What is inverse oflaplace-5/(s(50+20s)) ?

The answer to inverse oflaplace-5/(s(50+20s)) is -1/10 \H(t)+1/10 e^{-(5t)/2}

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inverse oflaplace-5/(s(50+20s))

inverse laplace

- \frac{5}{s ( 50 + 20 s )}

- \frac{1}{10} H (t) + \frac{1}{10} e^{- \frac{5 t}{2}}

Solution steps

L^{- 1} {- \frac{5}{s ( 20 s + 50 )}}

Take the partial fraction of

\frac{5}{s ( 50 + 20 s )} : \frac{1}{10 s} - \frac{1}{5 ( 2 s + 5 )}

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

L^{- 1} {\frac{1}{10 s}} : \frac{1}{10} H (t)

L^{- 1} {\frac{1}{5 ( 2 s + 5 )}} : \frac{1}{10} e^{- \frac{5 t}{2}}

= - \frac{1}{10} H (t) + \frac{1}{10} e^{- \frac{5 t}{2}}

\frac{d}{d x} (ln (\frac{1 - x ^{2}}{1 + x ^{2}}))

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{9}{2}}} d x

- 4 (t - 1)

\int_{0}^{2} e^{5 x} d x

x \to 0 lim (\frac{x + 1}{x})

What is inverse oflaplace-5/(s(50+20s)) ?
The answer to inverse oflaplace-5/(s(50+20s)) is -1/10 \H(t)+1/10 e^{-(5t)/2}