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Question: Obtain the inverse Laplace transform : \[\]\(\dfrac{{2s - 1}}{{{s^3} - s}}\)...

Obtain the inverse Laplace transform : $$$$2s−1s3−s\dfrac{{2s - 1}}{{{s^3} - s}}

Explanation

Solution

To find the inverse Laplace form we use the following notation:
f(t) = {L^{ - 1}}\left\\{ {F(s)} \right\\}
In the following question ,F(s)=2s−1s3−sF(s) = \dfrac{{2s - 1}}{{{s^3} - s}}

Complete step by step answer:
f(t) = {L^{ - 1}}\left\\{ {\dfrac{{2s - 1}}{{{s^3} - s}}} \right\\}
Now,
= {L^{ - 1}}\left\\{ {\dfrac{{2s - 1}}{{{s^3} - s}}} \right\\} \\\ = {L^{ - 1}}\left\\{ {\dfrac{{2s - 1}}{{s({s^2} - 1)}}} \right\\} \\\ = {L^{ - 1}}\left\\{ {\dfrac{{2s - 1}}{{s(s - 1)(s + 1)}}} \right\\} \\\
Using partial fraction ;
As+Bs−1+Cs+1=2s−1s(s−1)(s+1) A(s−1)(s+1)+Bs(s+1)+Cs(s−1)s(s−1)(s+1)=2s−1s(s−1)(s+1) A(s2−1)+B(s2+s)+C(s2−s)s(s−1)(s+1)==2s−1s(s−1)(s+1)  \dfrac{A}{s} + \dfrac{B}{{s - 1}} + \dfrac{C}{{s + 1}} = \dfrac{{2s - 1}}{{s(s - 1)(s + 1)}} \\\ \dfrac{{A(s - 1)(s + 1) + Bs(s + 1) + Cs(s - 1)}}{{s(s - 1)(s + 1)}} = \dfrac{{2s - 1}}{{s(s - 1)(s + 1)}} \\\ \dfrac{{A({s^2} - 1) + B({s^2} + s) + C({s^2} - s)}}{{s(s - 1)(s + 1)}} = = \dfrac{{2s - 1}}{{s(s - 1)(s + 1)}} \\\

Comparing the coefficients of s2{s^2}:
A+B+C=0−(i)A + B + C = 0 - (i)
Comparing the coefficient of ss:
B−C=2−(ii)B - C = 2 - (ii)
Comparing the coefficients of constants:
\-A=−1 A=1−(iii)  \- A = - 1 \\\ A = 1 - (iii) \\\
From (i)(i) and (ii)(ii)
B=C+2 A+B+C=0 A+C+2+C=0 A+2C=−2   B = C + 2 \\\ A + B + C = 0 \\\ A + C + 2 + C = 0 \\\ A + 2C = - 2 \\\ \\\
From (iii)(iii)
1+2c=−2 2c=−2−1 2c=−3 c=−32  1 + 2c = - 2 \\\ 2c = - 2 - 1 \\\ 2c = - 3 \\\ c = \dfrac{{ - 3}}{2} \\\
Now,
B=C+2 B=−32+2 B=12  B = C + 2 \\\ B = \dfrac{{ - 3}}{2} + 2 \\\ B = \dfrac{1}{2} \\\
Hence;
= {L^{ - 1}}\left\\{ {\dfrac{1}{s} + \dfrac{1}{2}(\dfrac{1}{{s - 1}}) - \dfrac{3}{2}(\dfrac{1}{{s + 1}})} \right\\} \\\ = {L^{ - 1}}\left\\{ {\dfrac{1}{s}} \right\\} + \dfrac{1}{2}{L^{ - 1}}\left\\{ {\dfrac{1}{{s - 1}}} \right\\} - \dfrac{3}{2}{L^{ - 1}}\left\\{ {\dfrac{1}{{s + 1}}} \right\\} \\\ \\\
Since, the standard formula is {L^{ - 1}}\left\\{ {\dfrac{1}{{s - a}}} \right\\} = {e^{at}}
=e0t+12e1t−32e−1t =1+12et−32e−t  = {e^{0t}} + \dfrac{1}{2}{e^{1t}} - \dfrac{3}{2}{e^{ - 1t}} \\\ = 1 + \dfrac{1}{2}{e^t} - \dfrac{3}{2}{e^{ - t}} \\\

Note: Laplace transform is a mathematical tool mostly used in the engineering field to transform an equation from one form to another. It is also an integral transform that converts a function of a real variable (often time) to a function of a complex variable.