What is inverse oflaplace (10s+8)/((s^3)) ?

The answer to inverse oflaplace (10s+8)/((s^3)) is 10t+4t^2

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inverse oflaplace (10s+8)/((s^3))

inverse laplace

\frac{10 s + 8}{( s ^{3} )}

10 t + 4 t^{2}

Solution steps

L^{- 1} {\frac{10 s + 8}{( s ^{3} )}}

Expand

= L^{- 1} {\frac{10 s}{s ^{3}} + \frac{8}{s ^{3}}}

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)}

= 10 L^{- 1} {\frac{s}{s ^{3}}} + 8 L^{- 1} {\frac{1}{s ^{3}}}

L^{- 1} {\frac{s}{s ^{3}}} : t

L^{- 1} {\frac{1}{s ^{3}}} : \frac{t ^{2}}{2}

= 10 t + 8 \cdot \frac{t ^{2}}{2}

Refine

10 t + 8 \frac{t ^{2}}{2} : 10 t + 4 t^{2}

= 10 t + 4 t^{2}

What is inverse oflaplace (10s+8)/((s^3)) ?
The answer to inverse oflaplace (10s+8)/((s^3)) is 10t+4t^2