What is inverse oflaplace (4s)/(s^2-16) ?

The answer to inverse oflaplace (4s)/(s^2-16) is 2e^{-4t}+2e^{4t}

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inverse oflaplace (4s)/(s^2-16)

inverse laplace

\frac{4 s}{s ^{2} - 16}

2 e^{- 4 t} + 2 e^{4 t}

Solution steps

L^{- 1} {\frac{4 s}{s ^{2} - 16}}

Take the partial fraction of

\frac{4 s}{s ^{2} - 16} : \frac{2}{s + 4} + \frac{2}{s - 4}

= L^{- 1} {\frac{2}{s + 4} + \frac{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{2}{s + 4}} + L^{- 1} {\frac{2}{s - 4}}

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

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

= 2 e^{- 4 t} + 2 e^{4 t}

\frac{d}{d x} (sin (x + \frac{π}{2}) - cos (x))

1 - 3 x^{2}

12 x^{3} - 3 y x^{4} + 1 \frac{d y}{d x} = 0

\int_{0}^{4} 2^{s} d s

y^{^{'}} + 3 y = e^{- 4 t}

What is inverse oflaplace (4s)/(s^2-16) ?
The answer to inverse oflaplace (4s)/(s^2-16) is 2e^{-4t}+2e^{4t}