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

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

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

inverse laplace

\frac{1}{s ^{2} - 2 s + 5}

e^{t} \frac{1}{2} sin (2 t)

Solution steps

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

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

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

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 - 1 ) ^{2} + 4} : a = 1, F (s) = \frac{1}{s ^{2} + 4}

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

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

= e^{t} \frac{1}{2} sin (2 t)

\frac{8 + x f ( x )}{x}

x^{2} - 2 x + 1

x \to 0 + lim (\frac{5}{x ^{2}})

(x + y)^{2} d x + (2 x y + x^{2} - 8) d y = 0

f (x) = 3 x^{2} + 5 x - \frac{1}{x ^{2}} + e^{2}

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