What is inverse oflaplace 1/s-2/((s+3)^2-9) ?

The answer to inverse oflaplace 1/s-2/((s+3)^2-9) is (2\H(t)+e^{-6t})/3

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inverse oflaplace 1/s-2/((s+3)^2-9)

inverse laplace

\frac{1}{s} - \frac{2}{( s + 3 ) ^{2} - 9}

\frac{2 H ( t ) + e ^{- 6 t}}{3}

Solution steps

L^{- 1} {\frac{1}{s} - \frac{2}{( s + 3 ) ^{2} - 9}}

Take the partial fraction of

\frac{2}{( s + 3 ) ^{2} - 9} : \frac{1}{3 s} - \frac{1}{3 ( s + 6 )}

= L^{- 1} {\frac{1}{s} - \frac{1}{3 s} + \frac{1}{3 ( s + 6 )}}

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}{s}} - L^{- 1} {\frac{1}{3 s}} + L^{- 1} {\frac{1}{3 ( s + 6 )}}

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

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

L^{- 1} {\frac{1}{3 ( s + 6 )}} : \frac{1}{3} e^{- 6 t}

= H (t) - \frac{1}{3} H (t) + \frac{1}{3} e^{- 6 t}

Simplify

H (t) - \frac{1}{3} H (t) + \frac{1}{3} e^{- 6 t} : \frac{2 H ( t ) + e ^{- 6 t}}{3}

= \frac{2 H ( t ) + e ^{- 6 t}}{3}

\int \frac{2 x ^{2} - 1}{x ^{3} - x} d x

\int \frac{1}{x ^{4} + y ^{2}} d y

\int 7 e^{- 3 x} d x

\frac{\partial}{\partial x} (arcsin (7 x yz))

x \to - 5 lim (\frac{3}{x - 5})

What is inverse oflaplace 1/s-2/((s+3)^2-9) ?
The answer to inverse oflaplace 1/s-2/((s+3)^2-9) is (2\H(t)+e^{-6t})/3