What is inverse oflaplace (1/s)((10)/(s+5)) ?

The answer to inverse oflaplace (1/s)((10)/(s+5)) is 2\H(t)-2e^{-5t}

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inverse oflaplace (1/s)((10)/(s+5))

inverse laplace

(\frac{1}{s}) (\frac{10}{s + 5})

2 H (t) - 2 e^{- 5 t}

Solution steps

L^{- 1} {(\frac{1}{s}) (\frac{10}{s + 5})}

Expand

\frac{1}{s} \frac{10}{s + 5} : \frac{10}{s ^{2} + 5 s}

= L^{- 1} {\frac{10}{s ^{2} + 5 s}}

Take the partial fraction of

\frac{10}{s ^{2} + 5 s} : \frac{2}{s} - \frac{2}{s + 5}

= L^{- 1} {\frac{2}{s} - \frac{2}{s + 5}}

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{2}{s}} - L^{- 1} {\frac{2}{s + 5}}

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

L^{- 1} {\frac{2}{s + 5}} : 2 e^{- 5 t}

= 2 H (t) - 2 e^{- 5 t}

\frac{3 x + 4}{2 x - 1}

\frac{d}{d x} (4 x^{- \frac{1}{3}})

y (6 y^{2} - x - 1) d x + 2 x d y = 0

\frac{d}{d x} (\frac{sin ( x )}{cos ^{3} ( x )})

f (x) = \frac{2}{x + 5}, at x = 4

What is inverse oflaplace (1/s)((10)/(s+5)) ?
The answer to inverse oflaplace (1/s)((10)/(s+5)) is 2\H(t)-2e^{-5t}