What is inverse oflaplace (s+3)/(s^2+6s+10) ?

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

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inverse oflaplace (s+3)/(s^2+6s+10)

inverse laplace

\frac{s + 3}{s ^{2} + 6 s + 10}

e^{- 3 t} cos (t)

Solution steps

L^{- 1} {\frac{s + 3}{s ^{2} + 6 s + 10}}

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

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

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

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

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

\frac{d}{d x} (5^{x} + 5 cos (\frac{5 x}{2}))

\int \frac{5 + t + t ^{2}}{t} d t

x \frac{d y}{d x} = y - \frac{1}{y}

x \to - 5 - lim (\frac{∣ x + 5 ∣}{x + 5})

\int x^{5} e^{- x^{6}} d x

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