What is inverse oflaplace (80)/(s^2+40s) ?

The answer to inverse oflaplace (80)/(s^2+40s) is 2\H(t)-2e^{-40t}

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inverse oflaplace (80)/(s^2+40s)

inverse laplace

\frac{80}{s ^{2} + 40 s}

2 H (t) - 2 e^{- 40 t}

Solution steps

L^{- 1} {\frac{80}{s ^{2} + 40 s}}

Take the partial fraction of

\frac{80}{s ^{2} + 40 s} : \frac{2}{s} - \frac{2}{s + 40}

= L^{- 1} {\frac{2}{s} - \frac{2}{s + 40}}

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{2}{s + 40}}

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

L^{- 1} {\frac{2}{s + 40}} : 2 e^{- 40 t}

= 2 H (t) - 2 e^{- 40 t}

What is inverse oflaplace (80)/(s^2+40s) ?
The answer to inverse oflaplace (80)/(s^2+40s) is 2\H(t)-2e^{-40t}