What is inverse oflaplace ((e^{-s}))/(s^2-6s+9) ?

The answer to inverse oflaplace ((e^{-s}))/(s^2-6s+9) is \H(t-1)e^{3(t-1)}(t-1)

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inverse oflaplace ((e^{-s}))/(s^2-6s+9)

inverse laplace

\frac{( e ^{- s} )}{s ^{2} - 6 s + 9}

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

Solution steps

L^{- 1} {\frac{( e ^{- s} )}{s ^{2} - 6 s + 9}}

Factor

s^{2} - 6 s + 9 : (s - 3)^{2}

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

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{1}{( s - 3 ) ^{2}}} (t - 1)

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

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

y^{^{'}} = \frac{2 y}{t} + t

x^{- 1}

\int \frac{1 - u}{1 + u} d u

x \to 0 lim (2 sin (x) - 1)

\int (\frac{ln ( x )}{x ^{2}}) d x

What is inverse oflaplace ((e^{-s}))/(s^2-6s+9) ?
The answer to inverse oflaplace ((e^{-s}))/(s^2-6s+9) is \H(t-1)e^{3(t-1)}(t-1)