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

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

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

inverse laplace

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

- \frac{1}{4} H (t) + \frac{1}{4} e^{2 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= - \frac{1}{4} H (t) + \frac{1}{4} e^{2 t}

\frac{d}{d x} (sin (x) + arcsin (x))

\int \frac{x}{sin ^{2} ( x ^{2} )} d x

12 x^{3} - 3 y^{2} x^{4} + 1 (\frac{d y}{d x}) = 0

\int_{0}^{\frac{1}{2}} (2 x - 1)^{- 3} d x

\int sin^{2} (a x) d x

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