The inverse Laplace Transform of $\frac{s^2}{(s-3)^3}$ can be written as $\frac{e^{3t}}{2}[At^2 + Bt + C]$. The values of $A$, $B$ and $C$, respectively are:
Options:
(A) 3, 5 and 7
(B) 2, 10 and 12
(C) 10, 12 and 4
(D) 9, 12 and 2
Official Answer Key:
D
Step-by-Step Solution & Explanation
Let $F(s) = \frac{s^2}{(s-3)^3}$. By the first shifting theorem, let $s - 3 = p \implies s = p + 3$.
$F(s) = \frac{(p+3)^2}{p^3} = \frac{p^2 + 6p + 9}{p^3} = \frac{1}{p} + \frac{6}{p^2} + \frac{9}{p^3}$.
Taking the inverse Laplace transform with respect to $p$:
$\mathcal{L}^{-1}\left\{ \frac{1}{p} + \frac{6}{p^2} + \frac{9}{p^3} \right\} = 1 + 6t + \frac{9}{2}t^2$.
Multiplying by $e^{3t}$:
$f(t) = e^{3t} \left( 1 + 6t + \frac{9}{2}t^2 \right) = \frac{e^{3t}}{2} [9t^2 + 12t + 2]$.
Comparing with $\frac{e^{3t}}{2}[At^2 + Bt + C]$:
$A = 9$, $B = 12$, $C = 2$. Option (D).
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