}}{6} + \frac{e^{-5t}}{30})$$

これらをまとめると

$$y(t) = 2{\cal L}^{-1}\{\frac{1}{s(s^2 + 4s -5)}\} - {\cal L}^{-1}\{e^{-\pi s}\frac{3s^2 + 2}{s(s^2 + 1)(s^2 + 4s -5)} \}$$

$$= 2(-\frac{1}{5} + \frac{e^{t}}{6} + \frac{e^{-5t}}{30})$$

$$- u_{\pi}(t)(-\frac{2}{5} - \frac{3\cos{(t-\pi)}}{26} + \frac{\sin{... ...pi)}}{13} + \frac{5e^{t-\pi}}{12} + \frac{77e^{-5(t-\pi)}}{780}) \ \framebox{終}$$

(d) 両辺にラプラス変換を施すと，
$\left\{\begin{array}{l} {\cal L}\{y_{1}^{\prime}\} + {\cal L}\{y_{2}\} = 0\\ {\cal L}\{y_{1}\} + {\cal L}\{y_{2}^{\prime}\} = 0 \end{array}\right .$
$\left\{\begin{array}{l} sY_{1}(s) - y_{1}(0) + Y_{2}(s) = 0\\ Y_{1}(s) + sY_{2}(s) - y_{2}(0) = 0 \end{array}\right .$
$\left\{\begin{array}{l} sY_{1}(s) - 2 + Y_{2}(s) = 0\\ Y_{1}(s) + sY_{2}(s) = 0 \end{array}\right .$
$\left\{\begin{array}{l} sY_{1}(s) + Y_{2}(s) = 2\\ Y_{1}(s) + sY_{2}(s) = 0 \end{array}\right .$
Cramerの公式を用いて $Y_{1}(s),Y_{2}(s)$について解くと
$Y_{1}(s) = \frac{\left\vert\begin{array}{cc} 2 & 1\\ 0 & s \end{array}\right \... ...s \end{array}\right \vert} = \frac{2s}{s^2 - 1} = \frac{1}{s+1} + \frac{1}{s-1}$
$Y_{2}(s) = \frac{\left\vert\begin{array}{cc} s & 2\\ 1 & 0 \end{array}\right \... ...s \end{array}\right \vert} = \frac{-2}{s^2 - 1} = \frac{1}{s+1} - \frac{1}{s-1}$
これより $y_{1}(t) = {\cal L}^{-1}\{\frac{1}{s+1} + \frac{1}{s-1} \} = e^{-t} + e^{t}$
$y_{2}(t) = {\cal L}^{-1}\{\frac{1}{s+1} - \frac{1}{s-1} \} = e^{-t} - e^{t}$

(e) 両辺にラプラス変換を施すと，

$$\left\{\begin{array}{l} {\cal L}\{y_{1}^{\prime}\} - {\cal L}\{y_... ...{y_{2}^{\prime}\} - {\cal L}\{y_{2}\} = {\cal L}\{e^{2t}\} \end{array}\right.$$

$$\left\{\begin{array}{l} sY_{1}(s) - y_{1}(0) - (sY_{2}(s) - y_{2}... ...Y_{1}(s) + sY_{2}(s) - y_{2}(0) - Y_{2}(s) = \frac{1}{s-2} \end{array}\right.$$

$$\left\{\begin{array}{l} sY_{1}(s) - (sY_{2}(s) - 1) - Y_{2}(s) = ... ...1} \\ Y_{1}(s) + sY_{2}(s) - 1 - Y_{2}(s) = \frac{1}{s-2} \end{array}\right.$$

$$\left\{\begin{array}{l} sY_{1}(s) - (s + 1)Y_{2}(s) = -\frac{1}{s... ...(s) + (s - 1)Y_{2}(s) = \frac{1}{s-2} + 1 = \frac{s-1}{s-2} \end{array}\right.$$

ここでCramerの公式を用いて $Y_{1}(s),Y_{2}(s)$を求めると

$$Y_{1}(s) = \frac{\left\vert\begin{array}{cc} -\frac{s}{s-1} & -(s... ...rt} = \frac{-s + \frac{s^2 - 1}{(s-2)}}{s^2 +1} = \frac{2s - 1}{(s-2)(s^2 + 1)}$$

$$Y_{2}(s) = \frac{\left\vert\begin{array}{cc} s & -\frac{s}{s-1}\... ...2} + \frac{s}{s-1}}{s^2 +1} = \frac{(s^2 - s)(s-1)+s(s-2)}{(s-2)(s-1)(s^2 + 1)}$$

となる．次に逆変換を用いて $y_{1}(t), y_{2}(t)$を求める．

$$y_{1}(t) = {\cal L}^{-1}\{\frac{2s-1}{(s-2)(s^2 +1)} \} = {\cal L}^{-1}\{\frac{A}{(s-2)} + \frac{Bs+C}{s^2 + 1} \}$$

$$A = \frac{2s-1}{s^2 + 1}\mid_{s=2} = \frac{3}{5}$$

$$Bs + C\mid_{s = i} = \frac{2s-1}{s-2}\mid_{s=i} = \frac{2i-1}{i-2} = \frac{(2i-1)(i+2)}{-5} = \frac{-4+3i}{-5} \Rightarrow$$

$$B = -\frac{3}{5}, C = \frac{4}{5}$$

これより，

$$y_{1}(t) = {\cal L}^{-1}\{\frac{3/5}{(s-2)} + \frac{-3s/5+4/5}{s^2 + 1} \} = \frac{3}{5}e^{2t} -\frac{3}{5}\cos{t} + \frac{4}{5}\sin{t}$$

次に$y_{2}$を求める．

$$y_{2}(t) = {\cal L}^{-1}\{\frac{(s^2 - s)(s-1)+s(s-2)}{(s-2)(s-1)... ...)} \} = {\cal L}^{-1}\{\frac{A}{s-2} + \frac{B}{s-1} + \frac{Cs+D}{s^2 + 1} \}$$

$$A = \frac{(s^2 - s)(s-1)+s(s-2)}{(s-1)(s^2 + 1)}\mid_{s=2} = \frac{2}{5}, B = \frac{(s^2 - s)(s-1)+s(s-2)}{(s-2)(s^2 + 1)}\mid_{s=1} = \frac{1}{2}$$

$$Cs+D\mid_{s = i} = \frac{(s^2 - s)(s-1)+s(s-2)}{(s-2)(s-1)}\mid_{s=i}$$

$$= \frac{(-1-i)(i-1) + i(i-2)}{(i-2)(i-1)} = \frac{1-2i}{1-3i} = \frac{7+i}{10}$$

$$\Rightarrow C = \frac{1}{10}, D = \frac{7}{10}$$

これより，

$$y_{2}(t) = {\cal L}^{-1}\{\frac{2/5}{s-2} + \frac{1/2}{s-1} + \frac{s/10+7/10}{s^2 + 1}\}$$

$$= \frac{2}{5}e^{2t} + \frac{1}{2}e^{t} + \frac{1}{10}\cos{t} + \frac{7}{10}\sin{t} \ \framebox{終}$$

(f) 両辺にラプラス変換を施すと，

$$\left\{\begin{array}{l} {\cal L}\{y_{1}^{\prime\prime}\} + 2{\cal... ...}\} + 2{\cal L}\{y_{1}^{\prime}\} + {\cal L}\{y_{2}\} = 0 \end{array}\right .$$

微分法則を用いて書き直すと

$$\left\{\begin{array}{l} s^2 Y_{1}(s) - sy_{1}(0) - y_{1}^{\prime}... ..._{2}^{\prime}(0) + 2(sY_{1}(s) - y_{1}(0)) + Y_{2}(s) = 0 \end{array}\right .$$

整理すると

$$\left\{\begin{array}{l} (s^2 + 1)Y_{1}(s) + 2sY_{2}(s) = \frac{1}{s^2 + 1} \\ 2sY_{1}(s) + (s^2 + 1)Y_{2}(s) = 0 \end{array}\right .$$

となる．ここでCramerの公式を用いて $Y_{1}(s),Y_{2}(s)$を求めると

$$Y_{1}(s) = \frac{\left\vert\begin{array}{cc} \frac{1}{s^2 + 1} & ... ...} s^2 + 1 & 2s\\ 2s & s^2 + 1 \end{array}\right \vert} = \frac{1}{(s^2 -1)^2}$$

$$Y_{2}(s) = \frac{\left\vert\begin{array}{cc} s^2 + 1 & \frac{1}{s... ...s\\ 2s & s^2 + 1 \end{array}\right \vert} = \frac{-2s}{(s^2 -1)^{2}(s^2 + 1)}$$

を得る．次に逆変換を用いて $y_{1}(t), y_{2}(t)$を求める．

$$y_{1}(t) = {\cal L}^{-1}\{ \frac{1}{(s^2 -1)^2}\} = {\cal L}^{-1}\{\frac{A}{(s-1)} + \frac{B}{(s-1)^2} + \frac{C}{s+1} + \frac{D}{(s+1)^2} \}$$

$$B = \frac{1}{(s+1)^2}\mid_{s=1} = \frac{1}{4}, A = (\frac{1}{(s+1)^2})^{\prime}\mid_{s=1} = -\frac{1}{4}$$

$$D = \frac{1}{(s-1)^2}\mid_{s=-1} = \frac{1}{4}, C = (\frac{1}{(s-1)^2})^{\prime}\mid_{s=-1} = \frac{1}{4}$$

より

$$y_{1}(t) = {\cal L}^{-1}\{\frac{-1/4}{(s-1)} + \frac{1/4}{(s-1)^2} + \frac{1/4}{s+1} + \frac{1/4}{(s+1)^2}$$

$$= \frac{1}{4}e^{t} + \frac{1}{4}te^{t} + \frac{1}{4}e^{-t} + \frac{1}{4}te^{-t}$$

となる．次に

$$y_{2}(t) = {\cal L}^{-1}\{ \frac{-2s}{(s^2 -1)^{2}(s^2 + 1)}\}$$

$$= {\cal L}^{-1}\{\frac{A}{(s-1)} + \frac{B}{(s-1)^2} + \frac{C}{s+1} + \frac{D}{(s+1)^2} + \frac{Es + F}{s^2 + 1} \}$$

$$B = \frac{-2s}{(s+1)^{2}(s^2 +1)}\mid_{s=1} = -\frac{1}{4}, A = (\frac{-2s}{(s+1)^{2}(s^2 +1)})^{\prime}\mid_{s=1} = \frac{1}{4}$$

$$D = \frac{-2s}{(s-1)^{2}(s^2 +1)}\mid_{s=-1} = \frac{1}{4}, C = (\frac{-2s}{(s-1)^{2}(s^2 +1)}^{\prime})\mid_{s=-1} = \frac{1}{4}$$

$$Es+F\mid_{s=i} = \frac{-2s}{(s^2 - 1)^2}\mid_{s=i} = \frac{-2i}{4}$$

$$\Rightarrow E = \frac{1}{2} , F = 0$$

より

$$y_{2}(t) = {\cal L}^{-1}\{\frac{-1/4}{(s-1)} + \frac{-1/4}{(s-1)^2} + \frac{1/4}{s+1} + \frac{1/4}{(s+1)^2} + \frac{s/2 }{s^2 + 1}$$

$$= \frac{1}{4}e^{t} + \frac{1}{4}te^{t} + \frac{1}{4}e^{-t} + \frac{1}{4}te^{-t} + \frac{1}{2}\cos{t} \ \framebox{終}$$