What is inverse oflaplace (2s+1)/(s(2s+10)) ?

The answer to inverse oflaplace (2s+1)/(s(2s+10)) is 1/10 \H(t)+9/10 e^{-5t}

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inverse oflaplace (2s+1)/(s(2s+10))

inverse laplace

\frac{2 s + 1}{s ( 2 s + 10 )}

\frac{1}{10} H (t) + \frac{9}{10} e^{- 5 t}

Solution steps

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

Take the partial fraction of

\frac{2 s + 1}{s ( 2 s + 10 )} : \frac{1}{10 s} + \frac{9}{10 ( s + 5 )}

= L^{- 1} {\frac{1}{10 s} + \frac{9}{10 ( 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{9}{10 ( s + 5 )}}

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

L^{- 1} {\frac{9}{10 ( s + 5 )}} : \frac{9}{10} e^{- 5 t}

= \frac{1}{10} H (t) + \frac{9}{10} e^{- 5 t}

f (x) = 3 e^{2 x^{2} + 4 x}

x - 3

\int x^{22} e^{x^{23}} d x

y = (3 - x^{2}) (x^{3} - 2 x + 1)

\frac{d x}{d t} = \frac{1 - x}{k}

What is inverse oflaplace (2s+1)/(s(2s+10)) ?
The answer to inverse oflaplace (2s+1)/(s(2s+10)) is 1/10 \H(t)+9/10 e^{-5t}