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# Finding solution to the Initial Value Problem 2

DIFFEQ-ERET1B

Find the solution to the initial value problem:

$$X'=\left(\begin{array}{c} 0 & -8 \\\ 2 & 0 \end{array}\right)X, \ X(0)=\left(\begin{array}{c} 2 \\\ -1 \end{array}\right)$$

A

$X=\left(\begin{array}{c} 2 \\\ -1 \end{array}\right)\cos 4t+\left(\begin{array}{c} 2 \\\ 1 \end{array}\right)\sin 4t$

B

$X=\left(\begin{array}{c} 2 \\\ 1 \end{array}\right)\cos 4t+\left(\begin{array}{c} 2 \\\ -1 \end{array}\right)\sin 4t$

C

$X=\left(\begin{array}{c} 2 \\\ 0 \end{array}\right)\cos 4t+\left(\begin{array}{c} 0 \\\ -1 \end{array}\right)\sin 4t$

D

$X=\left(\begin{array}{c} 2 \\\ -1 \end{array}\right)\cos 4t-\left(\begin{array}{c} 2 \\\ 1 \end{array}\right)\sin 4t$

E

None of the above