Johnsy's question over Facebook about an Inverse Laplace Transform

[Prove It]

Find the Inverse Laplace Transform of $\displaystyle \begin{align*} F(s) = \frac{4s^3 + 5s^2 + 57s + 45}{\left( s^2 + 9 \right) ^2} \end{align*}$

To start with, apply partial fractions...

$\displaystyle \begin{align*} \frac{A\,s + B}{ s^2 + 9} + \frac{C\,s + D}{ \left( s^2 + 9 \right) ^2 } &\equiv \frac{ 4s^3 + 5s^2 + 57s + 45}{ \left( s^2 + 9 \right) ^2 } \\ \left( A\,s + B \right) \left( s^2 + 9 \right) + C\,s + D &\equiv 4s^3 + 5s^2 + 57s + 45 \\ A\,s^3 + 9A\,s + B\,s^2 + 9B + C\,s + D &\equiv 4s^3 + 5s^2 + 57s + 45 \\ A\,s^3 + B\,s^2 + \left( 9A + C \right) \, s + 9B + D &\equiv 4s^3 + 5s^2 + 57s + 45 \end{align*}$

Thus $\displaystyle \begin{align*} A = 4, B = 5, C = 21, D = 0 \end{align*}$, giving

$\displaystyle \begin{align*} F(s) &= \frac{4s + 5}{ s^2 + 9 } + \frac{21s}{ \left( s^2 + 9 \right) ^2} \\ &= \frac{4s}{s^2 + 9 } + \frac{5}{s^2 + 9} + \frac{21s}{ \left( s^2 + 9 \right) ^2 } \\ &= 4 \left( \frac{s}{s^2 + 3^2 } \right) + \frac{5}{3} \left( \frac{3}{s^2 + 3^2} \right) + \frac{7}{2} \left[ -\frac{\mathrm{d}}{\mathrm{d}s} \left( \frac{3}{s^2 + 3^2} \right) \right] \end{align*}$

and thus when we take the Inverse Laplace Transform we have

$\displaystyle \begin{align*} f(t) = 4\cos{ \left( 3t \right) } + \frac{5}{3} \sin{ \left( 3t \right) } + \frac{7}{2} t\sin{ \left( 3t \right) } \end{align*}$

 

[jvicens]

where we have used the following property of the Inverse Laplace Transform:

$\displaystyle \mathcal{L}^{-1} \left[ \frac{1}{ \left( s^2 + a^2 \right) ^n} \right] = \frac{\left( -1 \right) ^{n-1}}{\left( n-1 \right) !} t^{n-1} \sin{ \left( at \right) } $