What is inverse oflaplace 4/(s^2(s+1)) ?

The answer to inverse oflaplace 4/(s^2(s+1)) is -4\H(t)+4t+4e^{-t}

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inverse oflaplace 4/(s^2(s+1))

inverse laplace

\frac{4}{s ^{2} ( s + 1 )}

- 4 H (t) + 4 t + 4 e^{- t}

Solution steps

L^{- 1} {\frac{4}{s ^{2} ( s + 1 )}}

Take the partial fraction of

\frac{4}{s ^{2} ( s + 1 )} : - \frac{4}{s} + \frac{4}{s ^{2}} + \frac{4}{s + 1}

= L^{- 1} {- \frac{4}{s} + \frac{4}{s ^{2}} + \frac{4}{s + 1}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= - L^{- 1} {\frac{4}{s}} + L^{- 1} {\frac{4}{s ^{2}}} + L^{- 1} {\frac{4}{s + 1}}

L^{- 1} {\frac{4}{s}} : 4 H (t)

L^{- 1} {\frac{4}{s ^{2}}} : 4 t

L^{- 1} {\frac{4}{s + 1}} : 4 e^{- t}

= - 4 H (t) + 4 t + 4 e^{- t}

n \to \infty lim (\frac{3 ^{n + 1}}{( n ) !})

y^{^{'}} = \frac{x ^{3} - 2 y}{x}

\int t^{2} + e^{t} d t

\int x^{3} e^{4 x} d x

\int_{0}^{π} 9 sin^{2} (4 x) d x

What is inverse oflaplace 4/(s^2(s+1)) ?
The answer to inverse oflaplace 4/(s^2(s+1)) is -4\H(t)+4t+4e^{-t}