What is inverse oflaplace 4/(s^2(s+2))+1/s ?

The answer to inverse oflaplace 4/(s^2(s+2))+1/s is 2t+e^{-2t}

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inverse oflaplace 4/(s^2(s+2))+1/s

inverse laplace

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

2 t + e^{- 2 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

Refine

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

= 2 t + e^{- 2 t}

7 lo g_{7} (x)

f (x) = e^{6 x}

x \to - 2 - lim (x + 2)

\frac{7 s + 53}{s ^{2} + 10 s + 61}

\int cos (2 x) cos (3 x) d x

What is inverse oflaplace 4/(s^2(s+2))+1/s ?
The answer to inverse oflaplace 4/(s^2(s+2))+1/s is 2t+e^{-2t}