2.4 已知描述系统的微分方程和初始状态如下，试求其零输入响应、零状态响应和全(1)y''(t)+4y'(t)+3y(t)=f(t),y'(0_(-))=1,f(t)=varepsilon(t)(2)y''(t)+4y'(t)+4y(t)=f'(t)+3f(t),y(0_(-))=1,y'(0_(-))=2,f(t)=e^-tvarepsilon(t)(3)y''(t)+2y'(t)+2y(t)=f'(t),y(0_(-))=0,y'(0_(-))=1,f(t)=varepsilon(t)

(1) **零输入响应**：特征根为-1和-3，解得 $y_i(t) = (2e^{-t} - e^{-3t})u(t)$。 **零状态响应**：拉普拉斯变换后解得 $y_f(t) = \left(\frac{1}{3} - \frac{1}{2}e^{-t} + \frac{1}{6}e^{-3t}\right)u(t)$。 (2) **零输入响应**：特征根为-2（二重），解得 $y_i(t) = (1 + 4t)e^{-2t}u(t)$。 **零状态响应**：拉普拉斯变换后解得 $y_f(t) = [2e^{-t} - (2 + t)e^{-2t}]u(t)$。 (3) **零输入响应**：特征根为-1±i，解得 $y_i(t) = e^{-t} \sin t u(t)$。 **零状态响应**：拉普拉斯变换后解得 $y_f(t) = e^{-t} \sin t u(t)$。 \[ \boxed{ \begin{array}{ll} (1) & y_i(t) = (2e^{-t} - e^{-3t})u(t), \quad y_f(t) = \left(\frac{1}{3} - \frac{1}{2}e^{-t} + \frac{1}{6}e^{-3t}\right)u(t); \\ (2) & y_i(t) = (1 + 4t)e^{-2t}u(t), \quad y_f(t) = \left[2e^{-t} - (2 + t)e^{-2t}\right]u(t); \\ (3) & y_i(t) = e^{-t} \sin t u(t), \quad y_f(t) = e^{-t} \sin t u(t). \end{array} } \]