5.6 解答

$\displaystyle {\cal L}\{y(t)\} = {\cal L}\{e^{-t}\} + 2{\cal L}\{\int_{0}^{t}e^{-3\tau}y(t- \tau)d\tau \}$

$\displaystyle Y(s)$	$\displaystyle =$	$\displaystyle \frac{1}{s+1} + 2{\cal L}\{e^{-3t}\ast y(t)\}$
	$\displaystyle =$	$\displaystyle \frac{1}{s+1} + 2{\cal L}\{e^{-3t}\}{\cal L}\{y(t)\}$
	$\displaystyle =$	$\displaystyle \frac{1}{s+1} + 2(\frac{1}{s+3})Y(s)$

$\displaystyle Y(s)(1 - \frac{2}{s+3}) = -\frac{1}{s+1}$

$\displaystyle Y(s) = \frac{s+3}{(s+1)^2}$

$\displaystyle y(t) = {\cal L}^{-1}\{\frac{s+3}{(s+1)^2}\} = {\cal L}^{-1}\{\frac{A}{s+1} + \frac{B}{(s+1)^2}\}$

$\displaystyle B = (s+3)\mid_{s=-1} = 2, A = (s+3)^{\prime}\mid_{s=-1} = 1$

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

$\displaystyle {\cal L}\{y(t)\} = {\cal L}\{\cos{t}\} + {\cal L}\{\int_{0}^{t}y(t-\tau)e^{-2\tau} d\tau \}$

$\displaystyle Y(s)$	$\displaystyle =$	$\displaystyle \frac{s}{s^2 +1} + {\cal L}\{y(t)\ast e^{-2t}\}$
	$\displaystyle =$	$\displaystyle \frac{s}{s^2 +1} + {\cal L}\{y(t)\}{\cal L}\{e^{-2t}\}$
	$\displaystyle =$	$\displaystyle \frac{s}{s^2 +1} + Y(s)(\frac{1}{s+2})$

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

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

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

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

$\displaystyle Bs+C\mid_{s=i}$	$\displaystyle =$	$\displaystyle \frac{s(s+2)}{s+1}\mid_{s=i} = \frac{-1+2i}{i+1}$
	$\displaystyle =$	$\displaystyle \frac{1 + 3i}{2}$
	$\displaystyle \Rightarrow$	$\displaystyle B = \frac{3}{2}, \ C = \frac{1}{2}$

$\displaystyle y(t) = {\cal L}^{-1}\{\frac{-1/2}{s+1} + \frac{3s/2 -1/2}{s^2 + 1}\} = -\frac{1}{2}e^{-t} + \frac{3}{2}\cos{t} + \frac{1}{2}\sin{t} \ \framebox{終}$

$\displaystyle {\cal L}\{y(t)\} = {\cal L}\{2t+1\} + {\cal L}\{\int_{0}^{t}y(t-\tau)e^{-\tau} d\tau \}$

$\displaystyle Y(s)$	$\displaystyle =$	$\displaystyle \frac{2}{s^2} + \frac{1}{s} + {\cal L}\{y(t)\ast e^{-t}\}$
	$\displaystyle =$	$\displaystyle \frac{s+2}{s^2} + {\cal L}\{y(t)\}{\cal L}\{e^{-t}\}$
	$\displaystyle =$	$\displaystyle \frac{s+2}{s^2} + Y(s)(\frac{1}{s+1})$

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

$\displaystyle Y(s) = \frac{(s+1)(s+2)}{s^3}$

$\displaystyle y(t) = {\cal L}^{-1}\{\frac{(s+1)(s+2)}{s^3}\} = {\cal L}^{-1}\{\frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} \}$

$\displaystyle C = (s+1)(s+2)\mid_{s=0} = 2, B = ((s+1)(s+2))^{\prime}\mid_{s=0} = 3$

$\displaystyle A = \frac{((s+1)(s+2))^{\prime\prime}}{2}\mid_{s=0} = 1$

$\displaystyle y(t) = {\cal L}^{-1}\{\frac{1}{s} + \frac{3}{s^2} + \frac{2}{s^3} \} =1 + 3t + t^2 \ \ \framebox{終}$