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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

\frac{d}{d x} (5 x tan (x))

y^{^{'}} = t + t y^{2}, y (2) = 0

\frac{d}{d x} (\frac{12}{x ^{3}})

\frac{d}{d x} (\frac{x - 9 x x}{x})

f (x) = \frac{x}{5}

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