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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

x \to 4 - lim (c x^{2} + 9 x)

\int 1 + \frac{4 x ^{2}}{( 1 - x ^{2} ) ^{2}} d x

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = - 7

\frac{\partial}{\partial y} (3.25 yz cos (x yz))

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

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