What is inverse oflaplace (s^3+s+6)/(s^4(s+2)) ?

The answer to inverse oflaplace (s^3+s+6)/(s^4(s+2)) is 1/4 \H(t)+t/2-(t^2)/2+(t^3)/2-1/4 e^{-2t}

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inverse oflaplace (s^3+s+6)/(s^4(s+2))

inverse laplace

\frac{s ^{3} + s + 6}{s ^{4} ( s + 2 )}

\frac{1}{4} H (t) + \frac{t}{2} - \frac{t ^{2}}{2} + \frac{t ^{3}}{2} - \frac{1}{4} e^{- 2 t}

Solution steps

L^{- 1} {\frac{s ^{3} + s + 6}{s ^{4} ( s + 2 )}}

Take the partial fraction of

\frac{s ^{3} + s + 6}{s ^{4} ( s + 2 )} : \frac{1}{4 s} + \frac{1}{2 s ^{2}} - \frac{1}{s ^{3}} + \frac{3}{s ^{4}} - \frac{1}{4 ( s + 2 )}

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

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

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

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

L^{- 1} {\frac{3}{s ^{4}}} : \frac{t ^{3}}{2}

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

= \frac{1}{4} H (t) + \frac{t}{2} - \frac{t ^{2}}{2} + \frac{t ^{3}}{2} - \frac{1}{4} e^{- 2 t}

\int_{7}^{8} \frac{x}{x ^{2} - 3} d x

x \to 7 + lim (3 x + 5)

x

\frac{d y}{d t} = 2 y - 3 e^{- t}

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

What is inverse oflaplace (s^3+s+6)/(s^4(s+2)) ?
The answer to inverse oflaplace (s^3+s+6)/(s^4(s+2)) is 1/4 \H(t)+t/2-(t^2)/2+(t^3)/2-1/4 e^{-2t}