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

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

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

inverse laplace

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

H (t - 2) (- \frac{3}{2} e^{t - 2} + \frac{3}{2} e^{3 (t - 2)})

Solution steps

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

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

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

ln (100 sin^{2} (x))

f (x) = \frac{4 x ^{2} tan ( x )}{sec ( x )}

x \to - 2 lim (- 0.8)

x d y - y d x = (2 x^{2} - 3) d x

\frac{3}{x ^{2} - x - 2}

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