The state equation of a second-order linear system is given by, \[\mathop x\limits^ \cdot \left( t \right) = Ax\left( t \right),\,x\left( 0 \right) = {x_0}\]
For \[{x_0} = \left[ \begin{array}{l} 1\\ - 1 \end{array} \right],\,x\left( t \right) = \left[ \begin{array}{l} {e^{ - t}}\\ - {e^{ - t}} \end{array} \right]\]       and for \[{x_0} = \left[ \begin{array}{l} 0\\ 1 \end{array} \right],\,x\left( t \right) = \left[ \begin{array}{l} {e^{ - t}} - {e^{ - 2t}}\\ - {e^{ - t}} + {e^{ - 2t}} \end{array} \right].\]
When \[{x_0} = \left[ \begin{array}{l} 3\\ 5 \end{array} \right],\,x\left( t \right)\]    is

A. \[\left[ \begin{array}{l} - 8{e^{ - t}} + 11{e^{ - 2t}}\\ 8{e^{ - t}} - 22{e^{ - 2t}} \end{array} \right]\]

B. \[\left[ \begin{array}{l} 11{e^{ - t}} - 8{e^{ - 2t}}\\ - 11{e^{ - t}} + 16{e^{ - 2t}} \end{array} \right]\]

C. \[\left[ \begin{array}{l} 3{e^{ - t}} - 5{e^{ - 2t}}\\ - 3{e^{ - t}} + 10{e^{ - 2t}} \end{array} \right]\]

D. \[\left[ \begin{array}{l} - 5{e^{ - t}} + 6{e^{ - 2t}}\\ - 5{e^{ - t}} + 6{e^{ - 2t}} \end{array} \right]\]

Answer: Option B