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}

t an g e n t y = \frac{1}{4 - x}

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d er i v a t i v e - \frac{7}{x ^{2}}

(x^{2} + 4) d y = (2 x - 8 x y) d x

\int - \frac{2}{πn} cos (\frac{πn}{2} x) d x

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