例2给定系统 =-MX(0)=0x(0)=x0 解:由方程=-M得 i=-Mt+Cl x=-oMt+Ct+C2 2 由初始条件可知:C1=0C2=x x=-Mt x==Mt- +o x=2(x0-x)M=-2M(x-x0)例2 给定系统 解:由方程 可得: 由初始条件可知: x = −M x = −Mt + C1 x = − Mt + C t + C 1 2 2 1 2 C1 = 0 C2 = x0 x = −M x(0) x = 0 x (0) = 0 x = −Mt x = − Mt + x 1 2 2 0 x (x x)M M(x x ) 2 = 2 0 − = −2 − 0