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

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

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

inverse laplace

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

H (t) - e^{- 2 t} - e^{- 2 t} \cdot 2 t

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{s}} : H (t)

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

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

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

\int \frac{1}{x ^{2} - 6 x + 8} d x

\int cos (x) cos (2 x) d x

ln (x + 5)

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

f (x) = x^{2} + 12

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