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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

3^{1 - 2 n}

\frac{\partial}{\partial x} (3 x y - x^{2} y^{2} + 23)

\frac{d y}{d x} = \frac{x}{e ^{y}}

3 x^{2} - 2

f (x) = - 9.8 x^{2} + 2 x + 1

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