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

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

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

inverse laplace

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

e^{- 2 t} sin (3 t)

Solution steps

L^{- 1} {\frac{3}{( 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{3}{( s + 2 ) ^{2} + 9} : a = - 2, F (s) = \frac{3}{s ^{2} + 9}

= e^{- 2 t} L^{- 1} {\frac{3}{s ^{2} + 9}}

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

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

\int \frac{1}{9 sin ^{2} ( θ )} d θ

\int_{0}^{π} sin^{5} (\frac{x}{2}) d x

t \to π lim (sin (\frac{t}{6}))

\frac{d y}{d x} = 2 y 16 - x^{2}, y (0) = 1

\frac{d}{d x} (ln (16 - x^{2}))

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