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

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

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

inverse laplace

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

H (t) + e^{- t}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= H (t) + e^{- t}

x \to 0 lim (x^{3} + x^{2})

f (x) = 3 x \cdot ln (x)

n \to \infty lim (3 n)

f (x) = x + 1, at x = 3

sin (3 x^{2} + 5)

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