What is inverse oflaplace (2/(2s+1))-(1/s) ?

The answer to inverse oflaplace (2/(2s+1))-(1/s) is e^{-t/2}-\H(t)

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inverse oflaplace (2/(2s+1))-(1/s)

inverse laplace

(\frac{2}{2 s + 1}) - (\frac{1}{s})

e^{- \frac{t}{2}} - H (t)

Solution steps

L^{- 1} {(\frac{2}{2 s + 1}) - (\frac{1}{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}{2 s + 1}} - L^{- 1} {\frac{1}{s}}

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

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

= e^{- \frac{t}{2}} - H (t)

(y^{2} + x y) d x - x^{2} d y = 0

\int (\frac{1}{t}) d t

\frac{d}{d x} (5 + 8 x^{\frac{3}{2}})

x \to \infty lim (x \cdot (- 2))

\frac{d}{d x} (5 sin (5 x))

What is inverse oflaplace (2/(2s+1))-(1/s) ?
The answer to inverse oflaplace (2/(2s+1))-(1/s) is e^{-t/2}-\H(t)