What is inverse oflaplace (s^2)/((s^2+4)^2) ?

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

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inverse oflaplace (s^2)/((s^2+4)^2)

inverse laplace

\frac{s ^{2}}{( s ^{2} + 4 ) ^{2}}

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

Solution steps

L^{- 1} {\frac{s ^{2}}{( s ^{2} + 4 ) ^{2}}}

Take the partial fraction of

\frac{s ^{2}}{( s ^{2} + 4 ) ^{2}} : \frac{1}{s ^{2} + 4} - \frac{4}{( s ^{2} + 4 ) ^{2}}

= L^{- 1} {\frac{1}{s ^{2} + 4} - \frac{4}{( s ^{2} + 4 ) ^{2}}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{1}{s ^{2} + 4}} - L^{- 1} {\frac{4}{( s ^{2} + 4 ) ^{2}}}

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

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

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

Simplify

\frac{1}{2} sin (2 t) - \frac{1}{4} (sin (2 t) - 2 t cos (2 t)) : \frac{sin ( 2 t ) + 2 t cos ( 2 t )}{4}

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

(2 y + 3 x) d x = - x d y

\frac{d}{d x} (\frac{12}{x ^{2}})

y^{^{''}} + y^{^{'}} + 3 y = 3 t^{2}

\int \frac{49 x ^{2} - 1}{x} d x

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