In Exercises 17-21, use Laplace transforms to solve the given initial-value problems.>\( \left\{\begin{array}{l}y^{\prime \prime \prime}+8 y=0 \\ y(0)=0, \quad y^{\prime}(0)=1 \\ y^{\prime \prime}(0)=0\end{array}\right. \)

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s^{3} L\{y\}-s+8 L\{y\}=0, soL\{y\}=\frac{s}{s^{3}+8}=\frac{s}{(s+2)\left(s^{2}-2 s+4\right)}=\frac{A}{s+2}+\frac{B s+C}{s^{2}-2 s+4} Adding these fractions and equating coefficients of powers of s with the numerator s, we obtain A=-1 / 6, B=1 / 6, and C=1 / 3. ThusL\{y\}=\frac{1}{6}\left(\frac{-1}{s+2}+\frac{s+2}{s^{2}-2 s+4}\right)=\frac{1}{6}\left(\frac{-1}{s+2}+\frac{(s-1)+3}{(s-1)^{2}+3}\right)so we havey=\frac{1}{6}\left(-e^{-2 t}+e^{t} \cos (\sqrt{3} t)+\sqrt{3} e^{t} \sin (\sqrt{3} t)\right) . ...