What is inverse oflaplace (e^{-3s})/(s^2+6s+13) ?

The answer to inverse oflaplace (e^{-3s})/(s^2+6s+13) is \H(t-3)e^{-3(t-3)} 1/2 sin(2(t-3))

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inverse oflaplace (e^{-3s})/(s^2+6s+13)

inverse laplace

\frac{e ^{- 3 s}}{s ^{2} + 6 s + 13}

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

Solution steps

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

Apply inverse transform rule:

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

then

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

Where

H (t)

is Heaviside step function

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

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

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

f (x) = 13

\int \frac{1 + u}{- 2 u - 1 - u ^{2}} d u

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

\int \frac{x ^{2} + 8 x - 3}{x ^{3} + 3 x} d x

y^{^{''}} + 5 y^{^{'}} + 6 y = 108 x^{2}

What is inverse oflaplace (e^{-3s})/(s^2+6s+13) ?
The answer to inverse oflaplace (e^{-3s})/(s^2+6s+13) is \H(t-3)e^{-3(t-3)} 1/2 sin(2(t-3))