The inverse Laplace transform of Fs -s -3-s2 - 2s -1 for t - 0 is

F(s) =s +3(s^2 +2s +1) = s+3(s+1)^2 =s(s+1)^2 +3(s +1)^2 =(s+1) 1(s+1)^2+3(s+1)^2 =s+1(s+1)^2 1(s +1)^2 +3(s +1)^2 =1s+1 +2(s +1)^2 ∵ f(t) = L^ 1 [1s+1 + 2(s ...

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The inverse L...

Question

The inverse Laplace transform of $F (s) = \frac{s + 3}{s^{2} + 2 s + 1}$ for $t \geq 0$ is

e^{- t} + 3 t e^{- t}

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e^{- t} + 4 t e^{- t}

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e^{- t} + 2 t e^{- t}

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3 t e^{- t}

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Solution

The correct option is C $e^{- t} + 2 t e^{- t}$
$F (s) = \frac{s + 3}{(s^{2} + 2 s + 1)} = \frac{s + 3}{(s + 1)^{2}} = \frac{s}{(s + 1)^{2}} + \frac{3}{(s + 1)^{2}}$

$= \frac{(s + 1) - 1}{(s + 1)^{2}} + \frac{3}{(s + 1)^{2}} = \frac{s + 1}{(s + 1)^{2}} - \frac{1}{(s + 1)^{2}} + \frac{3}{(s + 1)^{2}}$

$= \frac{1}{s + 1} + \frac{2}{(s + 1)^{2}}$

$∵ f (t) = L^{- 1} [\frac{1}{s + 1} + \frac{2}{(s + 1)^{2}}]$

$= e^{- t} + 2 t e^{- t}$

Note : $L^{- 1} [\frac{1}{s + a}] = e^{- a t}$

$L^{- 1} [\frac{1}{(s + a)^{n}}] = \frac{e^{- a t} t^{n - 1}}{(n - 1)!}$

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The inverse Laplace transform of F(s)=s+3s2+2s+1 for t≥0 is

The correct option is C e−t+2te−tF(s)=s+3(s2+2s+1)=s+3(s+1)2=s(s+1)2+3(s+1)2 =(s+1)−1(s+1)2+3(s+1)2=s+1(s+1)2−1(s+1)2+3(s+1)2 =1s+1+2(s+1)2 ∵f(t)=L−1[1s+1+2(s+1)2] =e−t+2te−t Note : L−1[1s+a]=e−at L−1[1(s+a)n]=e−attn−1(n−1)!

The inverse Laplace transform of $F (s) = \frac{s + 3}{s^{2} + 2 s + 1}$ for $t \geq 0$ is