What is inverse oflaplace 1/(s^2(s+1.1)) ?

The answer to inverse oflaplace 1/(s^2(s+1.1)) is -0.82644…\H(t)+(10t)/(11)+0.82644…e^{-1.1t}

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inverse oflaplace 1/(s^2(s+1.1))

inverse laplace

\frac{1}{s ^{2} ( s + 1.1 )}

- 0.82644 \dots H (t) + \frac{10 t}{11} + 0.82644 \dots e^{- 1.1 t}

Solution steps

L^{- 1} {\frac{1}{s ^{2} ( s + 1.1 )}}

Take the partial fraction of

\frac{1}{s ^{2} ( s + 1.1 )} : - \frac{0.82644 \dots}{s} + \frac{10}{11 s ^{2}} + \frac{0.82644 \dots}{s + 1.1}

= L^{- 1} {- \frac{0.82644 \dots}{s} + \frac{10}{11 s ^{2}} + \frac{0.82644 \dots}{s + 1.1}}

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{0.82644 \dots}{s}} + L^{- 1} {\frac{10}{11 s ^{2}}} + L^{- 1} {\frac{0.82644 \dots}{s + 1.1}}

L^{- 1} {\frac{0.82644 \dots}{s}} : 0.82644 \dots H (t)

L^{- 1} {\frac{10}{11 s ^{2}}} : \frac{10 t}{11}

L^{- 1} {\frac{0.82644 \dots}{s + 1.1}} : 0.82644 \dots e^{- 1.1 t}

= - 0.82644 \dots H (t) + \frac{10 t}{11} + 0.82644 \dots e^{- 1.1 t}

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

\int x x^{2} - 5 d x

x \to - 1 - lim (arccos (x))

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

\frac{d}{d x} (32 x)

What is inverse oflaplace 1/(s^2(s+1.1)) ?
The answer to inverse oflaplace 1/(s^2(s+1.1)) is -0.82644…\H(t)+(10t)/(11)+0.82644…e^{-1.1t}