What is inverse oflaplace (s+4)/((s+2)^2) ?

The answer to inverse oflaplace (s+4)/((s+2)^2) is e^{-2t}+e^{-2t}*2t

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inverse oflaplace (s+4)/((s+2)^2)

inverse laplace

\frac{s + 4}{( s + 2 ) ^{2}}

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

Solution steps

L^{- 1} {\frac{s + 4}{( s + 2 ) ^{2}}}

Take the partial fraction of

\frac{s + 4}{( 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{d y}{d x} = - \frac{x}{y e ^{x^{2}}}

y = x^{2} - 1.52, y = - x^{2} + 1.52

\frac{d}{d x} (cos (x y^{2}))

\frac{\partial}{\partial y} (2 x^{2} y + 2 y + 5)

x \to 0 lim (x + y)

What is inverse oflaplace (s+4)/((s+2)^2) ?
The answer to inverse oflaplace (s+4)/((s+2)^2) is e^{-2t}+e^{-2t}*2t