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

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

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

inverse laplace

\frac{2 s + 1}{s ^{2} + 2 s + 1}

2 e^{- t} - e^{- t} t

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= 2 e^{- t} - e^{- t} t

\frac{d}{d x} (m e^{m x})

x \to 0 lim (x 1 + \frac{1}{2})

\frac{\partial}{\partial v} (\frac{u - v}{2})

n = 0 \sum \infty \frac{n + 5}{5 ^{n}}

y^{'} = 0.1 (5 - y) y

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