Answer
$x(t)=0.35cos(2\sqrt{5}t)$
Work Step by Step
$k(0.25)=25$
$k=100$
$5r^2+100=0$
$r=\pm2\sqrt{5}i$
$x(t)=c_{1}cos(2\sqrt{5}t)+c_{2}sin(2\sqrt{5})$
$x(0)=0.35$
$c_{1}=0.35$
$x'(t)=-2\sqrt{5}(0.35)sin(2\sqrt{5}t)+2\sqrt{5}c_{2}cos(2\sqrt{5}t)$
$x'(0)=0$
$c_{2}=0$
$x(t)=0.35cos(2\sqrt{5}t)$
