What is inverse oflaplace (15)/((s+10)(s+11)) ?

The answer to inverse oflaplace (15)/((s+10)(s+11)) is 15e^{-10t}-15e^{-11t}

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inverse oflaplace (15)/((s+10)(s+11))

inverse laplace

\frac{15}{( s + 10 ) ( s + 11 )}

15 e^{- 10 t} - 15 e^{- 11 t}

Solution steps

L^{- 1} {\frac{15}{( s + 10 ) ( s + 11 )}}

Take the partial fraction of

\frac{15}{( s + 10 ) ( s + 11 )} : \frac{15}{s + 10} - \frac{15}{s + 11}

= L^{- 1} {\frac{15}{s + 10} - \frac{15}{s + 11}}

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{15}{s + 10}} - L^{- 1} {\frac{15}{s + 11}}

L^{- 1} {\frac{15}{s + 10}} : 15 e^{- 10 t}

L^{- 1} {\frac{15}{s + 11}} : 15 e^{- 11 t}

= 15 e^{- 10 t} - 15 e^{- 11 t}

\int_{0}^{2.8} 0.2 x d x

\frac{4 π r ^{3}}{3}

\frac{x ^{3}}{x + 1}

\int \frac{1}{x ^{p}} d x

\frac{d}{d x} (- 4 x e^{- 2 x^{2}})

What is inverse oflaplace (15)/((s+10)(s+11)) ?
The answer to inverse oflaplace (15)/((s+10)(s+11)) is 15e^{-10t}-15e^{-11t}