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

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

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inverse oflaplace e^{-2s} 1/((s+1)^2)

inverse laplace

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

H (t - 2) e^{- t + 2} (t - 2)

Solution steps

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

Apply inverse transform rule:

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

then

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

Where

H (t)

is Heaviside step function

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

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

= H (t - 2) e^{- (t - 2)} (t - 2)

= H (t - 2) e^{- t + 2} (t - 2)

y^{^{'}} = \frac{x ^{2}}{y}

\int \frac{8}{( x - 2 ) ( x ^{2} + 4 )} d x

\int e^{6 x} - 4 d x

\int \frac{y}{x ^{2} + 1} d x

\frac{d}{d x} (2000 - 4 x)

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