What is inverse oflaplace 1/(2s^2+8s+10) ?

The answer to inverse oflaplace 1/(2s^2+8s+10) is 1/2 e^{-2t}sin(t)

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inverse oflaplace 1/(2s^2+8s+10)

inverse laplace

\frac{1}{2 s ^{2} + 8 s + 10}

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

Solution steps

L^{- 1} {\frac{1}{2 s ^{2} + 8 s + 10}}

Expand

\frac{1}{2 s ^{2} + 8 s + 10} : \frac{1}{2} \cdot \frac{1}{( s + 2 ) ^{2} + 1}

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

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)}

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

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

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

\int \frac{1}{2} (x + y) e^{- x - y} d x

\frac{d}{d x} (\frac{1}{2} x^{\frac{4}{5}})

x \to 3 - lim (3 + x)

\frac{d}{d x} (\frac{4}{e ^{x} + e ^{- x}})

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

What is inverse oflaplace 1/(2s^2+8s+10) ?
The answer to inverse oflaplace 1/(2s^2+8s+10) is 1/2 e^{-2t}sin(t)