What is inverse oflaplace 1/(1+s)*8/s ?

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

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inverse oflaplace 1/(1+s)*8/s

inverse laplace

\frac{1}{1 + s} \cdot \frac{8}{s}

8 H (t) - 8 e^{- t}

Solution steps

L^{- 1} {\frac{1}{s + 1} \cdot \frac{8}{s}}

Expand

\frac{1}{1 + s} \frac{8}{s} : \frac{8}{s + s ^{2}}

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

Take the partial fraction of

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

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

L^{- 1} {\frac{8}{s}} : 8 H (t)

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

= 8 H (t) - 8 e^{- t}

\int \frac{e ^{u}}{( 1 - e ^{u} ) ^{2}} d u

\frac{d y}{d x} = y^{2} (4 - e^{x})

3 y^{^{''}} + 5 y^{^{'}} - 12 y = 0

\frac{d}{d y} (- 2 e^{x} + 4 x cos (y))

\int 2 x - 8 y d x

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