What is inverse oflaplace 1/((s-2)^2-4) ?

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

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inverse oflaplace 1/((s-2)^2-4)

inverse laplace

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

- \frac{1}{4} H (t) + \frac{1}{4} e^{4 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= - \frac{1}{4} H (t) + \frac{1}{4} e^{4 t}

y^{^{''''}} - 2 y^{^{''}} + y = 0

\frac{d}{d x} (3 cos^{2} (x) - 6 sin (x))

\frac{\partial}{\partial x} (arcsin (3 x))

\frac{d ^{2}}{d x ^{2}} (\frac{x}{x - 1})

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