What is inverse oflaplace 1/((s^2+1)(s^2+4)) ?

The answer to inverse oflaplace 1/((s^2+1)(s^2+4)) is 1/3 sin(t)-1/6 sin(2t)

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inverse oflaplace 1/((s^2+1)(s^2+4))

inverse laplace

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

\frac{1}{3} sin (t) - \frac{1}{6} sin (2 t)

Solution steps

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

Take the partial fraction of

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 4 )} : \frac{1}{3 ( s ^{2} + 1 )} - \frac{1}{3 ( s ^{2} + 4 )}

= L^{- 1} {\frac{1}{3 ( s ^{2} + 1 )} - \frac{1}{3 ( s ^{2} + 4 )}}

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}{3 ( s ^{2} + 1 )}} - L^{- 1} {\frac{1}{3 ( s ^{2} + 4 )}}

L^{- 1} {\frac{1}{3 ( s ^{2} + 1 )}} : \frac{1}{3} sin (t)

L^{- 1} {\frac{1}{3 ( s ^{2} + 4 )}} : \frac{1}{6} sin (2 t)

= \frac{1}{3} sin (t) - \frac{1}{6} sin (2 t)

x (\frac{d y}{d x}) = \frac{( 1 - 4 y ^{2} )}{3 y}

x x + x^{2} x

\frac{\partial}{\partial x} (\frac{5}{x})

x \to 0 lim ((\frac{1}{2})^{x})

\frac{d}{d t} (C e^{3 t} - e^{2})

What is inverse oflaplace 1/((s^2+1)(s^2+4)) ?
The answer to inverse oflaplace 1/((s^2+1)(s^2+4)) is 1/3 sin(t)-1/6 sin(2t)