Answer
$x(t)=0.6 e^{-4t} \sin 4t$
Work Step by Step
Substitute the given data in the differential equation:
$2\dfrac{d^2 x}{dt^2}+16\dfrac{dx}{dt}+64 x=0$
The auxiliary equation is: $2r^2+16r+64=0$
$\implies r= -4 \pm 4i$
Given: $x'=2.4$(velocity)
Then, we have $x= e^{-4t}(c_1 \cos 4t +c_2 \sin 4t)$
At $t=0$, we have $x(0)= e^{0}(c_1 \cos 0 +c_2 \sin 0)$
$\implies c_1=0$ (simplify)
and $x'(t) = -e^{-4t}(c_1 \cos 4t +c_2 \sin 4t)+e^{-4t}(-4c_1 \sin 4t +4c_2 \cos 4t)$
when $t=0$
$x'(0) = -e^{0}((0) \cos 0 +c_2 \sin 0)+e^{0}(-4c_1 \sin 0 +4c_2 \cos 0) \implies c_2=0.6$ (simplify)
Hence, we get $x(t)=e^{-4t}[(0.6) \sin 4t]=0.6 e^{-4t} \sin 4t$
