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)

\int \frac{x ^{4} - 3 x ^{2} + 5}{x ^{4}} d x

x \to - 2 lim (\frac{x + 2}{2 x})

\int x sin^{2} (x^{2}) d x

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

x \to 0 lim (\frac{60}{x ^{4}})

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