Question

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

Answer 1

Hint: In this question, we need to determine the inverse Laplace transform of the given function $ \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 $ \dfrac{{2s - 1}}{{{s^3} - s}} $ can be re-written as:
 $ \dfrac{{2s - 1}}{{{s^3} - s}} = \dfrac{{2s - 1}}{{s({s^2} - 1)}} - - - - (i) $ .
Now, using the arithmetic identity $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ in the above function where $ a = s $ and $ b = 1 $ .
Hence, equation (i) can be re-written as:
 $
  \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:
 $
  \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
 $
  \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
 $
  \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 = {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) \;
  $
On solving the above three equations, we get $ A = 1 $ .
Now we substitute the value of A in the equation $ A + B + C = 0 $ we get
 $
  1 + B + C = 0 \\
   \Rightarrow B + C = - 1 - - (d) \;
  $
Now we add equation (b) with the equation (d), hence we get
 $
  B - C = 2 \\
  \underline {B + C = - 1} \\
  2B = 1 \\
  B = \dfrac{1}{2} \;
  $
Hence we get the value of $ B = \dfrac{1}{2} $ .
Now substitute the value of B in the equation (d), we get
 $
  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 = \dfrac{1}{2}{\text{ and }}C = - \dfrac{3}{2} $ .
Substituting these values in the equation (iii), we get
 $
  \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
\[{L^{ - 1}}\left[ {\dfrac{{2s - 1}}{{{s^3} - s}}} \right] = {L^{ - 1}}\left[ {\dfrac{1}{s} + \dfrac{1}{{2\left( {s - 1} \right)}} + \dfrac{3}{{2\left( {s + 1} \right)}}} \right]\]
As, the Laplace transform are distributive in nature so, we can write the above equation as;
 $ {L^{ - 1}}\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}}\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 $ \dfrac{{2s - 1}}{{{s^3} - s}} $ is $ \left( {1 + \dfrac{1}{2}{e^t} + \dfrac{3}{2}{e^{ - t}}} \right) $ .
So, the correct answer is “ $ \dfrac{{2s - 1}}{{{s^3} - s}} $ is $ \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}}\left[ {\dfrac{1}{s}} \right] = u(s) $ where, u(s) is the unit step function.
 $ {L^{ - 1}}\left[ {\dfrac{1}{{s + a}}} \right] = {e^{ - at}} $