What is inverse oflaplace (-3e^{(-s2)})/((s^2-5s+6)) ?

The answer to inverse oflaplace (-3e^{(-s2)})/((s^2-5s+6)) is \H(t-2)(3e^{2(t-2)}-3e^{3(t-2)})

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inverse oflaplace (-3e^{(-s2)})/((s^2-5s+6))

inverse laplace

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

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

Solution steps

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

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

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

x \to 0 + lim (x^{3} \cdot ln (x))

ln (x + 8 + x^{2})

\frac{1}{2 x ^{2}}

(1 - t^{2}) y^{^{''}} + 2 t y^{^{'}} - 2 y = 0

f (x) = x^{2} - 3 x + 5, at x = - 2

What is inverse oflaplace (-3e^{(-s2)})/((s^2-5s+6)) ?
The answer to inverse oflaplace (-3e^{(-s2)})/((s^2-5s+6)) is \H(t-2)(3e^{2(t-2)}-3e^{3(t-2)})