What is inverse oflaplace 4/s+6/(s^2)-1/(s+8) ?

The answer to inverse oflaplace 4/s+6/(s^2)-1/(s+8) is 4\H(t)+6t-e^{-8t}

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inverse oflaplace 4/s+6/(s^2)-1/(s+8)

inverse laplace

\frac{4}{s} + \frac{6}{s ^{2}} - \frac{1}{s + 8}

4 H (t) + 6 t - e^{- 8 t}

Solution steps

L^{- 1} {\frac{4}{s} + \frac{6}{s ^{2}} - \frac{1}{s + 8}}

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{4}{s}} + L^{- 1} {\frac{6}{s ^{2}}} - L^{- 1} {\frac{1}{s + 8}}

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

L^{- 1} {\frac{6}{s ^{2}}} : 6 t

L^{- 1} {\frac{1}{s + 8}} : e^{- 8 t}

= 4 H (t) + 6 t - e^{- 8 t}

\frac{d}{d x} (\frac{cos ( x ) e ^{x}}{tan ( x )})

y^{^{'''}} + 8 y^{^{''}} - 12 y^{^{'}} - 144 y = 0

\frac{d}{d x} (sin (π f (x)))

ln (x + 8)

\int_{0}^{2} x + 1 d x

What is inverse oflaplace 4/s+6/(s^2)-1/(s+8) ?
The answer to inverse oflaplace 4/s+6/(s^2)-1/(s+8) is 4\H(t)+6t-e^{-8t}