- #1
Prove It
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To start with, let's work on the Partial Fraction decomposition.
$\displaystyle \begin{align*} \frac{A\,s + B}{s^2 - 16s +128} + \frac{C\,s + D}{s^2 + 16s + 128} &\equiv \frac{3s}{s^4 + 16\,384} \\ \frac{ \left( A\,s + B \right) \left( s^2 + 16s + 128 \right) + \left( C\,s + D \right) \left( s^2 - 16s + 128 \right) }{ \left( s^2 - 16s + 128 \right) \left( s^2 + 16s + 128 \right) } &\equiv \frac{3s}{s^4 + 16\,384} \\ \left( A\,s + B \right) \left( s^2 + 16s + 128 \right) + \left( C\,s + D \right) \left( s^2 - 16s + 128 \right) &\equiv 3s \end{align*}$
Unfortunately with so many variables it is very difficult to eliminate variables, so your method to try to expand, equate coefficients of like powers of s is a good one. In fact, if we were to attempt to use the "eliminating variables by substitution" method, we would have to use complex numbers in this case (which we're allowed to do, it's just not pretty...
$\displaystyle \begin{align*} \left( A\,s + B \right) \left( s^2 + 16s + 128 \right) + \left( C\,s + D \right) \left( s^2 - 16s + 128 \right) &\equiv 3s \\ A\,s^3 + 16A\,s^2 + 128A\,s + B\,s^2 + 16B\,s + 128B + C\,s^3 - 16C\,s^2 + 128C\,s + D\,s^2 - 16D\,s + 128D &\equiv 3s \\ \left( A + C \right) \, s^3 + \left( 16A + B - 16C + D \right) \, s^2 + \left( 128A + 16B + 128C - 16D \right) \, s + 128B + 128D &= 3s \end{align*}$
and so we get the system of equations:
$\displaystyle \begin{align*} A + C &= 0 \\ 16A + B - 16C + D &= 0 \\ 128A + 16B + 128C - 16D &= 3 \\ 128B + 128D &= 0 \end{align*}$
Subtracting 16 times the first equation from the second, and 128 times the first equation from the third, we get
$\displaystyle \begin{align*} A + C &= 0 \\ B - 32C + D &= 0 \\ 16B - 16D &= 3 \\ 128B + 128D &= 0 \end{align*}$
Subtracting 16 times the second equation from the third, and 128 times the second equation from the fourth, we get
$\displaystyle \begin{align*} A + C &= 0 \\ B - 32C + D &= 0 \\ 512C - 32D &= 3 \\ 4096C &= 0 \end{align*}$
Clearly $\displaystyle \begin{align*} C = 0 \end{align*}$ and from there it can be seen that $\displaystyle \begin{align*} A = 0, D = -\frac{3}{32} \end{align*}$ and $\displaystyle \begin{align*} B = \frac{3}{32} \end{align*}$. So the partial fraction decomposition is
$\displaystyle \begin{align*} \frac{3}{32 \left( s^2 - 16s + 128 \right) } - \frac{3}{32 \left( s^2 + 16s + 128 \right) } \end{align*}$
which is consistent with what your online calculator gave.Now as for the actual inverse transform, you have made a small mistake when completing the square, it should be
$\displaystyle \begin{align*} \frac{3}{32 \left[ \left( s - 8 \right) ^2 + 64 \right] } - \frac{3}{32 \left[ \left( s + 8 \right) ^2 + 64 \right] } \end{align*}$
and so now applying the shifts:
$\displaystyle \begin{align*} \mathcal{L} \left\{ \frac{3}{32 \left[ \left( s - 8 \right) ^2 + 64 \right] } - \frac{3}{32 \left[ \left( s + 8 \right) ^2 + 64 \right] } \right\} &= \frac{3}{32} \mathcal{L} \left\{ \frac{1}{ \left( s - 8 \right) ^2 + 8^2 } \right\} - \frac{3}{32} \mathcal{L} \left\{ \frac{1}{ \left( s + 8 \right) ^2 + 8^2 } \right\} \\ &= \frac{3}{32}e^{8t} \mathcal{L} \left\{ \frac{1}{s^2 + 8^2} \right\} - \frac{3}{32} e^{-8t} \mathcal{L} \left\{ \frac{1}{s^2 + 8^2} \right\} \\ &= \frac{3}{256}e^{8t} \mathcal{L} \left\{ \frac{8}{s^2 + 8^2} \right\} - \frac{3}{256}e^{-8t} \mathcal{L} \left\{ \frac{8}{s^2 + 8^2} \right\} \\ &= \frac{3}{256}e^{8t} \sin{ \left( 8t \right) } - \frac{3}{256} e^{-8t} \sin{ \left( 8t \right) } \\ &= \frac{3}{128} \sin{ \left( 8t \right) } \left( \frac{1}{2} e^{8t} - \frac{1}{2}e^{-8t} \right) \\ &= \frac{3}{128}\sin{ \left( 8t \right) } \sinh{ \left( 8t \right) } \end{align*}$