What is inverse oflaplace s/((s^2+16)^2) ?

The answer to inverse oflaplace s/((s^2+16)^2) is (tsin(4t))/8

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inverse oflaplace s/((s^2+16)^2)

inverse laplace

\frac{s}{( s ^{2} + 16 ) ^{2}}

\frac{t sin ( 4 t )}{8}

Solution steps

L^{- 1} {\frac{s}{( s ^{2} + 16 ) ^{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} + 16 ) ^{2}} : F (s) = \frac{1}{( s ^{2} + 16 ) ^{2}}

= \frac{d}{d t} (L^{- 1} {\frac{1}{( s ^{2} + 16 ) ^{2}}}) + L^{- 1} {\frac{1}{( s ^{2} + 16 ) ^{2}}} (0)

L^{- 1} {\frac{1}{( s ^{2} + 16 ) ^{2}}} : \frac{1}{128} (sin (4 t) - 4 t cos (4 t))

\frac{d}{d t} (\frac{1}{128} (sin (4 t) - 4 t cos (4 t))) = \frac{t sin ( 4 t )}{8}

Evaluate

\frac{1}{128} (sin (4 t) - 4 t cos (4 t))

at point

0 = 0

= \frac{t sin ( 4 t )}{8}

4 y^{^{''}} + 6 y^{^{'}} + 7 y = 0

\frac{d}{d x} (sin (e^{2 x y}) + e^{s i n (2 x y)})

t an g e n t y = 4 x + 33, (4, 7)

\int (2 x)^{\frac{1}{2}} d x

\int x^{2} 1 + (2 x)^{2} d x

What is inverse oflaplace s/((s^2+16)^2) ?
The answer to inverse oflaplace s/((s^2+16)^2) is (tsin(4t))/8