What is inverse oflaplace 4/(s^2(s+2))+3/(s+2) ?

The answer to inverse oflaplace 4/(s^2(s+2))+3/(s+2) is -\H(t)+2t+4e^{-2t}

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inverse oflaplace 4/(s^2(s+2))+3/(s+2)

inverse laplace

\frac{4}{s ^{2} ( s + 2 )} + \frac{3}{s + 2}

- H (t) + 2 t + 4 e^{- 2 t}

Solution steps

L^{- 1} {\frac{4}{s ^{2} ( s + 2 )} + \frac{3}{s + 2}}

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{3}{s + 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{2}{s ^{2}}} + L^{- 1} {\frac{1}{s + 2}} + L^{- 1} {\frac{3}{s + 2}}

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

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

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

L^{- 1} {\frac{3}{s + 2}} : 3 e^{- 2 t}

= - H (t) + 2 t + e^{- 2 t} + 3 e^{- 2 t}

Refine

- H (t) + 2 t + e^{- 2 t} + 3 e^{- 2 t} : - H (t) + 2 t + 4 e^{- 2 t}

= - H (t) + 2 t + 4 e^{- 2 t}

\frac{d}{d x} (arctan (\frac{x}{4 - 16 - x ^{2}}))

x \to 1 lim (\frac{2 x}{6 x - 10})

\frac{d}{d x} (60)

x x (9.27)

\int \frac{2 x ^{2} + 3}{x ( x - 1 ) ^{2}} d x

What is inverse oflaplace 4/(s^2(s+2))+3/(s+2) ?
The answer to inverse oflaplace 4/(s^2(s+2))+3/(s+2) is -\H(t)+2t+4e^{-2t}