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

The answer to inverse oflaplace 1/(s^2+2*0.5*1s+1) is e^{-t/2} 2/(sqrt(3))sin((sqrt(3)t)/2)

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inverse oflaplace 1/(s^2+20.51s+1)

inverse laplace

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

e^{- \frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2})

Solution steps

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

\frac{1}{s ^{2} + 20.51 s + 1} = \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{4}}

= L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{3}{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 + \frac{1}{2} ) ^{2} + \frac{3}{4}} : a = - \frac{1}{2}, F (s) = \frac{1}{s ^{2} + \frac{3}{4}}

= e^{- \frac{1}{2} t} L^{- 1} {\frac{1}{s ^{2} + \frac{3}{4}}}

L^{- 1} {\frac{1}{s ^{2} + \frac{3}{4}}} : \frac{2}{3} sin (\frac{3 t}{2})

= e^{- \frac{t}{2}} \frac{2}{3} sin (\frac{3 t}{2})

\int_{0}^{1} (1 - x^{2}) d x

f (x) = e^{3 x} cos (2 x)

x \to + 2 lim (2 x + 3)

f (x) = \frac{2}{x}, at x = \frac{1}{9}

t \to + 0 lim (\frac{sin ( 7 t )}{t})

What is inverse oflaplace 1/(s^2+2*0.5*1s+1) ?
The answer to inverse oflaplace 1/(s^2+2*0.5*1s+1) is e^{-t/2} 2/(sqrt(3))sin((sqrt(3)t)/2)