What is inverse oflaplace 1/((1+s^2)(s^2+3)) ?

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

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inverse oflaplace 1/((1+s^2)(s^2+3))

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\int_{0}^{1} ∣ x ∣ d x

f (x) = x + x, at x = 1

\int_{0}^{1} 2^{- θ} d θ

x \to 0 lim (3 x - 4)

\frac{d}{d y} (cos (x y))

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