演習問題

1. 次の初期値問題を解け．
$(a) \ y^{\prime\prime} + 4y^{\prime} - 5y = e^{t}, \ y(0) = 0, \ y^{\prime}(0) = 1$
$(b) \ y^{\prime\prime} + 4y^{\prime} + 4y = e^{-t}, \ y(0) = 1, \ y^{\prime}(0) = 1$
$(c) \ y^{\prime\prime} + 4y^{\prime} - 5y = f(t), \ y(0) = 0, \ y^{\prime}(0) = 0$

$$f(t) = \left\{\begin{array}{rl} 2,&0 < t < \pi\\ \cos{t},&t > \pi \end{array} \right.$$

$(d) \ \left\{\begin{array}{l} y_{1}^{\prime} + y_{2} = 0 \\ y_{1} + y_{2}^{\prime} = 0, \ y_{1}(0) = 2, \ y_{2}(0) = 0 \end{array}\right.$
$(e) \ \left\{\begin{array}{l} y_{1}^{\prime} - y_{2}^{\prime} - y_{2} = -e^{t} ... ...}^{\prime} - y_{2} = e^{2t}, \ y_{1}(0) = 0, \ y_{2}(0) = 1 \end{array}\right.$
$(f) \ \left\{\begin{array}{l} y_{1}^{\prime\prime} + 2y_{2}^{\prime} + y_{1} = ... ...}(0) = y_{1}^{\prime}(0) = y_{2}(0) = y_{2}^{\prime}(0) = 0 \end{array}\right.$