What is inverse oflaplace ((s+3))/(s^2+2s+1) ?

The answer to inverse oflaplace ((s+3))/(s^2+2s+1) is e^{-t}+e^{-t}*2t

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inverse oflaplace ((s+3))/(s^2+2s+1)

inverse laplace

\frac{( s + 3 )}{s ^{2} + 2 s + 1}

e^{- t} + e^{- t} \cdot 2 t

Solution steps

L^{- 1} {\frac{( s + 3 )}{s ^{2} + 2 s + 1}}

Take the partial fraction of

\frac{s + 3}{s ^{2} + 2 s + 1} : \frac{1}{s + 1} + \frac{2}{( s + 1 ) ^{2}}

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

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

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

= e^{- t} + e^{- t} \cdot 2 t

4 y^{^{''}} - 16 y^{^{'}} + 18 y = 0

\frac{d}{d x} (\frac{25 x ^{2} - 1}{x + 3})

\int_{0}^{1} (e^{- 0.1 x}) d x

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

z \to 0 lim (z^{3} e^{\frac{1}{z}})

What is inverse oflaplace ((s+3))/(s^2+2s+1) ?
The answer to inverse oflaplace ((s+3))/(s^2+2s+1) is e^{-t}+e^{-t}*2t