What is inverse oflaplace (1-e^{-2s})/(6s^2) ?

The answer to inverse oflaplace (1-e^{-2s})/(6s^2) is t/6-1/6 \H(t-2)(t-2)

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inverse oflaplace (1-e^{-2s})/(6s^2)

inverse laplace

\frac{1 - e ^{- 2 s}}{6 s ^{2}}

\frac{t}{6} - \frac{1}{6} H (t - 2) (t - 2)

Solution steps

L^{- 1} {\frac{1 - e ^{- 2 s}}{6 s ^{2}}}

Expand

= L^{- 1} {\frac{1}{6 s ^{2}} - \frac{e ^{- 2 s}}{6 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)}

= \frac{1}{6} L^{- 1} {\frac{1}{s ^{2}}} - \frac{1}{6} L^{- 1} {\frac{e ^{- 2 s}}{s ^{2}}}

L^{- 1} {\frac{1}{s ^{2}}} : t

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

= \frac{1}{6} t - \frac{1}{6} H (t - 2) (t - 2)

Refine

\frac{1}{6} t - \frac{1}{6} H (t - 2) (t - 2) : \frac{t}{6} - \frac{1}{6} H (t - 2) (t - 2)

= \frac{t}{6} - \frac{1}{6} H (t - 2) (t - 2)

What is inverse oflaplace (1-e^{-2s})/(6s^2) ?
The answer to inverse oflaplace (1-e^{-2s})/(6s^2) is t/6-1/6 \H(t-2)(t-2)