Question

Find the inverse laplace transform for 1/s(s+a) ?

Answer 1

Answer:

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Answer 2

Given,

$\[\frac{1}{{s\left( {s + a} \right)}}\]$

Solution,

$\[ = \frac{1}{{s\left( {s + a} \right)}}\]$

Calculate the partial fraction,

$\[ = \frac{A}{s} + \frac{B}{{s + a}}\]$

$\[ = \frac{{As + Aa + Bs}}{{s + a}}\]$--------(1)

$\[\begin{array}{l}A + B = 0\\aA = 1\\ \Rightarrow A = \frac{1}{a}\\ \Rightarrow B = - \frac{1}{a}\end{array}\]$

Put the values in equation 1,

$\[ = \frac{1}{{as}} - \frac{1}{{a\left( {s + a} \right)}}\]$

Calculate the inverse Laplace transform,

$\[ = {L^{ - 1}}\left( {\frac{1}{{as}}} \right) - {L^{ - 1}}\left( {\frac{1}{{a\left( {s + a} \right)}}} \right)\]$

$\[ = \frac{1}{a}{L^{ - 1}}\left( {\frac{1}{s}} \right) - \frac{1}{a}{L^{ - 1}}\left( {\frac{1}{{\left( {s + a} \right)}}} \right)\]$

$\[ = \frac{1}{a} \times 1 - \frac{1}{a} \times {e^{ - at}}\]$

$\[ = \frac{1}{a}\left( {1 - {e^{ - at}}} \right)\]$

Hence the inverse Laplace transform is $\[\frac{1}{a}\left( {1 - {e^{ - at}}} \right)\]$