Answer
$x(t)=-\frac{1}{5}e^{-6t}+1\frac{1}{5}e^{-t}$
Work Step by Step
$k(0.5)=6$
$k=12$
$2\frac{d^2x}{dt^2}+14\frac{dx}{dt}+12x=0$
$r^2+7r+6=0$
$r=-6$ and $r=-1$
$y=c_{1}e^{-6t}+c_{2}e^{-t}$
$x(0)=1$
$y=c_{1}+c_{2}=1$
$x'(0)=0$
$y'=-6c_{1}e^{-6t}-c_{2}e^{-t}$
$-6c_{1}-c_{2}=0$
Use the two equations to find
$c_{1}=-\frac{1}{5}$
$c_{2}=1\frac{1}{5}$
$x(t)=-\frac{1}{5}e^{-6t}+1\frac{1}{5}e^{-t}$
