What is inverse oflaplace 6/(s^2+9) ?

The answer to inverse oflaplace 6/(s^2+9) is 2sin(3t)

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inverse oflaplace 6/(s^2+9)

inverse laplace

\frac{6}{s ^{2} + 9}

2 sin (3 t)

Solution steps

L^{- 1} {\frac{6}{s ^{2} + 9}}

= L^{- 1} {2 \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)}

= 2 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)

= 2 sin (3 t)

x \to 3 - lim (\frac{2}{( x - 3 )})

\frac{\partial}{\partial u ( v )} (\frac{e ^{v}}{u ( v ) + v ^{9}})

0.3 x^{3} 5^{x}

5 x^{2} - x

What is inverse oflaplace 6/(s^2+9) ?
The answer to inverse oflaplace 6/(s^2+9) is 2sin(3t)