What is inverse oflaplace 2/(s^2+9) ?

The answer to inverse oflaplace 2/(s^2+9) is 2/3 sin(3t)

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inverse oflaplace 2/(s^2+9)

inverse laplace

\frac{2}{s ^{2} + 9}

\frac{2}{3} sin (3 t)

Solution steps

L^{- 1} {\frac{2}{s ^{2} + 9}}

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

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{3}{s ^{2} + 3 ^{2}}} = sin (3 t)

= \frac{2}{3} sin (3 t)

\int ln (a t + 1 + (a t)^{2}) d t

\int \frac{x ^{3}}{( 4 + x ^{2} ) ^{2}} d x

\int \frac{1}{x ^{2} - 9 x + 18} d x

\int \frac{25 - x ^{2}}{x ^{2}} d x

y = \frac{x ^{10} - 5 x ^{6} - 12 x}{x ^{6}}

What is inverse oflaplace 2/(s^2+9) ?
The answer to inverse oflaplace 2/(s^2+9) is 2/3 sin(3t)