What is inverse oflaplace 1/(s^2+3s+100) ?

The answer to inverse oflaplace 1/(s^2+3s+100) is e^{-(3t)/2} 2/(sqrt(391))sin((sqrt(391)t)/2)

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inverse oflaplace 1/(s^2+3s+100)

inverse laplace

\frac{1}{s ^{2} + 3 s + 100}

e^{- \frac{3 t}{2}} \frac{2}{391} sin (\frac{391 t}{2})

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 3 s + 100}}

\frac{1}{s ^{2} + 3 s + 100} = \frac{1}{( s + \frac{3}{2} ) ^{2} + \frac{391}{4}}

= L^{- 1} {\frac{1}{( s + \frac{3}{2} ) ^{2} + \frac{391}{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{3}{2} ) ^{2} + \frac{391}{4}} : a = - \frac{3}{2}, F (s) = \frac{1}{s ^{2} + \frac{391}{4}}

= e^{- \frac{3}{2} t} L^{- 1} {\frac{1}{s ^{2} + \frac{391}{4}}}

L^{- 1} {\frac{1}{s ^{2} + \frac{391}{4}}} : \frac{2}{391} sin (\frac{391 t}{2})

= e^{- \frac{3 t}{2}} \frac{2}{391} sin (\frac{391 t}{2})

\int cos^{4} (x) sin^{4} (x) d x

y = tan (4 x^{2})

x + y = 8

y^{^{''}} - 8 y^{^{'}} + 17 y = 0, y (0) = 4, y^{^{'}} (0) = - 1

What is inverse oflaplace 1/(s^2+3s+100) ?
The answer to inverse oflaplace 1/(s^2+3s+100) is e^{-(3t)/2} 2/(sqrt(391))sin((sqrt(391)t)/2)