What is inverse oflaplace 1/(s^2+0.5s+1) ?

The answer to inverse oflaplace 1/(s^2+0.5s+1) is 1.03279…e^{-0.25t}sin(0.96824…t)

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inverse oflaplace 1/(s^2+0.5s+1)

inverse laplace

\frac{1}{s ^{2} + 0.5 s + 1}

1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 0.5 s + 1}}

\frac{1}{s ^{2} + 0.5 s + 1} = \frac{1}{( s + 0.25 ) ^{2} + 0.9375}

= L^{- 1} {\frac{1}{( s + 0.25 ) ^{2} + 0.9375}}

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 + 0.25 ) ^{2} + 0.9375} : a = - 0.25, F (s) = \frac{1}{s ^{2} + 0.9375}

= e^{- 0.25 t} L^{- 1} {\frac{1}{s ^{2} + 0.9375}}

L^{- 1} {\frac{1}{s ^{2} + 0.9375}} : \frac{1}{0.96824 \dots} sin (0.96824 \dots t)

Simplify

e^{- 0.25 t} \frac{1}{0.96824 \dots} sin (0.96824 \dots t) : 1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

= 1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

x \to 1 lim ((x - 1)^{l n (x)})

t \to - 2 lim (\frac{t ^{4} - 8}{2 t ^{2} - 3 t + 8})

\frac{d y}{d t} - 2 t y = t

\frac{d y}{d x} = - 10 x y

\int t sin (t) cos (t) d t

What is inverse oflaplace 1/(s^2+0.5s+1) ?
The answer to inverse oflaplace 1/(s^2+0.5s+1) is 1.03279…e^{-0.25t}sin(0.96824…t)