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

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

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

inverse laplace

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

2 e^{- t} t

Solution steps

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

Expand

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

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

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

= 2 e^{- t} t

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