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

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

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

inverse laplace

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

e^{- t} t

Solution steps

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

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

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s + 1 ) ^{2}} : a = - 1, F (s) = \frac{1}{s ^{2}}

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

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

= e^{- t} t

\int sin^{4} (5 x) d x

2 y y^{^{'}} + 6 = y^{2} + 6 x, y (0) = 7

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

\int \frac{x}{( x ^{2} + 4 ) ^{2}} d x

- \frac{11}{s ^{4}}

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