What is inverse oflaplace (3e^{(-2s)})/((s^2+2s+5)) ?

The answer to inverse oflaplace (3e^{(-2s)})/((s^2+2s+5)) is \H(t-2) 3/2 e^{-t+2}sin(2(t-2))

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inverse oflaplace (3e^{(-2s)})/((s^2+2s+5))

inverse laplace

\frac{3 e ^{(- 2 s)}}{( s ^{2} + 2 s + 5 )}

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

Solution steps

L^{- 1} {\frac{3 e ^{(- 2 s)}}{( s ^{2} + 2 s + 5 )}}

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 - 2) L^{- 1} {\frac{3}{s ^{2} + 2 s + 5}} (t - 2)

L^{- 1} {\frac{3}{s ^{2} + 2 s + 5}} : \frac{3}{2} e^{- t} sin (2 t)

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

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

\int \frac{3 x ( 2 )}{( x )} d x

\frac{d}{d x} (\frac{ln ( x )}{x})

\int_{0}^{1} \frac{5}{1 + t ^{2}} d t

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

ln (x^{2} - 1)

What is inverse oflaplace (3e^{(-2s)})/((s^2+2s+5)) ?
The answer to inverse oflaplace (3e^{(-2s)})/((s^2+2s+5)) is \H(t-2) 3/2 e^{-t+2}sin(2(t-2))