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)

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{e ^{3 x}}{x ^{2}}

\frac{d y}{d x} = \frac{10 x ^{2}}{1 + x ^{3}}

x = 15 - y^{2}, x = y^{2} - 3

\int_{0}^{\frac{π}{8}} arctan (2 x) d x

\frac{\partial}{\partial w} (w arctan (uv))

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)