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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\int \frac{sec ^{2} ( x )}{csc ( x )} d x

\frac{d}{d x} (lo g_{10} (4 x))

\int \frac{1}{t - \frac{1}{2}} d t

e^{- y} sin (x) + y^{^{'}} = 0

\frac{d}{d x} (2 x x + 1)

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