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

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

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

inverse laplace

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

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

Solution steps

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

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

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

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

\int csc^{2} (5 y) d y

y = \frac{1}{x}, y = \frac{1}{x ^{2}}, x = 3

\frac{d}{d x} (x sin (x))

\int \frac{x e ^{x}}{( x + 1 ) ^{2}} d x

\frac{d}{d x} ((5 x^{2} - 8 x) e^{x})

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