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

The answer to inverse oflaplace 1/(s^2+8) is 1/(2sqrt(2))sin(2sqrt(2)t)

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inverse oflaplace 1/(s^2+8)

inverse laplace

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

\frac{1}{2 2} sin (22 t)

Solution steps

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

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

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

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{2 2}{s ^{2} + ( 2 2 ) ^{2}}} = sin (22 t)

= \frac{1}{2 2} sin (22 t)

\int \frac{3}{1 + x} d x

\frac{d}{d x} (e^{x^{3 + 6 x^{2}}})

\frac{d y}{d t} = \frac{y}{1 + t} - \frac{y}{t} + t^{2} (1 + t)

\frac{\partial}{\partial x} ((13 x - 1) (3 y - 1))

y = ln (3 x - 7)

What is inverse oflaplace 1/(s^2+8) ?
The answer to inverse oflaplace 1/(s^2+8) is 1/(2sqrt(2))sin(2sqrt(2)t)