If the Mass in the Previous Problem Is Pulled Down $x = 0.02 e ^ { - 2 t } + 0.04 t e ^ { - 2 t }$

Multiple Choice

If the mass in the previous problem is pulled down two centimeters and released, the solution for the position is

A) $x = 0.02 e ^ { - 2 t } + 0.04 t e ^ { - 2 t }$
B) $x = 2 e ^ { - 2 t } + 4 t e ^ { - 2 t }$
C) $x = 0.02 e ^ { 2 t } - 0.04 t e ^ { 2 t }$
D) $x = e ^ { - 2 t } \sin t$
E) $x = 0.02 e ^ { - 2 t } \cos t$