What is inverse oflaplace (s+2)/(s^2+4s+5) ?

The answer to inverse oflaplace (s+2)/(s^2+4s+5) is e^{-2t}cos(t)

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inverse oflaplace (s+2)/(s^2+4s+5)

inverse laplace

\frac{s + 2}{s ^{2} + 4 s + 5}

e^{- 2 t} cos (t)

Solution steps

L^{- 1} {\frac{s + 2}{s ^{2} + 4 s + 5}}

\frac{s + 2}{s ^{2} + 4 s + 5} = \frac{s + 2}{( s + 2 ) ^{2} + 1}

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

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

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

= e^{- 2 t} cos (t)

y^{^{''}} + 25 y = t sin (4 t)

\frac{\partial}{\partial w} (\frac{w}{u})

x \to 0 lim (\frac{e ^{x} - x - 1}{x})

8^{x}

\frac{d}{d x} (\frac{cot ( x )}{1 - sin ( x )})

What is inverse oflaplace (s+2)/(s^2+4s+5) ?
The answer to inverse oflaplace (s+2)/(s^2+4s+5) is e^{-2t}cos(t)