What is inverse oflaplace 3/((s+1)^2) ?

The answer to inverse oflaplace 3/((s+1)^2) is e^{-t}*3t

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inverse oflaplace 3/((s+1)^2)

inverse laplace

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

e^{- t} \cdot 3 t

Solution steps

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

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

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

= e^{- t} \cdot 3 t

n = 1 \sum \infty \frac{4 ^{n}}{2 ^{n}}

x y - 3 y^{2} = 8, (14, 4)

\int x e^{- x} d x

\frac{d}{d x} (\frac{2 x}{ln ( x )})

\frac{d}{d x} (\frac{x ^{3}}{1 - x ^{3}})

What is inverse oflaplace 3/((s+1)^2) ?
The answer to inverse oflaplace 3/((s+1)^2) is e^{-t}*3t