What is inverse oflaplace 1/(s^2-6s+9) ?

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

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inverse oflaplace 1/(s^2-6s+9)

inverse laplace

\frac{1}{s ^{2} - 6 s + 9}

e^{3 t} t

Solution steps

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

\frac{1}{s ^{2} - 6 s + 9} = \frac{1}{( s - 3 ) ^{2}}

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s - 3 ) ^{2}} : a = 3, F (s) = \frac{1}{s ^{2}}

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

L^{- 1} {\frac{1}{s ^{2}}} : t

= e^{3 t} t

\int 22 arctan (x) d x

cos (x - 7 x)

n = 0 \sum \infty 6^{n} x^{n}

n = 1 \sum \infty \frac{4}{n}

\int \frac{3}{2 x - 1} d x

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