正解例

1. $$\begin{eqnarray*} \mbox{$\cal L$}[u'+2v] & = & \mbox{$\cal L$}[u']+2\mbox{$\cal... ... L$}[u]+2(s\mbox{$\cal L$}[v]-v(0))\\ & = & U(s)+2sV(s)-2v(0) \end{eqnarray*}$$
   
   
   
   
   

2. 方程式の両辺をラプラス変換し、$u(0)=0$, $v(0)=-2$ を用いると、 (1) より
   
   $$\left\{\begin{array}{lc} sU+2V=0 & \hspace{6zw}(3) \\ \d... ...ac{2}{s^2}+\frac{2}{s}-4 & \hspace{6zw}(4) \end{array}\right.$$
   
   
   となる ( $\mbox{$\cal L$}[2x+2]=2/s^2+2/s$)。よって (3)$\times s-$(4) より
   $$\begin{eqnarray*} (s^2-1)U & = & -\frac{2}{s^2}-\frac{2}{s}+4 =\frac{4s^2-2s-... ...c{1}{s+1}\\ & = & \mbox{$\cal L$}[-1-\mbox{\large\it e}^{-x}] \end{eqnarray*}$$
   
   
   
   
   
   ゆえに
   
   $$\left\{\begin{array}{l} u=2+2x-2\mbox{\large\it e}^{-x}\\ v=-1-\mbox{\large\it e}^{-x} \end{array}\right.$$