The Solution of the Previous Problem That Satisfies the Initial $X ( 0 ) = \left( \begin{array} { l } 0 \\0\end{array} \right)$

Multiple Choice

The solution of the previous problem that satisfies the initial condition $X ( 0 ) = \left( \begin{array} { l } 0 \\0\end{array} \right)$ is

A) $\left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { - t } + \left( \begin{array} { l } 3 \\2\end{array} \right) e ^ { 4 t } / 16 + \left( \begin{array} { c } t / 4 - 19 / 16 \\- t / 2 + 7 / 8\end{array} \right)$
B) $\left(\begin{array}{c}1 \\-1\end{array}\right) e^{-t} / 16+\left(\begin{array}{l}3 \\2\end{array}\right) e^{4 t}+\left(\begin{array}{c}t / 4+19 / 16 \\-t / 2+7 / 8\end{array}\right)$
C) $\left( \begin{array} { c } 1 \\- 1\end{array} \right) e ^ { - t } + \left( \begin{array} { l } 3 \\2\end{array} \right) e ^ { 4 t } / 16 + \left( \begin{array} { c } t / 4 + 19 / 16 \\- t / 2 + 7 / 8\end{array} \right)$
D) $\left(\begin{array}{c}1 \\-1\end{array}\right) e^{-t} / 16+\left(\begin{array}{l}3 \\2\end{array}\right) e^{4 t}+\left(\begin{array}{c}t / 4-19 / 16 \\-t / 2+7 / 8\end{array}\right)$
E) none of the above