What is inverse oflaplace 1/((s^2+6s+10)) ?

The answer to inverse oflaplace 1/((s^2+6s+10)) is e^{-3t}sin(t)

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inverse oflaplace 1/((s^2+6s+10))

inverse laplace

\frac{1}{( s ^{2} + 6 s + 10 )}

e^{- 3 t} sin (t)

Solution steps

L^{- 1} {\frac{1}{( s ^{2} + 6 s + 10 )}}

\frac{1}{s ^{2} + 6 s + 10} = \frac{1}{( s + 3 ) ^{2} + 1}

= L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 1}}

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

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

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

= e^{- 3 t} sin (t)

\frac{d}{d t} (e^{2 t} sin (4 t))

2 x^{2} e^{x}

\frac{d}{d x} (x^{2} y^{2} z^{2})

(9.8) (5)

\int \frac{1}{25 + 4 x ^{2}} d x

What is inverse oflaplace 1/((s^2+6s+10)) ?
The answer to inverse oflaplace 1/((s^2+6s+10)) is e^{-3t}sin(t)