What is inverse oflaplace ((s+1))/((s^2+2s+9)) ?

The answer to inverse oflaplace ((s+1))/((s^2+2s+9)) is e^{-t}cos(2sqrt(2)t)

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inverse oflaplace ((s+1))/((s^2+2s+9))

inverse laplace

\frac{( s + 1 )}{( s ^{2} + 2 s + 9 )}

e^{- t} cos (22 t)

Solution steps

L^{- 1} {\frac{( s + 1 )}{( s ^{2} + 2 s + 9 )}}

\frac{s + 1}{s ^{2} + 2 s + 9} = \frac{s + 1}{( s + 1 ) ^{2} + 8}

= L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 8}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{s + 1}{( s + 1 ) ^{2} + 8} : a = - 1, F (s) = \frac{s}{s ^{2} + 8}

= e^{- t} L^{- 1} {\frac{s}{s ^{2} + 8}}

L^{- 1} {\frac{s}{s ^{2} + 8}} : cos (22 t)

= e^{- t} cos (22 t)

x = 3 y, x = 2 + y^{2}

6 u^{- 1}

x \to 0 lim (\frac{( e ^{x} )}{x})

38 x e^{x^{2}}

\frac{\partial}{\partial y} (5 - 2 x^{2} - y^{2})

What is inverse oflaplace ((s+1))/((s^2+2s+9)) ?
The answer to inverse oflaplace ((s+1))/((s^2+2s+9)) is e^{-t}cos(2sqrt(2)t)