What is inverse oflaplace (8s)/(3(s^2-16)^2) ?

The answer to inverse oflaplace (8s)/(3(s^2-16)^2) is -(e^{-4t}t)/6+(e^{4t}t)/6

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inverse oflaplace (8s)/(3(s^2-16)^2)

inverse laplace

\frac{8 s}{3 ( s ^{2} - 16 ) ^{2}}

- \frac{e ^{- 4 t} t}{6} + \frac{e ^{4 t} t}{6}

Solution steps

L^{- 1} {\frac{8 s}{3 ( s ^{2} - 16 ) ^{2}}}

Take the partial fraction of

\frac{8 s}{3 ( s ^{2} - 16 ) ^{2}} : - \frac{1}{6 ( s + 4 ) ^{2}} + \frac{1}{6 ( s - 4 ) ^{2}}

= L^{- 1} {- \frac{1}{6 ( s + 4 ) ^{2}} + \frac{1}{6 ( 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}{6 ( s + 4 ) ^{2}}} + L^{- 1} {\frac{1}{6 ( s - 4 ) ^{2}}}

L^{- 1} {\frac{1}{6 ( s + 4 ) ^{2}}} : \frac{e ^{- 4 t} t}{6}

L^{- 1} {\frac{1}{6 ( s - 4 ) ^{2}}} : \frac{e ^{4 t} t}{6}

= - \frac{e ^{- 4 t} t}{6} + \frac{e ^{4 t} t}{6}

3 y + x^{4} y^{^{'}} = 2 x y

(2, 2), (- 2, - 2)

x \to \infty lim (- 3^{x})

\int \frac{x ^{2} + 1}{x ( x ^{2} + 3 )} d x

What is inverse oflaplace (8s)/(3(s^2-16)^2) ?
The answer to inverse oflaplace (8s)/(3(s^2-16)^2) is -(e^{-4t}t)/6+(e^{4t}t)/6