What is inverse oflaplace 1/(s^2+9) ?

The answer to inverse oflaplace 1/(s^2+9) is 1/3 sin(3t)

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inverse oflaplace 1/(s^2+9)

inverse laplace

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

\frac{1}{3} sin (3 t)

Solution steps

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

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

\frac{d}{d x} (4 x - \frac{3}{x})

\frac{\partial}{\partial x} (ln (x - 2 y))

\frac{\partial}{\partial x} (sin (\frac{x}{y}))

\frac{d}{d t} (\frac{ln ( 2 t - 1 )}{t ^{2} - 1})

\frac{\partial}{\partial x} (2 x^{2} - 3 x y + z - 4)

What is inverse oflaplace 1/(s^2+9) ?
The answer to inverse oflaplace 1/(s^2+9) is 1/3 sin(3t)