What is inverse oflaplace 2/(s^3)+6/(s^2)-3/s ?

The answer to inverse oflaplace 2/(s^3)+6/(s^2)-3/s is t^2+6t-3\H(t)

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inverse oflaplace 2/(s^3)+6/(s^2)-3/s

inverse laplace

\frac{2}{s ^{3}} + \frac{6}{s ^{2}} - \frac{3}{s}

t^{2} + 6 t - 3 H (t)

Solution steps

L^{- 1} {\frac{2}{s ^{3}} + \frac{6}{s ^{2}} - \frac{3}{s}}

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{2}{s ^{3}}} + L^{- 1} {\frac{6}{s ^{2}}} - L^{- 1} {\frac{3}{s}}

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

L^{- 1} {\frac{6}{s ^{2}}} : 6 t

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

= t^{2} + 6 t - 3 H (t)

y 1 - x^{2} y^{^{'}} - x 1 - y^{2} = 0

\frac{d P}{d t} = P (9 - P) - \frac{81}{4}

\int \frac{3 - 4 x ^{3} + 6 x ^{6}}{x ^{6}} d x

\frac{d y}{d x} = 1 - e^{x - y + 8}

- y^{^{'''}} + 2 y^{^{''}} + 3 y^{^{'}} + 3 = 0

What is inverse oflaplace 2/(s^3)+6/(s^2)-3/s ?
The answer to inverse oflaplace 2/(s^3)+6/(s^2)-3/s is t^2+6t-3\H(t)