What is inverse oflaplace (-4e^{(-2s)})/((s^2-6s+8)) ?

The answer to inverse oflaplace (-4e^{(-2s)})/((s^2-6s+8)) is \H(t-2)(2e^{2(t-2)}-2e^{4(t-2)})

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inverse oflaplace (-4e^{(-2s)})/((s^2-6s+8))

inverse laplace

\frac{- 4 e ^{(- 2 s)}}{( s ^{2} - 6 s + 8 )}

H (t - 2) (2 e^{2 (t - 2)} - 2 e^{4 (t - 2)})

Solution steps

L^{- 1} {\frac{- 4 e ^{(- 2 s)}}{( s ^{2} - 6 s + 8 )}}

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{- 4}{s ^{2} - 6 s + 8}} (t - 2)

L^{- 1} {\frac{- 4}{s ^{2} - 6 s + 8}} : 2 e^{2 t} - 2 e^{4 t}

= H (t - 2) (2 e^{2 (t - 2)} - 2 e^{4 (t - 2)})

y^{^{'}} = t y

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

\frac{\partial}{\partial x} ((ln (1 + e^{(x)}))^{2})

t \to 0 lim (\frac{2 e ^{t} - 2}{t})

\int 2 + 2 cos (x) d x

What is inverse oflaplace (-4e^{(-2s)})/((s^2-6s+8)) ?
The answer to inverse oflaplace (-4e^{(-2s)})/((s^2-6s+8)) is \H(t-2)(2e^{2(t-2)}-2e^{4(t-2)})