Answer:
Explanation:
[tex]\dot x (t) - ax(t)=bu(t)\\\\e^{-at}\dot x = e^{-at}ax = \frac{d}{dt} (e^{-at})=e^{-at}bu\\\\\int\limits^t_0 \frac{d}{d\tau}(e^{-a\tau}x(\tau))d\tau=e^{-at}x(t)-x(0)=\int\limits^t_0 e^{-a\tau}bu(\tau))d\tau\\\\x(t)=e^{at}x(0)+ \int\limits^t_0 e^{-a(t-\tau)}bu(\tau))d\tau[/tex]
Similarly,
[tex]e^{-At}\dot x(t) -e^{-At}Ax(t) = \frac{d}{dt} (e^{-At}x(t))=e^{-At}Bu(t)\\\\\int\limits^t_0 \frac{d}{d\tau}(e^{-A\tau}x(\tau))d\tau=e^{-At}x(t)-e^{-A.0}x(0)=\int\limits^t_0 e^{-A\tau}Bu(\tau))d\tau\\\\since\:\:\:e^{-A.0}=I \:\:\: and\:\:\:[e^{-At}]^{-1}=e^{At}\\\\x(t)=e^{At}x(0)+ e^{At}\int\limits^t_0 e^{-A\tau}Bu(\tau))d\tau\\\\x(t)=e^{At}x(0)+ \int\limits^t_0 e^{A(t-\tau)}Bu(\tau))d\tau[/tex]
For initial condition x(0) = 0
[tex]x(t)= \int\limits^t_0 e^{A(t-\tau)}Bu(\tau))d\tau[/tex]
