What is inverse oflaplace (400)/(s(s+2)^3) ?

The answer to inverse oflaplace (400)/(s(s+2)^3) is 50\H(t)-50e^{-2t}-e^{-2t}*100t-e^{-2t}*100t^2

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inverse oflaplace (400)/(s(s+2)^3)

inverse laplace

\frac{400}{s ( s + 2 ) ^{3}}

50 H (t) - 50 e^{- 2 t} - e^{- 2 t} \cdot 100 t - e^{- 2 t} \cdot 100 t^{2}

Solution steps

L^{- 1} {\frac{400}{s ( s + 2 ) ^{3}}}

Take the partial fraction of

\frac{400}{s ( s + 2 ) ^{3}} : \frac{50}{s} - \frac{50}{s + 2} - \frac{100}{( s + 2 ) ^{2}} - \frac{200}{( s + 2 ) ^{3}}

= L^{- 1} {\frac{50}{s} - \frac{50}{s + 2} - \frac{100}{( s + 2 ) ^{2}} - \frac{200}{( s + 2 ) ^{3}}}

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{50}{s}} - L^{- 1} {\frac{50}{s + 2}} - L^{- 1} {\frac{100}{( s + 2 ) ^{2}}} - L^{- 1} {\frac{200}{( s + 2 ) ^{3}}}

L^{- 1} {\frac{50}{s}} : 50 H (t)

L^{- 1} {\frac{50}{s + 2}} : 50 e^{- 2 t}

L^{- 1} {\frac{100}{( s + 2 ) ^{2}}} : e^{- 2 t} \cdot 100 t

L^{- 1} {\frac{200}{( s + 2 ) ^{3}}} : e^{- 2 t} \cdot 100 t^{2}

= 50 H (t) - 50 e^{- 2 t} - e^{- 2 t} \cdot 100 t - e^{- 2 t} \cdot 100 t^{2}

\frac{\partial}{\partial x} (6 x^{\frac{y}{z}})

\frac{d}{d x} (x^{2} - 2 x e^{- x} + e^{- 2 x})

\int_{0}^{2} ∣ x ∣ d x

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

f (x) = x^{2} + 3 x - 3, at x = 1

What is inverse oflaplace (400)/(s(s+2)^3) ?
The answer to inverse oflaplace (400)/(s(s+2)^3) is 50\H(t)-50e^{-2t}-e^{-2t}*100t-e^{-2t}*100t^2