步骤 1:利用齐次解关系建立矩阵方程
由齐次解公式 x(t)=Φ(t)x(0),代入两组已知数据:
[e−2t−e−2t]=Φ(t)[1−1],[2e−t−e−t]=Φ(t)[2−1]
将两组列向量横向拼合成 2×2 矩阵:
[e−2t−e−2t2e−t−e−t]=Φ(t)[1−12−1]
步骤 2:求初始状态矩阵的逆矩阵
令 X(0)=[1−12−1],计算其行列式:
det[X(0)]=1×(−1)−2×(−1)=−1+2=1=0
求逆矩阵:
X−1(0)=11[−11−21]=[−11−21]
步骤 3:矩阵乘法求解 Φ(t)
Φ(t)=[e−2t−e−2t2e−t−e−t][1−12−1]−1=[e−2t−e−2t2e−t−e−t][−11−21]=[e−2t(−1)+(2e−t)(1)(−e−2t)(−1)+(−e−t)(1)e−2t(−2)+(2e−t)(1)(−e−2t)(−2)+(−e−t)(1)]=[2e−t−e−2t−e−t+e−2t2e−t−2e−2t−e−t+2e−2t]
步骤 4:反求系统矩阵 A(考研拓展)
对 Φ(t) 求导:
dtdΦ(t)=[−2e−t+2e−2te−t−2e−2t−2e−t+4e−2te−t−4e−2t]
代入 t=0:
A=dtdΦ(t)t=0=[−2+21−2−2+41−4]=[0−12−3]
Discussion
Comments
Thoughts, corrections, and follow-up notes are welcome here.