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# Finding the Solution to the Initial Value Problem 6

DIFFEQ-GQKE5L

The solution to the initial value problem $X'(t)=AX(t)$ where $A$ is given by $A=\left( \begin{array}{cc} 2 & 1 \\\ 1 & 2 \end{array} \right)$ and $X(0)=\left( \begin{array}{c}2 \\\ 2 \end{array} \right)$ is:

A

$X(t)=\left( \begin{array} {c} 2e^{t }\\\ -2e^{t} \end{array} \right)$

B

$X(t)=\left( \begin{array} {c} 2e^{3t} \\\ 2e^{3t} \end{array} \right)$

C

$X(t)=\left( \begin{array}{c} e^{t}+e^{3t} \\\ e^{t}-e^{3t} \end{array} \right)$

D

$X(t)=\left( \begin{array}{c} e^{t}-e^{3t} \\\ e^{t}+e^{3t} \end{array} \right)$