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

question 18

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

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