What is inverse oflaplace 2/(5(s+1)^2-80) ?

The answer to inverse oflaplace 2/(5(s+1)^2-80) is 1/20 e^{3t}-1/20 e^{-5t}

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inverse oflaplace 2/(5(s+1)^2-80)

inverse laplace

\frac{2}{5 ( s + 1 ) ^{2} - 80}

\frac{1}{20} e^{3 t} - \frac{1}{20} e^{- 5 t}

Solution steps

L^{- 1} {\frac{2}{5 ( s + 1 ) ^{2} - 80}}

Take the partial fraction of

\frac{2}{5 ( s + 1 ) ^{2} - 80} : \frac{1}{20 ( s - 3 )} - \frac{1}{20 ( s + 5 )}

= L^{- 1} {\frac{1}{20 ( s - 3 )} - \frac{1}{20 ( s + 5 )}}

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}{20 ( s - 3 )}} - L^{- 1} {\frac{1}{20 ( s + 5 )}}

L^{- 1} {\frac{1}{20 ( s - 3 )}} : \frac{1}{20} e^{3 t}

L^{- 1} {\frac{1}{20 ( s + 5 )}} : \frac{1}{20} e^{- 5 t}

= \frac{1}{20} e^{3 t} - \frac{1}{20} e^{- 5 t}

x \to 2 lim (\frac{1 - x ^{2}}{1 - x})

x \to - 3 + lim (x x + 3)

y = ∣ x ∣, y = x^{2} - 2

x \to 0 lim ((e x - 1) cot (2 x))

\frac{d}{d x} (- sin (4 x) \cdot 4)

What is inverse oflaplace 2/(5(s+1)^2-80) ?
The answer to inverse oflaplace 2/(5(s+1)^2-80) is 1/20 e^{3t}-1/20 e^{-5t}