Question

Find the Laplace transform: \[\dfrac{1}{{(s + 2)({s^2} + 4s + 9)}}\]

Answer 1

Hint: Here in this question, we have to find the Laplace transform of the given function. We solve the given function first by the partial fractions and then we determine the constant values for the partial fractions and then we apply the inverse Laplace transform formula and hence we determine the solution for the given function.

Complete step by step solution:
Consider the given function
  \[\dfrac{1}{{(s + 2)({s^2} + 4s + 9)}}\]
This can be written as
  \[ \Rightarrow \dfrac{1}{{(s + 2)({s^2} + 4s + 9)}} = \dfrac{A}{{(s + 2)}} + \dfrac{{Bs + C}}{{{s^2} + 4s + 9}}\]
Take the LCM in the RHS we get
  \[ \Rightarrow \dfrac{1}{{(s + 2)({s^2} + 4s + 9)}} = \dfrac{{A({s^2} + 4s + 9) + Bs + C(s + 2)}}{{(s + 2)({s^2} + 4s + 9)}}\]
Hence on simplifying we have
  \[ \Rightarrow 1 = A{s^2} + 4As + 9A + B{s^2} + 2Bs + Cs + 2C\]
Equate the coefficient of \[{s^2}\] , we have
  \[ \Rightarrow 0 = A + B\]
  \[ \Rightarrow A = - B\] --- (1)
Equate the coefficient of \[s\] , we have
  \[ \Rightarrow 0 = 4A + 2B + C\]
  \[ \Rightarrow C = - 4A - 2B\] ----(2)
Now equate the coefficients of the constant we get
  \[ \Rightarrow 1 = 9A + 2C\] ----- (3)
Substitute the (2) in the (3) we get
  \[ \Rightarrow 1 = 9A + 2( - 4A - 2B)\]
On simplifying we get
  \[ \Rightarrow 1 = 9A - 8A - 4B\] --- (4)
Substitute the (1) in (4) we get
  \[ \Rightarrow 1 = - 9B + 8B - 4B\]
On simplifying we get
  \[ \Rightarrow 1 = - 5B\]
  \[ \Rightarrow B = - \dfrac{1}{5}\]
Therefore from the equation (1) we have
  \[ \Rightarrow A = \dfrac{1}{5}\]
From the equation (3) we have
  \[ \Rightarrow 1 = \dfrac{9}{5} + 2C\]
Take \[\dfrac{9}{5}\] to LHS we get
  \[ \Rightarrow 1 - \dfrac{9}{5} = 2C\]
  \[ \Rightarrow - \dfrac{4}{5} = 2C\]
On simplifying the above equation we get
  \[ \Rightarrow C = - \dfrac{2}{5}\]
Therefore the given equation is written as
  \[ \Rightarrow \dfrac{1}{{(s + 2)({s^2} + 4s + 9)}} = \dfrac{1}{{5(s + 2)}} - \dfrac{s}{{5({s^2} + 4s + 9)}} - \dfrac{2}{{5({s^2} + 4s + 9)}}\]
On applying the inverse Laplace transform to the above function
  \[ \Rightarrow \dfrac{1}{5}{L^{ - 1}}\left( {\dfrac{1}{{(s + 2)}}} \right) - \dfrac{1}{5}{L^{ - 1}}\left( {\dfrac{s}{{({s^2} + 4s + 4) + 5}}} \right) - \dfrac{2}{5}{L^{ - 1}}\left( {\dfrac{1}{{({s^2} + 4s + 4) + 5}}} \right)\]
On simplifying we get
  \[ \Rightarrow \dfrac{1}{5}{L^{ - 1}}\left( {\dfrac{1}{{(s + 2)}}} \right) - \dfrac{1}{5}{L^{ - 1}}\left( {\dfrac{s}{{{{(s + 2)}^2} + {{(\sqrt 5 )}^2}}}} \right) - \dfrac{2}{5}{L^{ - 1}}\left( {\dfrac{1}{{{{(s + 2)}^2} + {{(\sqrt 5 )}^2}}}} \right)\]
  \[ \Rightarrow \dfrac{1}{5}{L^{ - 1}}\left( {\dfrac{1}{{(s + 2)}}} \right) - \dfrac{1}{5}{L^{ - 1}}\left( {\dfrac{{s + 2 - 2}}{{{{(s + 2)}^2} + {{(\sqrt 5 )}^2}}}} \right) - \dfrac{2}{5}{L^{ - 1}}\left( {\dfrac{1}{{{{(s + 2)}^2} + {{(\sqrt 5 )}^2}}}} \right)\]
  \[ \Rightarrow \dfrac{1}{5}{L^{ - 1}}\left( {\dfrac{1}{{(s + 2)}}} \right) - \dfrac{1}{5}{L^{ - 1}}\left( {\dfrac{{s + 2}}{{{{(s + 2)}^2} + {{(\sqrt 5 )}^2}}}} \right) + \dfrac{2}{5}{L^1}\left( {\dfrac{1}{{{{(s + 2)}^2} + {{(\sqrt 5 )}^2}}}} \right) - \dfrac{2}{5}{L^{ - 1}}\left( {\dfrac{1}{{{{(s + 2)}^2} + {{(\sqrt 5 )}^2}}}} \right)\]
On simplifying we get
  \[ \Rightarrow \dfrac{1}{5}{L^{ - 1}}\left( {\dfrac{1}{{(s + 2)}}} \right) - \dfrac{1}{5}{L^{ - 1}}\left( {\dfrac{{s + 2}}{{{{(s + 2)}^2} + {{(\sqrt 5 )}^2}}}} \right)\]
On applying the formula we get
  \[ \Rightarrow \dfrac{1}{5}{e^{ - 2t}} - \dfrac{1}{5}{e^{ - 2t}}\cos \left( {\sqrt 5 t} \right)\]
On taking common the above inequality is written as
  \[ \Rightarrow \dfrac{{{e^{ - 2t}}}}{5}\left( {1 - \cos \left( {\sqrt 5 t} \right)} \right)\]
Therefore the Laplace transform of \[\dfrac{1}{{(s + 2)({s^2} + 4s + 9)}}\] is \[\dfrac{{{e^{ - 2t}}}}{5}\left( {1 - \cos \left( {\sqrt 5 t} \right)} \right)\]
So, the correct answer is “ \[\dfrac{{{e^{ - 2t}}}}{5}\left( {1 - \cos \left( {\sqrt 5 t} \right)} \right)\] ”.

Note: While solving the problems related to the Laplace transform we must know the Laplace transform formulas. Here we have used the Laplace formula, that is \[{L^{ - 1}}\left( {\dfrac{1}{{(s - a)}}} \right) = {e^{at}}\] and \[{L^{ - 1}}\left( {\dfrac{1}{{{{(s - a)}^2} + {b^2}}}} \right) = {e^{at}}\cos bt\] , hence we can obtain the required solution for the given question.