Answer
$x=0.6 e^{-4t} \sin 4t$
Work Step by Step
Plug the given values in the given equation:
$2\dfrac{d^2 x}{dt^2}+16\dfrac{dx}{dt}+64 x=0$
Write the auxiliary equation: $2r^2+16r+64=0$
This gives: $r= -4 \pm 4i$
We are given that $x'=2.4$
Here, $x= e^{-4t}(c_1 \cos 4t +c_2 \sin 4t)$
$x(0)= e^{0}(c_1 \cos 0 +c_2 \sin 0)$
This gives: $c_1=0$
and $x'(t) = -e^{-4t}(c_1 \cos 4t +c_2 \sin 4t)+e^{-4t}(-4c_1 \sin 4t +4c_2 \cos 4t)$
Thus, we have $x'(0) = -e^{0}((0) \cos 0 +c_2 \sin 0)+e^{0}(-4c_1 \sin 0 +4c_2 \cos 0)$
Then, $c_2=0.6$
Hence, $x(t)=e^{-4t}((0) \cos 4t +(0.6) \sin 4t)=0.6 e^{-4t} \sin 4t$
