What is inverse oflaplace 1/(s^2+3s) ?

The answer to inverse oflaplace 1/(s^2+3s) is 1/3 \H(t)-1/3 e^{-3t}

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inverse oflaplace 1/(s^2+3s)

inverse laplace

\frac{1}{s ^{2} + 3 s}

\frac{1}{3} H (t) - \frac{1}{3} e^{- 3 t}

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 3 s}}

Take the partial fraction of

\frac{1}{s ^{2} + 3 s} : \frac{1}{3 s} - \frac{1}{3 ( s + 3 )}

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

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

L^{- 1} {\frac{1}{3 s}} : \frac{1}{3} H (t)

L^{- 1} {\frac{1}{3 ( s + 3 )}} : \frac{1}{3} e^{- 3 t}

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

\frac{d}{d x} ((2 x)^{8})

x \to 0 + lim (6 - ln (x))

n = 3 \sum \infty \frac{1}{n + 2}

y^{^{'}} (x) = \frac{3 x ^{2}}{3 y ^{2} - 7}, y (1) = 0

f (x) = 3 x + 4, (7, 5)

What is inverse oflaplace 1/(s^2+3s) ?
The answer to inverse oflaplace 1/(s^2+3s) is 1/3 \H(t)-1/3 e^{-3t}