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

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

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

inverse laplace

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

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

Solution steps

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

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

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

\frac{d}{d x} (x^{2} + 2 x - 3)

\int \frac{arctan ( 9 x )}{1 + 81 x ^{2}} d x

\int \frac{300}{x ^{2} 100 - x ^{2}} d x

\frac{\partial}{\partial x} (cos^{5} (x^{6} y^{7}))

y = 10 (3 x + 1) (1 - 5 x)

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