- #1
Prove It
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It's not entirely obvious what to do with this question, as the denominator does not easily factorise. However, if we realize that $\displaystyle \begin{align*} s^4 + 40\,000 = \left( s^2 \right) ^2 + 200^2 \end{align*}$ it's possible to do a sneaky completion of the square...
$\displaystyle \begin{align*} \left( s^2 \right) ^2 + 200^2 &= \left( s^2 \right) ^2 + 400\,s^2 + 200^2 - 400\,s^2 \\ &= \left( s^2 + 200 \right) ^2 - 400\,s^2 \\ &= \left( s^2 + 200 \right) ^2 - \left( 20\,s \right) ^2 \\ &= \left( s^2 - 20\,s + 200 \right) \left( s^2 + 20\,s + 200 \right) \end{align*}$
and thus it's now possible to perform partial fractions
$\displaystyle \begin{align*} \frac{A\,s + B}{s^2 - 20\,s + 200 } + \frac{C\,s + D}{s^2 + 20\,s + 200} &\equiv \frac{s^2 + 200}{\left( s^2 - 20\,s + 200 \right) \left( s^2 + 20\,s + 200 \right) } \\ \left( A\,s + B \right) \left( s^2 + 20\,s + 200 \right) + \left( C\,s + D \right) \left( s^2 - 20\,s + 200 \right) &\equiv s^2 + 200 \\ \left( A + C \right) \,s ^3 + \left( 20\,A + B - 20\,C + D \right) \, s^2 + \left( 200\,A + 20\,B + 200\,C - 20\,D \right) \, s + 200\,\left( B + D \right) &\equiv s^2 + 200 \end{align*}$
so it can be seen that $\displaystyle \begin{align*} A + C = 0 , \, \, 20\,A + B - 20\,C + D = 1 , \, \, 200\,A + 20\,B + 200\,C - 20\,D = 0 \textrm{ and } B+ D = 1 \end{align*}$, so solving the system gives
$\displaystyle \begin{align*} \left[ \begin{matrix} 1 & 0 & \phantom{-}1 & \phantom{-}0 & 0 \\ 20 & 1 & -20 & \phantom{-}1 & 1 \\ 200 & 20 & \phantom{-}200 & -20 & 0 \\ 0 & 1 & \phantom{-}0 & \phantom{-}1 & 1 \end{matrix} \right] \end{align*}$
apply R2 - 20R1 to R2 and R3 - 200R1 to R3 and we have
$\displaystyle \begin{align*} \left[ \begin{matrix} 1 & 0 & \phantom{-}1 & \phantom{-}0 & 0 \\ 0 & 1 & -40 & \phantom{-}1 & 1 \\ 0 & 20 & \phantom{-}0 & -20 & 0 \\ 0 & 1 & \phantom{-}0 & \phantom{-}1 & 1 \end{matrix} \right] \end{align*}$
apply R3 - 20R2 to R3 and R4 - R2 to R4 and we have
$\displaystyle \begin{align*} \left[ \begin{matrix} 1 & 0 & \phantom{-}1 & \phantom{-}0 & \phantom{-}0 \\ 0 & 1 & -40 & \phantom{-}1 & \phantom{-}1 \\ 0 & 0 & 800 & -40 & -20 \\ 0 & 0 & \phantom{-}40 & \phantom{-}0 & \phantom{-}0 \end{matrix} \right] \end{align*}$
and thus
$\displaystyle \begin{align*} 40\,C = 0 \implies C = 0 \end{align*}$
$\displaystyle \begin{align*} 800\,C - 40\,D = -20 \implies D = \frac{1}{2} \end{align*}$
$\displaystyle \begin{align*} B - 40\,C + D = 1 \implies B = \frac{1}{2} \end{align*}$
$\displaystyle \begin{align*} A + C = 0 \implies A = 0 \end{align*}$
So the partial fraction decomposition is
$\displaystyle \begin{align*} \frac{1}{2\,\left( s^2 - 20\,s + 200 \right) } + \frac{1}{2\,\left( s^2 + 20\,s + 200 \right) } &\equiv \frac{s^2 + 200}{\left( s^2 - 20\,s + 200 \right) \left( s^2 + 20\,s + 200 \right) } \end{align*}$
So moving on to the Inverse Laplace Transform now...
$\displaystyle \begin{align*} \mathcal{L}^{-1}\,\left\{ \frac{-3\,\left( s^2 + 200 \right) }{ s^2 + 40\,000 } \right\} &= -\frac{3}{2}\,\mathcal{L}^{-1} \,\left\{ \frac{1}{s^2 - 20\,s + 200 } + \frac{1}{s^2 + 20\,s + 200 } \right\} \\ &= -\frac{3}{2}\,\mathcal{L}^{-1}\,\left\{ \frac{1}{s^2 - 20\,s + \left( -10 \right) ^2 - \left( -10 \right) ^2 + 200 } + \frac{1}{s^2 + 20\,s + 10^2 - 10^2 + 200 } \right\} \\ &= -\frac{3}{2}\,\mathcal{L}^{-1} \, \left\{ \frac{1}{ \left( s - 10 \right) ^2 + 100 } + \frac{1}{ \left( s + 10 \right) ^2 + 100 } \right\} \\ &= -\frac{3}{2}\,\mathrm{e}^{10\,t}\,\mathcal{L}^{-1}\,\left\{ \frac{1}{s^2 + 10^2} \right\} - \frac{3}{2}\,\mathrm{e}^{-10\,t}\,\mathcal{L}^{-1}\,\left\{ \frac{1}{s^2 + 10^2} \right\} \\ &= -\frac{3}{2}\,\mathrm{e}^{-10\,t} \,\sin{ \left( 10\,t \right) } - \frac{3}{2}\,\mathrm{e}^{10\,t} \,\sin{ \left( 10\,t \right) } \\ &= -3\sin{ \left( 10\,t \right) } \left[ \frac{1}{2}\,\left( \mathrm{e}^{-10\,t} + \mathrm{e}^{10\,t} \right) \right] \\ &= -3\sin{ \left( 10\,t \right) } \cosh{ \left( 10\,t \right) } \end{align*}$
It's not entirely obvious what to do with this question, as the denominator does not easily factorise. However, if we realize that $\displaystyle \begin{align*} s^4 + 40\,000 = \left( s^2 \right) ^2 + 200^2 \end{align*}$ it's possible to do a sneaky completion of the square...
$\displaystyle \begin{align*} \left( s^2 \right) ^2 + 200^2 &= \left( s^2 \right) ^2 + 400\,s^2 + 200^2 - 400\,s^2 \\ &= \left( s^2 + 200 \right) ^2 - 400\,s^2 \\ &= \left( s^2 + 200 \right) ^2 - \left( 20\,s \right) ^2 \\ &= \left( s^2 - 20\,s + 200 \right) \left( s^2 + 20\,s + 200 \right) \end{align*}$
and thus it's now possible to perform partial fractions
$\displaystyle \begin{align*} \frac{A\,s + B}{s^2 - 20\,s + 200 } + \frac{C\,s + D}{s^2 + 20\,s + 200} &\equiv \frac{s^2 + 200}{\left( s^2 - 20\,s + 200 \right) \left( s^2 + 20\,s + 200 \right) } \\ \left( A\,s + B \right) \left( s^2 + 20\,s + 200 \right) + \left( C\,s + D \right) \left( s^2 - 20\,s + 200 \right) &\equiv s^2 + 200 \\ \left( A + C \right) \,s ^3 + \left( 20\,A + B - 20\,C + D \right) \, s^2 + \left( 200\,A + 20\,B + 200\,C - 20\,D \right) \, s + 200\,\left( B + D \right) &\equiv s^2 + 200 \end{align*}$
so it can be seen that $\displaystyle \begin{align*} A + C = 0 , \, \, 20\,A + B - 20\,C + D = 1 , \, \, 200\,A + 20\,B + 200\,C - 20\,D = 0 \textrm{ and } B+ D = 1 \end{align*}$, so solving the system gives
$\displaystyle \begin{align*} \left[ \begin{matrix} 1 & 0 & \phantom{-}1 & \phantom{-}0 & 0 \\ 20 & 1 & -20 & \phantom{-}1 & 1 \\ 200 & 20 & \phantom{-}200 & -20 & 0 \\ 0 & 1 & \phantom{-}0 & \phantom{-}1 & 1 \end{matrix} \right] \end{align*}$
apply R2 - 20R1 to R2 and R3 - 200R1 to R3 and we have
$\displaystyle \begin{align*} \left[ \begin{matrix} 1 & 0 & \phantom{-}1 & \phantom{-}0 & 0 \\ 0 & 1 & -40 & \phantom{-}1 & 1 \\ 0 & 20 & \phantom{-}0 & -20 & 0 \\ 0 & 1 & \phantom{-}0 & \phantom{-}1 & 1 \end{matrix} \right] \end{align*}$
apply R3 - 20R2 to R3 and R4 - R2 to R4 and we have
$\displaystyle \begin{align*} \left[ \begin{matrix} 1 & 0 & \phantom{-}1 & \phantom{-}0 & \phantom{-}0 \\ 0 & 1 & -40 & \phantom{-}1 & \phantom{-}1 \\ 0 & 0 & 800 & -40 & -20 \\ 0 & 0 & \phantom{-}40 & \phantom{-}0 & \phantom{-}0 \end{matrix} \right] \end{align*}$
and thus
$\displaystyle \begin{align*} 40\,C = 0 \implies C = 0 \end{align*}$
$\displaystyle \begin{align*} 800\,C - 40\,D = -20 \implies D = \frac{1}{2} \end{align*}$
$\displaystyle \begin{align*} B - 40\,C + D = 1 \implies B = \frac{1}{2} \end{align*}$
$\displaystyle \begin{align*} A + C = 0 \implies A = 0 \end{align*}$
So the partial fraction decomposition is
$\displaystyle \begin{align*} \frac{1}{2\,\left( s^2 - 20\,s + 200 \right) } + \frac{1}{2\,\left( s^2 + 20\,s + 200 \right) } &\equiv \frac{s^2 + 200}{\left( s^2 - 20\,s + 200 \right) \left( s^2 + 20\,s + 200 \right) } \end{align*}$
So moving on to the Inverse Laplace Transform now...
$\displaystyle \begin{align*} \mathcal{L}^{-1}\,\left\{ \frac{-3\,\left( s^2 + 200 \right) }{ s^2 + 40\,000 } \right\} &= -\frac{3}{2}\,\mathcal{L}^{-1} \,\left\{ \frac{1}{s^2 - 20\,s + 200 } + \frac{1}{s^2 + 20\,s + 200 } \right\} \\ &= -\frac{3}{2}\,\mathcal{L}^{-1}\,\left\{ \frac{1}{s^2 - 20\,s + \left( -10 \right) ^2 - \left( -10 \right) ^2 + 200 } + \frac{1}{s^2 + 20\,s + 10^2 - 10^2 + 200 } \right\} \\ &= -\frac{3}{2}\,\mathcal{L}^{-1} \, \left\{ \frac{1}{ \left( s - 10 \right) ^2 + 100 } + \frac{1}{ \left( s + 10 \right) ^2 + 100 } \right\} \\ &= -\frac{3}{2}\,\mathrm{e}^{10\,t}\,\mathcal{L}^{-1}\,\left\{ \frac{1}{s^2 + 10^2} \right\} - \frac{3}{2}\,\mathrm{e}^{-10\,t}\,\mathcal{L}^{-1}\,\left\{ \frac{1}{s^2 + 10^2} \right\} \\ &= -\frac{3}{2}\,\mathrm{e}^{-10\,t} \,\sin{ \left( 10\,t \right) } - \frac{3}{2}\,\mathrm{e}^{10\,t} \,\sin{ \left( 10\,t \right) } \\ &= -3\sin{ \left( 10\,t \right) } \left[ \frac{1}{2}\,\left( \mathrm{e}^{-10\,t} + \mathrm{e}^{10\,t} \right) \right] \\ &= -3\sin{ \left( 10\,t \right) } \cosh{ \left( 10\,t \right) } \end{align*}$
