What is inverse oflaplace (8s^2+12s+7)/(s^2+s) ?

The answer to inverse oflaplace (8s^2+12s+7)/(s^2+s) is 8\delta(t)+7\H(t)-3e^{-t}

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inverse oflaplace (8s^2+12s+7)/(s^2+s)

inverse laplace

\frac{8 s ^{2} + 12 s + 7}{s ^{2} + s}

8 δ (t) + 7 H (t) - 3 e^{- t}

Solution steps

L^{- 1} {\frac{8 s ^{2} + 12 s + 7}{s ^{2} + s}}

Take the partial fraction of

\frac{8 s ^{2} + 12 s + 7}{s ^{2} + s} : 8 + \frac{7}{s} - \frac{3}{s + 1}

= L^{- 1} {8 + \frac{7}{s} - \frac{3}{s + 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} {8} + L^{- 1} {\frac{7}{s}} - L^{- 1} {\frac{3}{s + 1}}

L^{- 1} {8} : 8 δ (t)

L^{- 1} {\frac{7}{s}} : 7 H (t)

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

= 8 δ (t) + 7 H (t) - 3 e^{- t}

\frac{d}{d x} (ln ((2 x + 5)^{3}))

x \to \infty lim (∣ x - 5 ∣)

\frac{\partial}{\partial x} (\frac{1}{x + t + 1})

y^{^{'}} = \frac{2 x}{y + 1}

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

What is inverse oflaplace (8s^2+12s+7)/(s^2+s) ?
The answer to inverse oflaplace (8s^2+12s+7)/(s^2+s) is 8\delta(t)+7\H(t)-3e^{-t}