What is inverse oflaplace 2/(s(s^2+15s+50)) ?

The answer to inverse oflaplace 2/(s(s^2+15s+50)) is 1/25 \H(t)-2/25 e^{-5t}+1/25 e^{-10t}

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inverse oflaplace 2/(s(s^2+15s+50))

inverse laplace

\frac{2}{s ( s ^{2} + 15 s + 50 )}

\frac{1}{25} H (t) - \frac{2}{25} e^{- 5 t} + \frac{1}{25} e^{- 10 t}

Solution steps

L^{- 1} {\frac{2}{s ( s ^{2} + 15 s + 50 )}}

Take the partial fraction of

\frac{2}{s ( s ^{2} + 15 s + 50 )} : \frac{1}{25 s} - \frac{2}{25 ( s + 5 )} + \frac{1}{25 ( s + 10 )}

= L^{- 1} {\frac{1}{25 s} - \frac{2}{25 ( s + 5 )} + \frac{1}{25 ( s + 10 )}}

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}{25 s}} - L^{- 1} {\frac{2}{25 ( s + 5 )}} + L^{- 1} {\frac{1}{25 ( s + 10 )}}

L^{- 1} {\frac{1}{25 s}} : \frac{1}{25} H (t)

L^{- 1} {\frac{2}{25 ( s + 5 )}} : \frac{2}{25} e^{- 5 t}

L^{- 1} {\frac{1}{25 ( s + 10 )}} : \frac{1}{25} e^{- 10 t}

= \frac{1}{25} H (t) - \frac{2}{25} e^{- 5 t} + \frac{1}{25} e^{- 10 t}

\int x ln (y + 1) d y

n = 0 \sum \infty (\frac{A}{A + 1})^{n}

f (x) = e^{9 x}

f (x) = 3 x^{2} + 4 x, at x = - 2

What is inverse oflaplace 2/(s(s^2+15s+50)) ?
The answer to inverse oflaplace 2/(s(s^2+15s+50)) is 1/25 \H(t)-2/25 e^{-5t}+1/25 e^{-10t}