What is inverse oflaplace 8/(s-1)-3/s ?

The answer to inverse oflaplace 8/(s-1)-3/s is 8e^t-3\H(t)

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inverse oflaplace 8/(s-1)-3/s

inverse laplace

\frac{8}{s - 1} - \frac{3}{s}

8 e^{t} - 3 H (t)

Solution steps

L^{- 1} {\frac{8}{s - 1} - \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{8}{s - 1}} - L^{- 1} {\frac{3}{s}}

L^{- 1} {\frac{8}{s - 1}} : 8 e^{t}

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

= 8 e^{t} - 3 H (t)

(2 x^{2}) + (x - 8)

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

x \to 0 lim (\frac{x}{x ^{2} + 4 x})

n = 1 \sum \infty (1 - \frac{3}{n})^{n}

\int (e^{\frac{x}{2}}) d x

What is inverse oflaplace 8/(s-1)-3/s ?
The answer to inverse oflaplace 8/(s-1)-3/s is 8e^t-3\H(t)