What is inverse oflaplace (s+3)/((s+3)^2+4) ?

The answer to inverse oflaplace (s+3)/((s+3)^2+4) is e^{-3t}cos(2t)

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inverse oflaplace (s+3)/((s+3)^2+4)

inverse laplace

\frac{s + 3}{( s + 3 ) ^{2} + 4}

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

Solution steps

L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 4}}

Apply inverse transform rule:

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

then

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

For

\frac{s + 3}{( s + 3 ) ^{2} + 4} : a = - 3, F (s) = \frac{s}{s ^{2} + 4}

= e^{- 3 t} L^{- 1} {\frac{s}{s ^{2} + 4}}

L^{- 1} {\frac{s}{s ^{2} + 4}} : cos (2 t)

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

\frac{d}{d x} (\frac{1}{2} cos (\frac{x}{2}))

y^{^{'}} = y^{2} + y

\frac{d}{d x} ((3 x + 2)^{6})

\int_{- 7}^{7} \frac{1}{x} d x

x \to 2 lim (3 ln (x - 1))

What is inverse oflaplace (s+3)/((s+3)^2+4) ?
The answer to inverse oflaplace (s+3)/((s+3)^2+4) is e^{-3t}cos(2t)