What is inverse oflaplace 3/s+(24)/(s^4) ?

The answer to inverse oflaplace 3/s+(24)/(s^4) is 3\H(t)+4t^3

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inverse oflaplace 3/s+(24)/(s^4)

inverse laplace

\frac{3}{s} + \frac{24}{s ^{4}}

3 H (t) + 4 t^{3}

Solution steps

L^{- 1} {\frac{3}{s} + \frac{24}{s ^{4}}}

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{3}{s}} + L^{- 1} {\frac{24}{s ^{4}}}

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

L^{- 1} {\frac{24}{s ^{4}}} : 4 t^{3}

= 3 H (t) + 4 t^{3}

\frac{\partial}{\partial y} (4 x^{3} y^{2})

\frac{1}{( s - 4 ) ^{3}}

\frac{d}{d x} (6 x^{2} - 8 x)

[0, 5], 5 x, \frac{5}{x}, x = 5

\frac{d}{d s} (ln ((s^{8} x^{7} e^{- s x^{8}})^{n}))

What is inverse oflaplace 3/s+(24)/(s^4) ?
The answer to inverse oflaplace 3/s+(24)/(s^4) is 3\H(t)+4t^3