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 ( s + 1 ) ( 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}

(- 1, - 2), (5, 4)

y \frac{d y}{d x} = x e^{- y^{2}}

\frac{5 x}{x - 2}

\int \frac{( x ^{2} - x + 12 )}{x ^{3} + 2 x} d x

\int_{0}^{\infty} e^{- u} d u

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}