Question

what is the inverse Laplace transform of 1 upon x square in bracket X - 1
​

Answer 1

SOLUTION

TO DETERMINE

The inverse Laplace transform of

$\displaystyle \sf{ \frac{1}{ {s}^{2}(s - 1) } }$

EVALUATION

Here the given function is

$\displaystyle \sf{ \frac{1}{ {s}^{2}(s - 1) } }$

Let

$\displaystyle \sf{ \frac{1}{ {s}^{2}(s - 1) } = \frac{a}{s} + \frac{b}{ {s}^{2} } + \frac{c}{s - 1} }$

$\displaystyle \sf{ \implies \frac{1}{ {s}^{2}(s - 1) } = \frac{as(s - 1) + b(s - 1) + c {s}^{2} }{ {s}^{2}(s - 1) } }$

Comparing both sides we get

a + c = 0 , b - a = 0 , - b = 1

Which gives

a = 1 , b = - 1 , c = - 1

Thus we get

$\displaystyle \sf{ \frac{1}{ {s}^{2}(s - 1) } = \frac{1}{s} - \frac{1}{ {s}^{2} } - \frac{1}{s - 1} }$

Taking inverse Laplace transform we get

$\displaystyle \sf{{L}^{ - 1} \bigg[ \frac{1}{ {s}^{2}(s - 1) }\bigg] = {L}^{ - 1} \bigg[\frac{1}{s} - \frac{1}{ {s}^{2} } - \frac{1}{s - 1} \bigg]}$

$\displaystyle \sf{ \implies \: {L}^{ - 1} \bigg[ \frac{1}{ {s}^{2}(s - 1) }\bigg] = {L}^{ - 1} \bigg[\frac{1}{s} \bigg] - {L}^{ - 1} \bigg[\frac{1}{ {s}^{2} }\bigg] - {L}^{ - 1} \bigg[\frac{1}{s - 1} \bigg]}$

$\displaystyle \sf{ \implies \: {L}^{ - 1} \bigg[ \frac{1}{ {s}^{2}(s - 1) }\bigg] = 1 - t - {e}^{ t} }$

FINAL ANSWER

$\displaystyle \sf{ \: {L}^{ - 1} \bigg[ \frac{1}{ {s}^{2}(s - 1) }\bigg] = 1 - t - {e}^{ t} }$

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