What is inverse oflaplace 1/(s^2)-(48)/(s^5) ?

The answer to inverse oflaplace 1/(s^2)-(48)/(s^5) is t-2t^4

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inverse oflaplace 1/(s^2)-(48)/(s^5)

inverse laplace

\frac{1}{s ^{2}} - \frac{48}{s ^{5}}

t - 2 t^{4}

Solution steps

L^{- 1} {\frac{1}{s ^{2}} - \frac{48}{s ^{5}}}

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}{s ^{2}}} - L^{- 1} {\frac{48}{s ^{5}}}

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

L^{- 1} {\frac{48}{s ^{5}}} : 2 t^{4}

= t - 2 t^{4}

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

x = \frac{( y - 2 ) ^{2}}{3}, y = 8 - x

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

\int 4 x^{2} - x + 1 d x

\frac{d}{d x} (arctan (sin (ln (x^{3}))))

What is inverse oflaplace 1/(s^2)-(48)/(s^5) ?
The answer to inverse oflaplace 1/(s^2)-(48)/(s^5) is t-2t^4