What is inverse oflaplace (-5e^{(-3s)})/((s^2-5s+6)) ?

The answer to inverse oflaplace (-5e^{(-3s)})/((s^2-5s+6)) is \H(t-3)(5e^{2(t-3)}-5e^{3(t-3)})

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inverse oflaplace (-5e^{(-3s)})/((s^2-5s+6))

inverse laplace

\frac{- 5 e ^{(- 3 s)}}{( s ^{2} - 5 s + 6 )}

H (t - 3) (5 e^{2 (t - 3)} - 5 e^{3 (t - 3)})

Solution steps

L^{- 1} {\frac{- 5 e ^{(- 3 s)}}{( s ^{2} - 5 s + 6 )}}

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 - 3) L^{- 1} {\frac{- 5}{s ^{2} - 5 s + 6}} (t - 3)

L^{- 1} {\frac{- 5}{s ^{2} - 5 s + 6}} : 5 e^{2 t} - 5 e^{3 t}

= H (t - 3) (5 e^{2 (t - 3)} - 5 e^{3 (t - 3)})

\int \frac{2 x ^{2} - 5}{x ^{4} - 5 x ^{2} + 6} d x

\int (2 x^{2} + 2 x - 3)^{10} (2 x + 1) d x

\frac{\partial}{\partial x} (5 x^{7} cos (y^{8}))

x \to 0 - lim (\frac{7}{x} - \frac{7}{∣ x ∣})

\int_{0}^{1} \frac{1}{1 - x ^{2}} d x

What is inverse oflaplace (-5e^{(-3s)})/((s^2-5s+6)) ?
The answer to inverse oflaplace (-5e^{(-3s)})/((s^2-5s+6)) is \H(t-3)(5e^{2(t-3)}-5e^{3(t-3)})