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

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

Explanation

Solution

Hint : In this question, we need to determine the inverse Laplace transform of the given function 2s−1s3−s\dfrac{{2s - 1}}{{{s^3} - s}} . For this, we will use the properties of the inverse Laplace function after redefining the given function in the simplest form. We also use partial fractions methods to resolve the s terms.

Complete step by step solution:
The given function 2s−1s3−s\dfrac{{2s - 1}}{{{s^3} - s}} can be re-written as:
2s−1s3−s=2s−1s(s2−1)−−−−(i)\dfrac{{2s - 1}}{{{s^3} - s}} = \dfrac{{2s - 1}}{{s({s^2} - 1)}} - - - - (i) .
Now, using the arithmetic identity a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) in the above function where a=sa = s and b=1b = 1 .
Hence, equation (i) can be re-written as:
2s−1s3−s=2s−1s(s2−1) =2s−1s(s−1)(s+1)−−−−(ii)  \dfrac{{2s - 1}}{{{s^3} - s}} = \dfrac{{2s - 1}}{{s({s^2} - 1)}} \\\ = \dfrac{{2s - 1}}{{s(s - 1)(s + 1)}} - - - - (ii) \\\
Now, applying partial fraction method to simplify the given equation as:
2s−1s3−s=2s−1s(s−1)(s+1) =As+Bs−1+Cs+1−−−−(iii)  \dfrac{{2s - 1}}{{{s^3} - s}} = \dfrac{{2s - 1}}{{s(s - 1)(s + 1)}} \\\ = \dfrac{A}{s} + \dfrac{B}{{s - 1}} + \dfrac{C}{{s + 1}} - - - - (iii) \\\
From the equation (iii), we can write
2s−1s3−s=As+Bs−1+Cs+1 =A(s−1)(s+1)+Bs(s+1)+Cs(s−1)s3−s  \dfrac{{2s - 1}}{{{s^3} - s}} = \dfrac{A}{s} + \dfrac{B}{{s - 1}} + \dfrac{C}{{s + 1}} \\\ = \dfrac{{A(s - 1)(s + 1) + Bs(s + 1) + Cs(s - 1)}}{{{s^3} - s}} \\\
Simplifying the numerator of the above equation, we get
2s−1s3−s=A(s−1)(s+1)+Bs(s+1)+Cs(s−1)s3−s =As2−A+Bs2+Bs+Cs2−Css3−s =s2(A+B+C)+s(B−C)−As3−s−−−−(iv)  \dfrac{{2s - 1}}{{{s^3} - s}} = \dfrac{{A(s - 1)(s + 1) + Bs(s + 1) + Cs(s - 1)}}{{{s^3} - s}} \\\ = \dfrac{{A{s^2} - A + B{s^2} + Bs + C{s^2} - Cs}}{{{s^3} - s}} \\\ = \dfrac{{{s^2}\left( {A + B + C} \right) + s\left( {B - C} \right) - A}}{{{s^3} - s}} - - - - (iv) \\\
Now, comparing the numerator of the equation (iv), we get
2s−1=s2(A+B+C)+s(B−C)−A2s - 1 = {s^2}\left( {A + B + C} \right) + s\left( {B - C} \right) - A
So, we can write
A+B+C=0−−(a) B−C=2−−(b) \-A=−1−−(c)   A + B + C = 0 - - (a) \\\ B - C = 2 - - (b) \\\ \- A = - 1 - - (c) \;
On solving the above three equations, we get A=1A = 1 .
Now we substitute the value of A in the equation A+B+C=0A + B + C = 0 we get
1+B+C=0 ⇒B+C=−1−−(d)   1 + B + C = 0 \\\ \Rightarrow B + C = - 1 - - (d) \;
Now we add equation (b) with the equation (d), hence we get
B−C=2 B+C=−1‾ 2B=1 B=12   B - C = 2 \\\ \underline {B + C = - 1} \\\ 2B = 1 \\\ B = \dfrac{1}{2} \;
Hence we get the value of B=12B = \dfrac{1}{2} .
Now substitute the value of B in the equation (d), we get
B+C=−1 ⇒12+C=−1 ⇒C=−1−12 ⇒C=−32   B + C = - 1 \\\ \Rightarrow \dfrac{1}{2} + C = - 1 \\\ \Rightarrow C = - 1 - \dfrac{1}{2} \\\ \Rightarrow C = - \dfrac{3}{2} \;
Hence, we get the value of the constants as
A=1;B=12 and C=−32A = 1;B = \dfrac{1}{2}{\text{ and }}C = - \dfrac{3}{2} .
Substituting these values in the equation (iii), we get
2s−1s3−s=As+Bs−1+Cs+1 =1s+12s−1−−32s+1−−−−(v)  \dfrac{{2s - 1}}{{{s^3} - s}} = \dfrac{A}{s} + \dfrac{B}{{s - 1}} + \dfrac{C}{{s + 1}} \\\ = \dfrac{1}{s} + \dfrac{{\dfrac{1}{2}}}{{s - 1}} - \dfrac{{ - \dfrac{3}{2}}}{{s + 1}} - - - - (v) \\\
Applying inverse Laplace transform to both sides of the equation (v), we get
L−1[2s−1s3−s]=L−1[1s+12(s−1)+32(s+1)]{L^{ - 1}}\left[ {\dfrac{{2s - 1}}{{{s^3} - s}}} \right] = {L^{ - 1}}\left[ {\dfrac{1}{s} + \dfrac{1}{{2\left( {s - 1} \right)}} + \dfrac{3}{{2\left( {s + 1} \right)}}} \right]
As, the Laplace transform are distributive in nature so, we can write the above equation as;
L−1[2s−1s3−s]=L−1[1s]+12L−1[1s−1]+32L−1[1s+1]{L^{ - 1}}\left[ {\dfrac{{2s - 1}}{{{s^3} - s}}} \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]
Using the properties of the inverse Laplace transform in the above equation, we get
L−1[2s−1s3−s]=L−1[1s]+12L−1[1s−1]+32L−1[1s+1] =1+et2+32e−t  {L^{ - 1}}\left[ {\dfrac{{2s - 1}}{{{s^3} - s}}} \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] \\\ = 1 + \dfrac{{{e^t}}}{2} + \dfrac{3}{2}{e^{ - t}} \\\
Hence, the inverse Laplace transform of the function 2s−1s3−s\dfrac{{2s - 1}}{{{s^3} - s}} is (1+12et+32e−t)\left( {1 + \dfrac{1}{2}{e^t} + \dfrac{3}{2}{e^{ - t}}} \right) .
So, the correct answer is “ 2s−1s3−s\dfrac{{2s - 1}}{{{s^3} - s}} is (1+12et+32e−t)\left( {1 + \dfrac{1}{2}{e^t} + \dfrac{3}{2}{e^{ - t}}} \right) ”.

Note : It is worth noting down here that the partial fraction plays a very important role in simplifying the function that will be suitable to apply the direct properties of the inverse Laplace transform. Some of the properties for the inverse Laplace transform which are used here in this question are:
L−1[1s]=u(s){L^{ - 1}}\left[ {\dfrac{1}{s}} \right] = u(s) where, u(s) is the unit step function.
L−1[1s+a]=e−at{L^{ - 1}}\left[ {\dfrac{1}{{s + a}}} \right] = {e^{ - at}}