What is inverse oflaplace 3+1/s ?

The answer to inverse oflaplace 3+1/s is 3\delta(t)+\H(t)

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inverse oflaplace 3+1/s

inverse laplace

3 + \frac{1}{s}

3 δ (t) + H (t)

Solution steps

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

L^{- 1} {3} : 3 δ (t)

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

= 3 δ (t) + H (t)

\int_{o}^{1} 42 x (1 - x)^{5} d x

\frac{d}{d x} (a^{2} + x^{2})

x \to 1 lim (\frac{1 + x}{1 - x ^{2}})

\int_{0}^{- \frac{π}{6}} x cos (3 x) d x

y = 8 x + 7

What is inverse oflaplace 3+1/s ?
The answer to inverse oflaplace 3+1/s is 3\delta(t)+\H(t)