What is inverse oflaplace (s+11)/((s-10)(s+4)) ?

The answer to inverse oflaplace (s+11)/((s-10)(s+4)) is 3/2 e^{10t}-1/2 e^{-4t}

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inverse oflaplace (s+11)/((s-10)(s+4))

inverse laplace

\frac{s + 11}{( s - 10 ) ( s + 4 )}

\frac{3}{2} e^{10 t} - \frac{1}{2} e^{- 4 t}

Solution steps

L^{- 1} {\frac{s + 11}{( s - 10 ) ( s + 4 )}}

Take the partial fraction of

\frac{s + 11}{( s - 10 ) ( s + 4 )} : \frac{3}{2 ( s - 10 )} - \frac{1}{2 ( s + 4 )}

= L^{- 1} {\frac{3}{2 ( s - 10 )} - \frac{1}{2 ( s + 4 )}}

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{3}{2 ( s - 10 )}} - L^{- 1} {\frac{1}{2 ( s + 4 )}}

L^{- 1} {\frac{3}{2 ( s - 10 )}} : \frac{3}{2} e^{10 t}

L^{- 1} {\frac{1}{2 ( s + 4 )}} : \frac{1}{2} e^{- 4 t}

= \frac{3}{2} e^{10 t} - \frac{1}{2} e^{- 4 t}

2 e^{- 2 t} cos (7 t) - 5 e^{- 2 t} sin (7 t)

\frac{d}{d x} (\frac{e ^{x}}{x})

n = 0 \sum \infty n + 3

4 sec^{7} (2 - 4 x)

x \to 0 + lim (arcsin (2 x - 1))

What is inverse oflaplace (s+11)/((s-10)(s+4)) ?
The answer to inverse oflaplace (s+11)/((s-10)(s+4)) is 3/2 e^{10t}-1/2 e^{-4t}