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Henry's question via email about an Inverse Laplace Transform

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
inv laplace.png

It's not entirely obvious what to do with this question, as the denominator does not easily factorise. However, if we realise 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...

$\displaystyle \begin{align*} \mathcal{L}^{-1}\,\left\{ \frac{-3\,\left( s^2 + 200 \right) }{ s^2 + 40\,000 } \right\} &= -\frac{3}{2}\,\mathcal{L}^{-1} \,\left\{ \frac{1}{s^2 - 20\,s + 200 } + \frac{1}{s^2 + 20\,s + 200 } \right\} \\ &= -\frac{3}{2}\,\mathcal{L}^{-1}\,\left\{ \frac{1}{s^2 - 20\,s + \left( -10 \right) ^2 - \left( -10 \right) ^2 + 200 } + \frac{1}{s^2 + 20\,s + 10^2 - 10^2 + 200 } \right\} \\ &= -\frac{3}{2}\,\mathcal{L}^{-1} \, \left\{ \frac{1}{ \left( s - 10 \right) ^2 + 100 } + \frac{1}{ \left( s + 10 \right) ^2 + 100 } \right\} \\ &= -\frac{3}{2}\,\mathrm{e}^{10\,t}\,\mathcal{L}^{-1}\,\left\{ \frac{1}{s^2 + 10^2} \right\} - \frac{3}{2}\,\mathrm{e}^{-10\,t}\,\mathcal{L}^{-1}\,\left\{ \frac{1}{s^2 + 10^2} \right\} \\ &= -\frac{3}{2}\,\mathrm{e}^{-10\,t} \,\sin{ \left( 10\,t \right) } - \frac{3}{2}\,\mathrm{e}^{10\,t} \,\sin{ \left( 10\,t \right) } \\ &= -3\sin{ \left( 10\,t \right) } \left[ \frac{1}{2}\,\left( \mathrm{e}^{-10\,t} + \mathrm{e}^{10\,t} \right) \right] \\ &= -3\sin{ \left( 10\,t \right) } \cosh{ \left( 10\,t \right) } \end{align*}$
 

Joppy

Well-known member
MHB Math Helper
Mar 17, 2016
256
I'm confused. Who is Collin's and Henry and all the others?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
I'm confused. Who is Collin's and Henry and all the others?
My students, they email me asking for help, and I direct them here...