What is inverse oflaplace 5/(4(s-3)^2-64) ?

The answer to inverse oflaplace 5/(4(s-3)^2-64) is -5/32 e^{-t}+5/32 e^{7t}

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inverse oflaplace 5/(4(s-3)^2-64)

inverse laplace

\frac{5}{4 ( s - 3 ) ^{2} - 64}

- \frac{5}{32} e^{- t} + \frac{5}{32} e^{7 t}

Solution steps

L^{- 1} {\frac{5}{4 ( s - 3 ) ^{2} - 64}}

Take the partial fraction of

\frac{5}{4 ( s - 3 ) ^{2} - 64} : - \frac{5}{32 ( s + 1 )} + \frac{5}{32 ( s - 7 )}

= L^{- 1} {- \frac{5}{32 ( s + 1 )} + \frac{5}{32 ( s - 7 )}}

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{5}{32 ( s + 1 )}} + L^{- 1} {\frac{5}{32 ( s - 7 )}}

L^{- 1} {\frac{5}{32 ( s + 1 )}} : \frac{5}{32} e^{- t}

L^{- 1} {\frac{5}{32 ( s - 7 )}} : \frac{5}{32} e^{7 t}

= - \frac{5}{32} e^{- t} + \frac{5}{32} e^{7 t}

\frac{d}{d x} (ln (\frac{x + 2}{x ^{2}}))

\frac{d}{d x} (x^{2} y^{4})

x \to 2 lim (x^{2} - 5)

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

x \to 0 lim (\frac{1 - e ^{- x}}{x})

What is inverse oflaplace 5/(4(s-3)^2-64) ?
The answer to inverse oflaplace 5/(4(s-3)^2-64) is -5/32 e^{-t}+5/32 e^{7t}