What is inverse oflaplace 1/(s^2+6s+13) ?

The answer to inverse oflaplace 1/(s^2+6s+13) is e^{-3t} 1/2 sin(2t)

inverse oflaplace 1/(s^2+6s+13)

inverse laplace

\frac{1}{s ^{2} + 6 s + 13}

e^{- 3 t} \frac{1}{2} sin (2 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 6 s + 13}}

\frac{1}{s ^{2} + 6 s + 13} = \frac{1}{( s + 3 ) ^{2} + 4}

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

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

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

= e^{- 3 t} \frac{1}{2} sin (2 t)

d er i v a t i v e f (x) = x - 4

\frac{d}{d x} (\frac{1}{2 x ^{3}})

\frac{\partial}{\partial r} (r e^{s})

\int \frac{4 x ^{3} - x ^{2} + 4 x + 2}{( x ^{2} + 1 ) ^{2}} d x

x \to 6 - lim (\frac{1}{( x - 6 ) ^{2}})

What is inverse oflaplace 1/(s^2+6s+13) ?
The answer to inverse oflaplace 1/(s^2+6s+13) is e^{-3t} 1/2 sin(2t)