What is inverse oflaplace ((s+3))/(s^2+6s+13) ?

The answer to inverse oflaplace ((s+3))/(s^2+6s+13) is e^{-3t}cos(2t)

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inverse oflaplace ((s+3))/(s^2+6s+13)

inverse laplace

\frac{( s + 3 )}{s ^{2} + 6 s + 13}

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

Solution steps

L^{- 1} {\frac{( s + 3 )}{s ^{2} + 6 s + 13}}

\frac{s + 3}{s ^{2} + 6 s + 13} = \frac{s + 3}{( s + 3 ) ^{2} + 4}

= 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)

\int \frac{13 x ^{3}}{9 + x ^{2}} d x

\int (3 x^{4} + 5 x - 6) d x

x \to 0 lim (\frac{2 sin ( 2 x )}{x})

\frac{d}{d x} (x^{9} ln (5 x))

y y^{^{'}} + 3 y = x^{2} y

What is inverse oflaplace ((s+3))/(s^2+6s+13) ?
The answer to inverse oflaplace ((s+3))/(s^2+6s+13) is e^{-3t}cos(2t)