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{\partial}{\partial y} (- 2 y)

y = 2 x + 3

\frac{\partial}{\partial x} (x - z)

\frac{d}{d x} (- x - 5)

\frac{d}{d x} (tan (3 x) - 3 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)