$$y′′′′ + 2y′′ + y = \sin x$$ $$y(0) = y′(0) = y′′(0) = y′′′(0)= 0$$
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?
On
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}
On
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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 $\quad\ds{\mu = 1}$: $$ \color{#66f}{\large\int_{\gamma - \infty\ic}^{\gamma + \infty\ic} {\expo{st} \over \pars{s^{2} + 1}^{3}}\,{\dd s \over 2\pi\ic} =-\,\frac{1}{8} \left[-\left(t^2-3\right) \sin\left(t\right)-3 t \cos\left(t\right)\right]} $$
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 $$ \sin2\theta=2\sin\theta\cos\theta. $$