Solving an ODE using Laplace Transforms $$y′′′′ + 2y′′ + y = \sin x$$ 
  $$y(0) = y′(0) = y′′(0) = y′′′(0)= 0$$
After solving I got $y(s)=\dfrac1{(s^2 + 1)^3}$ for which I am unable to find the inverse Laplace transform. Please let me know if what I have done is correct and how to proceed further?
 A: Yes, you have done it correctly. Now, notice that
$$
\mathcal{L}[\sin at]=\frac{a}{s^2+a^2}
$$
and
$$
\mathcal{L}[\sin at-at\cos at]=\frac{2a^3}{(s^2+a^2)^2},
$$
then rewrite
$$
\frac{1}{(s^2+1)^3}=\frac12\cdot\frac{1}{s^2+1}\cdot\frac{2}{(s^2+1)^2}.
$$
Taking the inverse Laplace transform and using the convolution theorem, we obtain
\begin{align}
\mathcal{L}^{-1}\left[\frac{1}{(s^2+1)^3}\right]&=\frac12\mathcal{L}^{-1}\left[\frac{1}{s^2+1}\cdot\frac{2}{(s^2+1)^2}\right]\\
&=\frac12\int_0^t\sin(t-\tau)[\sin \tau-\tau\cos \tau]\ d\tau\\
&=\frac12\int_0^t(\sin t\cos\tau-\cos t\sin\tau)(\sin \tau-\tau\cos \tau)\ d\tau.
\end{align}
The rest part should be easy using integration by parts and the following identities
$$
\sin^2\theta=\frac{1-\cos2\theta}{2},
$$
$$
\cos^2\theta=\frac{1+\cos2\theta}{2},
$$
and
$$
\sin2\theta=2\sin\theta\cos\theta.
$$
A: The inverse Laplace transform in question is a reduction of the more general form
given by
\begin{align}
\mathcal{L}^{-1}\left\{ \frac{1}{(s^{2}+a^{2})^{\nu}} \right\} = \frac{ \sqrt{\pi} t^{\nu - 1/2} J_{\nu-1/2}(at)}{(2a)^{\nu-1/2} \Gamma(\nu)}
\end{align}
where $J_{\mu}(z)$ is the Bessel function of the first kind. In the case of $a=1$ and $\nu = 3$ it is seen that
\begin{align}
\mathcal{L}^{-1}\left\{ \frac{1}{(s^{2}+1)^{3}} \right\} &= \frac{ \sqrt{\pi} t^{5/2} J_{5/2}(t)}{(2)^{7/2}} \\
&= \frac{1}{8} \left[ (3-t^{2}) \sin(t) - 3 t \cos(t) \right] 
\end{align}
A: $\newcommand{\+}{^{\dagger}}
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With $\quad\ds{\gamma > 0\,,\ \mu > 0}$:

\begin{align}
&\color{#66f}{\large\int_{\gamma - \infty\ic}^{\gamma + \infty\ic}
{\expo{st} \over s^{2} + \mu}\,{\dd s \over 2\pi\ic}}
={\expo{\ic\root{\mu}t} \over 2\ic\root{\mu}}+
{\expo{-\ic\root{\mu}t}  \over -2\ic\root{\mu}}
={\sin\pars{\root{\mu}t} \over \root{\mu}}
\end{align}
Take derivatives respect of $\ds{\mu}$:

\begin{align}
&\color{#66f}{\large\int_{\gamma - \infty\ic}^{\gamma + \infty\ic}
{\expo{st} \over \pars{s^{2} + \mu}^{2}}\,{\dd s \over 2\pi\ic}}
={\sin\pars{\root{\mu}t} \over 2\mu^{3/2}} - {t\cos\pars{\root{\mu}t} \over 2\mu}
\end{align}

\begin{align}
&\color{#66f}{\large\int_{\gamma - \infty\ic}^{\gamma + \infty\ic}
{\expo{st} \over \pars{s^{2} + \mu}^{3}}\,{\dd s \over 2\pi\ic}}
\\[3mm]&=-\,\half\bracks{%
\frac{-\frac{t^2 \sin \left(\sqrt{\mu } t\right)}{4 \mu }-\frac{t \cos \left(\sqrt{\mu } t\right)}{4 \mu ^{3/2}}}{\sqrt{\mu }}+\frac{3 \sin \left(\sqrt{\mu } t\right)}{4 \mu ^{5/2}}-\frac{t \cos \left(\sqrt{\mu } t\right)}{2 \mu ^2}}
\end{align}

Set $\quad\ds{\mu = 1}$:
  $$
\color{#66f}{\large\int_{\gamma - \infty\ic}^{\gamma + \infty\ic}
{\expo{st} \over \pars{s^{2} + 1}^{3}}\,{\dd s \over 2\pi\ic}
=-\,\frac{1}{8} \left[-\left(t^2-3\right) \sin\left(t\right)-3 t \cos\left(t\right)\right]}
$$

