What is inverse oflaplace (10)/(s(s+1)) ?

The answer to inverse oflaplace (10)/(s(s+1)) is 10\H(t)-10e^{-t}

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inverse oflaplace (10)/(s(s+1))

inverse laplace

\frac{10}{s ( s + 1 )}

10 H (t) - 10 e^{- t}

Solution steps

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

Take the partial fraction of

\frac{10}{s ( s + 1 )} : \frac{10}{s} - \frac{10}{s + 1}

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

L^{- 1} {\frac{10}{s}} : 10 H (t)

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

= 10 H (t) - 10 e^{- t}

What is inverse oflaplace (10)/(s(s+1)) ?
The answer to inverse oflaplace (10)/(s(s+1)) is 10\H(t)-10e^{-t}