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

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

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inverse oflaplace S/(((S^2+9)^2))

inverse laplace

\frac{S}{( ( S ^{2} + 9 ) ^{2} )}

\frac{t sin ( 3 t )}{6}

Solution steps

L^{- 1} ⎩ ⎨ ⎧ \frac{S}{( ( S ^{2} + 9 ) ^{2} )} ⎭ ⎬ ⎫

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{S}{( S ^{2} + 9 ) ^{2}} : F (S) = \frac{1}{( S ^{2} + 9 ) ^{2}}

= \frac{d}{d t} (L^{- 1} {\frac{1}{( S ^{2} + 9 ) ^{2}}}) + L^{- 1} {\frac{1}{( S ^{2} + 9 ) ^{2}}} (0)

L^{- 1} {\frac{1}{( S ^{2} + 9 ) ^{2}}} : \frac{1}{54} (sin (3 t) - 3 t cos (3 t))

\frac{d}{d t} (\frac{1}{54} (sin (3 t) - 3 t cos (3 t))) = \frac{t sin ( 3 t )}{6}

Evaluate

\frac{1}{54} (sin (3 t) - 3 t cos (3 t))

at point

0 = 0

= \frac{t sin ( 3 t )}{6}

\int 3 e^{3 x} sin (e^{3 x}) d x

f (x) = x x

(4 x^{2} + 2 x y) \frac{d y}{d x} + (12 x y + 2 y^{2}) = 0

\int_{\frac{1}{2}}^{1} sin (2 π x) d x

y = - x^{2} + 4, y = 2 x + 1

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