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

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

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

inverse laplace

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

e^{- 2 t} - e^{- 2 t} \cdot 2 t

Solution steps

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

Take the partial fraction of

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

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

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

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

= e^{- 2 t} - e^{- 2 t} \cdot 2 t

\frac{\partial}{\partial x} (y e^{x} - x e^{y})

x \to 0 + lim (\frac{1}{e ^{x} + 1})

[3 ln (x) - 12 sin (x)]

\frac{\partial}{\partial x} (z e^{x^{2} - y^{2}})

y^{^{''}} - 4 y^{^{'}} - 4 y = e^{2 x}

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