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

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

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

inverse laplace

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

e^{- 3 t} \cdot 9 t

Solution steps

L^{- 1} {\frac{3 ^{2}}{( 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{3 ^{2}}{( s + 3 ) ^{2}} : a = - 3, F (s) = \frac{3 ^{2}}{s ^{2}}

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

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

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

\int 1 + cos (t) d t

y = 2 x^{3}, (- 1, - 2)

\int tan (y) d y

\int \frac{x ^{6} e ^{x} - 4 x ^{5}}{x ^{6}} d x

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