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

The answer to inverse oflaplace (e^{-2s})/(s^2-2s+5) is \H(t-2)e^{t-2} 1/2 sin(2(t-2))

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

inverse laplace

\frac{e ^{- 2 s}}{s ^{2} - 2 s + 5}

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

Solution steps

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

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

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

\frac{\partial}{\partial x} ((7 x + 3 y)^{5})

\frac{\partial}{\partial y} ((- x^{4} + 7 y^{2} + 3 y)^{6})

\frac{d}{d x} (x^{2} - 2 y)

\frac{d y}{d x} = 7 x cos^{2} (y)

\frac{d}{d x} (\frac{1}{2 x + 8})

What is inverse oflaplace (e^{-2s})/(s^2-2s+5) ?
The answer to inverse oflaplace (e^{-2s})/(s^2-2s+5) is \H(t-2)e^{t-2} 1/2 sin(2(t-2))