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}

f (x) = \frac{2 x}{1 + x ^{2}}, at x = 4

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x} + 2 e^{x} + e^{- x}

\frac{\partial}{\partial x} (x^{3})

\int 2 + sin (x) d x

\int_{0}^{1} \frac{1}{x ^{1.2}} d x

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}