What is inverse oflaplace 9/((s+3)^2) ?

The answer to inverse oflaplace 9/((s+3)^2) is e^{-3t}*9t

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inverse oflaplace 9/((s+3)^2)

inverse laplace

\frac{9}{( s + 3 ) ^{2}}

e^{- 3 t} \cdot 9 t

Solution steps

L^{- 1} {\frac{9}{( s + 3 ) ^{2}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{9}{( s + 3 ) ^{2}} : a = - 3, F (s) = \frac{9}{s ^{2}}

= e^{- 3 t} L^{- 1} {\frac{9}{s ^{2}}}

L^{- 1} {\frac{9}{s ^{2}}} : 9 t

= e^{- 3 t} \cdot 9 t

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What is inverse oflaplace 9/((s+3)^2) ?
The answer to inverse oflaplace 9/((s+3)^2) is e^{-3t}*9t