What is inverse oflaplace 1/(s^4+s^3) ?

The answer to inverse oflaplace 1/(s^4+s^3) is \H(t)-t+(t^2)/2-e^{-t}

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inverse oflaplace 1/(s^4+s^3)

inverse laplace

\frac{1}{s ^{4} + s ^{3}}

H (t) - t + \frac{t ^{2}}{2} - e^{- t}

Solution steps

L^{- 1} {\frac{1}{s ^{4} + s ^{3}}}

Take the partial fraction of

\frac{1}{s ^{4} + s ^{3}} : \frac{1}{s} - \frac{1}{s ^{2}} + \frac{1}{s ^{3}} - \frac{1}{s + 1}

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

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

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

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

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

= H (t) - t + \frac{t ^{2}}{2} - e^{- t}

x \to - \infty lim (\frac{8 x ^{2} + 7 x ^{\frac{1}{3}}}{16 x ^{4} + 6})

(cos (t))^{^{'}}

\frac{d}{d x} ((x e^{2 x} + x^{3})^{6})

\int \frac{1}{2 x - x ^{2}} d x

What is inverse oflaplace 1/(s^4+s^3) ?
The answer to inverse oflaplace 1/(s^4+s^3) is \H(t)-t+(t^2)/2-e^{-t}