What is inverse oflaplace 1/((s^2-4s+5)) ?

The answer to inverse oflaplace 1/((s^2-4s+5)) is e^{2t}sin(t)

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inverse oflaplace 1/((s^2-4s+5))

inverse laplace

\frac{1}{( s ^{2} - 4 s + 5 )}

e^{2 t} sin (t)

Solution steps

L^{- 1} {\frac{1}{( s ^{2} - 4 s + 5 )}}

\frac{1}{s ^{2} - 4 s + 5} = \frac{1}{( s - 2 ) ^{2} + 1}

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

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

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

L^{- 1} {\frac{1}{s ^{2} + 1}} : sin (t)

= e^{2 t} sin (t)

\frac{d}{d x} (sin (3 x) + 3 x cos (3 x))

\frac{d}{d x} (x - e^{- x})

\int (3 x + 2)^{2.4} d x

\frac{\partial}{\partial x} (x^{6} y - 2 x^{5} y^{2})

- 2 x^{4} + 5 x^{2} - 3 (1)

What is inverse oflaplace 1/((s^2-4s+5)) ?
The answer to inverse oflaplace 1/((s^2-4s+5)) is e^{2t}sin(t)