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

The answer to inverse oflaplace 6/((s^2+9)^2) is 1/9 (sin(3t)-3tcos(3t))

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inverse oflaplace 6/((s^2+9)^2)

inverse laplace

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

\frac{1}{9} (sin (3 t) - 3 t cos (3 t))

Solution steps

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

= L^{- 1} {\frac{1}{9} \cdot \frac{2 \cdot 3 ^{3}}{( s ^{2} + 3 ^{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}{9} L^{- 1} {\frac{2 \cdot 3 ^{3}}{( s ^{2} + 3 ^{2} ) ^{2}}}

Use Inverse Laplace Transform table:

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

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

= \frac{1}{9} (sin (3 t) - 3 t cos (3 t))

x \to 2 + lim (\frac{7}{x - 2})

\frac{d y}{d x} + (\frac{7}{x}) y = x + 2

\int_{- 2}^{3} \frac{1}{1 + 9 t ^{2}} d t

What is inverse oflaplace 6/((s^2+9)^2) ?
The answer to inverse oflaplace 6/((s^2+9)^2) is 1/9 (sin(3t)-3tcos(3t))