What is inverse oflaplace (0.2)/((10s^3+s^2)) ?

The answer to inverse oflaplace (0.2)/((10s^3+s^2)) is -2\H(t)+0.2t+2e^{-t/(10)}

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inverse oflaplace (0.2)/((10s^3+s^2))

inverse laplace

\frac{0.2}{( 10 s ^{3} + s ^{2} )}

- 2 H (t) + 0.2 t + 2 e^{- \frac{t}{10}}

Solution steps

L^{- 1} {\frac{0.2}{( 10 s ^{3} + s ^{2} )}}

Take the partial fraction of

\frac{0.2}{10 s ^{3} + s ^{2}} : - \frac{2}{s} + \frac{0.2}{s ^{2}} + \frac{20}{10 s + 1}

= L^{- 1} {- \frac{2}{s} + \frac{0.2}{s ^{2}} + \frac{20}{10 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{2}{s}} + L^{- 1} {\frac{0.2}{s ^{2}}} + L^{- 1} {\frac{20}{10 s + 1}}

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

L^{- 1} {\frac{0.2}{s ^{2}}} : 0.2 t

L^{- 1} {\frac{20}{10 s + 1}} : 2 e^{- \frac{t}{10}}

= - 2 H (t) + 0.2 t + 2 e^{- \frac{t}{10}}

What is inverse oflaplace (0.2)/((10s^3+s^2)) ?
The answer to inverse oflaplace (0.2)/((10s^3+s^2)) is -2\H(t)+0.2t+2e^{-t/(10)}