What is inverse oflaplace (4e^{(-s)})/((s^2+4s+13)) ?

The answer to inverse oflaplace (4e^{(-s)})/((s^2+4s+13)) is \H(t-1) 4/3 e^{-2(t-1)}sin(3(t-1))

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inverse oflaplace (4e^{(-s)})/((s^2+4s+13))

inverse laplace

\frac{4 e ^{(- s)}}{( s ^{2} + 4 s + 13 )}

H (t - 1) \frac{4}{3} e^{- 2 (t - 1)} sin (3 (t - 1))

Solution steps

L^{- 1} {\frac{4 e ^{(- s)}}{( s ^{2} + 4 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 - 1) L^{- 1} {\frac{4}{s ^{2} + 4 s + 13}} (t - 1)

L^{- 1} {\frac{4}{s ^{2} + 4 s + 13}} : \frac{4}{3} e^{- 2 t} sin (3 t)

= H (t - 1) \frac{4}{3} e^{- 2 (t - 1)} sin (3 (t - 1))

What is inverse oflaplace (4e^{(-s)})/((s^2+4s+13)) ?
The answer to inverse oflaplace (4e^{(-s)})/((s^2+4s+13)) is \H(t-1) 4/3 e^{-2(t-1)}sin(3(t-1))