What is inverse oflaplace 8/((s-4)*(s+4)^2) ?

The answer to inverse oflaplace 8/((s-4)*(s+4)^2) is 1/8 e^{4t}-1/8 e^{-4t}-e^{-4t}t

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inverse oflaplace 8/((s-4)*(s+4)^2)

inverse laplace

\frac{8}{( s - 4 ) \cdot ( s + 4 ) ^{2}}

\frac{1}{8} e^{4 t} - \frac{1}{8} e^{- 4 t} - e^{- 4 t} t

Solution steps

L^{- 1} {\frac{8}{( s - 4 ) ( s + 4 ) ^{2}}}

Take the partial fraction of

\frac{8}{( s - 4 ) ( s + 4 ) ^{2}} : \frac{1}{8 ( s - 4 )} - \frac{1}{8 ( s + 4 )} - \frac{1}{( s + 4 ) ^{2}}

= L^{- 1} {\frac{1}{8 ( s - 4 )} - \frac{1}{8 ( s + 4 )} - \frac{1}{( s + 4 ) ^{2}}}

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}{8 ( s - 4 )}} - L^{- 1} {\frac{1}{8 ( s + 4 )}} - L^{- 1} {\frac{1}{( s + 4 ) ^{2}}}

L^{- 1} {\frac{1}{8 ( s - 4 )}} : \frac{1}{8} e^{4 t}

L^{- 1} {\frac{1}{8 ( s + 4 )}} : \frac{1}{8} e^{- 4 t}

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

= \frac{1}{8} e^{4 t} - \frac{1}{8} e^{- 4 t} - e^{- 4 t} t

x \to 0 lim (\frac{( x ^{2} - 2 x )}{x})

\frac{\partial}{\partial x} (4 y e^{5 x})

n = 0 \sum \infty \frac{x}{2 ^{n}}

\int_{0}^{3} \frac{1}{( x - 3 ) ^{3}} d x

x \to 0 lim (5 + x^{2})

What is inverse oflaplace 8/((s-4)*(s+4)^2) ?
The answer to inverse oflaplace 8/((s-4)*(s+4)^2) is 1/8 e^{4t}-1/8 e^{-4t}-e^{-4t}t