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

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

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inverse oflaplace 1/(s^2+2)

inverse laplace

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

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

Solution steps

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

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

Use Inverse Laplace Transform table:

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

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

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

\int \frac{x ^{2}}{x ^{2} - 2 x + 1} d x

\frac{1}{8 x ^{2}}

n = 1 \sum \infty 3^{n}

\int \frac{1 + 2 x}{x} d x

n \to \infty lim (3)

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