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

The answer to inverse oflaplace (10)/(s*(s+1)*(s+10)) is \H(t)-10/9 e^{-t}+1/9 e^{-10t}

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

inverse laplace

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

H (t) - \frac{10}{9} e^{- t} + \frac{1}{9} e^{- 10 t}

Solution steps

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

Take the partial fraction of

\frac{10}{s ( s + 1 ) ( s + 10 )} : \frac{1}{s} - \frac{10}{9 ( s + 1 )} + \frac{1}{9 ( s + 10 )}

= L^{- 1} {\frac{1}{s} - \frac{10}{9 ( s + 1 )} + \frac{1}{9 ( s + 10 )}}

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{10}{9 ( s + 1 )}} + L^{- 1} {\frac{1}{9 ( s + 10 )}}

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

L^{- 1} {\frac{10}{9 ( s + 1 )}} : \frac{10}{9} e^{- t}

L^{- 1} {\frac{1}{9 ( s + 10 )}} : \frac{1}{9} e^{- 10 t}

= H (t) - \frac{10}{9} e^{- t} + \frac{1}{9} e^{- 10 t}

2 t

\frac{\partial}{\partial x} (cos (2 x + 6 y))

\int t cos (a t) d t

\int tan^{4} (x se) c^{6} x d x

\frac{d}{d x} (9 e^{1 - 3 x})

What is inverse oflaplace (10)/(s*(s+1)*(s+10)) ?
The answer to inverse oflaplace (10)/(s*(s+1)*(s+10)) is \H(t)-10/9 e^{-t}+1/9 e^{-10t}