What is inverse oflaplace (3+(4/s))/(s+5) ?

The answer to inverse oflaplace (3+(4/s))/(s+5) is (4\H(t)+11e^{-5t})/5

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inverse oflaplace (3+(4/s))/(s+5)

inverse laplace

\frac{3 + ( \frac{4}{s} )}{s + 5}

\frac{4 H ( t ) + 11 e ^{- 5 t}}{5}

Solution steps

L^{- 1} {\frac{3 + ( \frac{4}{s} )}{s + 5}}

Expand

= L^{- 1} {\frac{3}{s + 5} + \frac{\frac{4}{s}}{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)}

= 3 L^{- 1} {\frac{1}{s + 5}} + 4 L^{- 1} {\frac{\frac{1}{s}}{s + 5}}

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

L^{- 1} {\frac{\frac{1}{s}}{s + 5}} : \frac{1}{5} H (t) - \frac{1}{5} e^{- 5 t}

= 3 e^{- 5 t} + 4 (\frac{1}{5} H (t) - \frac{1}{5} e^{- 5 t})

Simplify

3 e^{- 5 t} + 4 (\frac{1}{5} H (t) - \frac{1}{5} e^{- 5 t}) : \frac{4 H ( t ) + 11 e ^{- 5 t}}{5}

= \frac{4 H ( t ) + 11 e ^{- 5 t}}{5}

\frac{d}{d x} (x + x y)

n = 1 \sum \infty \frac{5}{n + 1}

\int_{0}^{1} (3 t - 1)^{50} d t

x \to \infty lim (\frac{x}{4})

x \to 0 lim (\frac{x}{x ^{2} - 3 x})

What is inverse oflaplace (3+(4/s))/(s+5) ?
The answer to inverse oflaplace (3+(4/s))/(s+5) is (4\H(t)+11e^{-5t})/5