What is inverse oflaplace (-4e^{(-s)})/((s^2-4s+3)) ?

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

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inverse oflaplace (-4e^{(-s)})/((s^2-4s+3))

inverse laplace

\frac{- 4 e ^{(- s)}}{( s ^{2} - 4 s + 3 )}

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

Solution steps

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

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 + 3}} (t - 1)

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

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

y^{^{'}} = (t + y)^{2} - 1, y (2) = 4

\int x (x - 1)^{3} d x

\frac{d}{d x} (\frac{x}{81 + x ^{2}})

\frac{\partial}{\partial x} ((9 x + 2 y)^{6})

x \to 0 lim (x^{4} cos (\frac{3}{x}))

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