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)

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