What is inverse oflaplace 4/(s^2)-(10)/s ?

The answer to inverse oflaplace 4/(s^2)-(10)/s is 4t-10\H(t)

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inverse oflaplace 4/(s^2)-(10)/s

inverse laplace

\frac{4}{s ^{2}} - \frac{10}{s}

4 t - 10 H (t)

Solution steps

L^{- 1} {\frac{4}{s ^{2}} - \frac{10}{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{4}{s ^{2}}} - L^{- 1} {\frac{10}{s}}

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

L^{- 1} {\frac{10}{s}} : 10 H (t)

= 4 t - 10 H (t)

\frac{d}{d x} (\frac{1}{1 + e ^{- x}})

h \to 0 - lim (\frac{2 ^{h} - 1}{h})

x \to 0 lim (- \frac{5}{x})

\int \frac{x + 5}{( x + 3 ) ( x - 1 )} d x

\frac{d}{d x} (15 ln (\frac{x}{80}))

What is inverse oflaplace 4/(s^2)-(10)/s ?
The answer to inverse oflaplace 4/(s^2)-(10)/s is 4t-10\H(t)