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

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

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

inverse laplace

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

e^{- t} t

Solution steps

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

n = 0 \sum \infty z^{2 n + 1}

\int (7 cos (x) + 2 sin (x)) d x

y^{^{''}} + 4 y = sin (2 t) cos (2 t)

\frac{( e ^{- s} )}{s ^{5}}

x^{2} + y^{2} = 2

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