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

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

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

inverse laplace

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

e^{- 2 t} \frac{1}{3} sin (3 t)

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s + 2 ) ^{2} + 9} : a = - 2, F (s) = \frac{1}{s ^{2} + 9}

= e^{- 2 t} L^{- 1} {\frac{1}{s ^{2} + 9}}

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

= e^{- 2 t} \frac{1}{3} sin (3 t)

\frac{\partial}{\partial x} (2 x - 3 y + 5)

\frac{\partial}{\partial x} (e^{x^{2 - y^{2}}})

\int_{0}^{\frac{π}{2}} cos^{3} (21 x) d x

t \to 0 lim (\frac{e ^{2 t} - 1}{t})

y = ln (30) + 3 0^{x + 4}

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