What is inverse oflaplace (2s^3+3s^2-2s)/(s^3) ?

The answer to inverse oflaplace (2s^3+3s^2-2s)/(s^3) is 2\delta(t)+3-2t

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inverse oflaplace (2s^3+3s^2-2s)/(s^3)

inverse laplace

\frac{2 s ^{3} + 3 s ^{2} - 2 s}{s ^{3}}

2 δ (t) + 3 - 2 t

Solution steps

L^{- 1} {\frac{2 s ^{3} + 3 s ^{2} - 2 s}{s ^{3}}}

Take the partial fraction of

\frac{2 s ^{3} + 3 s ^{2} - 2 s}{s ^{3}} : 2 + \frac{3 s - 2}{s ^{2}}

= L^{- 1} {2 + \frac{3 s - 2}{s ^{2}}}

Expand

= L^{- 1} {\frac{3 s}{s ^{2}} - \frac{2}{s ^{2}} + 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} {2} + 3 L^{- 1} {\frac{s}{s ^{2}}} - 2 L^{- 1} {\frac{1}{s ^{2}}}

L^{- 1} {2} : 2 δ (t)

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

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

= 2 δ (t) + 3 \cdot 1 - 2 t

Refine

2 δ (t) + 31 - 2 t : 2 δ (t) + 3 - 2 t

= 2 δ (t) + 3 - 2 t

m (t) = \frac{4 t + 11}{7 t ^{2} - 3 t + 5}

\frac{d}{d x} (\frac{x}{x + \frac{y}{x}})

\int cos^{8} (x) sin (x) d x

\int cos (5 t) d t

x \to 1 - lim (4 x)

What is inverse oflaplace (2s^3+3s^2-2s)/(s^3) ?
The answer to inverse oflaplace (2s^3+3s^2-2s)/(s^3) is 2\delta(t)+3-2t