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

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

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

inverse laplace

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

\frac{5}{2} e^{- t} sin (2 t)

Solution steps

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

Expand

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

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 4}} : e^{- t} \frac{1}{2} sin (2 t)

= 5 e^{- t} \frac{1}{2} sin (2 t)

Refine

5 e^{- t} \frac{1}{2} sin (2 t) : \frac{5}{2} e^{- t} sin (2 t)

= \frac{5}{2} e^{- t} sin (2 t)

\frac{d}{d x} (ln (1 + 4 x^{2}))

\int_{- 1}^{3} - x^{2} d x

\frac{d}{d x} (\frac{6}{5 x ^{3}})

\int sin^{2} (\frac{x}{2}) cos (\frac{x}{2}) d x

f (x) = (6 x + 5) (x^{3} - 2)

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