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

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

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

inverse laplace

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

\frac{t sin ( 2 t )}{4}

Solution steps

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

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

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

\frac{d}{d t} (\frac{1}{16} (sin (2 t) - 2 t cos (2 t))) = \frac{t sin ( 2 t )}{4}

Evaluate

\frac{1}{16} (sin (2 t) - 2 t cos (2 t))

at point

0 = 0

= \frac{t sin ( 2 t )}{4}

\frac{\partial}{\partial x} (e^{x c o s (y)} sin (x y))

(x^{2} - 4)^{2}

\int \frac{9 x ^{3} - x ^{2} + 45 x + 5}{x ^{3} + 5 x} d x

x \to 2 lim (x^{4} - 16)

\int 5 t e^{t} d t

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