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

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

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

inverse laplace

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

- 3 e^{- 3 t} t

Solution steps

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

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

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

= - 3 e^{- 3 t} t

17 e^{x}

\frac{\partial}{\partial t} (e^{- t})

\int cos (3 x) sin (x) d x

(9 x^{6} + 4 x^{3})^{4}

8 \cdot e^{(- 2 t)} \cdot t \cdot cos (3 t) \cdot cos (t)

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